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Table of Contents1. DC and AC Voltages .............................................................................................................. 1

DC Voltage Sources ........................................................................................................... 1AC Voltage Sources ........................................................................................................... 2Measuring voltages using a multimeter ................................................................................ 3

2. Current and Resistors ............................................................................................................ 4Introduction ........................................................................................................................ 4Resistors in series .............................................................................................................. 4Resistors in parallel ............................................................................................................ 5Creating a voltage divider using resistors ............................................................................ 5Measuring current using a multimeter .................................................................................. 6Measuring resistance using a multimeter ............................................................................. 7

3. Power dissipation ................................................................................................................... 8Introduction ........................................................................................................................ 8Power dissipation in a resistor ............................................................................................ 8Power dissipation of series-connected resistors ................................................................... 8Power dissipation of parallel-connected resistors ................................................................. 9

4. Capacitors ............................................................................................................................ 10Introduction ...................................................................................................................... 10The impedance of a capacitor ........................................................................................... 10Phase shift ....................................................................................................................... 11Relation between voltage and current ................................................................................ 11Frequency filters ............................................................................................................... 12ESR ................................................................................................................................. 12Timer circuits .................................................................................................................... 13Types of capacitors .......................................................................................................... 14

5. Diodes ................................................................................................................................. 15Introduction ...................................................................................................................... 15An AC Voltage Rectifier .................................................................................................... 15LEDs ............................................................................................................................... 16Zener Diodes ................................................................................................................... 16Testing diodes using a multimeter ..................................................................................... 17

6. Bipolar Transistors ................................................................................................................ 19Introduction ...................................................................................................................... 19The transistor as a switch ................................................................................................. 19The Darlington ................................................................................................................. 20The transistor as an amplifier ............................................................................................ 20Testing transistors using a multimeter ................................................................................ 22

7. Project: A simple adjustable DC power supply ....................................................................... 24The transformer ................................................................................................................ 24The diagram .................................................................................................................... 24

8. Differential Amplifier .............................................................................................................. 26Typical example ............................................................................................................... 26DC Current Source ........................................................................................................... 26

9. Operational Amplifier ............................................................................................................ 28Introduction ...................................................................................................................... 28Opamp used as an amplifier ............................................................................................. 28Opamp used as a threshold switch ................................................................................... 29Opamp used as a voltage-controlled current source ........................................................... 30Diode and transistor tester ................................................................................................ 31

10. Lab Power Supply .............................................................................................................. 33The diagram .................................................................................................................... 33Voltage feedback .............................................................................................................. 33Current limitation .............................................................................................................. 34Protective circuitry ............................................................................................................ 34Choosing your components ............................................................................................... 35


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Assembly ......................................................................................................................... 3511. Heatsink ............................................................................................................................. 38

Introduction ...................................................................................................................... 38Calculations ..................................................................................................................... 38

12. Power Amplifiers ................................................................................................................ 40Introduction ...................................................................................................................... 40Emitter follower ................................................................................................................ 40Balance amplifier .............................................................................................................. 40Bias ................................................................................................................................. 41Darlingtons ....................................................................................................................... 42Opamp ............................................................................................................................. 42

13. Inductors ............................................................................................................................ 44Introduction ...................................................................................................................... 44The impedance of an inductor .......................................................................................... 44Relation between voltage and current ................................................................................ 44Frequency filters ............................................................................................................... 45Quality factor .................................................................................................................... 46

14. Decibels (dB) ..................................................................................................................... 47Power and Voltage Ratios ................................................................................................ 47Reference-related dBs ...................................................................................................... 47

15. Vocal Eliminator ................................................................................................................. 48Introduction ...................................................................................................................... 48Schematic ........................................................................................................................ 48Choosing components ...................................................................................................... 48Testing ............................................................................................................................. 49Assembly ......................................................................................................................... 49

16. Symmetric power supply ..................................................................................................... 51Introduction ...................................................................................................................... 51Up to about 25mA ............................................................................................................ 51Up to 1A .......................................................................................................................... 52

17. JFETs ................................................................................................................................ 54Introduction ...................................................................................................................... 54JFET amplifier .................................................................................................................. 54JFET current source ......................................................................................................... 55

18. MOSFETs .......................................................................................................................... 56Introduction ...................................................................................................................... 56MOSFET amplifier ............................................................................................................ 57Dual-gate MOSFETs ......................................................................................................... 57

19. LC filters ............................................................................................................................ 59Introduction ...................................................................................................................... 59High pass filter ................................................................................................................. 59Band pass filter ................................................................................................................ 60Band pass filter with a smaller band .................................................................................. 62Resonance ....................................................................................................................... 63Tuned circuit .................................................................................................................... 63

20. Miscellaneous filters ........................................................................................................... 65Introduction ...................................................................................................................... 65Coupled filters .................................................................................................................. 65Twin T-filter ...................................................................................................................... 66Bridged T-filters ................................................................................................................ 67

21. Frequency-independant voltage devider ............................................................................... 68Introduction ...................................................................................................................... 68HF probe ......................................................................................................................... 68Attenuator in an oscilloscope ............................................................................................ 69

22. DIACs, SCRs and TRIACs .................................................................................................. 70DIAC ................................................................................................................................ 70SCR ................................................................................................................................ 70TRIAC .............................................................................................................................. 71


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23. TV Deflection Circuit ........................................................................................................... 73Warning ........................................................................................................................... 73The picture tube ............................................................................................................... 73Deflection coils ................................................................................................................. 73Deflection circuit ............................................................................................................... 73Linearity correction ........................................................................................................... 75S-Correction ..................................................................................................................... 75EW-Correction .................................................................................................................. 76Practical Examples ........................................................................................................... 76

24. Automatic volume control .................................................................................................... 78Introduction ...................................................................................................................... 78Schematic ........................................................................................................................ 78Choosing components ...................................................................................................... 79Power supply ................................................................................................................... 79

A. Calculating RMS value ......................................................................................................... 80B. Inside semiconductors .......................................................................................................... 81

Inside a diode .................................................................................................................. 81P type and N type semiconductors ............................................................................ 81Joining P and N together .......................................................................................... 81Zener diodes ............................................................................................................ 82Varicap diodes ......................................................................................................... 82

Inside a bipolar transistor ................................................................................................. 82Joining three layers of P and N together .................................................................... 82Applying voltages across a transistor ......................................................................... 82

Inside a JFET .................................................................................................................. 83Joining three layers of P and N together .................................................................... 83Applying voltages across a JFET .............................................................................. 83

Inside a MOSFET ............................................................................................................ 84Joining three layers of P and N together .................................................................... 84Applying some voltages across an enhancement MOSFET ......................................... 84Depletion MOSFETs ................................................................................................. 84

C. Buffer capacitor ................................................................................................................... 85Calculation of the value of a buffer capacitor ..................................................................... 85ESR ................................................................................................................................. 85

D. Amplifier stability .................................................................................................................. 88Introduction ...................................................................................................................... 88Stability ............................................................................................................................ 88

E. Complex math ..................................................................................................................... 89Calculations on a capacitor and resistor in series. .............................................................. 89Calculations on an inductor and resistor in series. .............................................................. 90Calculations on a capacitor and inductor in series. ............................................................. 90Calculations on a tuned LC circuit. .................................................................................... 90

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Chapter 1. DC and AC VoltagesDC Voltage Sources

DC stands for: Direct Current. DC voltage sources have a positive and a negative terminal. Thesymbol of a DC voltage source is

An example of a DC voltage source is a battery or a DC power supply.

To increase the voltage, you can connect multiple voltage sources in series:

The total voltage will be the sum of each voltage source. So when you connect two 1.5V batteries inseries, you'll measure a total voltage of 3 volts.

By the way, most designers don't draw voltage sources in their schematics; they just draw theterminals:

The third drawing is the most common. The horizontal line at the bottom is the ground symbol.Ground is not always the negative terminal. Many audio devices for example use a so calledsymmetric power supply. Symmetric power supplies consist of two DC voltage sources connected toeach other:

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In these cases ground is the 'middle point' where both sources are connected. Ground is always thereference point. This means that all the voltages in the design or description are always with respectto ground. In other words: the black wire (connected to the COM bus) of your voltage meter shouldalways be connected to ground, as shown in the picture above. In this case voltage meter M1 reads+9V and M2 reads -9V. And this immediately explains why it is called a symmetric power supply.

Later in this course you will learn what symmetric power supplies are used for and are we going tobuild one.

AC Voltage Sources

AC stands for: Alternating Current. AC voltage sources don't have a positive and a negative terminal:the polarity reverses in time. Take a look at the picture below.

In this picture SW1 is a switch. In the drawn position, A is connected to C and therefore to the positiveside of the DC voltage source. B is connected to E and therefore to the negative terminal of the 9Vsource. When you toggle switch SW1, the polarity will be reversed: A will be connected - via D - tothe negative end of the voltage source, and B will be connected to the positive end. Now imagine thatsomeone toggles switch SW1 frequently. The signal at terminals A and B will then be an AC voltage.

The top value of an AC voltage is called the amplitude. In this case, the amplitude is 9Vt. The voltagebetween the two tops is called the top-top value; in this case 18Vtt.

If we toggle SW1 forth and back in exactly 1 second, we create a 1Hertz signal. Hertz is the unit offrequency: the number of times a signal repeats itself in one second. Hertz is usually abbreviated toHz. The time it takes for a signal to repeat itself is called the period time, symbol T; in this case T = 1s. A 10Hz signal means that the signal repeats itself 10 times per second; in that case T=0.1s. So:

T = 1/f and f = 1/T

It is common practice to use symbols in capitals for DC signals and lower case symbols for ACsignals. For example VA would mean the DC voltage at point A, and iR4 would mean the AC currentflow in resistor R4.

Examples of AC voltage sources are: a microphone, a house outlet, and the speaker terminals of anamplifier.

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An AC voltage source doesn't really have a symbol of it's own. They are usually drawn as one or twoterminals with a ~ sign. If only one terminal is drawn, the other one is connected to ground.

Some AC voltage sources have their own symbols, e.g. a microphone:

Measuring voltages using a multimeterMost digital multimeters look like this:

1 = Display, 2 = Function switch, 3 = Transistor socket (optional), 4...6 = Test lead jacks

If you want to measure DC voltages set the function switch to the DC voltage range you want to use.For example, if you want to measure the voltage across a 9V battery, set the switch to 20V DC. Ifyou have no idea what to expect, set the function switch to the highest DC range available and workdown.

Having done that, we can connect the test leads. Mulimeters usually come with two test leads: a blackone and a red one. To measure voltages, you need to connect the black test lead to the COM jackand the red lead to the V/Ω jack. Connect the other ends of the test leads to the source or load undermeasurement. In case of a 9V battery, connect the black wire to the negative and red wire to thepositive terminal of the battery. If you swap the test leads, you will read a negative value.

If you want to measure AC voltages, set the function switch to the proper AC voltage range. Connectthe test leads to the device-under-test. Swapping test leads makes no difference (of course!).

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Chapter 2. Current and ResistorsIntroductionWhen you connect the terminals of a voltage source to each other, you create a short circuit. Thismeans a high current flow. To limit the current flow, you can use a resistor. The symbol of a resistor is:

Voltage, current and resistance are related to each other as follows:

R = V/I

V is the voltage across the resistor [unit: volts, or V]; I is the current in the resistor [unit: amperes, orA]; R is the resistance [unit: ohms, or Ω].

Example: Imagine you connect a 1000Ω (or 1kΩ) resistor to a 9V battery. In that case, the current inthe resistor (and in the battery of course!) will be: I = V/R = 9V / 1000Ω = 9mA (milli-amps).

You can't buy resistors of any value. You can choose from a series of resistors, e.g. the E12 series.The E12 series has the following values: 10, 12, 15, 18, 22, 27, 33, 39, 47, 56, 68, 82. If you wantother values, you may select one from another (more expensive) series, or create one by connectingmultiple resistors in series or parallel.

Resistors in series

Now we'll connect 3 resistors in series with the battery (see picture above). What will be the totalresistance of R1, R2 and R3?

The voltage across R1 (V1) equals to: V1 = I∙R1. And V2 = I∙R2, and V3 = I∙R3.

We know that V1 + V2 + V3 = Vbat, so:

Vbat = I∙R1 + I∙R2 + I∙R3 = I∙(R1 + R2 + R3).

This tells us that the total resistance of resistors is series equals to R1 + R2 + R3 + ..., or:

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In this case, the total resistance is 3kΩ. The current I will be: 9V / 3k = 3mA.

Resistors in parallel

The picture on the left shows a DC voltage source connected with 3 parallel-connected resistors. Thequestion is again: what is the total resistance?

The current in R1 (I1) equals to: I1 = Vbat/R1. And I2 = Vbat/R2, and V3 = Vbat/R3. The total current Itotequals I1 + I2 + I3, so:

Itot = Vbat/R1 + Vbat/R2 + Vbat/R3.

This proves that the total resistance of parallel connected resistors equals to:

1/Rtot = 1/R1 + 1/R2 + 1/R3 + ... or:

In this case, the total resistance is 333Ω. The total current will be 3 ∙ 9mA = 27mA.

Creating a voltage divider using resistors

Take a look at the picture on the right. We see three series connected resistors. We've alreadylearned that the total resistance is 3k. So the current I will be 9V / 3k = 3mA. The voltage at point


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B, VB, equals 1k∙3mA = 3V. (Do you still remember what is meant by 'voltage at point B'? It means:connect the red wire of the volt meter to point B and the black wire to ground.)

The general way of calculating the voltage across a resistor in a series connection is:

I = Vsource / Rtotal, and Vres = I∙R. So:

There are three ways to calculate the voltage at point A:

1. The total resistance of R2 and R3 is 2k, so VA = 2k∙3mA = 6V.

2. The voltage across each resistor is 3V, so VA = 6V.

3. Using the equation above: VA = 9V∙(2k/3k) = 6V.

Does this mean that you can connect your 3V portable cassette player to point B? Well, of courseyou could, but don't expect it to work! The player acts like a resistor of, say, 50 ohms. That resistor isparallel connected with R3, resulting in a resistance of 47.6 ohms. So VB will drop to 9V∙(47.6/2047.6)= 0.2V. And that will never be enough for your player.

Conclusion: If you design a voltage divider, don't forget to take the load into account!

Measuring current using a multimeterMost digital multimeters look like this:


If you want to measure DC current, set the function switch to the DC current range you want to use.For example, if you expect to measure 1mA, set the switch to 2mA DC. If you have no idea what toexpect, set the function switch to the highest DC range available and work down.

Having done that, we can connect the test leads. Mulimeters usually come with two test leads: a blackone and a red one. To measure current, you need to connect the black test lead to the COM jackand the red lead to the A jack. Connect the other ends of the test leads in series with the load undermeasurement. If the current flows from red to black, you will read a positive value. Otherwise, a minussign appears in the display.

If you want to measure AC current, set the function switch to the proper AC current range. Connectthe test leads in series with device-under-test. Swapping test leads makes no difference (of course!).


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Note: many meters have a separate jack for measuring high current. Usually the A jack measuresup to 200mA. The separate jack will be labeled '20A'. This jack only works when the function switchhas been set to 20A. Warning: the 20A jack is usually unfused! Overload may seriously damage yourmultimeter.

Tip: if you want to measure the current flow in a component, you'll have to connect the meter in serieswith that component. This means you may need to unsolder one end of that component. If the samecurrent also flows in a resistor, you can simply measure the voltage across that resistor and calculatethe current.

After current measurement, disconnect the leads from the meter. If you forget this and want tomeasure voltages again, you may cause disasterous shorts!

Measuring resistance using a multimeterIf you want to measure resistance, set the function switch to the resistance range you want to use.For example, if you expect the resistance to be 1kΩ, set the switch to 2kΩ. If you have no idea what toexpect, set the function switch to the highest resistance range available and work down.

Having done that, we can connect the test leads. Connect the black test lead to the COM jack andthe red lead to the V/Ω jack. Connect the other ends of the test leads across the resistance undermeasurement.

Please note that in-cicuit measurement may lead to wrong results, since there may be othercomponents parallel-connected to the resistance. It is also a good idea to make sure that thevoltage across the resistance is 0V before starting resistance measurement. Also make sure that theequipment-under-test has been turned off!

Interesting links: Circuit Fantasia's electronics course

http://www.circuit-fantasia.com/my-students/ske2004/classes/current-ske-1.htm

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Chapter 3. Power dissipationIntroductionWhen a current flows in a component, that component will heat up. This process is called powerdissipation and is measured in Watts. The power dissipation in a device can be calculated very easily:

P = V ∙ I

Power dissipation in a resistorLet's calculate the power dissipation in a 100 ohms resistor connected to a 9V battery.

The voltage across the resistor will be 9V. The current is 9V/100ohms = 90mA. So the powerdissipation will be: 9V ∙ 90mA = 810mW.

It is very important to calcuate the power dissipation in the components in your design. A regularresistor has a maximum dissipation rating of 0.25W (= 250mW). If you would have used such aresistor in the example above, it would have blown. A 1W resistor is a good choise.

Since it's so important, let's create an equation with which we can easily calculate the powerdissipation in a resistor. We know:

(1) P = V ∙ I (2) V = I ∙ R (3) I = V / R

Substituting (2) in (1) and (3) in (1) respectively results in:

P = I2 ∙ R P = V2 / R

With these equations you can easily calculate the power dissipation when you connect a DC voltagesource to a resistor. But what will the power dissipation be if you connect an AC voltage source toa resistor? In that case, simply substitute V and I by the so called RMS values vRMS and iRMS. RMSstands for Root Mean Square. The RMS value is defined as the DC equivalent that provides the samepower as the original waveform. Let's approximate the RMS value of a 1Vt/1Hz sinusoidal signal: v= sin(2∙π∙f∙t) = sin(2∙π∙t). We take 4 samples: at 0s, 0.25s, 0.5s and at 0.75s. The values are 0, 1, 0,and -1. Next, calculate the square of each value: 0, 1, 0, and 1. The mean value of these squares is (0+ 1 + 0 + 1)/4 = 2/4 = 0.5. Finally, calculate the square root of the mean of the squares: √0.5 = 0.707V.So the approximated RMS value of a 1Vt sinusoidal signal is 0.707V. Of course, the approximationis more accurate if you take more samples. Using some math, you can prove that vRMS = A/√2 (for asinusoidal signal). Using this equation we can calculate the RMS value of the signal of our example:vRMS=1/√2. = 0.707V.

Using the theory above, we can calculate the power dissipation of a 100ohms resistor connected to a9Vt sinusoidal signal: P=v2

RMS/R = A2/2R = 81/200 = 0.401W.

Power dissipation of series-connected resistorsIf you don't have a 1W resistor, and you still want to perform the experiment above, you may connectfour 25ohms resistors in series. We've already learned that resistors in series act like a voltagedevider: the voltage across each resistor is 9V/4 = 2.25V. The current is still 90mA since the totalresistance is the same. So each resistor dissipates 2.25V ∙ 90mA = 0.20W. (Of couse we could alsouse one of our 'easy' equations: P = I2 ∙ R = (90mA)2 ∙ 25 = 0.20W.)

Be carefull: always take resistors with the same resistance. Of course you could also create a100ohms resistor with three 33ohms resistors and one 1ohm resistor in series, but you're gonna smellsome smoke! Which resistor(s) will blow? The 1ohm resistor because it's the smallest? Let's see.

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Since we know the current is 90mA, we use the equation P = I2 ∙ R = (90mA)2 ∙ 1 = 8.1mW. The 1ohmresistor will survive! The power dissipation of each 33ohms resistor will be (90mA)2 ∙ 33 = 0.27W. Itmay take some time, but you certainly will loose three resistors!

Power dissipation of parallel-connected resistorsAnother way to create your own high wattage resistor is to connect multiple resistors in parallel. Let'screate a 100ohms resistor with four parallel-connected 400ohms resistors and connect it to a 9Vbattery. The voltage across each resistor is 9V. So the power dissipation of each resistor is P = V2 / R= 92 / 400 = 0.20W.

Again: always use resistors with the same resistance.

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Chapter 4. CapacitorsIntroductionA capacitor consists of two metal plates with a thin insulator in between, as its symbol shows:

What will happen if we connect a DC voltage source (battery) to a capacitor?

The positive side of the battery attracts the electrons in the top plate of the capacitor. This plate willbecome positively charged. Because the insulator is very thin, the top plate will attract the electronsin the bottom plate. The gaps these electrons leave behind, will be filled up with the electrons fromthe negative end of the battery. So it seems if the current flows right through the capacitor, as ifthere were no insulator at all. But of course, this can't continue for ever. Eventually, there will be noelectrons left on the top plate, and no room for more electrons on the bottom plate. The capacitor isnow completely charged, and the current flow will stop.

Now let's swap the terminals of the battery. The positive terminal of the battery will attract theelectrons on the bottom plate of the capacitor and the negative end of the battery will emit electrons tofill in the gaps on the top plate. This process will continue until the capacitor is charged again.

If we continually swap the terminals of the battery, there will be a continuous current flow. In otherwords: a capacitor conducts AC currents, but blocks DC currents.

The capacitance depends on the size of the plates and the matial between them. This materialis called the dielectric and reduces the electric field between the plates. This will increase thecapacitance.

The capacitance can be calculated with: C = εA/d, where ε is the dielectric constant, A the area of oneplate and d the distance between the plates. Since we can buy capacitors in any electronics show,we'll seldomly need this equation.

The unit of capacity is Farad, symbol F. This unit is usually far too large; uF (micro Farad), nF (nanoFarad), and pF (pico Farad) are more common. 1F = 1000000uF, 1uF = 1000nF, 1nF = 1000pF.

The impedance of a capacitorThe impedance of a component is the resistance of that component for AC voltages. The symbol forresistance is R; the symbol for impedance is X. The impedance of a capacitor is not zero; it depends

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on the capacity (size of the plates) and the frequency of the signal (number of polarity changes (forthand back) per second). The impedance can be calculated using the following equation.

f is the frequency in Hertz; C is the capacitance in Farad

Example: We have a 1nF capacitor and connect it to a 50Hz AC voltage source. Calculate theimpedance of the capacitor.

XC = 1/(2∙π∙50∙10-9) = 3.18MΩ.

Phase shift

When the voltage across a certain resistor increases, the current flow in that resistor will also increase(and visa versa). This is not true for a capacitor. We already saw in the introduction that if a capacitoris fully charged (so the voltage across it has reached its maximum), the current flow stops. Thecurrent will have its maximum value when the capacitor is empty. Let's look what happens if weconnect a capacitor to a sinusoidal voltage source.

We connected a capacitor to a 1kHz voltage source. The green curve shows the voltage across thecapacitor and the blue curve shows the current flow. We see that the current reaches the top value1/4 period before the voltage. Since 1/4 period of a sine wave equals 90 degrees, we say that thecurrent leads the voltage by 90 degrees, because the current reaches its top value before the voltagedoes. We can also say that the voltage lags the current by 90 degrees.

Relation between voltage and current

The following equation can be used for any current i(t):

It clearly shows the 90 degrees phase shift if i(t) is sinusoidal: the integral of a sine is a (-)cosine. Wealso see that, if we charge a capacitor with a constant current I, the voltage across it will increaselinearly: v(t) = v(0) + (1/C)∙I∙t.

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Frequency filters

Take a look on the diagram above. Assume that the voltage source supplies a 1V/10kHz signal (thismeans: the amplitude is 1V and the frequency is 10kHz = 10000Hz).

The impedance of capacitor C will be XC = 1/(2∙π∙104∙10-6) = 15,9Ω. The output voltage (voltageacross capacitor C) will be 1V∙(XC/ZR+C), where ZR+C is the total impedance of R and C. Becausea capacitor causes a phase shift in the current flow, we cannot just state that ZR+C = R + XC. Usingsome complex math we can prove that:

ZR+C = √(R2+XC2).

In our case ZR+C = √(1k2+15.92) = 1000.13Ω. So the output voltage becomes 1V∙(15.9/1000.13) =0.0159V.

Now assume that the voltage source supplies a 1V/10Hz signal. The impedance of capacitor Cwill then be XC = 1/(2∙π∙10∙10-6) = 15,9kΩ. The output voltage will be 1V∙(XC/(ZR+C)) = 1V∙(15.9k/

√(1k2+15.9k2)) = 0.998V. So we created a very simple frequency filter with just a resistor and acapacitor.

In this case we created a so called low pass filter (LPF) since it passes low frequency signals andsuppresses high frequency signals. If you swap R and C, you create a high pass filter (HPF).

Let's calculate the cut-off frequency of our filter. The cut-off frequency is the frequency at which R=XC=> R = 1/(2∙π∙f∙C) =>

In our case f = 1/(2∙π∙103∙10-6) = 159Hz.

ESREvery capacitor has a certain series resistance. This resistance is not only caused by the leads, butalso by the metal plates and the dielectric the capacitor is made of. The sum of these resistances iscalled ESR, Equivalent Series Resistance. This resistance will not always remain the same, but mayincrease due to aging.

When will the ESR bother us? Of couse this depends on how large the ESR is and the application inwhich the capacitor is used. Assume the ESR of capacitor C in the filter above is 10Ω. At very highfrequencies the output voltage will not be 0V, but 1V∙(10/1010) = 10mV. In most cases, this will not beany problem. However, if resistor R were also 10Ω, the output voltage would have been 0.5V!

We can also expect ESR problems when large charge and discharge currents flow though thecapacitor. Remember, a large current means a large voltage across the series resistance. This mayeven heat up the capacitor. If a capacitor heats up, the ESR may increase. This will heat up the

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capacitor even more, and so on. Eventually (and this may take months) the capacitor will be ready forthe dumpster. Troubleshooting can be a pain; a simple capacitance meter uses small currents and willtherefore not notice that the ESR has increased.

How can we measusre the ESR? The are special ESR meters available for this purpose, but theseare pretty expensive. Most of the time we only need an indication. We can connect the capacitorto a power supply via a known resistor R and a switch. If the switch is open, the voltage acrossthe capacitor and ESR will be 0V. On the momen the switch is closed, the capacitor is still empty.The voltage we measure across the capacitor is therefore equal to the voltage across its ESR. Ifthat voltage is equal to half the supply voltage, the ESR must be equal to the known resistor R. Ofcourse: the lower the voltage, the lower the ESR must be. The disadvantage of this method is that thepower supply must be able to deliver the current peak. Moreover, we must also include the internalresistance of the power supply in our calculations. That's why we often use the opposite method: wecharge a capacitor to a certain voltage and then discharge it via a known resistor. Of couse: the higherthe voltage at the moment of discharge, the lower the ESR must be. Please find below a picture ofboth methods. Resistor R is 10Ω. The supply voltage is 1V.

At t=0, the voltage across the ESR is about 0.34V. So ESR/(R+ESR)=0.34 => ESR=R(0.34/(1-0.34))= 10(0.34/0.66) = 5.2Ω.

At t=100us, the capacitor is discharged via the same resistor R. The voltage immediately drops to0.66V.

So R/(ESR+R)=0.66 => ESR=R((1-0.66)/0.66) = 10(0.34/0.66) = 5.2Ω.

Timer circuitsNow we'll exchange the AC voltage source for a 1V DC voltage source. Since the frequency is 0Hz,XC is infinite, so there will be no current flow. That's true, but not for the first period of time afterconnecting the voltage source as we already saw in the introduction of this chapter.

Assume that capacitor C is completely discharged: VC=0 => VR=1V. So the current flow in resistor Rwill be 1mA. Having nowhere else to go to, this current will flow 'in' the capacitor, charging it. While thecapacitor is charging, the voltage across it raises, leaving less voltage for resistor R. This means thatthe current flow decreases. Suppose that after T seconds, the capacitor is half full: VC=0.5V. In thatcase VR=0.5V => IR = IC = 0.5mA. So after 2T seconds, the capacitor will not be completely chargedsince the current flow isn't 1mA anymore. To calculate the voltage at any given time, use the followingequation.

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VB is the voltage of the DC voltage source. t is the time in seconds since the capacitor was connectedto the voltage source. e is Euler's constant (2.7182818).

When t=RC, -t/(RC) will be -1 and VC = 0.63V, so the capacitor will be 63% full. This time is referred toas the 'RC time'.

RC circuits are often used in timers, for example in a simple burglar alarm:

When you enter your own house, you don't want the alarm to go off immediately; you want to havesome time to switch it off. In the circuit above you have R∙C = 100k∙100u = 10 seconds to do that.After 10 seconds the voltage across the capacitor will raise above 0.63V, and a switch will closecausing the flash light to give alarm.

Types of capacitorsThere are generally two types of capacitors: polarized and bipolar. Polarized capacitors have apositive and a negative terminal; bipolar capacitors don't. In polarized capacitors the insulatorbetween the plates is usually an electrolyte; hence the name electrolytic capacitors, or electro's. Theelectrolyte enables manufacturers to create large value capacitors with small dimensions. That's whyyou'll always see electro's with relatively large values: 1uF and above.

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Chapter 5. DiodesIntroductionA diode is a device that conducts the current in just one direction: the direction of the arrow in thediode symbol, which looks like this:

The most important parameters of a diode are: maximum forward current, forward voltage, maximumpower dissipation and reverse voltage.

The forward current is the current flow in the direction of the arrow of the diode symbol. This currentcauses a voltage across the diode: the forward voltage drop.

Each diode has a certain minimum voltage drop, called the knee voltage. The diode will not conductwhen the voltage across it is less than the knee voltage. The knee voltage of a generic silicon diode isabout 0.6V.

The power dissipation of a diode is the forward current multiplied by the forward voltage drop.

The reverse voltage is the voltage across a diode when it is reverse biased.

If you want to know how a diode works internally, you'll have to take a peek inside.

An AC Voltage RectifierSince diodes conduct current in only one direction, they can be used as an AC Voltage rectifier. Takea look at the picture below.

A triangular AC voltage is connected to the input terminals of the rectifier. The output voltage will bemeasured across resistor R1. When the upper input terminal is positive, there will be a current flow inthe diode and the resistor. This current causes a voltage across R1. Assume the peak voltage is (plusand minus) 9V, and the forward voltage of the diode is 0.7V. The peak current will then be (9V - 0.7V)/1k = 8.3mA. The maximum power dissipation of the diode will be 0.7∙8.3mA = 5.8mW.

When the voltage at the upper input terminal becomes negative, the diode is reverse biased blockingthe current flow. Since the diode has a very large resistance, all the voltage will be across the diode.This should not exceed the maximum reverse voltage.

So if you want to perform this experiment, you'll need a diode with the following requirements:the maximum forward current must be 8.3mA or higher; the maximum power dissipation must be5.8mW or higher; and the maximum reverse voltage must be 9V or higher. Any small signal diodewill meet these requirements. The resistor can be a regular 0.25W resistor since the maximum powerdissipation is (8.3mA)2∙1k = 69mW.

The circuit above is called a half wave rectifier, since the ouput contains only the positive half of theinput. The circuit below shows a full wave rectifier.

Diodes

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This circuit works as follows. When the input signal is positive, the currents flows from the upperterminal, via diode D1, resistor R1, and diode D3 to the lower terminal. When the input signal isnegative, the currents flows from the lower terminal, via diode D2, resistor R1, and diode D4 to theupper terminal. Notice that the current always flows in two diodes: either D1 and D3, or D2 and D4.This means that the output voltage will always be about 1.4 volts (two 'forward voltage drops') lessthan the input voltage.

The circuit D1...D4 is called a bridge rectifier. When you look at a bridge rectifier, you'll probably seesomething imprinted like 'B80C5000/3300'. The number after the 'B' indicates the maximum (reverse)voltage, in this case 80V. The number after the 'C' indicates the maximum peak/continuous (forward)current in mA. In this case the maximum peak current is 5A and the maximum continuous current is3.3A. Smaller bridge rectifiers only indicate the maximum voltage and current, e.g. 'B40C800'.

LEDsThe abbreviation LED stands for Light Emitting Diode. LEDs consume less power than light bulbs,and have a much longer life time: about 100000 hours. A regular LED needs a current flow of10...20mA, and has a forward voltage drop of 1.5 to 2 volts, depending on the color.

With the circuit below, you can test and experiment with LEDs.

Question: What will be a good value for R1? Assume that the voltage across LED D1 is 2 volts, andwe want a current flow of 15mA.

Answer: The voltage across R1 will be 9V-2V = 7V. The current flow in R1 will also be 15mA. So R1should have a value of 7V/15mA = 467Ω. From the E12 series, 470Ω is a good choice.

Zener DiodesA zener diode in conducting state acts like a normal diode. It's the reverse voltage that distinguishes azener diode from a regular diode. Take a look at the picture below.

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In this picture you see a reverse connected zener diode. The 'value' of a zener diode is given in volts;this is the reverse voltage. But a zener diode doesn't blow when the voltage tends to get higher. Azener diode stabilizes the voltage at the reverse voltage. So the voltage across the zener diode in thepicture above will always be 4.7 volts, even when the battery voltage increases.

Again, we need to calculate the value of R1. Unfortunately, it's difficult to say what's the ideal currentflow in a zener diode. (Yep, altough the diode is reverse biased, there is a current flow!) In most cases5mA is fine. Since the voltage across R1 will be 9V-4.7V = 4.3V, a good value of R1 is 4.3V/5mA =860Ω.

A zener diode manufacturer publishes the maximum power dissipation of a zener diode. 0.4 or0.5W is a very common value for a small zener diode. Using this characteristic, we can calculate theminimum value of R1: Assume we use a 0.4W zener in the design above. Since the voltage acrossthe zener is 4.7V, the maximum current flow is 0.4W/4.7V = 85mA. The voltage across R1 will be 9V- 4.7V = 4.3V. So the minimum value for R1 is 4.3V/85mA = 51Ω. So a good value of R1 ranges from51 to 860Ω. 820Ω is a good choice. Note however that the calculations above only count for a zenerwithout a load. Take a look at the picture below.

In this design zener D1 has a 50Ω load (R2). Again, we'll calculate a proper value for R1. Since thevoltage across the load R2 is always 4.7V, the current flow in R2 will always be 4.7V/50 = 94mA.The current flow in D1 should be between 5 and 85mA. So the current flow in R1 ranges from 99 to179mA. The voltage across R1 is always 4.3V, so the resistance should be between 24 and 43Ω. 39Ωmay be a good choice. In that case, the power dissipation is 4.32/39 = 0.47W. So you'd better take a1W resistor!

But what should we do if the 50Ω load can be detached, e.g. because it's an external load? With theload connected, the maximum value of R1 is 43Ω, but without the load the minimum value is 51Ω!

The answer is simple: use a higher wattage zener diode, e.g. 1.3W. In that case, the maximumcurrent flow in D1 is 1.3W/4.7V = 276mA. This means, without the load connected, a minimum valueof R1 of 4.3V/276mA = 16Ω. Now we have an overlapping range of values for R1 from which you maychoose one. Again, a 39Ω/1W resistor is a good choice.

Testing diodes using a multimeterMost digital multimeters look like this:

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If you want to test a diode, set the function switch to "diode test".

Next, connect the test leads. Mulimeters usually come with two test leads: a black one and a red one.Connect the black test lead to the COM jack and the red lead to the V/Ω jack. Connect the other endsof the test leads across the diode. Connect the black wire to the cathode and red wire to the anode.The display should now read about 0.6V (600mV). If you swap the test leads, the display will indicatean overflow.

Note that in-circuit testing may lead to wrong results, since other components may be parallel-connected to the diode. Also make sure that the equipment-under-test has been turned off!

Later in this course we'll build a nice device for testing diodes (and transistors).

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Chapter 6. Bipolar TransistorsIntroductionBipolar transistors are amplifying devices and can also be used as switches. There are two types:NPN and PNP. See the picture below for a typical circuit.

A bipolar transistor has three terminals: Base, Collector and Emitter. In case of a NPN transistor, asmall current flows from B to E (IB) causes a larger current flow from C to E (IC). The ratio IC/IB iscalled the current gain, symbol hFE.

Inside a transistor, there's a diode between B and E and between B and C, so VBE,max and VBC,maxare about 0.6V to 0.7V.

Let's assume for example RB = 1M, RL = 1k, VS = 9V, hFE = 300 and VBE = 0.6V.

The voltage across RB will be VS-VBE=8.4V, so IB=8.4/1M=8.4μA. IC=IB∙hFE=8.4μA∙300=2.52mA. Sothe voltage across RL will be 2.52V.

The transistor as a switchWhat will happen if in the example above, RB=100k instead of 1M?

IB=8.4/100k=84μA. You may expect that IC will be 84μA∙300=25.2mA, but that isn't possible since thevoltage across RL would be 25.2V which is more than VS. IC,max in this circuit is VS/RL=9/1k=9mA.So even if IB=84μA, IC will be 9mA. IC/IB=107, which is less than hFE. In such a case, when IC/IB < hFE,we say that the transistor has become saturated and can be considered as a closed switch (betweenC and E).

Take a look at the diagram below.

You see a battery operated clock, operating at 3V. This clock has an alarm function: at a preset time,you hear a chime. Imagine that you don't want to hear a chime, but that you want to switch someother equipment on, for example a radio operating a 9V. This radio has an internal resistance of 100Ω.At alarm time, the output voltage of the clock is 3V. VBE=0.6V. hFE=100

What would be a proper value for RB? A large value may not saturate the transistor; a small resistormay overload the output stage of the alarm circuitry of the clock.

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IC,max=9/100=90mA. IC/IB < hFE => IB > IC/hFE => IB > 90mA/100 = 0.9mA. The voltage across RBequals 3-0.6=2.4V. This means RB < 2.4V/0.9mA = 2.7k. To be on the safe side, 2.2k would be agood value. IB will then be 2.4/2k2 = 1.09mA.

Please note that this will only work if the "ground" of the clock (the minus terminal of its battery) isconnected to the ground of our little switch (the emitter).

The DarlingtonIf the clock in the diagram above is a wrist watch, even a 1.09mA current may overload its alarmcircuitry. In that case, you may use two transistors as shown in the diagram below. This is called adarlington.

A darlington can be considered as a single transistor with the following characteristics:

VBE,darlington=VBE1+VBE2

IB,darlington=IB1. IC1= hFE1∙IB1. IC2=hFE2∙IB2=hFE2∙IE1=hFE2∙(hFE1+1)∙IB1.IC,darlington=IC1+IC2=hFE1∙IB1+hFE2∙(hFE1+1)∙IB1=(hFE1+(hFE1+1)∙hFE2)∙IB1. hFE,darlington=IC,darlington/IB,darlington=hFE1+(hFE1+1)∙hFE2. hFE1>>1 => hFE,darlington=hFE1+hFE1∙hFE2. hFE1∙hFE2>>hFE1 =>hFE,darlington=hFE1∙hFE2.

Let's now recalculate a proper value for RB. Assume T1's current gain is 300 and T2's current gain is100.

IC,darlington = 9/100=90mA. hFE,darlington=300∙100=30000. VBE,darlington=1.2V.

The voltage across RB equals 3V-1.2V=1.8V => RB < 1.8V/3µA = 600k. 560k is a safe value.IB,darlington will be 1.8/560k = 3.21µA.

The transistor as an amplifierAs discussed in the first part of this chapter, a transistor is an excellent amplifier. The picture belowshows an example.

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The input signal is connected to the amplifier via C1. C1 prevents a DC current flow in R1 and theinput signal source, e.g. a microphone. DC currents may destroy the microphone (unless it's anelectret; a type of microphone with a built-in amplifier).

The characteristics of transistor T1 are: hFE=100 and VBE=0.6V.

Assume we want to connect an end amplifier with a 10k input resistance to the OUT terminal. Formaximum power transfer, the output resistance of our amplifier must be equal to the input resistanceof the end amplifier. The output impedance of an amplifier is defined as vOUT/iOUT. In our case vOUT= vRC and iOUT = iRC. So the output impedance of this amplifier is vRC/iRC = RC. So RC = 10k. Forstability reasons VRE must be VS/5. Since VS = 9V, VRE must be 1.8V. This means that the voltage atthe OUT terminal can vary between 1.8 and 9V. So the maximum AC output voltage (vOUT,max) is 9-1.8=7.2Vtt. Obviously, this can only be arranged if the quiescent output voltage (when vIN=0) is exactlybetween 1.8 and 9V. This means VOUT=1.8+(9-1.8)/2=5.4V. We already know that RC=10k, so IC = (9-5.4V)/10k = 0.36mA. IE will also be 0.36mA, so RE=VRE/IE=1.8/0.36m=5k.

VB = VBE + VRE. VBE is always 0.6V, so vRE = vB. (Remember: AC voltages and currents are written inlower case letters.) When C1 is large enough, vIN = vB = vRE. VOUT = VS - VRC = 9V - VRC => vOUT = -vRC.

Knowing this, we can calculate the gain A of the amplifier, which is defined as: A = vOUT/vIN = -vRC/vRE=-(iC∙RC)/(iE∙RE). Since iC=iE (hFE is large enough to neglect iB), A=-(iC∙RC)/(iC∙RE) = -RC/RE.This means our amplifier's gain is -10k/5k =-2.

VR1 = VS-VBE-VRE = 9-0.6-1.8 = 6.6V. IR1=IC/hFE=0.36mA/100=3.6μA. R1 = 6.6V/3.6μA = 1.8M.

Unfortunately, transistors with the same type designation can have a wide range of hFE. For example,the hFE of a 2N3904 transistor ranges from 100 thru 300. The question is: will our amplifier stillfunction properly if hFE = 300? Let's see...

IB = (VS-VBE-VRE)/R1. VRE = IC∙RC = hFE∙IR1∙RE = hFE∙RE∙(VS-VBE-VRE)/R1 = hFE∙RE∙(VS-VBE)/R1 -hFE∙RE∙VRE/R1 =>

VRE+(hFE∙RE/R1)∙VRE = hFE∙RE∙(VS-VBE)/R1 => (1+300∙5k/1.8M)∙VRE = 300∙5k∙8.4/1.8M =>1.833∙VRE=7 => VRE = 7/1.833 = 3.8V.

IB = (8.4-3.8)/1.8M = 2.6μA. IC = 300∙2.6μA = 0.767mA. However, IC,max = VS/(RC+RE)=9/15k=0.6mA. So hFE∙IB > IC,max, which means that the transistor is saturated and thus acts like a closedswitch!

Solution: add an extra resistor which makes VRE (and therefore IC) independent of hFE:

Make sure IR1 >> IB => IR1≈IR2. Let's estimate proper values for R1 and R2.

VR2 = VBE+VRE = 0.6+1.8V = 2.4V. VR1 = VS-VR2 = 9-2.4 = 6.6V. So R1:R2=6.6:2.4. E.g. R1=33k andR2=12k. In that case IR1(=IR2) = 6.6V/33k = 0.2mA, which is much larger than IB.

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As mentioned before, the voltage gain of this amplifier is just (-)2. In many cases that will not beenough. You can easily increase the gain by adding an extra resistor and capacitor as shown in thepicture below: the most common transistor amplifier.

Capacitor C2 shorts RE2 for AC voltages. So for DC signals, RE = RE1 + RE2, and for AC signals,RE = RE1. If RE1 = 500Ω and RE2 = 4.5k, we have an amplifier with the same characteristics asabove, but the gain is 10k/500=20.

The impedance of C2 must be much smaller than RE1:

1/(2∙π∙fmin∙C2) « RE1 => C2 » 1/(2∙π∙fmin∙RE1) where fmin is the lowest frequency the amplifier mustbe able to handle.

For example: if fmin = 20Hz, C2 » 1/(2∙π∙20∙500) = 16μF. 47 or 100μF is a good choice.

The AC input resistance of the amplifier is approximately R1//R2 = 8.8k. So the impedance of C1 mustbe much less than 8.8k => C1 » 1/(2∙π∙20∙8.8k) = 0.9μF. 10μF is a good choose. The positive terminalof C1 must be connected to the amplifier, unless the input signal's DC component is larger than 2.4V.

Testing transistors using a multimeterMost digital multimeters look like this:


If your meter has a transistor socket, set the function switch to hFE and simply insert the transistorleads into the proper holes of the socket. To determine the type (NPN/PNP) and to locate the B, C andE leads, use the transistor's datasheet. The display will show the transistor's current gain (hFE).

If your meter doesn't have a hFE test, you can at least test the BE and CB diodes using the diode test.

http://hobby-electronics.info/links/icsearch.php

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Later in this course we'll build a nice device for testing transistors (and diodes).

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Chapter 7. Project: A simpleadjustable DC power supplyThe transformerIn this project we use all the components we've learned about in the previous lessons. The only newcomponent is a transformer. A transformer transforms high voltages to low voltages (or vice versa).It basically consists of two coils of wire wrapped around a soft-iron core. When you connect one ofthe coils to an AC voltage source, it produces an alternating magnetic field in the soft-iron core. Thismagnetic field also flows in the core of the second coil. This causes an alternating current flow in thesecond coil.

The coil connected to the source is called the primary coil; the second coil, connected to the load, iscalled the secondary coil.

The voltage ratio is equal to ratio of the number of turns of each coil:

vs:vp=Ns:Np

The current ratio is equal to: is:ip=Np:Ns

The diagramLet's take a look on the diagram below.

On the left, you see the aforementioned transformer. In this case, the transformer's output voltageis 15V AC. This AC voltage is rectified by bridge rectifier G1. The rectified voltage is smoothed bycapacitor C1. Without C1, the output would be just a rectified sine wave signal. If you would poweryour walkman with this voltage, you would hear a terrible 100Hz humming. The result of a computersimulation below shows what C1 does.

In this picture you see two signals: a rectified sine wave (the situation without C1), and the situationwith C1. At t=0, C1 is discharged, so VC1=0V. G1 will then charge C1 until t=T1. The top value of the

Project: A simple adjustableDC power supply

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rectified signal is 15V∙√2 - 2∙0.6V = 20V. After t=T1 C1 will be discharged by the load until t=T2. Then,everything will start all over again.

Vr is called the ripple voltage. Especially in audio equipment it should be as small as possible,because voltage ripple in the power lines means voltage ripple in the sound signals! You can reducethe ripple voltage by using larger capacitors. You can calaculate the value of the capacitor, but youcan also use the rule of thumb: 2000...5000uF (2...5mF) per ampere load current.

You might think that the maximum voltage across C1 will be 20V. But this is only true if the secondaryvoltage of the transformer is 15V. Unfortunately, this voltage depends on the load. The open linevoltage may be 18V or even more! Take this into consideration when buying a capacitor for C1, sinceall capacitors have a maximum voltage they can sustain.

The load-dependency of the output voltage of the transformer also explains the presence of zenerdiode D1. Without D1 the output voltage would depend on the load, and that is something we don'twant. Thanks to D1, the voltage across P1 and R2 is always 12V. So the voltage at the base of T1only depends on the position of P1. With P1 turned to the maximum position, VB=12V. The outputvoltage VE will be 12V-0.6V=11.4V. With P1 turned all the way down, VB=VR2. VE=VR2-0.6V. Of coursewe want VE to be 0V, so VR2=0.6V. We can now calculate R2: R2:P1=VR2:VP1=0.6:11.4, so R2=(0.6/11.4)∙10k = 526Ω. 470Ω is a good choise.

If we want 5mA current flow in D1, IR1=5mA+VD1/(P1+R2)=5mA+12/10470=6.15mA. R1=(Vtop-VD1)/IR1=(20-12)/6.15mA=1.3k. 1.2k is a good choise. This calculation assumes that T1's base current isvery small. And it should be, because a large current will cause a high voltage drop across the toppart of P1. This will reduce the base voltage and this the output voltage. And since the base currentdepends on the output current, the output voltage would depend on the output current. And we don'twant that. You may need to replace T1 with a darlington.

Zener diode D1 has another advantage: a small ripple voltage across C1 does not appear on theoutput terminals. Even if Vr=5V, the voltage across C1 will never drop below 15V and VD1 remains12V. This means that you can create a ripple free power supply without coffee table size capacitors!However, don't make C1 too small. Charging C1 causes a high current flow in the transformer. If C1 issmall, charging takes a relatively long time and might overheat the transformer.

When VC1=15V, IR1=(15-12)/1200 = 2.5mA. This leaves 2.5-1.15 = 1.35mA for D1. And this mightnot be enough for D1 to operate properly. If we want ID1 to be 5mA (and thus IR1 6.15mA), thenR1=(15-12)/6.15m = 488Ω. So R1 should be replaced with a 470Ω resistor. Let's now calculate thecurrent when VC1 reaches its top value (20V). IR1=(20-12)/470 = 17mA. So ID1=17-1.15 = 15.35mA.PD1=12∙17m=0.2W. So a 0.4W zener will survive.

Please keep in mind that the top value is only 20V when the transformer voltage is 15V. We alreadysaw that this voltage depends on the load current, and that the open line voltage can be 18V ormore. Always consider the worst-case scenario. Assume we measure an 18V open line voltage. Add10% to this, just to be safe. So we assume a top value of 20V∙√2 - 2∙0.6V = 27V. IR1=(27-12)/470= 32mA. PR1=(27-12)∙32m = 0.48W. R1 must therefore be a 1W resistor. ID1=32-1.15 = 30.85mA.PD1=12∙30.85m=0.37W. Although this is just below 0.4, D1 should better be replaced with a higherwattage zener.

Another method is enlarging C1, making the ripple voltage smaller. Assume that the ripple voltage hasbeen reduced to 2V. VC1,min is now 20-2=18V. R1=(18-12)/6.15m = 976Ω. We'll take an 820Ω resistor.When VC1=27V, IR1=(27-12)/820 = 18.3mA. PR1=(27-12)∙18.3m = 0.27W. An 1/3 of 1/2W resistor willnow be sufficient. ID1=18.3-1.15 = 17.15mA. PD1=12∙17.15m=0.21W. A 0.4W zener can handle thisvery easily.

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Chapter 8. Differential AmplifierTypical example

In the picture above, you see a typical schematic of a differential amplifier.

The DC current source I1 provides a continuous 1mA current flow.

Transistors T1 and T2 have the same electrical characteristics, e.g. hFE1=hFE2=100. Therefore, thequiescent emitter currents of T1 and T2 are the same: IE1=IE2=0.5mA. The voltage across R1 (andR2) will be: VR1=10k∙0.5mA=5V. So the voltage at the OUT terminal equals V1-VR2=9V-5V=4V.

If we inject a 1uA current in IN1, IE1 will raise by 1uA∙100=0.1mA, so IE1=0.6mA and IE2 willbe 0.4mA since the sum must be 1mA. VR2=0.4mA∙10k=4V and VOUT=9V-4V=5V. So we canwrite down the following (DC) formula for VOUT: VOUT=V1-VR2=9V-(0.5mA-IIN1∙hFE)∙R2=9V-5V+IIN1∙hFE∙R2=4V+IIN1∙hFE∙R2. When we omit the DC component we get this AC equation:vOUT=iIN1∙hFE∙R2.

Of course, we can also inject a current in IN2, resulting in: vOUT=-iIN2∙hFE∙R2 (Note the minus sign).Combine both equations and you get:

vOUT=(iIN1-iIN2)∙hFE∙R2

Hence the name differential amplifier.

All so called operational amplifiers are based on a differential amplifier. We'll take a closer lookon operational amplifiers in the next chapter. First, we have to find out how to create a DC currentsource.

DC Current Source

If you actually want to build a differential amplifier using the diagram in the previous section, you havea problem. You can buy the resistors, transistors and the battery in any shop, but where to buy a DCcurrent source? The answer is simple. Don't go looking for it, because no one sells one. We have tocreate one ourselves.

There are several ways to do this. We'll use a so called current mirror.

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Again, both transistors have the same characteristics. VR3=9V-0.6=8.4V, so IR3=8.4V/8.2k=1.02mA.Since the transistors have the same characteristics and VBE1=VBE2, both collector currents must bethe same: IC3=IC4=IR3=1.02mA.

You can obtain any current flow you like by adjusting R3.

To use the current source in the differential amplifier, connect the collector of T3 to the emitters of T1and T2.

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Chapter 9. Operational AmplifierIntroduction

The picture above shows the symbol on an operational amplifier, or opamp. Opamps are differentialamplifiers with a very large gain: VOUT=(VIN1-VIN2)∙A. Gain A is usually greater than 100000. Thismeans that VIN1-VIN2 must be very small: even 1mV would result in an output voltage of more than100V, which of course is impossible, because the source voltage is just 18V. In the next section we'llsee that the gain can be reduced by using resistors.

The input resistance of both input terminals is very high: »1GΩ. This means that there will be nocurrent flow in the input terminals of an opamp.

IN1 is called the non-invering input; IN2 is called the inverting input, because its signal is inverted: ifVIN2 increases, VOUT decreases and vice versa. The voltage at the non-inverting input is called Vp; thevoltage at the inverting input is called Vn. So in this case: Vp=VIN1 and Vn=VIN2.

Opamp used as an amplifierIn this section we'll learn the two easiest ways to reduce the huge gain of an opamp. First, we take alook at the non-inverting amplifier:

R1 and R2 make a voltage divider: Vn=VOUT∙R1/(R1+R2), so

VOUT=(Vp-Vn)∙A = (VIN-VOUT∙R1/(R1+R2))∙A = VIN∙A-VOUT∙A∙R1/(R1+R2) =>

VOUT+VOUT∙A∙R1/(R1+R2)=VIN∙A.

Since A is nearly infinite, term VOUT is negigible, so

VOUT∙A∙R1/(R1+R2)=VIN∙A =>

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VOUT=VIN∙(R1+R2)/R1

Remember the first equation of this section? Vn=VOUT∙R1/(R1+R2).

This is equal to: VOUT=Vn∙(R1+R2)/R1. And we've just proved that VOUT=VIN∙(R1+R2)/R1. This meansthat VIN=Vn, and since VIN=Vp, Vp must be equal to Vn!

This is always the case.

If an opamp is used as an amplifier: Vp=Vn

Note: the gain of this non-inverting amplifier is always greater than or equal to 1.

To create an amplifier with a gain less than 1 (an attenuator), we connect the input signal to R1,creating an inverting amplifier:

Since Vp is 0 and Vp=Vn, Vn=0. This means VOUT=VR2.

VR2=(VOUT-VIN)∙R2/(R1+R2) => VOUT=(VOUT-VIN)∙R2/(R1+R2) =>

VOUT=VOUT∙R2/(R1+R2)-VIN∙R2/(R1+R2) => VOUT-VOUT∙R2/(R1+R2)=-VIN∙R2/(R1+R2) =>

VOUT∙R2/(R1+R2)-VOUT=VIN∙R2/(R1+R2) => (R2/(R1+R2)-1)∙VOUT=VIN∙R2/(R1+R2) =>

-R1/(R1+R2)∙VOUT=VIN∙R2/(R1+R2) => -R1VOUT=R2∙VIN =>

VOUT=-(R2/R1)VIN

Note the minus sign: the input signal is inverted.

This amplifier becomes an attenuator if R1>R2.

Opamp used as a threshold switchSince VOUT=(Vp-Vn)∙A and A is infinite, it's easy to see that an opamp can be used as a thresholdswitch: if Vp>Vn then VOUT equals the positive source voltage; if Vp<Vn then VOUT equals the negativesource voltage. Take a look at the picture below.


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Assume P1 is in the middle position. Let's call the top section P1a and the bottom section P1b. In themiddle position P1a=P1b=5k => Vp=4.5V. If R1>R2, Vn<Vp and the lamp will be turned on. If R1<R2,Vn>Vp and the lamp will be turned off.

If R1 is a light dependent resistor (LDR), the circuit becomes a switch that turns the lamp onautomatically when it gets dark. LDRs have a high resistance in the dark and a lower resistancewhen light shines on it. So in the dark Vn<Vp and the lamp will be turned on. At dawn, Vn will becomegreater than Vp and the lamp will be turned off. Use P1 to change the threshold.

But what will happen if Vn=Vp? In that case, the lamp might switch on and off rapidly. Fortunately wecan prevent this by adding one resistor:

Feedback resistor R3 makes Vp dependent on the state of the switch. Assume again that P1 is inthe middle position. When the lamp is turned on, VOUT=9V, so R3 can be considered as parallel-connected to the top section of P1 (P1a): Vp=9V∙P1b/(P1b+[P1a//R3]) = 9V∙5k/(5k+[5k//10k]) = 5.4V.So the lamp will only be turned off when Vn>5.4V. In that case, VOUT=0V, so R3 can be consideredas parallel-connected to P1b: Vp=9V∙[P1b//R3]/(P1a+[P1b//R3]) = 9V∙[5k//10k]/(5k+[5k//10k]) = 3.6V.So the lamp will only be turned on again when Vn<3.6V. The difference 5.4V-3.6V=1.8V is called thehysteresis.

Opamp used as a voltage-controlled current source

The current flow in the collector and emitter will be the same. The current in the lamp and R1 willtherefore also be the same. The voltage across R1 depends on the current flowing in it; this voltagewill therefore also depend on the current flow in the lamp.

The opamp will try to keep Vn=Vp. This means that the current flow though the lamp depends on thevoltage at the non-inverting input of the opamp. Because R1 is 1 ohm, a 1V voltage will cause a 1Acurrent flow. The voltage across R1 will also be 1V, which leaves 11V for the lamp. R1 dissipates 1W.


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Diode and transistor tester

This gadget allows us to test diodes and transistors very easily. When testing diodes, it tells us whichterminal is the cathode and for transistors it shows whether it's an NPN- or PNP-transistor.

The circuitry around opamp U1.A make an oscillator. Directly after connecting the power supply, C1is still empty and thus the voltage across it will be 0V. The output will carry the supply voltage. Theoutput is also said to be 'high'. C1 is charged via R4. When the voltage over C1 exceeds the voltageat pin 3, the output voltage will be 0V, or 'low'. C1 will now discharge via R4 until the voltage is lessthan the voltage at pin 3. Now everything starts all over again. Resistor R3 provides the hysteresis.

U1.B's invertering input is connected to U1.A's non-inverting input and vice versa. So the voltages atpin 7 and at pin 1 are in antiphase. That is: if pin 1 is 'high', pin 7 is 'low' and vice versa.

Next, we connect our test diode to pins 1 and 3 of connector J1, making sure the cathode isconnected to pin 1. If U1.B's output is 'high' (and U1.A's output is 'low'), a current will flow from U1.Bvia the test diode, D2 and R7 to U1.A. LED D2 comes on. After the oscillator flipped the output of bothopamps, there is no path for the current to flow. After all, the test diode is in reverse. So actually D2blinks, but the frequency is so high that you won't notice that. When we swap the terminals of our testdiode, only D1lights. So the LEDs show us which side of the diode is the cathode. They also showif the diode is working properly: if both LEDs light, the test diode apparently conducts current in bothdirections. If neither of the LEDs light, the diode blocks current flow in both directions and we cantrash it as well.

Let's now try connecting a NPN-transistor to J1. The diagram above shows where the collector, baseand emitter go. If U1.A is 'high', current will flow via R5, the base and emitter of our transistor to U1.Bwhich will be 'low' of course. If the transistor functions properly, it will now switch on, allowing currentto flow from U1.A, via R6, D1, the collector and emitter of the transistor to U1.B. So, in case of a NPN-transistor, D1 comes on. If we connect a PNP-transistor D2 lights. And again: if both LEDs come on ofremain off, the transistor is busted.

To make it easier for you to assemble this nice device, a PCB has been disigned. The design can bedownloaded in JPEG, EPS and HPGL format.

The picture below shows which component goes where.

x_dttester-pcb.jpg

x_dttester-pcb.eps

x_dttester-pcb.plt


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J1 is a 5 pin socket. Of course 3 pins are enough to test a transistor, but this design allows us to testany transistor no matter if the base, collector or emitter is in the center. If we use a 3 pin socked, wemay need to bend the terminals of some transistors, which may result in shorts.

The power supply can be a 9V block battery of 4 AA batteries in series.

33

Chapter 10. Lab Power SupplyThe diagram

This is the schematic of the 30V/2A power supply I use in my own lab. Click here to see the fullsize image. It may look very complex, but it really isn't very difficult to understand: it uses only theknowledge we've learned in the previous lessons.

The top part looks like the power supply we built in a previous lesson: the transformer L1 transformsthe outlet voltage to a safe 30V, which is rectified by bridge rectifier G2 and smoothed by capacitorC5. Transistors T3 and T6 form a darlington transistor. This darlington replaces transistor T1 inLesson 7. However, the base voltage is not controlled by a simple potmeter, but by an 'electronicpotmeter' with voltage feedback. The advantage of this feedback is a load-independent outputvoltage.

Voltage feedbackImagine you connect your 'Lesson 7' power supply to a device that switches a lamp on and off. Thisdevice requires 4.75 ... 5.25V supply voltage to operate correctly. When the lamp is off, it draws just5mA. When the lamp is on, the current flow rises to, say, 1A. You connect the device to the powersupply with wires that have a total resistance of 1Ω. You use a volt meter to adjust P2 to an outputvoltage of 5V. After a while, the lamp switches on. The voltage drop over the wires will now be 1V,leaving just 4V for the device. This is not enough and the device shuts down, turning off the lamp.The device now receives 5V again and the lamp switches on. The supply voltage drops to 4V, and thelamp goes off, and so on, and so on...

The remedy for this is eighter using very thick wires or voltage feedback. When you want topermanently connect the device to a power supply, the first option is probably the best, especially

x_lps-l.jpg

x_lps-l.jpg

Lab Power Supply

34

when the wires are short. In this chapter however, we want to build a lab power supply. You mayalready have noticed R14. This is a 0.5Ω current sense resistor which will be discussed later in thischapter. This means that even if we use thick wires, the voltage drop will always be at least 0.5V perampere.

In the schematic above the feedback terminals are FB+ and FBGND. R19 and R16 devide the actualoutput voltage by 2 and R13 feeds it back to the non-inverting pin of opamp U2. The inverting pinof U2 is connected to P2 via R17. U3 is a cheap 15V voltage stabalizer. To keep it as stable aspossible, it is fed by separate transformer winding (or a separate transformer), bridge rectifier (G1)and smoothing electrolytic (C3). So Vp=0.5∙VOUT and Vn ranges from 0 ... 15V. Lesson 9 tought usthat if Vp > Vn, U2's output is 15V. T5 will be saturated and the base voltage of the driver darlingtomT3/T6 will be 0V. The load at the output terminals will discharge C2 until Vp < Vn. At that moment, U2'soutput will become 0V, T5 opens, and a current starts to flow in R4 and the bases of T3 and T6. Thiscurrent will saturate the darlington. A large current will charge C2 and feed the load. This will continueuntil Vp > Vn. At that point everything starts all over again. Again we see that opamp U2 tries to keepVp equal to Vn, just like an opamp amplifier. In matter of fact, U2 is used as an amplifier: it amplifiesthe voltages over P2 (=Vn) 2 times (because R19 and R16 devide the output voltage by 2). Since Vnranges from 0 ... 15V, VOUT will range from 0 ... 30V.

Without R21 and R22, we should not turn on the power supply until FB+ and FBGND were bothconnected! If FB+ and FBGND were disconnected, U2 might have thought that the output voltagewere too low, and a voltage of 40V or more might have appeared across the output terminals!

When T5 closes, there may still be up to 30V across C2. Since The base voltage of T3 is 0V, there willbe -30V between the base of T3 and the emitter of T6. However, T6 cannot withstand voltages lessthan -7V. For T3, VBE must be greater than -60V. R5 and R20 devide the -30V, so that T6 will survive.

Current limitationWhen the output current increases, the voltage across R14 will also increase. This voltage is amplifiedby the cicuitry around opamp U1. The gain can be controlled by potmeter P1. When U1's outputvoltage exceeds 0.6V, T2 switches on. This causes a current flow in T1, which also switches on. Thecurrent flow in R3 switches on T5 and the voltage at the output terminals will drop to 0V. Of course,the current flow will also be 0A, and VR14 will also be 0V. However, T1 also feeds T2 via R1, so theoutput voltage remains 0V until switch SW1 is closed.

WARNING: remove the load prior to closing SW1. The overload protection does not work as long asSW1 is closed!

You may wonder why the voltage across R14 is first devided by R9 and R10 and then amplified byU1. In the the worst case scenario, the OUT terminal is shorted to the GND terminal. This means thatthe full output voltage will be across R14. This voltage can be up to 30V. However the supply voltageof U1 is just 15V. If the non-inverting pin were directly connected to the GND terminal, opamp U1would be destroyed, because the input voltage of an opamp should never exceed its supply voltage.R9 and R10 make sure that Vp never exceeds 15V.

Capacitor C1 prevents current spikes triggering the overload protector.

Protective circuitryWhen you switch off the power supply, C5 and C3 will remain charged for a certain period of time.Since C5 is much larger than C3, it is very likely that C3 will be empty while C5 is still charged. If C3 isempty, U2 ceases to function, T5 will be open and the full voltage across C5 will appear on the outputterminals! The protective circuitry around T4 will prevent this. While U2 receives its supply voltageT4 remains closed and everything works as it should. However, when U2 loses its supply voltage, T4opens. In that case, T5 closes due to a current flow in R18 and D5. This will make sure that the outputvoltage remains 0V.

Lab Power Supply

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Choosing your components

When you want to build this power supply, you may encounter some difficulties while buying thecomponents. Or maybe you want to build one that has different characteristics.

For transformer L1, I use a 30V transformer with an extra 20V winding. If you can't find a transformerlike this, you can also use 2 transformers.

The maximum voltage across C5 will be 30V∙√2 - 1.4 = 41V. So I choose for G2 a B80C5000/3300.(80 = maximum voltage; 5000 = maximum peak current [mA]; 3300 = maximum continuous current[mA]). C5 should be able to sustain 50V.

The maximum power dissipation in R4 is 412/3k3 = 0.51W. Usually, you take the next value available,in this case 1W. However, in this case the voltage across R4 will not continuously be 41V, but just forshort periods of time. So I took a 0.5W resistor.

T6 is a high power transistor. Make sure you cool it using a proper heatsink. I also attached T6 to themetal case of my power supply. Its minimum current gain is 20, so the maximum current in T3 is 2A/20 = 0.10A. The maximum power dissipation in T3 is 41V∙0.10A=4.1W. According to the datasheet,the maximum power dissipation without heatsink is just 2W, so this transistor needs a little heatsink.N.B. Do NOT use a TIP41, since its VCE,max = 40V. Use a TIP41A, B or C.

T1 can be any PNP transistor. T2 can be any NPN transistor. T4 and T5 can be any transistor whereVCE,max >= 45V. So American users should take care when applying a 2N3904 here; the maximumVCE of this transistor is 40V. If anyone knows a good American replacement for it, please let me know.(I'm not sure if the BC-series are available in America.)

The maximum power dissipation in R14 is 32∙0.5 = 4.5W. The next available value is 5W.

For opamps U1 and U2 I used a CA3140. Do not use a cheap 741 or so; these are not suitable forthis job.

If you want more than 30V output voltage, you need to change more than just the transformer.You also need higher voltage versions of C2, C5, T4 and T5. You also need to change the voltagedeviders R19/R16 and (maybe) R9/R10. For example, if you want to build a 40V power supply, C2 =10uF/50V and C5 = 10000uF/63V. T4 and T5 should be replaced with a BC546. Without any change,the maximum voltage on the non-inverting pin of U1 is 40V/2.8 = 14.2V. Although this is less than 15V,you'd better replace R10 with a 2k2 resistor. When VFB+ = 40V, Vp,U1 should (still) be 15V. This meansthat R19/R16 should devide the voltage by 40/15 = 2.67: R19 = 56k and R16 = 33k.

If you need less than 30V, you only need a lower voltage transformer and different values for R19 andR16. For example, if VOUT,max = 20V, R19/16 should devide the feeback voltage by 20/15 = 1.33: R19= 18k and R16 = 56k.

Assembly

You can download the PCB (printed circuit board) in several formats here: JPEG, EPS and HPGL.There are many ways to create a PCB from a layout, but that's beyond the scope of this course.

The component layout looks like this:

http://hobby-electronics.info/mail.php

x_lps-pcb.jpg

x_lps-pcb.eps

x_lps-pcb.plt

Lab Power Supply

36

Components not mentioned on the board are: transformer L1, bridge rectifier G2, capacitor C5, C2and diode D1. These are mounted in the cabinet. The positive terminal of C5 is connected to pinVSIN on the board, and its negative terminal is connected to the VSGND pin. C2 and D1 are directlysoldered to the output terminals.

The 20V winding of the transformer is connected to the AC20A and AC20B pins on the PCB.

Transistor T6, potmeters P1 and P2, switch SW1 and LED D1 are mentioned on the PCB, but notmounted on it. I mounted T6 on the metal back pannel of the cabinet (electrically insulated!) and usedthree pieces of wire to connect it to the PCB. The other components are mounted on the front panel.

Pins OUT and GND are the output terminals of the power supply. FB+ and FBGND are the feedbackpins. I soldered them directly to the OUT and GND terminals.

It is highly recommended to use IC sockets for U1 and U2: soldering out a defective chip can be quitea pain.

When you're carefull not to mount diodes and electro's the other way around, you should have noproblems on this project. And if it doesn't work immediately, you have enough knowledge to deducethe problem!

Good luck!

Lab Power Supply

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A picture of the power supply from one of the course members.

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Chapter 11. HeatsinkIntroductionHeatsinks are used to prevent transistors from getting too hot. Air and plastic are a very bad heatconductors; the heat a transistor produces is not easily transferred to the surrounding air or plastic.Metal however is a very good heat conductor. That's why small power transistors have a small metalplate and large power transistors even have a full metal case. The larger the metal surface, the easierthe produced heat is transferred to the ambient. And that is exacly what a heatsink does: enlargingthe metal surface of a transistor.

This chapter describes how to calculate the size of the heatsink you need.

CalculationsThe ability to transfer heat is called the thermal conductance, but we always use the reciprocal value:the heat resistance. The thermal resistance (Rth) is measured in °C/W. If the Rth of a heatsink is 1°C/W, the temperature will rise 1°C per Watt power dissipation. So:

R=T/P

Doesn't this folmula look familiar? Yep, if you replace T with V and P with I, you get the formula forelectrical resistance. This anology makes it very easy to calculate the heak sink you need: replaceall heat producers with current sources and temperatures with voltages. Take a look at the diagrambelow.

This diagram shows transistor T6 of the lab power supply, a 2N3055 producing 60W at its junction.(The junction is the silicon chip inside the transistor.) The thermal resistance from junction to caseis 1.5°C/W. The ambient temperature is 25°C. Resistor RTHC-A is the thermal resistance of theheatsink. This is the resistance we want to calculate.

The maximum junction temperature of the 2N3055 is 200°C. The 'temperature drop' across RTHJ-Cis 60W∙1.5°C/W = 90°C. That leaves 200 - 90 - 25 = 85°C for the heatsink. So RTHC-A = 85°C/60W =1.4°C/W.

Let's now calculate the heatsink for T3, a TIP41A. The maximum junction temperature is 150°C. Thedatasheet says: when you keep the case temperature to 25°C, the transistor can dissipate 65W. Thismeans that RTHJ-C = 125°C/65W = 1,9°C/W. The transistor dissipates 4.1W, so the total resistancefrom junction to ambient may not exceed 125°C/4.1W = 30.5°C/W. So the thermal resistance of theheatsink should not exceed 30.5 - 1.9 = 28.6°C/W. A very small heatsink will suffice.

Important notes:

• When you attach a heatsink to a transistor, there will always be air pockets between the two metalplates. And air is a bad heat conductor. To reduce the thermal resistance between the transistorcase and the heatsink, you should use special heat conducting paste (heatsink compound).

• Bear in mind that one of the pins (base, collector or emitter) will be internally connecteded tothe metal case of a transistor. If you want to mount two transistors on one heatsink, make sureyou don't create shortcuts. For example, both the TIP41A and the 2N3055 have their collectors

Heatsink

39

connected to the case. In our lab supply, both collectors are connected anyway, so you can mountthese transistors on the same heatsink. But obviously, this is not always the case! Fortunately,you can buy insulators to put between the transistor and the heatsink. Note however that this willintroduce extra thermal resistance!

• In some catalogs you may find thermal resistances measured in K/W (Kelvin per Watt). This is thesame as °C/W: 1K/W = 1°C/W.

• Transistor datasheets can be found on the internet.

http://hobby-electronics.info/links/icsearch.php

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Chapter 12. Power AmplifiersIntroduction

During this course, we've already learned how to amplify small voltages using transistors andoperational amplifiers. The maximum output current of these amplifiers was very low and thereforenot suitable for driving a loudspeaker. In this lesson we will learn how to increase the maximum outputcurrent.

Emitter follower

We already know that transistors are made for this job. Take a look at the picture below.

RL is an 8Ω loudspeaker. VBE does not depend on the AC input voltage; it's always about 0.6V. Thismeans that for AC voltages vE = vB => vBE = 0 => vRL = vIN. Since the AC emitter voltage 'follows'the base voltage, this kind of amplifier is also called an 'emitter follower'. The picture next to theschematic above shows the output signal. As you can see, this amplifier works only for the positivehalf period of the input signal. During the negative half period, the base voltage drops below 0V, andT1 can be considered as an open switch. If we replace T1 with a PNP transistor (and connect itscollector to -15V), the amplifier works only for the negative half period of the input signal. So we mustcombine these two.

Balance amplifier

Ah, that's much better. But what's that horrible distortion at ground level? When vIN = 0V, VB1 = VB2 =0V (assumed R1 = R2). Since VRL is also 0V, VBE1 = VBE2 = 0V. This means that when VIN is between-0.6 and 0.6V, T1 and T2 both remain open. This usually happens when vIN crosses the 0V line.

Power Amplifiers

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That's why it is called 'cross-over distortion'. To get rid of this, we must make sure that the quiescentvoltages VBE1 = 0.6V and VBE2 = -0.6V. This can be simply accomplished by adding two diodes:

Bias

This looks much better. Let's now calculate the proper values for R1 and R2. Assume RL is an 8Wloudspeaker, VBE = 0.7V and hFE = 20. So u2

rms/RL = 8W => u2rms = 8W∙8Ω => urms = 8V => utop =

8V∙√2 = 11.3V => VRL,max = VE1,max = 11.3V => VB1,max = 11.3 + 0.7 = 12V => VR1 = VS - VB1 = 15 -12 = 3V. IRL,max = VRL,max/RL = 11.3/8 = 1.4A => IE1,max = 1.4A. IB1,max = IE1,max/(hFE+1) = 1.4A/21 =67mA = IR1 (IB»ID1). R1 = VR1/IR1 = 3V/67mA = 45Ω.

The quiescent diode current will be (2∙15V - 2∙0.7V)/(2∙45Ω) = 302mA. The power dissipation in R1and R2 will be I2∙R = 3.6W each! Needless to say, this is not a very economical amplifier. And tomake things even worse: every diode has a certain resistance. And even if this resistance is just 5Ω, itresults in a 302mA∙5Ω = 1.5V voltage drop. This means that the voltage across the diodes increasesfrom 0.7V to 2.2V! This means that VBE1 and VBE2 also increase. And if VBE increases, IC will alsoincrease. Result: the quiescent collector current will be very large!

Of course we could replace the diodes with resistors, making sure VBE will always be 0.7V. However,diodes have a major advantage: thermal stability. Transistor and diode parameters are temperature-dependent. If you keep VBE (or VD) constant, IC (ID) will increase with temperature. And vise versa:if you keep IC constant, VBE decreases with 2mV/°C. So if you use resistors to keep VBE 0.7V, anincrease in temperture will cause IC to rise. This may heat up the transistor, so IC increases evenmore, and so on... This is called thermal runaway. However, if you use diodes instead of resistorsand attach the diodes to the transistors (to make sure they have the same temperature) there will beno thermal runaway: if the transistor heats up, the diode also heats up, reducing VBE preventing andincrease of IC.

The only way to reduce the power dissipation in R1 and R2 and the voltage drop across D1 and D2,is to reduce the (quiescent) current flow in these components. And that is only possible with larger hFEvalues. So we need darlingtons!

Power Amplifiers

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Darlingtons

Assume hFE3 = hFE4 = 100. This means that the total hFE of each darlington is 2000. IRL,max = 1.4A =>IE1,max = 1.4A. IB1,max = IE1,max/(hFE+1) = 1.4A/2001 = 0.7mA = IR1. VB1,max = 11.3 + 2∙0.7 = 12.7V =>VR1 = VS - VB1 = 15 - 12.7 = 2.3V. R1 = VR1/IR1 = 2.3V/0.7mA = 3.3kΩ.

Let's now calculate the maximum power dissipation in T1 and T2. PT1 = VCE1∙IE1 = (VS - VRL)∙(VRL/

RL) = (VS∙VRL - V2RL)/RL. PT1 reaches its maximum value if dPT1/dVRL = VS - 2∙VRL = 0 => VRL = VS/

2 = 7.5V. PT1,max = (15∙7.5 - 7.52)/8 = 7W. Since T1 works only during the positive half periods, T1's(and T2's) maximum dissipation will be 3.5W.

Opamp

If you want to build an operational amplifier and a power amplifier into one device, you can do it likethis:

Power Amplifiers

43

You don't need diodes or resistors anymore to prevent cross-over distortion; the opamp takescare of this. Since the opamp tries to keep Vp equal to Vn, the voltage gain is 2. Variable resistor(potentiometer) P1 allows you to adjust the quiescent output voltage to exactly 0V.

An appropriate symmetric power supply can be found in another chapter.

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Chapter 13. InductorsIntroductionAn inductor consists of a coil of wire, as its symbol shows:

The image above shows two coils. Inductor L1 has no core; the wire of L2 is wrapped around a ferrite(soft iron) core.

What will happen if we connect a DC voltage source (battery) to an inductor?

As a child, you may have built an electromagnet by wrapping copper wire around a nail andconnecting it to a battery. Without knowing it, you actually created an inductor. Now you know whathappens if you connect an inductor to a battery: you end up with an electromagnet. Be careful though;small coils you buy in a store are made of very thin wire. If you connect it to a battery, the high currentmay overheat the inductor within a second.

Although electromagnets are very useful, things get really interesting if we connect an inductor to anAC voltage source. When you disconnect an inductor from its battery, it wants to keep its magneticfield. Reversing the magnetic field is therefore not an easy task. But that's exectly what an AC currentwants to do: it wants to reverse the magnetic field over and over again. The higher the frequency, theharder this will be. In other words: an inductor blocks AC currents, but conducts DC currents.

The unit of inductance is Henry, symbol H. In schematics, you usually find mH (miliHenry) and uH(microHenry).

The impedance of an inductorThe impedance of a component is the resistance of that component for AC voltages. The symbol forresistance is R; the symbol for impedance is X. The impedance of an inductor is not zero; it dependson the inductance (number of turns; core material) and the frequency of the signal (number of polaritychanges (forth and back) per second). The impedance can be calculated using the following equation.

f is the frequency in Hertz; L is the inductance in Henry

Example: We have a 1mH inductor and connect it to a 50Hz AC voltage source. Calculate theimpedance of the inductor.

XL = 2∙π∙50∙10-3 = 0.31Ω.

Relation between voltage and currentFor any voltage v(t) the following equation is true:

It shows a 90 degrees pahse shift if v(t) is sinusiodal: the integral of a sine is a (-)cosine.

Inductors

45

It also shows that, if we apply a constant voltage V across an inductor, the current flow in it willincrease linearly:

v(t) = v(0) + V∙t/L.

If we remove this voltage, the current flow will continue, provided of course there's a path. If there isno path for the current, a large voltage peak will appear across the inductor. Hence the diode across atransitor-controlled relay:

A relay consists of an electromagnet and a switch that closes when a current flows in theelectromagnet. This current is generally controlled by a transistor. When the transistor turns off, thereis no path for the coil current. The voltage peak across the relay may destroy the transistor. The diodewill prevent this. After turning off the relay, the coil current will flow in the diode.

Frequency filters

Take a look at the diagram on the right. Assume that the voltage source supplies a 1V/10kHz signal(this means: the amplitude is 1V and the frequency is 10kHz = 10000Hz).

The impedance of inductor L will be XL = 2∙π∙104∙10-3 = 62.8Ω. The output voltage (voltage acrossinductor L) will be 1V∙(XL/(ZR+L)), where ZR+L is the total impedance of R and L. Because an inductor,just like a capacitor, causes a phase shift in in the current flow, we cannot just state that ZR+L = R +XL. Using some complex math we can prove that:

ZR+L = √(R2+XL2).

In our case ZR+L = √(1k2+62.82) = 1002Ω. Thus, the output voltage is 1V∙(62.8/1002) = 0.0627V.

(By the way, an inductor causes a +90 degrees phase shift, while a capacitor causes a -90 degreesphase shift.)

Now assume that the voltage source supplies a 1V/10MHz signal. The impedance of inductor L willthen be XL = 2∙π∙107∙10-3 = 62.8kΩ. So ZR+L = √(1k2+62.8k2) = 62.8k. This is the same as XL, so

Inductors

46

the output voltage will be 1V. So we created a very simple frequency filter with just a resistor and ainductor.

In this case we created a so called high pass filter (HPF) since high frequency signals pass this filtereasily while frequency signals are suppressed. If you swap R and L, you create a low pass filter(LPF).

Let's calculate the cut-off frequency of our filter. The cut-off frequency is the frequency at which R = XL=> R = 2∙π∙f∙L =>

In our case f = 1k/(2∙π∙10-3) = 159kHz.

Quality factorIn the previous sections we only looked at ideal inductors. However, every inductors has a certainseries resistance. This is not only determined by the resistance of the wire, but also by the kindof material the wire has been wrapped around. This resistance looks very much like the ESR of acapacitor. But with inductors we don't talk about ESR, but about the quality factor Q, which is definedas:

Q = XL/rL

where rL is the series resistance. XL and rL are both frequency-dependant, but not to the same extent.Q is therefore also frequency-dependant.

Assume that the series resistance of the inductor in the previous section is 20 ohms at 159kHz. Thequality factor at 159kHz will then be 1k/20 = 50.

The Q plays an important role in determining the notch depth and the bandwidth of LC filters. This willbe discussed in a next lesson. This lesson will also teach us how to measure the Q without expensiveequipment.

47

Chapter 14. Decibels (dB)Power and Voltage RatiosIf you browse through the technical specs of your stereo amplifier or your boom box, you'll notice thatsome parameters are specified in dB. For example: S/N ratio: 70dB; channel separation: 60dB.

What does this mean?

dB is defined as 10∙log(P1/P2)

Since P1 = V12/R and P2 = V22/R, we can also write: dB = 10∙log(V12/V22) = 20∙log(V1/V2)

In case of a signal-to-noise ratio (S/N ratio or SNR), dB = 20∙log(Vsig/Vnoise)

Knowing this, we can calulate the Vsig:Vnoise ratio if the S/N ratio is 70dB:

70dB = 20∙log(Vsig/Vnoise) => log(Vsig/Vnoise) = 70/20 = 3.5 => Vsig/Vnoise = 103.5 = 3162. This meansthat the music signal is 3162 times stronger than the noise generated by the amplifier.

Channel separation indicates how much signal meant for the right channel is present in the leftchannel and vice versa. If the channel separation is 60dB, the wanted channel signal is 1060/20 =1000 times stronger than the unwanted channel signal. In other words: if you only listen to the leftloudspeaker, you will also hear the music that should only come from the right speaker. However, this'unwanted channel signal' is 1000 times weaker than the music coming from the right speaker.

In the previous lesson, we learnt that the attenuation of an RL filter is about 10 times per decade. Howmuch is this in dB?

Since we were looking at the voltage attenuation, the attenuation of an RL filter is 20∙log(10) = 20dBper decade.

Reference-related dBsIn case of reference-related dBs, dB is defined as 10∙log(P/Pref) or 20∙log(V/Vref). In the table belowyou find the most commonly used dBs.

dBV Vref = 1VRMS dBV = 20∙log(V/1VRMS)

dBW Pref = 1W dBW = 10∙log(P/1W)

dBj Vref = 1mVRMS dBj = 20∙log(V/1mVRMS)

dBm Pref = 1mW dBm = 10∙log(P/1mW)

dBu (=dBv) Vref = √0.6V = 0.775VRMS (0.6V is the voltage across a600 ohm resistor dissipating 1mW)

dBu = 20∙log(V/0.775VRMS)

Examples: 0dBV = 1V = 60dBj; 0dBW = 1W = 30dBm.

48

Chapter 15. Vocal EliminatorIntroduction

A vocal eliminator is a device that removes the vocals from a song. These devices are primarily usedby professional singers. Amateurs want to have one, but these eliminators are quite expensive. So thepoor amateurs always had to perform their soundmix show by simply singing louder than the artists onCD.

Well... not anymore! Simply build this vocal eliminator and amaze your friends!

Schematic

Vocal eliminators use the fact that only the music is recored in stereo; the vocals (of the lead singer)are recorded in mono. This means that we only have to subtract the left channel signal from the rightchannel signal. Unfortunately, this also eliminates nearly all low frequence signals (bass); these haveto be added afterwards. The schematic below does exactly that:

Opamps U1.A and U1.B buffer the input signals.

Opamp U1.D is used as a differential amplifier. It's a combination of an inverting and a non-invertingamplifier. Since R3 = R4 = R5 = R6, U1.D's output is RIN - LIN.

The inverting amplifier around opamp U1.C adds RIN and LIN. Its output is fed to a low pass filter(R11, C1, U2.A). The result is a signal that only contains the low frequency part of both the left andthe right channel.

This signal is added to RIN - LIN by opamp U2.B. Output resistors R14 and R15 make sure U2.Bsurvives an accidental short to ground of the output terminals.

Choosing components

Opamps:

You can just take any opamp you like. I used a TL084 for U1.A...U1.D and a TL082 for U2.A andU2.B. A single TL084 contains 4 opamps; a TL082 contains 2 opamps:

Vocal Eliminator

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Resistors

All resistors except R3...R6 are generic 1/4W resistors. R3...R6 should be 1% precision resistors tomake sure that U3's output is exactly RIN - LIN.

If you want to control the amount of bass signal, you may replace R12 with a variable resistor.

Power supply

The power supply has not been drawn in the schematic. It can be just a simple small symmetricpower supply. A +/- 6V supply is sufficient for the vocal eliminator, so, in the supply, TR1 can be a 15Vtransformer and U1 a 78(L)12 voltage regulator.

Testing

Connect RIN and LIN to the line out terminals of a CD player, cassette deck or whatever you useas the music source. Connect ROUT and LOUT to the line in terminals of an amplifier. If you havea mono amplifier, just connect either ROUT or LOUT. Note however that the music source must bestereo.

If you play a mono CD track or tape, RIN and LIN are the same, so RIN - LIN = 0. You can test this bytemporarily removing R12. You should hear no sound. If you disconnect LIN or RIN, you should hearthe music.

Having R12 still removed, play a stereo song. You should now hear the music, but no vocals.Otherwise, the vocals are recorded in stereo as well; try another song (preferably from another CD ortape).

If you hear that the vocals are gone, you can put R12 back in place. If you hear the vocals again, thelow pass filter may not have been properly dimensioned. Experiment a little with R11 and C1. If youcan't get it right, you may need to add a second filter (just between R11/C1 and the + pin of U2.A),making the attenuation 40dB per decade.

Assembly

You can download the PCB (printed circuit board) of the vocal eliminator in several formats here:JPEG, EPS and HPGL. There are many ways to create a PCB from a layout, but that's beyond thescope of this course.

The component layout looks like this:

x_ve-pcb.jpg

x_ve-pcb.eps

x_ve-pcb.plt

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The power supply has its own PCB.

You can mount both boards in one cabinet, or in separate cabinets. If you give the power supply itsown case, you can use it for other projects as well.

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Chapter 16. Symmetric power supplyIntroduction

Several projects and lessons need a symmetric power supply. This chapter describes several of them.

Up to about 25mA

G1 is a cheap B80C800 bridge rectifier.

The 7824 is a 24V/1A voltage regulator. It looks like this:

1=IN, 2=GND, 3=OUT.

You may also use a 78L24, a 24V/100mA voltage regulator. It looks like this:

1=OUT, 2=GND, 3=IN.

The resistors devide the 24V output voltage; the voltage across R2 (and R1) will be 12V. The opampbuffers this 12V voltage. The maximum power dissipation of the opamp is about 600mW, so makesure the output current doesn't exceed 600mW/24V = 25mA.


This is the component layout:

x_ops-pcb.jpg

x_ops-pcb.eps

x_ops-pcb.plt

Symmetric power supply

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After soldering every component in its place, connect a power cord to power pole connector J1, andmake sure the output voltage of the power supply is correct.

Up to 1A

G1 and G2 are both B40/C1500 bridge rectifiers.

The 7812 is a 12V voltage regulator; the 7912 is a -12V voltage regulator. They both look like this:

7812: 1=IN, 2=GND, 3=OUT. 7912: 1=GND, 2=IN, 3=OUT.

The voltage across C1 and C3 is about 21V. The output voltage of the regulators is +/- 12V. So thevoltage drop is 9V. Without a heatsink, the thermal resistance from junction to ambient is 65°C/W.The maximum junction temperature is 150°C. When the room temperature is 25°C, we don't needa heatsink when the power dissipation doesn't exceed (150-25)/65 = 1.9W. To be safe, start using aheatsink when the power dissipation exceeds 1.5W. This means it's safe not using a heatsink up to1.5W/9V = 166mA output current. Do NOT mount both regulators on the same heatsink without usingan insulator! Also make sure that the screws don't create shorts.


This is the component layout:

x_ps-pcb.jpg

x_ps-pcb.eps

x_ps-pcb.plt

Symmetric power supply

53

After soldering every component in its place, connect a power cord to power pole connector J1, andmake sure the output voltage of the power supply is correct.

54

Chapter 17. JFETsIntroductionThe acronym FET stands for Field Effect Transistor; the J stands for Junction.

JFETs are transistors with a very high input resistance. They have three terminals: Drain, Gate andSource. JFETs come in two flavours: N-channel and P-channel:

A JFET looks like a voltage-controlled current source; the current source between the Drain and theSource is controlled by the voltage across the Gate and the Source.

The ratio dID/dVGS is called the forward transfer admittance, symbol yfs.

To make an N-channel JFET work, the Gate voltage must always be less than the Drain and Sourcevoltages. That means that VGS must be negative. If VGS becomes more negative, ID will decrease.The voltage at which the drain current becomes zero, is called the pinch-off voltage.

A JFET only behaves like a voltage-controlled current source if VGD is less (more negative) thanthe pinch-off voltage. Otherwise, the JFET will behave like a voltage-controlled resistor. (For anexplanation, take a look inside a JFET.)

Let's take a look at the BF245A, an N-channel JFET. According to the datasheet, yfs = 3mA/V (or3mS, millisiemens). ID = 1mA if VGS = -1V. If VGS increases by 0.5V, ID will be 1 + 0.5∙3 = 2.5mA.

We can use this behaviour to create an amplifier.

JFET amplifier

The diagram above shows a very simple JFET amplifier. Let's make some DC calculations (no inputsignal). T1 is a BF245A. The datasheet tells us that the yfs is 3mA/V if -1<VGS<0. So we choose VGS =-0.5V. At that voltage, ID = 2.5mA. We want VOUT to be 6V, so RD = 6V/2.5mA = 2.4k.

JFETs

55

Since VG = 0V and VGS = -0.5V, VS must be 0.5V. So RS = 0.5V/2.5mA = 200Ω.

Next, we connect a 0.1Vt sinus signal to the IN termial. What will be the the amplitude of the outputsignal? A 0.1V change in VGS causes a 0.1V∙3mA/V = 0.3mA change in ID and thus a 0.3mA∙2.4k =0.8V change in the output signal. So the voltage gain is 8 (or actually -8; it's an inverting amplifier).

Capacitor CS makes sure that VS remains constant, so that vGS = vIN (for AC signals).

RI is usually 1M or so. It guarantees that VIN = 0V (DC), while the input resistance remains very high.

Just like the current gain (hFE) of a bipolar transistor may vary over a wide range, so may the forwardtransfer admittance of a JFET. In case of the BF245A, yFS may vary between 3 and 6.5mS. Far worseis the fact that the pinch-off voltage of a BF245A ranges from -0.25 to -8V. That means that at VGS=-0.5V, ID can be much less or greater than the 2.5mA mentioned above; that was just a typical value.Resistor RS is therefore often replaced by a current source.

JFET current source

We already saw that the JFET is a voltage-controlled current source. If VGS remains constant, so willID. The circuit above makes use of that behaviour. The transistor is a BF245B. VGS = 0V. According tothe datasheet ID will be 10mA. In this simple circuit we use it to supply an LED with a current flow thatis independant of the source voltage.

Note that we cannot use this current source in the JFET amplifier in the previous section. The JFET inthe current source has the same parameter variety as the one in the amplifier. That's no problem fordriving LEDs, but it is in an amplifier.

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Chapter 18. MOSFETsIntroduction

As said in the previous lesson, the acronym FET stands for Field Effect Transistor; MOS stands forMetal Oxide Silicon.

There are enhancement and depletion MOSFETs. Each type is available in an N-channel and P-channel flavour. This means that there are four types of MOSFETs available:

JFETs and MOSFETs have many properties in common:

• Both have a very high input resistance.

• Both have three terminals: Drain, Gate and Source. Some MOSFETs have an extra Bulk terminal.

• Both can look like a voltage-controlled current source; the current source between the Drain and theSource is controlled by the voltage across the Gate and the Source. The ratio dID/dVGS is called theforward transfer admittance, symbol yfs.

• Both can also look like a voltage-controlled resistor; VGD must be between 0V and the pinch-offvoltage.

JFETs and depletion MOSFETs have even more in common:

• At VGS=0V, the D-S channel is conductive. (Enhancement MOSFETs need a certain G-S voltagebefore a Drain current can flow.)

• To close an N-channel, VGS must be negative; increasing VGS will increase ID.

Of course, there are also differences between JFETs and MOSFETs. MOSFETs make perfectswitches. The channel resistance in 'on mode' is very low, generally less than 10 ohms. To switchan N-channel enhancement MOSFET on, simply apply a high enough voltage across the Gate andSource. You cannot do that with a JFET, because a JFET's VGS must be negative (N-channel); whenVGS is positive, the G-S diode will be forward-biased, dramatically decreasing the input resistance!

And although MOSFETs make perfect switches, they can be used in amplifiers as well.

MOSFETs

57

MOSFET amplifier

The diagram above shows a very simple MOSFET amplifier. T1 is an depletion N-channel MOSFET.Assume its datasheet tells us that the VGS threshold voltage is -2V, and that yFS=10mA/V at VGS=0V.If R1=470Ω, the gain will be 10mA/V∙470Ω = 4.7.

Dual-gate MOSFETsA dual-gate MOSFET consists of two MOSFETs in series:

Typical applications are:

• An amplifier with gain control.

The input signal is fed to G1. The voltage at G2 controls the gain, because it determines thethickness of the channel of the top MOSFET. The transfer characteristics below shows the draincurrent versus VG1 for different values of VG2.

MOSFETs

58

It clearly shows that yfs (=ID/VG1) depends on VG2. And, of course, this means that the gaindepends on VG2.

In an antenna amplifier, weak signals should be amplified enough to be processed by the nextstage. But stong signals shouldn't overload the next stage. We only need a simple circuit thatproduces a voltage of, say, 5V for weak signals and 1V for strong signals and feed that voltage toG2. That's how AGC (Automatic Gain Control) implemented in many receivers.

• An AM modulator.

An AM modulator is a device that varies the amplitude of a high frequency signal with a lowfrequency signal. This is simply done by connecting G1 to the HF signal and G2 to the LF signal.Like above, the gain will vary with VG2 and thus with the LF signal.

• HF amplifier.

The amplifier in the previous section works fine for low frequency signals, but it's not suitablefor amplifying antenna signals in a TV set. Due to the MOSFET's construction, the capacitancebetween Gate and Drain is relatively high, about 5pF for a small signal MOSFET. At 100MHz, thismeans an impedance of just 318Ω. But the capacitance between G1 and Drain of a dual-gateMOSFET can be as low as 20fF (1fF = 1 femto-Farad = 10-18F). At 100MHz, the impedance will be79.6kΩ. G2 if usually connected to the possitive supply voltage, or to an AGC voltage (see above).

59

Chapter 19. LC filtersIntroductionWe have already learned about filters consisting of a resistor and a capacitor and of a resistor and aninductor. In this lesson we will learn about filters consisting of a capacitor and an inductor.

High pass filterAs a start, we will look at the following two filters.

A previous lesson tought us that the filter on the left has a cut-off frequency of 159kHz. The XC ofthe 1n capacitor is 1k at 159kHz. So the cut-off freqency of the filter on the right is also 159kHz.However, at 10kHz, XC will be 15.9k. The output voltage will be 1V∙(XL/(ZL+C)). Again, we must usesome complex math to prove that:

ZL+C = |XL - XC|.

At 10kHz this will be |62.8 - 15.9k| = 15837.2. The output voltage will be 1V∙(62.8/15837.2) =0.00397V. This means that a CL filter suppresses unwanted freqencies much better than RL filters.Let's calculate how much better.

To do this we calculate the output voltage of both filters at one tenth of the cut-off frequency, i.e.15.9kHz. At this frequency, XC=10k and XL=100Ω.

Vout,RL=1V∙(XL/√(XL2+R2)) = 1V∙(100/1005k) = 0.0995V. This is 10 times less than the input voltage.

Vout,CL=1V∙(XL/|XL-XC|) = 1V∙(100/9.9k) = 0.0101V. This is 100 times less than the input voltage.

Calculating the output voltages at one hundreth of the cut-off frequency (i.e. 1.59kHz) gives:

Vout,RL=1V∙(XL/√(XL2+R2)) = 1V∙(10/1000.05) = 0.0099995V. This is 100 times less than the input

voltage.

Vout,CL=1V∙(XL/|XL-XC|) = 1V∙(10/99.99k) = 0.0001V. This is 10000 times less than the input voltage.

We see that the attenuation of an RL filter is 10 times (or 20dB) per decade, and the attenuation of aCL filter is 100 times (or 40dB) per decade!

The same is true for RC and LC filters.

Let's now calculate the cut-off frequency of an LC filter. We already know that XC=XL. It can now beeasily proven that:

LC filters

60

Filling in L=1mH and C=1nF gives f=159kHz (but we already knew that).

Band pass filter

The circuit above is called a band pass filter, because olnly a band of frquencies appears at theoutput. The capacitor stops all low frequencies and the inductor all high frequencies. The currentflow in C and L is the same. The phase shift in the voltage across the capacitor is +90 degrees; thephase shift in the voltage across the inductor is -90 graden. The phase shift between both voltagesis therefore 180 graden, and are thus in anti-phase! The output voltage reaches its maximum valueif the voltages across C and L are equal; they simply cancel each other, and the full input voltage willappear across R. If the voltages are equal, XC must be equal to XL. The previous section alreadyshowed us that in that case:

We have also found a way to determine the quality factor Q of the inductor. We already know thatQ=XL/rL. And rL can easily determined, because we have a simple voltage devider here, since thetotal impedance of C and L at resonance frequency is rL. The full procedure is:

1. Adjust the source frequency to the frequency you want to know the Q of;

2. Adjust C to maximum output voltage;

3. Calculate rL and Q.

Example: we want to determine the Q of an inductor at 159kHz.

1. Adjust the source frequency to 159kHz;

2. Adjust C to maximum output voltage. It appears to be 0.83 times the input voltage;

3. The voltage across rL is 0.17 times the input voltage. So the ratio rL:R is 0.17:0.83. This means thatrL is about 20 ohms. So: Q=1k/20=50.

Please note that we asumed that the internal resistance of the source is 0 ohms. In reality, that willnever be the case. Additionally, we need very accurate voltage meters and rL is usually very smallcausing high current flows. Later in this lesson we will discover a better method.

But let's first determine the bandwith of our filter.

LC filters

61

The bandwith is determined by the frequencies at which the output voltage is 3dB less than itsmaximum value. The maximum voltage is at -1.6dB, so we need to read the frequencies at whichthe output voltage is -4.6dB. The lowest frequency is about 150kHz and the highest 169kHz. So thebandwidth is about 169kHz-150kHz=19kHz. Fortunalely, we can also calculate the bandwidth:

B=f∙(R+rL)/XL

In our case: B=159kHz∙120/1k=19.1kHz.

(R+rL)/XL looks very much like 1/Q. XL/Rtotal is therefore also called the quality factor of the circuit. Sothe bandwidth can also be noted as:

B=f/Qc

In our example Qc=1k/120=8.3. So B=159kHz/8.3=19.1kHz.

When we swap R and C, we get a low pass filter:

At low frequencies, the output voltage is equal to the input voltage (so Vout/Vin=0dB). We may expectit remains this way until the cut-off frequency (159kHz) at which point it will decay by 40dB perdecade. However, we first see the capacitor voltage increasing to about 18.4dB. At this point, thevoltage is 8.3 times higher than the input voltage! This ratio is eual to Qc. By making R 0 ohms, thevoltage ratio will be equal to the inductor's Q! This is the second method to determine the Q of aninductor. We don't need very accurate voltage meters anymore, but the source's input resistance andthe (possible) high currents are still bothering us.

LC filters

62

Band pass filter with a smaller band

This circuit is also a band pass filter. High frequencies are shorted by the capacitor and lowfrequencies by the inductor.

It can be proven that:

Zr=Q∙XL

where Zr is the impedance at resonance freqency and Q is the quality factor of the inductor. Thisequation is an approximation, but the Q of most inductors is so large that it can almost always beused.

Some complex math proves that the impedance reaches its maximum value at:

Again, this is an approximation that assumes the series resistance of the inductor is so small that itdoesn't affect the frequency.

In our case f is 159kHz again. If rL is 20 ohms, Q is 1k/20=50. Zr will be 50∙1k=50k. The current andthe input voltage are in phase, so:

Vout,max/Vin=Zr/(R+Zr)=50k/1050k=0.0476.

The image below shows the frequency response.

The bandwidth is about 160.7kHz-157.5kHz=3.2kHz.

If R»Zr and Q is large enough, then: B=f/Q. In our case B=159kHz/50=3.18kHz.

LC filters

63

And this is finally the best way to determine the Q of an inductor:

1. Adjust the source frequency to the frequency you want to know the Q of;

2. Adjust C to maximum output voltage;

3. Turn the source frequency down until the output voltage decreased by 3dB (=factor 1.41);

4. Increase the source frequency until the output voltage is again 3dB less than the maximum voltage;

5. Subtract both frequencies. This is the bandwidth;

6. Calculate Q: Q=f/B

With this method, the internal resistance of the source doesn't matter, and we don't run the risk of ahigh current flow.

ResonanceWhat will happen if we attach a charged capacitor to an inductor? Of course, the capacitor willdischarge, causing a current to flow in the inductor. If the voltage across the capacitor has becomezero, there will still be a current flow in the inductor. This current will re-charge the capacitor.The voltage across it will become negative. When the indictor has transferred all its energy tothe capacitor, the capacitor will discharge again. Next, the inductor will charge the capacitor andeverything will start all over again. This is called oscillation. The frequency equals:

Tuned circuit

Ever wondered how on earth a radio (or a TV set) is capable of picking up the right channel out ofhundreds of channels available? One way of doing this using a tuned circuit. A tuned circuit consistsof a capacitor and a parallel-connected inductor. Usually, the capacitor can be tuned: you can changeits capacitance by turning the attached knob. Actually, this circuit is the band pass filter from aprevious section. So the bandwidth is equal to B=f/Q and the resonance frequency can be calculatedusing:

The notch must be small enough to block all other channels, but wide enough to allow the full signal ofthe selected channel to pass.

When the antenna picks up a signal close to this frequency, the tuned circuit will resonate on thatfrequency. Just look what happens if the antenna picks up a 159kHz radio signal:

LC filters

64

Another radio station broadcasts at 149kHz, just 10kHz below the resonant frequency. The imagebelow shows the signal for this frequency.

The difference is obvious. The detector (a circuit that demodulates a radio signal to audible sound) willonly work on the 159kHz signal; the 149kHz signal is simply too weak.

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Chapter 20. Miscellaneous filtersIntroductionIn this lesson we will learn about coupled filters and T-filters.

Coupled filtersFilters can be coupled both inductively and capacitively:

The top circuit is an inductively coupled filter. The two at the bottom are capacitively coupled filters.

In inductively coupled filters, the windings of the inductors are wrapped around the same core. Theratio VL2:VL1 (omitting C1 and C2) is called coefficient of coupling k. This coefficient has a majorinfluence of the frequency response:

From left to right, k increases. This will result in an increasing output voltage. When k reaches acertain value, the frequency response starts having two top values. A filters are critically coupledwhen this just doen't happen. This appears to be the case when k = 1/Q, in which Q is the qualityfactor of both inductors. Of course, both inductors need to have the same Q. If k < 1/Q; the filter isundercoupled and will be narrow banded. If k > 1/Q; the filter is overcoupled, and will be wide banded.

In capacitively coupled filters, k is deternined by capacitors.

The cirtuit on the left shows top coupling. Here, k = CT/√(C3∙C4).

The filter on the right shows foot coupling, where k = √(C6∙C7)/CF.

When we compare coupled filters with LC filters, we'll see that coupled filters are much steeper. In thenarrow banded LC-filter the difference in output voltage at 150kHz and 159kHz is about 15dB. In thecritically couple filter, this difference is about 35dB! Note however that the output voltage of a coupledfilter is much less than that of an LC filter.

Miscellaneous filters

66

The picture above is a part of the circuit diagram of a radio receiver.

The transistor amplifies the input voltage and the coupled filter picks out the wanted signal. You noticethat the manufacturer used a tapped inductor. The voltage at pin 9 will be less than at pin 8, whichreduces the current flow drawn by the next stage. This is important, because this current will changethe resonance frequency if it's too high.

Twin T-filter

The circuit above is called a twin T-filter. The current in the resistors is in phase with the voltage. Thephase shift in the current in the capacitors will be 180 degrees. At a certain frequency the currentamplitues will be same, but in anti-phase. This frequency will therefore be suppressed. The circuit istherefore called a band stop filter, or notch filter. The center frequency equals:

If R = 1k and C = 1n again, this frequency will be 159kHz. This picture below shows the frequencyresponse.

Miscellaneous filters

67

A twin T-filter with feedback can be found on my projects page.

Bridged T-filters

These are also band stop filters. Both filters have the same frequency response, which is evensteeper than that of a twin T-filter:

To calculate the central frequency, we can use the same equation as for simple LC filters:

At this frequency the output voltage will be 0V if R = Q∙XL/4.

http://hobby-electronics.info/projects/TwinTFilter.html

http://hobby-electronics.info/projects

68

Chapter 21. Frequency-independantvoltage deviderIntroduction

In one of the first lessons we have already learned about voltage deviders. These voltage devidersconsisted of resistors only, and were therefore frequency-independant. However, when we want toconnect a voltage devider to some measuring equipment, we'll have to take its input capacity intoaccount.

In this chapter the 'measuring equipment' is an oscilloscope, which input impedance can be seen asa 1MΩ resistor parallel connected to a 30pF capacitor; this is usually written as 1M//30pF. At 159Hzthe impedance of the 30pF capacitor is also 1MΩ. So the total impedance at 159Hz is just 500kΩ.The input-impedance is frequency-dependant. A 9MΩ resistor will not suffice to devide the voltage by10. This wordt for DC voltages, but at 159Hz the voltage will not be devided by 10, but by (9M+500k)/500k = 19!

We will first create an HF probe that doesn't have this problem. Next, we will discover how thisproblem is solved in the attenuator of an oscilloscope.

HF probe

The most simpele HF probe contains just a resistor and a variable capacitor:

RIN and CIN make the input impedance of the scope. RP and CP make the probe.

For DC voltages VRIN/VRP = RIN/RP. For high frequency voltages VRIN/VRP = XCIN/XCP = CP/CIN.To make the voltage devider frequency-independant RIN/RP must be equal to CP/CIN. So: CP =CIN∙RIN/RP.

In our case CP must be adjusted to 30p∙1M/9M = 3.33pF.

If you own a scope and HF probe, you already know that the probe has a little 'screw' that needs tobe adjusted. That screw is capacitor CP. CP can be adjusted properly by connecting the probe to asquare wave voltage. A square wave signal contains many sinusoidal signals. E.g. a 1kHz squarewave contains sinusoidal voltages of 1kHz, 3kHz, 5kHz, 7kHz, etcetera. If CP has been adjustedproperly, a square wave will appear on the scope's screen (left). All frequencies the square waveconsists of will be attentuated equally. If CP is too low, the higher frequencies will be attenuated toomuch (middle); if CP is set too high, the higher frequencies will not be attenuated enough (right).

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RP and RIN are in series, so the input resistance is 1M+9M=10M. CP and CIN are also in series, so:1/C = 1/CP+1/CIN = 1/3.33p+1/30p. Thus, the input capacity is 3pF. So the input impedance of theprobe is 10M//3pF.

Attenuator in an oscilloscopeOf course, the attenuator in an oscilloscope must also be frequency-independant. Moreover, the inputresistance and capacitance must always be the same. Otherwise, we need to re-adjust our HF probefor every attenuator.

The picture above shows two attenuators. The top one devides the voltage by 10, the bottom one by100. Let's first take a look at the top one.

R103 is parallel-connected to the input resistance (RIN) of the scope. So the total resistance is: 1/R =1/RIN+1/R103, so R = 100k. R102 is 900k, which means that the attenuation is 10, and that the inputresistance remains 1MΩ.

C103 is parallel-connected to the input capacity (CIN) of the scope. The total capacity will be 47p +30p = 77pF. VC102 must be adjusted to 77p∙100k/900k = 8.56pF. The new input capacity is 7.7pF. Tomake it 30pF again, VC101 must be set to 30p - 7.7p = 22.3pF.

For the bottom attenuator we can follow the same route. R105 makes RIN 10.1k//1M = 10k. R104 is990k, so the attenuation is 100. C105 makes CIN 82p + 30p = 112pF. VC104 must be set to 112p∙10k/990k = 1.13pF. The input capacity is now just 1.12pF. VC103 must therefore be adjusted to 30p -1.12p = 28.88pF.

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Chapter 22. DIACs, SCRs and TRIACsDIACA DIAC looks like a Diode for AC. If we increase the voltage across a DIAC (via a resistor), at acertain voltage it will start to conduct; the voltage across the DIAC becomes 0V. When we thendecrease the voltage, the DIAC remains conductive. The DIAC doen't close untile the current ceases.The picture belows may make things more clear.

This DIAC starts to conduct when the supply voltage reaches 32V. Even when the supply voltagebecomes less than 32V, the DIAC remains conductive. The DIAC only closes, when the supplyvoltages becomes 0V. It's obvious that the DIAC also works for negative voltages. Hence, a DIACdoen't have an anode and cathode like a normal diode.

SCRSCR stands for Sicilon Controlled Rectifier. An SCR looks like a transistor: a current flow in one ofthe terminals (the gate) makes the SCR conductive. The difference however is that the SCR remainsconductive when the gate current becomes zero. The SCR only closes when the current flow stops(just like a DIAC). Take a look at the picture below.

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By applying a voltage at the base of T1, T1 will turn on. This will provide base current to T2 wich willalso turn on. On turn, T2 provides base current to T1, so the SCR remains turned on.

The SCR only turns off in the voltage across anode and cathode becomes zero.

We can make a very simple dimmer switch using an SCR:

While experimenting, keep in mind that the whole circuit (including P1!) is connected to the mains!

Capacitor C1 is charged by R1 and P1. As soon as the voltage across C1 reaches the DIAC's break-down voltage, the SCR starts conductiong and will turn on the lamp. The SCR will remain turned onuntil the mains voltage becomes 0V. And then things start all over again. The larger P1, the longer ittakes before the voltage across the capacitor is high enough to make the DIAC and SCR turn on. Ahigher value of P1 will reduce the light output.

When we turn on the light, we sometimes here a 'pop' coming from the loudspeakers of a radio. Thiscirtuit turns on the light 100 or 120 times a second (depending on the mains frequency). R2 and C2reduce the amount of noise caused by the switching SCR.

TRIACThe name TRIAC actually stand for 'TRIode for AC'. And triode is the ancient name for transistor. But'SCR for AC' really is a better name. Just like an SCR, a TRIAC turns on by applying a gate currentand doesn't turn off until the voltage across the TRIAC becomes 0V.

A TRIAC is capable of conducting the positive and negative half period of an alternating current,provided that the gate current also alters its direction.

The circuit below is a very common switch.

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While experimenting, keep in mind that the whole circuit is connected to the mains!

U1 is an optocoupler. An optocoupler consists of an LED and a transistor. When the LED is turned on,the transistor will become conductive. This allows current to flow from the gate to A1. Now current willflow from A2 to A1 turning on the lamp.

This circuit does have a disadvantage: there can be no current flow from A1 to the gate; so the TRIACwill be off during every negative half cycle! Fortunately, this can easily be resolved:

Optocoupler U1 contains a TRIAC instead of a transistor. This allows gate current to flow in bothdirections. There are also optocouples available that contain a zero-crossing detector, e.g. theMOC3041. In that case R2 and C2 can be omitted.

One last remark: gate current always flows from or to A1. Therefore you cannot swap A1 and A2!

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Chapter 23. TV Deflection CircuitWarning

Caution! All information in this chapter is solely meant to explain the deflection circuitry in a TV set.It is not meant to teach you how to repair a TV. Leave this to the qualified service personel! Thereare dangerously voltages present in a TV (and especially in the deflection circuit), even if the plug ispulled!

The picture tube

The front of the tube is phosphor coated. This phosphor gives off light when hit by a high energyelectron. The electrons are emitted by electron guns, located in the neck of the tube. The guns emitmore electrons when they are warm. Hence the presence of the heating filament. That's the thing wecan see glowing in the neck of the tube.

When leaving the gun, the electrons don't have enough energy. To increase the energy, a high voltageis applied. This high voltage, which can be 25.000V or higher, remains on the tube, even if the TV hasbeen turned off! It may take a few days before the tube is discharged.

Deflection coils

Without any further actions, all electrons would hit the center of the screen. We can deflect theelectron beam by applying a magnetic field. This magnetic field is produced by deflection coils. Thereare vertical deflection coils, moving the electrons up and down, and horizontal deflection coils movingthe beam sideways.

The TV picture is refreshed 25 times per second (PAL format). So the frequency of the verticaldeflection coil current is 25Hz. The period time is 40ms. In that interval 625 horizontal lines arewritten. So the period time of the current in the horizontal deflection coils is 64us. This is also referredto as the scan time. It corresponds to a frequency of 15625Hz. This is the high pitched whine wesometing hear coming from a TV set.

Deflection circuit

TV Deflection Circuit

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Let's have a look at the horizontal deflection circuit.

When a positive voltage is applied to the base, the Horizontal Output Transistor (HOT) turns on.This means that the full power supply voltage will be across the deflection coil Ld. This will cause thecurrent in Ld to rise according to the law dI=(V/Ld)dt. In our case V = 150V and Ld = 1mH, so thecurrent in Ld will rise by 150000A per second! The electrons will now move from the center to theright. When the electrons arrive at the right end of the screen, the transistor must be turned off again.The on-time of the transistor is equal to half scan time. Or actually, it's somewhat less than half scantime, since there's also some time needed to move the electron beam from the right side to the leftfor the next line. This time is called the flyback time, and costs about 10us, leaving 54us to write oneline. This means that the HOT must be turned off after 54us/2 = 27us. The coil current has inceasedto (150V/1mH)27us=4.05A. After turning off the transistor, there will still be a current flow in Ld. Thiscurent cannot flow through the transistor anymore, nor can it flow through the reverse biased diode.The current can only flow into capacitor Cfb. When Ld has transferred all its energy to Cfb, the currentwill be 0.

Let's calculate the voltage across Cfb. The energy in a coil equals 0.5LI2. And for the energy in acapacitor we can write: E = 0.5CV2. Since all energy is tranferred from the coil to the capacitor, wecan write: 0.5LdILd

2 = 0.5VCfb2. The voltage across the capacitor will therefore be: VCfb,max = ILd√(Ld/

Cfb). Of course, we also need to add the 150V supply voltage: VCfb,max=150 + ILd√(Ld/Cfb) = 150 +4.05√(1m/10n) = 1431V.

The capacitor will now discharge through the coil. The current flow will be in opposite direction andwill therefore be negative. When Cfb has transferred all its energy to Ld, the voltage across Cfb willbe zero. At this point, the current in Ld will be -4.05A (assumed that all components are ideal). Ldwants to transfer its energy to Cfb again. The voltage across it will therefore be negative. This willcause the diode to become forward biased, so the voltage across it will be (almost) 0V. Again, a fixedvoltage exists across the coil, and the current will start to increase (become less negative) accordingto the law dI=(V/Ld)dt. By the time the current becomes 0, the HOT has been turned on again. So, thevoltage across the deflection coil remains 150V and everything starts all over again.

An appropriate value for Cfb can be easily calculated. We already saw that Cfb and Ld make a tunedcircuit. A half period takes 10us. So the period time is 20us. We know that f = 1/(2π√(Ld∙Cfb)), so T =2π√(Ld∙Cfb). Hence, Cfb = T2/(4π2Ld). In our case Cfb = 10nF.

If all components were ideal and the picture tube would be a perfect sphere, the deflection circuitwould be as simple as described above. Of course, reality is different.


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Linearity correction

We will first take a look at the linearity correction. The deflection coil is not ideal, but has a certain(wire) resistance. The voltage across this resistance rises as the current in it increases. So the voltageacross the actual coil is less than it should be. The current in the coil will not increase as fast as itshould. To prevent this, we need a negative resistance that compensates for the DC resistance ofthe deflection coil. Unfortunately, we cannot buy a negative resistor in a shop. But we can createone using an inductor with a ferrite core (Lsat). As the curent in this inductor increases, its core willbecome more and more magnetic, until its limit is reached. Because a (small) change in current doesnot affect the magnetic field anymore, the impedance has become zero. Of course it doesn't matterin which direction the current is flowing. By pre-megnetising the core, we'll end up with an inductorwhich inductance increases as the current becomes more negative and increases when the currentbecomes more positive. The voltage across Lsat equals VLsat=Lsat∙dI/dt. Because the inductance ofLsat is far less than the inductance of the deflection coil, dI/dt will remain (almost) the same. However,Lsat decreases when the current rises. This means that VLsat decreases when the current increases.So Lsat behaves like a negative resistance. It's obvious that when Lsat becomes defective, it must bereplaced with an original part from the manufacturer.

S-Correction

Let's see how we can get a nice picture on a flat (instead of a sphere-shaped) screen. The electronsare deflected too much to the left and to the right. This can be solved by replacing the voltage sourceby a capacitor. When the transistor is on, the current will now be drawn from the capacitor Cs. Thevoltage across it will decrease. This will cause the current to rise less rapidly (less than 150000A/s).When the transistor is off and the coil current is negative, the voltage across Cs will rise again. Andthat is exactly what we want. The coil current is now somewhat S-shaped. This correction is therefore


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referred to as the S-correction. We only need to take care that the average voltage across Cs remains150V. To assure that, we cannot just connect a voltage source across Cs. The capacitor is thereforecharged via the deflection coil. Choke Lp prevents shorting the voltage source when the HOT is on.

EW-Correction

We have now solved the problem that electrons are deflected too much to the left and to the right.But we have the same problem at the top and at the bottom of the screen, although not as obviousto the viewer. Especially TVs with large screens will have an East-West correction. To solve theproblem, we could vary the source voltage, but this will cause another problem. Lp is actually theprimary winding of a transformer. The secundary voltages must remain stable, so the voltage acrossLp must also remain the same. This can be accomplished by putting a 'dummy' deflection circuit isseries with the 'real' one. Both transistors switch on and off at the same time, so Lp still receives 150Vpulses. By varying the voltage across Cmod, we can change the voltage across Cs, because VCs=VB-VCmod. The maximum voltage across Cmod is usually 0.2VB, which equals 30V. Cfb and Cfb2 makea voltage devider. So the 30 volts can be obtained by making Cfb2 4 times larger than Cfb. In thatcase VCfb2 will be 1∙VB/(1+4) = 30V. Because the period time of both circuits must be the same,Cfb∙Ld = Cmod∙Lmod. So Lmod must be Ld/4. Nowadays, the voltage across Cmod is generated bya dedicated chip. Since both transistors must swich on and off at the same time, they can be replacedby one transistor. We now have a so called diode modulator used in most large TV sets.

Practical Examples

This is the deflection circuit of a small TV. It has no East-West correction.


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Transistor 7445 is the HOT. Capacitor 2451 is the Cs. It's obvious that choke Lp is part of atransformer.

Capacitor 2446 is the Cfb. Its peak voltage is 700V. That also shows that it must be a small TV.

This is a deflection circuit that does have an East-West correction.

Transistor Q503 is the HOT. D505 contains both diodes of the diode modulator. C524 is Cfb and C525is Cfs2. P501 represents the horizontal defelction coils. P502 is Lsat. C520 and C521 make the Cs.L501 is Lmod and C527 is Cmod.

By the way, this is not the deflection circuit of a TV, but of a computer monitor. This can be easilyconcluded from the fact that the scan time is just 3.3 + 15.2 = 18.5us, instead of 64us. The refreshfrequency is 1/11.75ms = 85Hz. That also explains why Cs consists of two capacitors in series. Atother frequencies, a MOSFET can be turned on, which connects C518 in parallel with C520. This willchange the value of Cs, providing a better picture quality for that frequency.

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Chapter 24. Automatic volume controlIntroductionAutomatic volume control means that the gain of an amplifier decreases when the input voltageincreases. This way, we can prevent overdriving the next stage.

Schematic

Opamp U1A is used as an amplifier. Its gain depends on the DS resistance of JFET T1. The lower thisresistance, the larger the gain of U1.

The resistance between T1's Drain and Source depends on VDS and VGS. To keep VDS as small aspossible, R1 and R3 will first devide the input voltage by 100. The voltage between Gate and Sourceis controlled by U2.

U2 amplifies U1A's output signal. To prevent a too strong low-frequency response, the signal is filteredby an HPF consisting of C2 and R7. U2's gain is determined by P1. D1 makes sure that only negativehalf cyles make it to the Gate, since the JFET can only pinched off by a negative Gate voltage. Thenegative half cycles are smoothed by C1. Please note that 'the positive' of C1 is connected to ground,because the voltage across it will always be negative!

A small input voltage will result in a small output voltage at U1A and U2. At T1's Gate, there will thusbe a small negative voltage. The DS resistance will therefore be low and U1A's gain will be high.

A larger input voltage will cause a more negative voltage at the Gate, resulting in a higher DSresistance, and a lower gain.

So U1A's gain is higher for small input voltages and smaller for large input voltages. And that isexactly what we want.

The minimum DS resistance of the chosen FET is about 150Ω. The maximum gain of U1A is thereforeabout 100. The maximum DS resistance is infinite, so the minumum gain is unity.

Assume that P1 has been set so that U2's gain is 10. Also assume that T1's pinch-off voltage is -1.5V.If the input voltage is 0V, U1A's gain will be 100, because the Gate voltage is also 0V.

If the input voltage is 200mVtop the output amplitude at U2 will initially be 200mV/100∙100∙10 = 2V.This voltage will charge C1. This will cause the Gate voltage to become more negative, increasingthe DS resistance and decreasing the gain of U1A. The output voltage at U2 will therefore decrease

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as well. C1 will be charged less and less rapidly. Eventually the Gate voltage will give U1A a gain atwhich the output voltage at U2 equals the Gate voltage.

What will happen if the Gate voltage is almost equal to the pinch-off voltage and the input voltage isincreased even more? The FET will never get pinched-off completely, because U1A's gain will then beunity and that will never be enough to maintain the Gate voltage at pinch-off voltage (unless the inputvoltage is increased to 15Vtop, but this circuit is not meant for that). So the output voltage at U2 willnot increase. This means that the output voltage at U1A cannot increase either. At the certain inputvoltage (depending on the position of P1 and the pinch-off voltage of the FET) the amplitude of theoutput voltage remains the same! So this circuit can prevent overdriving the next stage.

The circuit around U1B provides some extra gain.

Choosing componentsU1A and U1B must be low-noise opamps that have a low offset voltage. We need a low-noise opamp,because the input signal is devided by 100 by R1 and R3. So the input voltage at U1A is just a fewmV. A low offset voltage is needed because of the high gain. A generic cheap opamp will carry a largeDC component at its output. This DC voltage will also be amplified by U1B. A NE5532 perfect for thisjob. If we want to use a cheap opamp anyway (for example a TL082), we will have to add a 100ncapacitor between node pin 1/C2 and pin 5, and a 1MΩ resistor between pin 5 and ground. This willblock the DC component without attenuating the low-frequency signals too much.

U2 can be any generic opamp. A TL081 works perfectly.

JFET T1 is not very critical, as long as its pinch-off voltage isn't too low. A BF245A (pinch-off voltage:-1.5V) is a good choise. If we only have a BF245C (pinch-off voltage: -4.5V), then we can use it aswell. We'll just turn up U2's gain using P1.

Power supplyThis circuit draws just a few mA. Of course it needs a symmetrical power supply. We may use thesmall power supply discussed in a previous lesson. The supply voltage isn't very important, as long asthe opamps can sustain it. A NE5532 works reliably between +/-5V and 18V.

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Appendix A. Calculating RMS valueRMS stands for Root Mean Square, which shows what it is: the square root of the mean of thesquares. Let's assume we take n samples. The value of sample x will be v(x∙T/n) where T is theperiod of the signal. The RMS value is the square root of the average of the square of all samples (xranging from 0 thru n-1):

Next, we replace xT/n by t: t = xT/n. This means we also have to change the range of the sum sign:if x=0, t=0; if x=n-1, t=(n-1)T/n = T when n is very large. The calculation becomes more accurate if nnears infinite:

Multiply both the numerator and the denominator by the time step between each sample (which nearszero when n nears infinite), and we get the following equation:

In case of a sinusoidal signal, v(t) = A sin(2∙π∙f∙t) where A is the amplitude of the signal.

v2(t) = A2sin2(2∙π∙f∙t). You may have learned in high school that sin2(x) = 0.5(1-cos(2x)). So:

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Appendix B. Inside semiconductorsInside a diode

P type and N type semiconductors

Most diodes are made of silicon.

A silicon atom has four electrons in its outer orbit. Silicon atoms together form a crystal structure,leaving no "free electrons" moving around. Hence, pure silicon is an insulator.

Let's replace a few silicon atoms with a atoms that have just three electrons in the outer orbit. Eachatom forms a "hole" in the silicon lattice. It has now become attractive for electrons, just as if it werepositively charged. This material is called a P type semiconductor.

Of course we can also replace some silicon atoms with atoms that have five outer electrons. Thismaterial is called N type semiconductor, since it repells electrons.

Joining P and N together

What will happen if we join some P and N material together?

The P type material attracts electrons while the N side repells them. So at the junctions someelectrons from the N side will fill the holes on the P side, creating a depletion zone:

In this zone are no free holes and no free electrons: this zone is an insulator.

Let's see what happens if we apply a voltage across the PN material. First, we connect the positive tothe P side and the negative to the N side:

The negative lead of the voltage source pushes the electrons in the N material through the depletionzone, filling up the holes in the P type metarial. The P side is now negatively charged, and theelectrons will flow from the P side to the positive lead of the voltage source. The material has becomea conductor!

In our next experiment, we will reverse the voltage:

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The negative lead fills holes in the P side and the positive lead attracts free electrons in the N side.The result is that the depletion zone will become larger. In other words, the PN material is now aninsulator!

So the PN material forms a device that conducts current in only one direction: a diode.

Zener diodes

If you connect a diode reverse biased to a voltage source and increase the voltage, you'll notice thatat a certain voltage, current starts to flow. The voltage at which this happens is called the breakdownvoltage or zener voltage. As long as the current is kept within certain limits, breakdown will notdamage the diode.

The zener voltage depends on the amount of impurities (non-silicon atoms) in the lattice: the moreimpurities, the lower the voltage will be.

Varicap diodes

We already saw that the depletion zone is an insulator. The remaining part of the P and N materialdoes conduct electrical current. And what do we call a device that consists of two conductors with aninsulator in between? A capacitor!

We also saw that the depletion zone increases when you increase the reverse voltage. This meansthat the capacity depends on the voltage.

Although all diodes show this behaviour, there are diodes specially designed for this purpose. Theseare called varicaps. They are used in radio and TV tuners. The frequency can thus be controlled bythe reverse voltage across the diode.

Inside a bipolar transistor

Joining three layers of P and N together

When we look inside a diode, we see it consists of a P layer and an N layer. To create a transistor, weneed three layers. We have two possible combinations: N-P-N and P-N-P. Hence the names NPN andPNP transistor.

An NPN transistor looks like this:

Please note that the P layer between the two N layers is very thin. Just 2 micron or so.

Applying voltages across a transistor

Let's connect an NPN transistor to some voltage sources:

When VBE = 0V, there will be no current flow between C and E, since the CB diode is reverse biased.


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When we increase VBE, electrons will flow from E through the N layer and the P layer to B. As wehave noticed, the P layer is very thin. In no time the electrons flowing from B to E will have filled allthe holes in the P layer. The remaining electrons will be free electrons. The P layer now appears tobe an N layer. N type silicon always contains free electrons, so there is now a conductive path from Cto E! If IBE increases, the number of free electrons increases and so will ICE. And that is how a bipolartransistor works.

Inside a JFET


Like a bipolar transistor, a JFET consists of three layers of P and N silicon. However, these layers areconnected in a different way. The picture below shows the inside of an N-channel JFET.

The two P layers are connected to each other and form the Gate. The Source and the Drain are bothconnected to the N layer. That's why this is called an N-channel JFET.

Applying voltages across a JFET

Let's connect an N-channel JFET to some voltage sources:

While discussing the inside of a diode, we saw that the depletion zone between a PN junctionincreases if the reverse voltage increases. So if VGS becomes more negative, the depletion zonebetween the P layers and the N layer will become thicker, narrowing down the channel between theSource and the Drain. The voltage at which the channel is closed, is called the pinch-off voltage.

Now let's see what happens if VGS remains constant and VDS increases:

At first ID increases as well, just like the D-S channel were a resistor. However, increasing VDS alsomakes VGD more negative, narrowing down the channel. When VGD has reached a certain voltage,the pinch-off voltage, ID cannot increase any further. The FET has now become saturated.

So if VGD is between 0 and the pinch-off voltage, the S-D channel looks like a resistor; its resistancecan be controlled by VGS. If VGD becoms less than the pinch-off voltage, the S-D channel acts like acurrent source that can be controlled by VGS.


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We see that both VGS and VGD have a "pinch-off value". Since JFETs are symmetrical, the pinch-offvoltages are the same. You may even swap the Drain and the Source!

Inside a MOSFET


Like a JFET transistor, a MOSFET consists of three layers of P and N silicon, where one of the layersform a channel between the Source and the Drain. However, a MOSFET looks a bit different. Let'stake a look at the inside of an N-channel enhancement MOSFET:

The two N layers are connected to Source and the Drain. The Gate is connected to a layer of metal.Between that metal layer and the P layer, there a very thin film of insulating material (SiO2). The Player is connected to the Bulk terminal. In nearly all cases, the Bulk is internally connected to theSource. The metal layer and the P layer make a capacitor. Let's apply some voltages across thetransistor and see what happens.

Applying some voltages across an enhancement MOSFET

If VGS=0V, the D-S channel is closed, because there is always a reverse-biased PN junction.

If VGS>0V the metal layer becomes positively charged. The metal layer will now attract the electronsin the P layer. Thus, a layer of electrons is formed.

These electrons make the P layer close to the gate look like N silicon. And now the're a channel offree electrons between the Source and the Drain making current flow possible.

If VDS is small, the channel acts like a resistor, which resistance is controlled by VGS. However, if VDSincreases, the 'gate-bulk capacitor' will decrease at the Drain side. This will narrow down the channel.At a certain threshold voltage, the channel will be pinched off, and ID will remain constant.

Depletion MOSFETs

Depletion MOSFETs look very much like enhancement MOSFETs:

Please note the thin layer of N silicon near the gate. This means that even if VGS=0V, there alreadyis a conductive path between the Drain and Source. Increasing VGS makes the path wider. MakingVGS<0V narrows the channel down.

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Appendix C. Buffer capacitorCalculation of the value of a buffer capacitor

In the picture above you see that the discharge of buffer capacitor C1 starts at t=T1. Now assume thatT1=0. The rectified sine wave will now be a rectified cosine. At t=T2 VC1=-Vtopcos(2∙π∙f∙T2). (Note theminus sign. At t=T2, the original transformer output is negative.)

So cos(2∙π∙f∙T2)=-VC1/Vtop. This means

When you discharge a capacitor with current I, VC(t) = VC(0)-I∙t/C. In our case:

VC1(t) = VC1(0)-I∙t/C1, so at t=T2: VC1 = Vtop-I∙T2/C1. Since VC1=Vtop-Vr, we can also say Vtop-Vr =Vtop-I∙T2/C1. So Vr = I∙T2/C1. This means

If f=50Hz and Vtop=20V and we want a 2V ripple voltage, we need a 4.3mF capacitor per ampere loadcurrent.

Note: Make sure your calculator uses radians!

ESR

In the calculations above, we didn't take the ESR of the capacitor into account. However, the ESRplays an important role in power supplies due to the large charge and discharge currents. The ripplevoltage will increase by at least I∙ESR. This will be in the case where the capacitor can be fullycharged. However, that will never be the case. Calculating the ripple voltage for a certain ESR isnot an easy task. Since we often do not know the exact ESR value of a capacitor, we'd better usecomputer simpulation to view the results of a certain ESR:

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C = 4700uF; ESR = 0Ω

C = 4700uF; ESR = 1Ω

The frequency of the (non-rectified) input signal is 50Hz. The discharge current is 0.5A.

The first image shows the situation whithout ESR. The ripple voltage is about 0.9V. On the right, theoutput voltage is plotted when the ESR is 1Ω. The ripple voltage is now 1.8V! So the ESR added 0.9Vto the ripple voltage.

Let's see what happens if, due to aging, the ESR increases to 3Ω.

The ripple voltage has now grown to 3V. In audio applications, a heavily aged supply capacitor can beheard as a 100Hz humm.

Buffer capacitor

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If computer simulation shows that the ESR has to be impossibly small, you may connect multiplecapacitors in parallel:

This image shows the situation with two 2200uF capacitors each having an ESR of 1Ω. Although thetotal capacitance is less, the ripple voltage is less than the situation with one 4700uF capacitor: just1.3V. The increase of the ripple voltage (due to ESR) has dropped from 0.9V to 0.4V. This is one ofthe reasons you'll often see parallel-connected capacitors in power supplies. (Another reason may bethat there are no capacitors available with a higher capacitance.)

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Appendix D. Amplifier stabilityIntroductionEvery amplifier will produce a certain amount of heat. When a transistor heats up, its characteristicswill change. For example, the quiescent collector current will increase. This may heat up the transistoreven more, causing the quiescent current to increase, etcetera. This behaviour is called thermalrunaway. This may cause the amplifier to clip, and may destroy the transistor. This, of course, must beprevented.

StabilityAssume that a dT increase in temperature results in a dP increase in the transistor's powerdissipation, and that a power increase of dP results in a temperature increase of dT'. Everything willbe fine as long as dT' < dT.

We can calculate dT' by multiplying the power increase (dP) with the thermal resistance from junctionto ambient, so:

dT' = Rth,j-a ∙ dP

So an amplifier will be thermally stable if: Rth,j-a∙dP < dT, or: Rth,j-a∙dP/dT < 1

Let's take a look at a very simple amplifier:

Without input signal the transistor (NPN or PNP) will dissipate: P=VCE∙IC.

VCE=VS - IC∙RL => P=(VS - IC∙RL)∙IC=VS∙IC - IC2∙RL.

dP/dIC=VS - 2∙IC∙RL. => dP=(VS - 2∙IC∙RL)∙dIC

The amplifier will therefore be stable if:

Rth,j-a∙(VS - 2∙IC∙RL)∙dIC/dT < 1

This will always be true if: VS < 2∙IC∙RL => IC∙RL > VS/2

IC∙RL = VS - VCE => VS - VCE > VS/2 => VCE < VS/2

So, the amplifier will be stable if VCE is smaller than VS/2. For maximum output amplitude, thecollector voltage must be VS/2. If the emitter is grounded, VCE will be equal to VS/2 and the amplifiermay be thermally instable. Therefore a small resistor is connected between emitter and ground. Thevoltage across this resistor must be about VS/5. This is a compromise; less voltage may result intoinstability due to component spread and aging. A higher voltage will decrease the maximum outputamplitude and thus the efficiency.

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Appendix E. Complex mathCalculations on a capacitor and resistor in series.We already know how to calculate the impedance of a capacitor:

At 10kHz a 1nF capacitor has an impedance of 15.9k.

When we connect a 10k resistor in series with this capacitor, we may expect that the total impedancewill be 25.9k is. But that's not the case, because a capacitor causes a -90 degrees phase shift in thecurrent flow. The equation

doesn't show this. But what if we plot the impedance as a vector? The length will be the (absolute)impedance and the angle the phase shift:

This component has an absolute impedance of 1 ohm and causes a 45 degrees phase shift betweenvoltage and current.

A resistor doesn't cause a phase shift; this vector will be on the x-axis. a capacitor causes a -90degrees phase shift and will be on the (negative) y-axis. The total impedance of R and C is R + XC.However, we have to add vectors instead of plain numbers. Since we have 90 degrees angles, it'seasy to calculate the (absolute) impedance: we can just use Pythagoras' theorem. Zt = √(R2+XC

2).The phase shift is equal to arctan(-XC/R)

Wouldn't it be nice if we had a more simple way for saying: the impedance is x ohms and causes ay degrees fase shift? A 180 degrees phase shift is easy; in that case we could say: the impedanceequals -x ohms. A 180 degrees phase shift equal to multiplying by -1. Now suppose that a phaseshift of 90 degrees is equal to multiplying by j. A 180 degrees phase shift will be the same asmultiplying by j2. This means that j2 = -1. A negative square is only possible in so called 'complexmath'. (Mathematicians among us may be accustomed to use i instead of j. But we already use i as asymbol for current, so that's confusing.)

Every impedance can be written as: a + bj. Number a is called the real part and is plotted on the x-axis. Number b is called the imaginary part and is plotted on the y-axis.

A resistor doesn't cause a phase shift and is therefore purely real.

We know that a capacitor causes a -90 degrees phase shift; its impedance is therefore purelyimaginary, and can be written as:

Complex math

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As an example, we'll look again at our 10k resistor in series with a 15.9k capacitor. The total(complex) impedance is 10k - 15.9kj. The absolute value (|Z|) equals √(10k2 + 15.9k2) = 18.78k. Thephase shift it causes in the current (arg(Z)) is arctan(-15.9k/10k) = -57.8 degrees.

So, when we connect a resistor and capacitor in series: ZR+C = R - XCj.

The absolutie value is: |ZR+C| = √(R2+XC2).

The phase shift is: arg(ZR+C) = arctan(-XC/R).

Calculations on an inductor and resistor in series.

An inductor causes a 90 degrees phase shift. After reading the previous section, it's clear that thecomplex impedance can be written as:

So: |ZR+L| = √(R2+XL2) and arg(ZR+L) = arctan(XL/R).

As an example, we'll take a 10k resistor in series with a 15.9k inductor. The total (complex)impedance is 10k + 15.9kj. The absolute value is √(10k2 + 15.9k2) = 18.78k. The phase shift in thecurrent flow will be arctan(15.9k/10k) = 57.8 degrees.

Calculations on a capacitor and inductor in series.

From the previous sections we've learnt that:

So: ZL+C = XLj - XCj = (XL - XC)j. The impedance is purely imaginary.

|ZL+C| = |XL - XC|.

This means that the total impedance is less than the impedance of each single component. If XL andXC are equal, the impedance will be zero! This can also be explained in an other way. The current flowthrough both components is the same, while the voltage across one component has a 90 degreesphase shift and the voltage across the other component a -90 degrees phase shift. The phase shiftbetween the voltages across C and L will be 180 degrees. Because the absolute impedances are thesame, the voltage amplitudes will also be the same. So both voltages will 'rule each other out'. Thetotal voltage across L and C will be 0V. This means that the impedance is 0Ω.

Calculations on a tuned LC circuit.

We want to calculate for which frequency the tuned circuit reaches its maximum value.

Complex math

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The impedance of two parallel-connected components can be calculated using the equation:

From the previous sections we've already learned that

Substituting this in the first equation gives:

Since the nominator is frequency-independent, Z reaches its maximum if the denominator reaches 0,so:

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