Elecs 1_Diode Circuits.pptx

Units Associated with Basic Electrical Quantities

Diode CircuitsThe Transformer, The Half-Wave Rectifier, The Full-Wave Rectifier, The Bridge Rectifier, Peak Inverse Voltage and Surge Current, Clippers and Limiters, Clampers, Voltage Multipliers, The Choke Input Filter, The Capacitor Input Filter (4th 5th Week)1ObjectivesAt the end of this topic you should be able to:

Describe how a transformer works and list important transformer ratings.

Calculate the currents, voltages, and impedances of a transformer circuit. Identify the different types of transformer cores.

Explain the operation of a half-wave and full-wave rectifier.

Calculate the output voltage of half-wave and full-wave rectifiers.

ObjectivesAt the end of this topic you should be able to:

Explain the effect of a capacitor filter on the operation of half-wave and full-wave rectifiers.

List the characteristics of a light-emitting diode (LED).

List the forward- and reverse-bias characteristics of a zener diode.

Calculate the voltage and current values in a loaded zener voltage regulator.

Diode CircuitsTHE TRANSFORMER

The transformer is an important application of mutual inductance.

As shown, a transformer has a primary winding inductance LP connected to a voltage source that produces alternating current, and the secondary winding inductance LS is connected across the load resistance RL.


The purpose of the transformer is to transfer power from the primary, where the generator is connected, to the secondary, where the induced secondary voltage can produce current in the load resistance that is connected across LS.


Although the primary and secondary are not physically connected to each other, power in the primary is coupled into the secondary by the magnetic field linking the two windings.


The transformer is used to provide power for the load resistance RL, instead of connecting RL directly across the generator, whenever the load requires an ac voltage higher or lower than the generator voltage.


By having more or fewer turns in LS, compared with LP, the transformer can step up or step down the generator voltage to provide the required amount of secondary voltage.


Typical transformers are shown below.

Note that a steady dc voltage cannot be stepped up or down by a transformer because a steady current cannot produce induced voltage.



When the secondary has more turns than the primary, the secondary voltage is higher than the primary voltage and the primary voltage is said to be stepped up.

This principle is illustrated with a step-up ratio of 10100, or 1:10.


When the secondary has fewer turns, the voltage is stepped down.

In either case, the ratio is in terms of the primary voltage, which may be stepped up or down in the secondary winding.

These calculations apply only to iron-core transformers with unity coupling.


Air core transformers for rf circuits are generally tuned to resonance.

In this case, the resonance factor is considered instead of the turns ratio.

Diode CircuitsWith unity coupling between primary and secondary, the voltage induced in each turn of the secondary is the same as the self-induced voltage of each turn in the primary.

Therefore, the voltage ratio is in the same proportion as the turns ratio:

THE TRANSFORMER

Diode CircuitsThe transformer shown has a turns ratio, NP : NS, of 4:1. Therefore, the root-mean-square (rms) secondary voltage is calculated as shown: THE TRANSFORMER

Diode CircuitsTo calculate the peak secondary voltage, we proceed as shown:

The peak-to-peak value of the secondary voltage equals 2 x VS(pk) or 2 x 42.42-V = 84.84-Vp-p.

THE TRANSFORMER






It is important to note that power used by the secondary load, such as RL in the circuit, is supplied by the generator in the primary.


With current in the secondary winding, its magnetic field opposes the varying flux of the primary current.

The generator must then produce more primary current to maintain the self-induced voltage across LP and the secondary voltage developed in LS by mutual induction.


If the secondary current doubles, for instance, because the load resistance is reduced by one-half, the primary current will also double in value to provide the required power for the secondary.


Therefore, the effect of the secondary-load power on the generator is the same as though RL were in the primary, except that the voltage for RL in the secondary is stepped up or down by the turns ratio.


The current ratio is the inverse of the voltage ratio, that is, voltage step-up in the secondary means current step-down, and vice versa.


The secondary does not generate power but takes it from the primary.

Therefore, the current step-up or step-down is in terms of the secondary current IS, which is determined by the load resistance across the secondary voltage. Diode CircuitsTHE TRANSFORMER



As an aid in calculations, remember that the side with the higher voltage has the lower current. The primary and secondary V and I are in the same proportion as the number of turns in the primary and secondary.Diode CircuitsTHE TRANSFORMER

The circuit below illustrates a power transformer with two secondary windings L1 and L2.

There can be one, two, or more secondary windings with unity coupling to the primary as long as all the windings are on the same iron core.


Each secondary winding has induced voltage in proportion to its turns ratio with the primary winding, which is connected across the 120 V source.

The secondary winding L1 has a voltage step-up of 6:1, providing 720 V.


The 7200- load resistance R1, across L1, allows the 720 V to produce 0.1 A for I1 in this secondary circuit.

The power here is 720-V x 0.1-A = 72-W.


The other secondary winding L2 provides voltage step-down with the ratio 20:1, resulting in 6-V across R2.

The 0.6- load resistance in this circuit allows 10-A for I2.


Therefore, the power here is 6-V x 10-A, or 60-W.

Since the windings have separate connections, each can have its individual values of voltage and current.


The total power used in the secondary circuits is supplied by the primary.

In this example, the total secondary power is 132-W, equal to 72-W for P1 and 60-W for P2.


The power supplied by the 120-V source in the primary then is 72 + 60 = 132-W.

The primary current IP equals the primary power PP divided by the primary voltage VP. This is 132-W divided by 120-V, which equals 1.1-A for the primary current.


The same value can be calculated as the sum of 0.6-A of primary current providing power for L1 plus 0.5-A of primary current for L2, resulting in the total of 1.1 A as the value of IP.


This example shows how to analyze a loaded power transformer.

The main idea is that the primary current depends on the secondary load.

With more than one secondary, calculate each IS and PS.

Then add all PS values for the total secondary power, which equals the primary power.


The calculations can be summarized as follows:


Efficiency is defined as the ratio of power out to power in. Stated as a formula,

For example, when the power out in watts equals one-half the power in, the efficiency is one-half, which equals 0.5 100%, or 50%.

In a transformer, power out is secondary power, and power in is primary power.


Assuming zero losses in the transformer, power out equals power in and the efficiency is 100%.

Actual power transformers, however, have an efficiency slightly less than 100%.

The efficiency is approximately 80 to 90% for transformers that have high power ratings.

Transformers for higher power are more efficient because they require heavier wire, which has less resistance.


In a transformer that is less than 100% efficient, the primary supplies more than the secondary power.

The primary power that is lost is dissipated as heat in the transformer, resulting from I2R in the conductors and certain losses in the core material.

The R of the primary winding is generally about 10- or less for power transformers.

Diode CircuitsTHE TRANSFORMER: RATINGS

Like other components, transformers have voltage, current, and power ratings that must not be exceeded.

Exceeding any of these ratings will usually destroy the transformer. Diode CircuitsTHE TRANSFORMER: RATINGS

Manufacturers of transformers always specify the voltage rating of the primary and secondary windings.

Under no circumstances should the primary voltage rating be exceeded.

In many cases, the rated primary and secondary voltages are printed on the transformer.


Consider the transformer shown.

Its rated primary voltage is 120-V, and its secondary voltage is specified as 12.6012.6, which indicates that the secondary is center-tapped.

The notation 12.6012.6 indicates that 12.6-V is available between the center tap connection and either outside secondary lead.

The total secondary voltage available is 2 x 12.6 V or 25.2 V.


As illustrated, the black leads coming out of the top of the transformer provide connection to the primary winding.

The two yellow leads coming out of the bottom of the transformer provide connection to the outer leads of the secondary winding.

The bottom middle black lead connects to the center tap on the secondary winding.


Note that manufacturers may specify the secondary voltages of a transformer differently.

For example, the secondary may be specified as 25.2-V CT, where CT indicates a center-tapped secondary.

Another way to specify the secondary voltage would be 12.6 V each side of center.


Regardless of how the secondary voltage of a transformer is specified, the rated value is always specified under full-load conditions with the rated primary voltage applied.

A transformer is considered fully loaded when the rated current is drawn from the secondary.

When unloaded, the secondary voltage will measure a value that is approximately 5 to 10% higher than its rated value.


As an example, the secondary current is 2-A. If 120-V is connected to the primary and no load is connected to the secondary, each half of the secondary will measure somewhere between 13.2- and 13.9-V approximately.

However, with the rated current of 2-A drawn from the secondary, each half of the secondary will measure approximately 12.6-V.


As illustrated, the schematic diagram for the transformer below with the colors of each lead are identified for clarity.

Transformers can have more than one secondary winding, likewise, they can also have more than one primary winding with purpose to allow using the transformer with more than one value of primary voltage.


The illustrations shows a transformer with two separate primaries and a single secondary.

This transformer can be wired to work with a primary voltage of either 120- or 240-V.

For either value of primary voltage, the secondary voltage is 24-V.


Figure above shows the individual primary windings with phasing dots to identify those leads with the same instantaneous polarity.

Figure below shows how to connect the primary windings to 240 V.

Notice the connections of the leads with the phasing dots.


With this connection, each half of the primary voltage is in the proper phase to provide a series-aiding connection of the induced voltages.

Furthermore, the series connection of the primary windings provides a turns ratio NP / NS of 10:1, thus allowing a secondary voltage of 24-V.


For a 120-V input, notice the connection of the leads with the phasing dots.

When the primary windings are in parallel, the total primary current IP is divided evenly between the windings.

The parallel connection also provides a turns ratio NP /NS of 5:1, thus allowing a secondary voltage of 24-V.


Illustration shows a transformer that can operate with a primary voltage of either 120- or 440-V.

In this case, only one of the primary windings is used with a given primary voltage.


For example, if 120-V is applied to the lower primary, the upper primary winding is not used.

Conversely, if 440-V is applied to the upper primary, the lower primary winding is not used.


Manufacturers of transformers usually specify current ratings only for the secondary windings.

The reason is quite simple. If the secondary current is not exceeded, there is no possible way the primary current can be exceeded.

If the secondary current exceeds its rated value, excessive I2R losses will result in the secondary winding.


This will cause the secondary, and perhaps the primary, to overheat, thus eventually destroying the transformer.

The IR voltage drop across the secondary windings is the reason that the secondary voltage decreases as the load current increases.


The power rating of a transformer is the amount of power the transformer can deliver to a resistive load.

The power rating is specified in volt-amperes (VA) rather than watts (W) because the power is not actually dissipated by the transformer.

The product VA is called apparent power, since it is the power that is apparently used by the transformer.


The unit of apparent power is VA because the watt unit is reserved for the dissipation of power in a resistance.

Assume that a power transformer whose primary and secondary voltage ratings are 120-V and 25-V, respectively, has a power rating of 125-VA.

What does this mean? It means that the product of the transformers primary, or secondary, voltage and current must not exceed 125 VA.

If it does, the transformer will overheat and be destroyed.


The maximum allowable secondary current for this transformer can be calculated as

The maximum allowable primary current can be calculated as


With multiple secondary windings, the VA rating of each individual secondary may be given without any mention of the primary VA rating.

In this case, the sum of all secondary VA ratings must be divided by the rated primary voltage to determine the maximum allowable primary current.


In summary, you will never overload a transformer or exceed any of its maximum ratings if you obey two fundamental rules:

Never apply more than the rated voltage to the primary. Never draw more than the rated current from the secondary.


All transformers have a frequency rating that must be adhered to.

Typical frequency ratings for power transformers are 50-, 60-, and 400-Hz.

A power transformer with a frequency rating of 400-Hz cannot be used at 50- or 60-Hz because it will overheat.


However, many power transformers are designed to operate at either 50- or 60-Hz because many types of equipment may be sold in both Europe and the United States, where the power-line frequencies are 50- and 60-Hz, respectively.

Power transformers with a 400-Hz rating are often used in aircraft because these transformers are much smaller and lighter than 50- or 60-Hz transformers having the same power rating.

Diode CircuitsSINUSOIDAL INPUTS; HALF-WAVE RECTIFICATION

The diode analysis will now be expanded to include time-varying functions such as the sinusoidal waveform and the square wave.

There is no question that the degree of difficulty will increase, but once a few fundamental maneuvers are understood, the analysis will be fairly direct and follow a common thread.


The simplest of networks to examine with a time-varying signal appears in Fig. 2.43.


For the moment we will use the ideal model to ensure that the approach is not clouded by additional mathematical complexity.


Over one full cycle, defined by the period T, the average value (the algebraic sum of the areas above and below the axis) is zero.


The circuit shown, called a half-wave rectifier, will generate a waveform Vo that will have an average value of particular, use in the ac-to-dc conversion process.


When employed in the rectification process, a diode is typically referred to as a rectifier.

Its power and current ratings are typically much higher than those of diodes employed in other applications, such as computers and communication systems.


During the interval t = 0 T/2 in Fig. 2.43 the polarity of the applied voltage vi is such as to establish pressure in the direction indicated and turn on the diode with the polarity appearing above the diode.


Substituting the short-circuit equivalence for the ideal diode will result in the equivalent circuit shown, where it is fairly obvious that the output signal is an exact replica of the applied signal.


The two terminals defining the output voltage are connected directly to the applied signal via the short-circuit equivalence of the diode.


For the period T/2 T, the polarity of the input vi is as shown and the resulting polarity across the ideal diode produces an off state with an open-circuit equivalent.


The result is the absence of a path for charge to flow and vo = iR = (0)R = 0-V for the period T/2 T.


The input vi and the output vo were sketched together in Fig. 2.46 for comparison purposes.

The output signal vo now has a net positive area above the axis over a full period and an average value determined by


The process of removing one-half the input signal to establish a dc level is aptly called half-wave rectification.

The effect of using a silicon diode with VT = 0.7 V is demonstrated in the circuit for the forward-bias region.


The applied signal must now be at least 0.7 V before the diode can turn on.

For levels of vi less than 0.7 V, the diode is still in an open- circuit state and vo = 0-V as shown in the same figure.


When conducting, the difference between vo and vi is 0.7V and vo = vi - VT, as shown in the figure.

The net effect is a reduction in area above the axis, which naturally reduces the resulting dc voltage level.


For situations where Vm >> VT, the equation below can be applied to determine the average value with a relatively high level of accuracy.

In fact, if Vm is sufficiently greater than VT, it is often applied as a first approximation for Vdc.


Example 1:

Diode CircuitsSINUSOIDAL INPUTS; HALF-WAVE RECTIFICATIONExample 1: Solution



Diode CircuitsSINUSOIDAL INPUTS; HALF-WAVE RECTIFICATION PEAK INVERSE VOLTAGE

The Peak Inverse Voltage can be determined by the equation:

Diode CircuitsSINUSOIDAL INPUTS; FULL-WAVE RECTIFICATION BRIDGE NETWORK

The dc level obtained from a sinusoidal input can be improved 100% using a process called full-wave rectification.

The most familiar network for performing such a function appears, as shown, with its four diodes in a bridge configuration.


During the period t = 0 to T/2 the polarity of the input is as illustrated.

The resulting polarities across the ideal diodes are also shown to reveal that D2 and D3 are conducting while D1 and D4 are in the off state.


The net result is the configuration as shown, with its indicated current and polarity across R.

Since the diodes are ideal the load voltage is vo = vi, as shown in the same figure.


For the negative region of the input the conducting diodes are D1 and D4, resulting in the configuration shown.


The important result is that the polarity across the load resistor R is the same and establishing a second positive pulse, as shown.

Since the diodes are ideal the load voltage is vo = vi, as shown in the same figure.


Over one full cycle the input and output voltages will appear as shown.


Since the area above the axis for one full cycle is now twice that obtained for a half-wave system, the dc level has also been doubled, then


If silicon rather than ideal diodes are employed as, an application of Kirchhoffs voltage law around the conduction path would result in;


The peak value of the output voltage vo is therefore;


For situations where Vm >> 2VT, the following equation can be applied for the average value with a relatively high level of accuracy.


Then again, if Vm is sufficiently greater than 2VT, then the as a first approximation for Vdc is often applied.

Diode CircuitsSINUSOIDAL INPUTS; FULL-WAVE RECTIFICATION PEAK INVERSE VOLTAGE

The required PIV of each diode (ideal) can be determined from the illustration obtained at the peak of the positive region of the input signal.

PIV

+-Vm

Diode CircuitsSINUSOIDAL INPUTS; FULL-WAVE RECTIFICATION CENTER-TAPPED TRANSFORMER

A second popular full-wave rectifier appears below with only two diodes but requiring a center-tapped (CT) transformer to establish the input signal across each section of the secondary of the transformer.


During the positive portion of vi applied to the primary of the transformer, the network will appear as shown in Fig. 2.60. D1 assumes the short-circuit equivalent and D2 the open-circuit equivalent, as determined by the secondary voltages and the resulting current directions.


During the negative portion of the input the network appears as shown, reversing the roles of the diodes but maintaining the same polarity for the voltage across the load resistor R.

Diode CircuitsSINUSOIDAL INPUTS; FULL-WAVE RECTIFICATION PEAK INVERSE VOLTAGEThe net PIV for each diode for this full-wave rectifier is the maximum voltage for the secondary voltage and Vm as established by the adjoining loop will result in

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Diode Circuits

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Elecs 1_Diode Circuits.pptx