Ideal Diode

8/7/2019 Ideal Diode

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Ideal DiodeUp to this point in your career as an ECE student, youve been studying linear electrical components. Forexample, resistors (R), inductors (L), and capacitors (C) are ideally linear elements (and passive, of course).Linear Circuit ElementsWhat do we mean that a component is linear? To answer this question, recall there are just twoindependent qualities of electricity in electrical circuits. These are voltage and current.

A linear circuit element is one that linearly relates the voltage across that element to the current throughthe element.Linearity has a precise mathematical statement. If a quantity y is a function of another quantity x, asy= f( x) (1)Then y is linearly dependent on x ifmy= f( mx) (2)

where m is a constant. In other words, f is a linear function if when quantity x is multiplied by someconstant m results in the function simply being multiplied by m.

Here are a couple of examples of linear components in electrical circuits:

This is perhaps a bit tricker, but notice that differentiation is a linear operator: if v increases by a factor mthen i does as well.

Ideal Diodes

You will now learn about a new electrical circuit element, the diode. Diodes are made from two differenttypes of semiconducting materials that come together to form a junction:

The circuit symbol is

which is related to the physical markings on a typical diode as


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In stark contrast to resistors, inductors, and capacitors, the diode is a nonlinear element. For an ideal diode,the i-v characteristic curve is

It is apparent from this i-v characteristic curve that there are two distinct regions of operation of the idealdiode:

The ideal diode acts as an electronic valve allowing current in only one direction through the diode: in thedirection of the arrow in the circuit symbol.We will find this valve behavior very useful in some situations. For example, this is useful to preventdamage to an electronic device when the battery is inserted backwards, for example.

Applications of DiodesNow we will briefly consider a couple of applications for diodes. Well cover these in much more detail later.

Conversely, when vI < 0 , the ideal diode is off and there is no current through R.Therefore, v0 = 0.We have rectified the input voltage with this circuit.This process of rectification will work for any type of input signal, whether it is periodic (as shown above) ornot.


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Digital logic gate. Diodes and resistors together can be used to make rudimentary logic gates. For example:

Assuming the voltages are 0 V for low signals and 5 V for the high signals, then the circuit shown aboveis a two-input OR gate:o If vA = 5 V and vB = 0 or 5 V, then vY = 5 V.o If vB = 5 V and vA = 0 or 5 V, then vY = 5 V.for ideal diodes. This is an OR function Y=A+B.Why do we have the resistor in this circuit? Its a clamp down resistor and forces the voltage vY = 0 whenvA = vB = 0. (What if there was no R? Wouldnt the voltage be zero in this case as well?)One huge complication of diodes (or any nonlinear circuit element) in an electrical circuit is the use ofsuperposition in the analysis is disallowed! (The exception to this is if the analysis has been linearized foronly small amplitude signals. Well see this linearization throughout the course.)

Consequently, nonlinear circuit analysis is usually much more complicated than linear circuit analysis. Oneoften needs numerical analysis for solution, such as that provided by circuit simulation software.Example N1.1 Determine the voltage V and current I in the circuit below.

This is a nonlinear circuit, so a completely different analysis procedure is required than what youve used inthe past for linear circuits.One process you can use to solve this problem is to try (i.e. guess) different on/off combinations for the

diodes D1 and D2 until you achieve a physically plausible and self consistent solution.For example, if D1 is on and D2 is on, then V = 1 V and 3 V.It is simply impossible to have two different voltages simultaneously at a node. Voltages must be singlevalued at all nodes. We conclude that D1 and D2 cannot both be on simultaneously.Next, we try D1 off and D2 on. This leads to V = 3 V and I = 8 mA.This result is physically realistic and self consistent since with V = 3 V then the voltage drop across D1requires that it be off, which is what we have assumed.

Physical Operation of Diodes

Real diodes have a more complicated i-v characteristic curve than ideal diodes. As shown in the text fora silicon diode:


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The diode has three distinct regions of operation:

1. Forward bias note that when the diode is on, the voltage drop is approximately 0.6 V to 0.7 V for

a silicon diode.

2. Reverse bias in this region i = -IS , where IS is called the saturation current. For small signal

diodes, IS is often on the order of fA (10-15 A).

3. Breakdown in this region v VZK - for all I, where VZK is called the breakdown knee voltage. This

region of operation is useful in certain applications.

In the forward bias region of operation, it can be shown from first principals that

where

n = emission constant. Typically between 1 and 2.

VT = kT/q mV at room temperature (20C). Called the thermal voltage.

Notice the highly non-linear relationship between i and v in this equation.

which is true for operation in the reverse bias region.

Well now take a quick look at the basic semiconductor physics behind the pn junction, and then follow

this up with examples and applications.

PN Junction

Semiconductor junction diodes are made by joining two semiconductors together. A pn junction diode is

formed by joining a p-type semiconductor to an n-type semiconductor:

For a silicon diode, both the p and n regions are silicon, but in each of these regions, small amounts of

impurities have been added through a process called doping. To make p and n regions, we begin with a

silicon crystal as shown figure below. These atoms are held together by covalent bonds (sharing pairs of

electrons).


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At T = 0, the outermost electron (e-) of each atom is held in covalent bonds. No current is possible since

no electrons are available to contribute to conduction.

For T > 0, random thermal vibration provides enough energy for some of the e- to break their covalent

bonds.

These e- can contribute to conduction current.

Holes

When electrons are thermally excited out of covalent bonds, they leave a vacancy at the bond site, as

illustrated above in Fig. This is called a hole. Interestingly, holes can also contribute to conduction

current in a semiconductor material (see the figure below). This movement is usually much slower thane- so the mobility of holes is smaller.

Donors and Acceptors

The concentration of holes and free electrons can be changed in a silicon crystal by adding small

amounts of impurities called dopants. This is what makes electronic devices possible!

(1) To create holes, add acceptor dopants to the silicon. For such p-type semiconductors, the silicon is

doped with trivalent impurity elements such as boron. These impurity atoms displace silicon atoms

(having four electrons) with boron atoms (having three electrons).

Consequently, the regular silicon lattice has holes, or locations in the lattice that can accept a free

electron. This hole can also move through the lattice.


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(2) To create free electrons, add donor dopants For such n-type semiconductors, the silicon is doped

with pentavalent impurity elements such as phosphorus. These impurities displace silicon atoms with

phosphorous atoms (having five electrons). Consequently, one extra electron is available to move

through the silicon lattice.

Be aware that the entire p-type and n-type regions remain charge neutral at all times! The dopant atoms

are also charge neutral.

At room temperature, thermal ionization breaks some covalent bonds. In n-type materials we then have

free electrons while in p-type materials we have free holes. p type means positive charge carriers

predominate while n type means negative charge carriers predominate.

Depletion Region

Something very special occurs when we place p-type material in contact with n-type material. There now

appears to be an excess of holes in the p-type material and an excess of free electrons in the n type.

Through the mechanism of diffusion (random motion due to thermal agitation), excess holes will migrate

to the n-type region while excess free electrons will migrate to the p-type region.

More specifically, when the p- and n-type materials are placed in contact (forming a junction), two things

happen near the contact region:(1) Holes diffuse across the junction (diffuse because the hole concentration is higher in p type) into the

n-type region and recombine with majority electrons.

With this electron now gone, we have uncovered a


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positive charge from the dopant atom in the n-type region.

This forms a positively charged region.

(2) Similarly, the majority carriers in the n-type region (electrons) diffuse across the junction and

recombine with majority holes in the p-type region. This uncovers negative bound charge.

This contact region between the p and n regions now has a bound volume electric charge density. It is

called the depletion region. This may seem an unexpected name since only in this region is there a net

volume charge density (aka space charge)!

Reverse and Forward Biased Junction There are two important states for a pn junction, the reversed

biased and forward biased states:

(1) Reversed biased state:

An electric field E is created in the depletion region because of the uncovered charges near the

junction:

For the reversed biased state of the pn junction, the electric field produced by the battery Ebattery adds

to this electric field of the space charge E in the depletion region. This increases the width of the

depletion region.

Consequently, the majority carriers cannot flow through the region: holes in the p material are

opposed by E in the depletion region, as are electrons in the n material. Hence, little current flows (only

the drift current IS) unless the junction breaks down. This occurs when Ebattery is strong enough to

strip electrons from the covalent bonds of the atoms, which are then swept across the junction.

(2) Forward biased state:

When V is large enough so that Ebattery > E, then (i) holes are swept from the p to n regions, and (ii)

electrons are swept from the n to p regions. We now have current!


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DC Analysis of Diode Circuits

Well now move on to the DC analysis of diode circuits. Lets consider this very

simple diode circuit:

We will assume that the diode is forward biased. Using KVL VDD=IR+VD From the

characteristic equation for the diode

Assuming n, IS, and VT are known, we have two equations for the two unknown

quantities VDand I. Substituting (2) into (1):

which is a transcendental equation for VD. There is no simple analytical solution to

this equation. So how do we solve such a circuit problem? Over the next couple of

pages well mention five methods.

1. Graphical Analysis. Begin with the diode i-v characteristic curve:

2.

From (1), we can rearrange the equation in terms of I to also plot above. That is,

from (1)

which is an equation for a straight line ( y b mx = + ):

We call this straight line the load line.

Now, plot both of these curves on the same graph:


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The point where these two curves intersect is the simultaneous solution to the two

equations (1) and (2). This graphical method is an impractical solution method for

all but the simplest circuits. However, it is useful for a qualitative understanding of

these circuits. For example, what happens when:

(a) VDD increases?

(b) R increases?

(c)

3. Simulation packages. SPICE, Agilents Advanced Design System (ADS), etc. Here

is a simple example using ADS:

4.

3. Numerical methods. Use Mathematica, Matlab, Mathcad, etc.

Here is a simple example from Mathcad:


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4. Iterative analysis. See example 3.4 in the text.5. Approximate analysis. This is by far the most widely used approach for hand

calculations.

Approximate Diode Circuit Solutions

There is often a need for us to perform design with pencil and paper. Remember:

simulation packages dont design for you, they only analyze circuits. Theres a big

difference between design and analysis!

There are two very important approximate diode models that allow easier paper

designs:

1. Constant-Voltage-Drop (CVD) Model.

2. Piecewise Linear (PWL) Model.

Constant-Voltage-Drop (CVD) Model In this model, the characteristic curve is

approximated as:

In words, this model says that if the diode is forward biased, then the voltage drop

across the diode is VD. If not forward biased, the diode is then reversed biased and

the current is zero and VD can be any value < VD.

VD is often set to 0.7 V for silicon diodes, as shown above, while set to 0.2 V for

Schottky diodes, for example. The CVD circuit model for diodes is


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This is probably the most commonly used diode model for hand calculations.Example N3.1. Determine the current I in the circuit below using the CVD model and

assuming a silicon diode.

Using the CVD model, the equivalent approximate circuit is:

Assuming the diode is forward conducting (i.e., on) with VD = 0.7 V and using KVL

in this circuit:

The positive value of this current indicates our original assumption that the diode is

on is correct.

Since I is negative, then D must be reversed biased. This means our initial forward

conducting assumption was incorrect. Rather, in this situation I = 0 and VD = 0.5 V.

Piecewise Linear (PWL) Diode Model

This is a battery plus internal diode resistance model. It is one step better than the

CVD model by incorporating a slope to the interpolative line:

The finite slope to this curve means that the diode has a nonzero internal resistance,


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which we will label as rD. The equivalent circuit for the PWL diode model is then

Example N3.2. Determine the current I in the circuit below using the PWL diode

model shown in text Figure.

From Fig., we can determine VD0 and rD for the particular diode whose

characteristic equation is shown:

The equivalent circuit using the PWL model of the diode is then

Assuming the diode is on,, then by

KVL:

This is close to the 1.3 mA we computed in the last example using the cruder CVD

model. Again, the positive value of this current indicates that we made the correct

choice that the diode is on.

Whats the forward voltage drop across the diode?

Is this enough to turn the diode on? Yes, referring to the equivalent circuit above

Small-Signal Diode Model and Its Application

The diode analysis so far has focused only on DC signals. We must also consider the application of diodes


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in circuits with time varying signals. This analysis is also complicated by the nonlinear nature of the

diode. ts is often best left for circuit simulation packages. Conversely, small signal analysis of nonlinear

diode circuits can sometimes be done by hand. The concept behind small-signal operation is that a time

varying signal with small amplitude rides on a DC value that may or may not be large.

The analysis of the circuit is then divided into two parts: 1. DC bias 2. AC signal of small amplitude.

and the solutions are added together using superposition. For example:

where vd(t) is some time varying waveform, perhaps periodic such as a sinusoid or triangle signal. The

purpose of VD in this circuit is to set the operation of the diode about a point on the forward bias i-v

characteristic curve of the diode. This is called the quiescent point, or Q point, and the process of setting

these DC values is called biasing the diode.

The total voltage at any time t is the sum of the DC and AC Components d(t)=vd+vd(t) provided the AC

signal is small enough that the diode operates approximately in a linear fashion.

where ID is the DC diode current. We can series expand the exponential term using


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and if vd(t) is small enough so that truncate the series to two terms:

Substituting (3) in (2) gives

So, if vd(t) is small enough we can see from this last equation that iD is the sum (or superposition) of

two components: DC and AC signals. What weve done is to linearize the problem by limiting the AC

portion of vD to small values. The term T D nV I has units of ohms. It is called the diode smallsignal

resistance:

From a physical viewpoint, rd is the inverse slope of the tangent line at a particular bias point along the

haracteristic curve of the diode. Note that rd changes depending on the (DC) bias:

(Note that this rd is a fundamentally different quantity than rD used in the PWL model of the diode

discussed in the previous lecture.) The equivalent circuit for the small-signal operation of diodes is:

Because we have linearized the operation of the diode (by restricting the analysis to small AC signals),

we can use superposition to analyze the composite DC and AC signals. That is, signal analysis isperformed by eliminating all DC sources (short out DC voltage sources/open circuit DC current sources)

and replacing the diode with its small-signal resistance rd. This process is illustrated below:

Example N4.1 (Text example 3.6). For the circuit shown below, determine vD when


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The diode specifications are

0.7-V drop at 1 mAdc

n = 2.

As we discussed, for small AC signals we can separate the DC analysis from the AC (i.e., linearized). We

need to start with the DC bias. Assuming 0.7 D V V for a silicon diode the DC current is

Since 1 D I mA, then VD will be very close to the assumed value. At this DC bias, then the small-signal

resistance at the Q point is

We use this rd as the equivalent resistance in the small-signal model of the diode

The AC voltage across the diode is found from voltage division as

The corresponding phasor diode voltage is then

where the subscript p indicates a peak value and the pp subscript means a peak-to-peak value. Were

we justified in using a small-signal assumption for this problem?

which is much less than 2. So, yes, the small-signal assumption is valid here. As an aside, note that in

this circuit the ripple in the voltage has been reduced at the output. At the input, the ripple is 2/10=20%

of the DC component while at the output the ripple is 0.0107/0.7=1.5% of the DC component. See text

example 3.7 for another example of this ripple reduction.

Diode High Frequency ModelThis purely resistive AC model for the diode works well when the frequency of the AC signals is

sufficiently low. At high frequencies, we need to include the effects that arise due to these time varying

signals and the charge separation that exists in the depletion region and in the bulk p and n regions of

the diode under forward bias conditions.

Within the device and the depletion region there exists an electric field, as discussed in Lecture 2. For AC


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signals, this electric field is varying with time. As youve learned in electromagnetics, a time varying

electric field is a displacement current. The effects of a displacement current are modeled by equivalent

circuit capacitances:

We wont do anything with this effect now. This is presented primarily as an FYI. (However, later in the

course we will investigate this capacitive junction effect in transistors and how it affects the gain of

transistor amplifier circuits at high

Zener Diode

The very steep portion in the reverse biased i-v characteristic curve is called the breakdown region.

In this region the voltage across the diode remains nearly constant while the current varies (i.e., small

internal resistance). There are two physical mechanisms that can produce this behavior in the

breakdown region. One is the Zener effect in which the large electric field in the depletion region causes

electrons to be removed from the covalent bonds in the silicon. The second mechanism is the avalancheeffect in which charges that are accelerated to high speeds due to the large electric field in the depletion

region collide with atoms in the silicon lattice causing charges to be dislodged. In turn, these dislodged

charges have sufficient energy to liberate additional electrons. In other words, this avalanche effect is a

cascading, ionization process. Provided that the power dissipated in the diode is less than the maximum

rated, the diode is not damaged when operating in the breakdown region. In fact, Zener diodes are

designed to operate in this region. The circuit symbol for the Zener diode is

These diodes are usually operated in the reverse bias regime (i.e., breakdown region) so that IZ > 0 and

VZ > 0. An enlargement of this bre akdown region is shown in text


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The manufacturer specifies the VZ0 and test current IZT. One can design Zeners with a wide range of

voltages. The page below is from a Digikey catalog (www.digikey.com) and shows voltages from ranging

from 3.6 V to 200 V, for example.

The rated VZ at the specified IZT is listed for these Zener diodes. The circled component, for example,

has VZ = 8.2 V at IZT = 31 power is 1 W for this device. As the current deviates from the specified valueIZT, the voltage VZ also changes, though perhaps only by a small amount. The change in voltage VZ is

related to the change in the current IZ as

where rz is the incremental or dynamic resistance at the Q point and is usually a few Ohms to tens of

Ohms. See the datasheet for the particular devic e you are working with. Because of the nearly linear

relationships in the breakdown region, the reverse bias model of the Zener diode is

where vz=vz0+rzIz Applications of Zener Diodes

What are Zener diodes used for? Applications include: 1. Voltage overload protection. This circuit is from

the NorCal 40A radio that is built in EE 322 Electronics II Wireless Communication Electronics:


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2. Voltage regulation. See the figure below. An example of such a regulator circuit will be considered

next.

Example N6.1 (similar to text example 3.8). The Zener diode in the circuit below has the following

characteristics: 6.8-V rating at 5 mA, rz = 20 , and IZK = 0.2 mA. With

With these ratings

Note that the supply voltage can fluctuate by

1 V. Imagine this fluctuation is a random process rather

than a time periodic variation. Determine the following quantities:

(b) Find VO with no load and V+ at the nominal value. The equivalent circuit for the reverse bias

operation of the Zener diode is

From this circuit we calculate


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Therefore,

(c) Find the change in VO resulting from a 1 V change in V+. Using the circuit above the V+ = 11 V:

Similarly, with V+=9 V:

.

The ratio of the change in output voltage to the change in the source voltage is called the line regulation

of the regulator circuit. Its often expressed in units of mV/V. For this example and no load attached,

(d) Find the change in VO resulting from connecting a load of RL = 2 kohm with a nominal V+ = 10

V.Assuming that the diode is operating in the breakdown region:

Is this a reasonable value? Calculate IS:

So, yes, this is a reasonable value because IL < IS , as it must. From (1), Vo = rz VIz and since VIz =

-3.4 mA then

The ratio of the change in output voltage to the change in the load current ( V o/IL ) is called the load

regulation of the regulator circuit. Its often expressed in units of mV/mA. For this example,

(e) What is VO when RL = 0.5 kohm? Assume the diode is in breakdown. In this case,

Is this a reasonable value? No, because this value is greater than IS = 6.4 mA. Therefore, in this case

the Zener diode is not operating in the breakdown region. Also, the diode cant be forward biased.

Consequently, we conclude the diode must be operating in the reverse bias region.


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The equivalent circuit in this case is

From this circuit we calculate

This voltage is less than the breakdown voltage VZK, which is consistent with the reverse biased

assumption. (f) Determine the minimum RL for which the diode still remains in breakdown for all V+.

(We know from the results in parts (c) and (d) of this example that RL must lie between 500 and 2 k

when 10 V + = V.) Referring to Fig. 3.21, at the kneeI z=Izk=0.2 mA and Vz=Vzk~VZ0=6.7

If V+ = 9 V:

Therefore, IL = 4.6 mA-0.2 mA = 4.4 mA, so that

If V+ = 11 V:

Therefore, IL = 8.6 mA-0.2 mA = 8.4 mA, so that

The smallest load resistance that can be attached to this circuit and have the diode remain in breakdown

is RL = 1,522 . The reason is that for any smaller value when V+ = 9 V results in the diode leaving

breakdown and entering the reverse bias mode.


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Diode Rectifier Circuits(Half Cycle, Full Cycle, and Bridge).

We saw in the previous lecture that Zener diodes can be used in circuits that provide (1) voltage

overload protection, and (2) voltage regulation.

An important application of regular diodes is in rectification circuits. These circuits are used to convert

AC signals to DC in power supplies.

A block diagram of this process in a DC power supply is shown below

In this DC power supply, the first stage is a transformer:

An ideal transformer changes the amplitude of time varying voltages as

This occurs even though there is no direct contact between the input and output sections. This magic is

described by Faradays law:

By varying the ratio N2/N1 in (1) we can increase or decrease the output voltage relative to the input

voltage:

If N2> N1 , have a step-up transformer

If N2

For example, to convert wall AC at ~120 VRMS to DC at, say, 13.8 VDC, we need a step down

transformer with a ratio of:

We choose s v1 V5DC for a margin. For the remaining stages in this DC power supply:

Diode rectifier. Gives a unipolar voltage, but pulsating with time.

Filter. Smoothes out the pulsation in the voltage.

Regulator. Removes the ripple to produce a nearly pure DC voltage.

We will now concentrate on the rectification of the AC signal. Well cover filtering in the next lecture.


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Diode Rectification

We will discuss three methods for diode rectification:

1. Half-cycle rectification.

2. Full-cycle rectification.

3. Bridge rectification. (This is probably the most widely used.)

Half-Cycle Rectification

Weve actually already seen this circuit before in this class!

We will use the PWL model for the diode to construct the equivalent circuit for the rectifier:

From this circuit, the output voltage will be zero if Conversely, if vs(t ) >vds we can determine vO by

superposition of the two sources (DC and AC) in the circuit sources since we have linearized the diode:

Notice that were not making a small AC signal assumption here. Rather, we have used the assumption

of the PWL model to completely linearize this problem when ( ) 0 S D v t V > and then used

superposition of the two sources, which just happen to be DC and AC sources. (Consequently, we should

not use rd here.)The total voltage is the sum of the DC and AC components:

In many applications,

A sketch of this last result is shown in the figure below.

There are two important device parameters that must be considered when selecting rectifier diodes:


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1. Diode current carrying capacity.

2. Peak inverse voltage (PIV). This is the largest reverse voltage across the diode. The diode must be

able to withstand this voltage without shifting into breakdown. For the half-cycle rectifier with a periodic

waveform input having a zero average value

PIV = Vs

where Vs is the amplitude of vS.Full-Cycle Rectification

One disadvantage of half-cycle rectification is that one half of the source waveform is not utilized. No

power from the source will be converted to DC during these half cycles when the input waveform is

negative.

The full-cycle rectifier, on the other hand, utilizes both the positive and negative portions of the input

waveform. An example of a full-cycle rectifier circuit is:

Notice that the transformer has a center tap that is connected to ground.

On the positive half of the input cycle 0 S v > , which implies that D1 is on and D2 is off. Conversely,

on the negative half of the input cycle, 0 S v < which implies that D1 is off and D2 is on.

In both cases, the output current i0(t ) >0 and the output voltageV0(t)>0

While this full-cycle rectifier is a big improvement over the halfcycle, there are a couple of

disadvantages:

1 PIV = 2Vs - VD), which is about twice that of the half-cycle rectifier. This fact may require expensive or

hard-to-find diodes.

1 Requires twice as many transformer windings on the secondary as does the half-cycle rectifier.

Bridge Rectification


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The bridge rectifier uses four diodes connected in the famous bridge pattern:

Oftentimes these diodes can be purchased as a single, fourterminal device.

Note that the bridge rectifier does not require a center-tapped transformer, but uses four diodes

instead.

The operation of the bridge rectifier can be summarized as:1. When

are off:

2. 3. When4. are on:

The bridge rectifier is the most popular rectifier circuit. Advantages include:

1 PIV = Vs - VD), which is approximately the same as the half-cycle rectifier.

2 No center tapped transformer is required, as with the half cycle rectifier.

Peak Rectifiers


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The output of the rectifier circuits discussed in the last lecture is pulsating significantly with time. Hence,

its not useful as the output from a DC power supply.

One way to reduce this ripple is to use a filtering capacitor. Consider the half-cycle rectifier again, but

now add a capacitor in parallel with the load:

We expect that as soon as we turn on the source, the capacitor will charge up on + cycles of v Iand

discharge on the - cycles.

To smooth out the voltage, we need this discharge to occur slowly in time. This means we need to

choose C large enough to make this happen, presuming that R is a given quantity (the Thvenin

resistance of the rest of the circuit). The output voltage vO will then be a smoothed-out signal that

pulsates with time:

Notice the diode current and the capacitor voltage. They display behavior much different than what one

would find in an AC circuit.

Analysis of Peak Rectifier Circuits

Well require that RC T = , which means that the time constant of the RC circuit must be much greater

that the period

of the input sinusoidal signal:

Now, our quest is to approximately determine the ripple voltage

vrassuming t>>T


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When D is off, and assuming it is an ideal diode

[If D is not ideal then v0(t)~(v0-0.7)e-t/r.]

At the end of the discharge time, td, the output voltage equals

Substituting for vO from (1) at this time td leads to

This equation has the two unknowns Vr and td, assuming isknown. If we can determine td, then we

can find Vr. Finding tdcan be done numerically by equating (1) to the expression forthe input voltage

and solving for the time td when the two are equal as

This needs to be done numerically since (5) is a transcendentalequation.Alternatively, if t is small

compared to T (true when T , asassumed), then from (3)

Again, because T then we can truncate the series expansionof the exponential function to two terms

(see Lecture 4) giving

This simple equation gives the ratio of the ripple voltage to the peak voltage of the input sinusoidal

signal for the half-cyclerectifier. Its worth memorizing, or knowing how to derive.Often R and T are fixed

quantities. So from (7)

to obtain a small ripple voltage we need a large C in this case.

Conduction Interval


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Lastly, the conduction interval t is defined as the time intervalin which the diode is actually conducting

current. This timeperiod is sketched in the preceding two figures. The diode conducts current beginning

at time td and ending at T,within each period. Using equation (4) at time td

We expect the conduction interval to be small. So truncating theseries expansion of cosine to two terms,

(9) gives

The factor t is sometimes called the conduction angle, . For V r Vp this conduction angle (and

conduction interval) will besmall, as expected.

Discussion

To reiterate, the objective of the peak rectifier is to charge theshunt C when D is on, and slowly

discharge it during thosetimes when D is off.When does D conduct? During the t periods in the

previousfigure. Also seeNote that this peak rectifier is not a linear circuit. iD is a very complicated

waveform and not a sinusoid, as seen earlier in

There are no simple exact formulas for the solution tothis problem. The text only shows approximate

solutions for peak iD:

Example N8.1 (similar to text example 3.9). A half-cycle peakrectifier with R = 10 k is fed by a 60-Hz

sinusoidal voltagewith a peak amplitude of 100 V.(a) Determine C for a ripple voltage of 2 Vpp. From

(8):

or c=83.3ufFor a factor of safety of two, make C twice as large. Remember, a bigger C translates to smaller ripple.

(b) Determine the peak diode current. Using (11):

When specifying a diode for your circuit design, you would need to find one that could safely handle this

amount of current.Example N8.2. A half-cycle peak rectifier with R = 10 k is fedby a 60-Hz triangula

voltage with a peak amplitude of 100 V.

(a) Determine C for a ripple voltage of 2 Vpp. If you go back and look at the derivation of (8) youll find

that there were no approximations made that required a sinusoidal waveform. Consequently, (8) applies

to this triangular waveform as well, provided T . Hence, as before 83.3 C = F. (b) Determine the diode


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conduction time, t. Referring to this sketch of the region near the positive peak voltage for vI:

Because the rising portion of the waveform is a straight line:

To find t, equate

Therefore, for a triangular waveform

In this particular case,

Compare this time to a sinusoidal waveform:

This time is much longer than for the triangular waveform. Consequently, we would expect max D i for D

to be much larger for the triangular waveform than for the sinusoid!

Full-Cycle Peak Rectifiers

In a similar fashion, we can also add a shunt C to full cycle and bridge rectifiers to convert them to peak

rectifiers. For example, for a full-cycle peak rectifier:

The output voltage has less ripple than from a half-cycle peakrectifier (actually one half less ripple).


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The ripple frequency is twice that of a half-cycle peak rectifier. Using the same derivation procedure as

before with thehalf cycle, but with 2 T T . gives from (7)

Lastly, it can be shown that the id|max for the full-cycle peakrectifier:

is approximately one-half that of the half-cycle peak rectifierwhen Vr V p .

Limiting and Clamping Diode Circuits

Voltage Doubler. Special Diode Types .

Well finish up our discussion of diodes in this lecture by consider a few more applications. Well discuss

limiting and clamping circuits for diodes as well as voltage doubling circuits.

Voltage Limiting CircuitsThese types of circuits are used to cap voltages between preset limits. These are useful as voltage

protection circuitry or as signal conditioning. Examples of such circuits are shown in text

A simple signal conditioning example is a circuit with the following transfer function:


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Then one would see this output voltage vO for this particular input voltage v rI:

A circuit with ideal diodes can be designed to realize the abovetransfer function from a combination of

the concepts shownabove

Clamped Capacitor CircuitsAn idealized circuit of this type in shown below:

There are three important things to note about this circuit:

1. The ideal D keeps v0>0.

2. C charges only when v 1


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To demodulate the signal, one AC couples to give zero time average voltage (i.e., 0 VDC). The signal is then passed through a clamped capacitor circuit to give a

well-defined DC component, then through a low pass filter to extract the DC. This DC voltage is the time average value, which changes depending on the width of

the pulses (if the period is constant, as assumed).

(b) Combined clamped capacitor with peak rectifier. This is also called a voltage doubler circuit.

Ignoring the transient behavior when the input voltage is first applied, vD1 is:

This voltage is fed to a half-cycle peak rectifier yielding the output voltage:

Its obvious now why this is called a voltage doubler circuit.

Special Diode Types

1. Schottky barrier diode. Often just called a Schottky diode. (Used in Laboratory #1 and in the NorCal 40A in EE 322.) These are formed from a metal and an

n-doped semiconductor. The big d ifference from a silicon diode is a smaller forward-bias voltage drop of approximately 0.2 V. Also, because all conduction current

in a Schottky diode is carried by majority carriers (electrons) there is little to no junction capacitance due to the absence of minority carrier charge accumulation

in the vicinity of the depletion region. Because of this, one would expect the switching speeds of the Schottky diodes to be faster than silicon diodes, for example

2. Varactor. A reversed biased diode acting as a voltagecontrolled capacitance. (Used in the NorCal 40A in EE 322.) To understand the operation of the varactor,

recall that in the pn junction:

This separated charge region acts as a capacitance. As shown in the text, the junction capacitance can be expressed as

3. Photodiodes. This is a reversed biased pn junction illuminated by light: I t is readily apparent from this equation that as VR changes, so does Cj. (This model is

used in Spice.)


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When the pn junction is exposed to incident light in the correct frequency band(s), the incident photons can break covalent bonds in the depletion region thus

generating electron-hole pairs. These are swept away from the junction by the electric field in the depletion region with e- to the n region and holes to the p

region. Thus a reverse bias current has been generated. This is called a photocurrent. 4. Light Emitting Diode (LED). This is the reverse of the photodiode. In the

LED, a pn junction is forward biased:

When electron-hole recombination occurs, light can be given off in certain types of semiconductors such as GaAs.

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Ideal Diode