Practice English Speaking&Listening with: Lecture -27 Voltage Regulators

We were discussing regulator, voltage regulator and we had seen how reference, Zener reference

could be obtained and that could be fed as one input and the output voltage as another

input to the base emitter junction, so that this current, the error current, is going

to adjust the drop across the transistors so as to keep output constant at value of

V z minus V Gamma. That simple regulator, we had discussed in detail.

Now obviously, one limitation of that circuit is that V naught was not adjustable once V

z is fixed; and also, V naught was not exactly V z; it was V z minus V Gamma; and it was

dependent upon the base to emitter junction potential difference.

So, we would like to get rid of all these things and use an error amplifier and the

error amplifier we said could be an op-amp comparator or simple differential amplifier.

So now, we will use a simple differential amplifier which is a crude version of the

op-amp itself. So, this is the difference amplifier here and the voltage that is developed across the load

of the differential amplifier, that we will see is going to be taken to control the base...that

is collector emitter voltage drop of the transistor.

So earlier, this was compared with this and that became the input; and now this...and

portion of this...that is, this is being compared with, let us say, R 1 by R 1 plus R 2 times

V naught; portion of the V naught, output voltage. This...this is a comparator which

is nothing but a differential amplifier here. And let us say we are putting this as I naught

and this I naught could be the current source which we can derive; we know how.

So, this could be directly connected here. So essentially, we see here...either an op-amp

or a differential amplifier could be used. From here to here, there is no phase shift;

from here to here, there is a phase shift of 180 degree. So, this is minus terminal

of the op-amp. This is plus terminal of the op-amp. This is the output of the op-amp.

So, the output of the op-amp directly goes to the base. This structure, we had earlier

indicated as the basic structure, which will involve a voltage reference, a series pass

transistor and a comparator, and the load. So actually, the comparator is comparing the

voltage reference with portion of the output voltage. If therefore this is a negative feedback

structure, then, if this is V z, V z should be very nearly equal to V naught into R 1

by R 1 plus R 2.

If the gain of the...open loop gain of the amplifier is very high, then this voltage

should be same as this voltage.

So essentially, you can see that this particular structure is a very crude operational amplifier,

using a single supply. The input voltage itself is biasing this operational amplifier. So,

one end of the op-amp is connected to this end, supply; and the other one is connected

to ground; output goes to the base of the transistor.

So, you will see that this configuration is normally available as a voltage regulator

I C. The I C manufacturers have a differential amplifier and a series pass transistor and

a voltage reference as part of the I C. So, this is a crude picture of a voltage regulator

integrated circuit.

So now, let us see what exactly happens. V z is equal to V naught into R 1 by R 1 plus

R 2. So, V naught therefore essentially is equal to 1 plus R 2 by R 1 times V z. So,

depending upon V z, once V z is fixed, we can always find out what the R 2 by R 1 should

be, in order to get the required output as the voltage regulated.

So, this is normally put externally. This network is put externally. R 1 R 2 network

is put externally in order to fix the output voltage, regulated voltage, as what you want,

given V z. So, in every I C regulator also, V z is specified so that you can fix according

to your own requirement what the output voltage should be.

If this is the structure, that is, there in the...this thing...let us understand the structure

fully. This is V naught and this is taking a current of V naught by R L. That is the

load current; and this takes a current of...assuming that the transistor Beta is high, this will

take a current of V naught by R 1 plus R 2.

Obviously, R 1 plus R 2 should be chosen by us such that this V naught by R 1 plus R 2

forms a small part of the load current, because rest of the current should not be dominant

for efficiency, for maintaining good efficiency. So, this is made small compared to V naught

by R L; but this should be larger.

V naught by R 1 plus R 2 should be much less than R L. This is one requirement. Then we

have assumed that this current is V naught by R 1 plus R 2. That means this base current

is neglected. That means this should be greater than, much greater than the base current.

So, selection of R 1 plus R 2 is based on these requirements. It should be much greater

than base current and it should be much less than the load current; and R 1 by R 1 plus

R 2 - ratio is fixed depending upon the output voltage. So, R 1 and R 2 get uniquely fixed

based on this.

Next, this current is nothing but V naught by R L plus V naught by R 1 plus R 2. Now,

that current divided by Beta; that by...that is the current delivered by the base of this.

Let us see what this current is.

This is at V i, we know this potential; and this is V naught, this is at V naught, plus

V Gamma. So, the current in R c, this current is V i minus V naught minus V Gamma by R c.

So, this current keeps fluctuating as V i is changing. One thing is this current has

to be much greater than this current. This current is the one which is split between

this and the transistor.

So, V i minus V naught minus V Gamma by R c; this current should be much greater than

V naught by Beta R L. Of course, this current can be ignored.

Now, by how much greater?- in the sense, this current should be capable of supplying this

current and also there should be some current left out for this transistor to be in the

active region.

That means this current should not also go too high or too small. It should be high enough

so that it can give some current to this and some current to this, so that this current...transistor

does not go off. If this keeps on increasing, this current remains the constant because

V naught is constant. Rest of the current should go into this and this current can keep

on increasing. So, what will happen? This transistor will conduct more and more. And

ultimately, it will carry only a current of I naught. That means it is capable of driving

this transistor to half state. So, the...in the working of this regulator, we have to

understand clearly the limitations on input voltage.

Now, for certain load here and output voltage, this current is constant. Let us say, this

current is 100 milliamperes and Beta is 100, let us say. So, the base current is a constant

requirement of 1 milliampere. So, this current has to be greater than 1 milliampere. Let

us say it is 2 milliamperes for a given input voltage.

So, this 1 milliampere will go into this. Therefore, I naught has to be greater than

1 milliampere so that some current is left out for this transistor to remain in the active

region. So, as V i keeps on increasing, this current will keep on increasing; and this

current will keep on increasing and this will keep on decreasing.

So, for a certain limiting value of V i, this transistor will stop...the differential amplifier

will stop functioning, because it will reach the limit of its current switching. Again,

when V i is kept on decreasing, this current will go to zero. That is the other one. So,

because of this limitation itself, we will have a V i max and V i minimum that we can

determine from these equations. So this current should be much greater than this and it should

be less than I naught. This is the requirement.

So, as far as this structure is concerned, we can as well put a current source here instead

of R c so that the open loop gain of this differential amplifier is high. Otherwise,

the gain of this is simply g m R c by 2. The gain of this is g m R c by 2. g m is 1 over

R e of the transistors. So, all these things you can assume, as long as the current division

is I naught by 2 and I naught by 2. That I naught by 2 division occurs only for a specific

V i situation. Otherwise, these two currents will be different and the g m of the thing

is 1 over R e 1 plus R e 2 and the gain is that g m into R c by 2; R e 1 and R e 2 will

depend upon the currents sharing in this.

So, all these things you have to remember because it depends upon the operating point.

The small signal definition of the equivalent circuit is based on the operating point and

it is not necessary that this should operate at all times with equal division of current

here. So, this is a D C amplifier actually. Voltage regulator is using a D C amplifier

for regulation purposes; and therefore, the operating point keeps on shifting as far as

this is concerned.

So, we will perhaps work out an example of this regulator and see for ourselves all the

limitations that can come about in this circuit. And, as far as the output impedance is concerned,

since this is using an op-amp in the negative feedback structure, it is much lower than

the earlier output impedance which was nothing but R e; and R e plus R z by Beta plus 1.

Here it will be much lower than that; it will be lower than the earlier impedance by the

loop gain.

So, we will work out an example and show the various parameters associated with the regulator.

So now, consider Example 22. For the voltage regulator circuit shown here, determine all

the important parameters which we have already defined earlier. So, this is very illustrative

of what exactly happens in a typical voltage regulator circuit.

V i is the applied unregulated voltage which let us say, keeps changing from 20 to 30 volts.

This is the typical variation possible because normally, this unregulated voltage is derived

from the power line and that much variation occurs in the power line and you are deriving

from rectifier, transformer rectifier, filter and unregulated voltage which is going to

supply to this.

So, that is varying from 20 to 30. You want a regulated supply voltage of, let us say

in this case, 14 volts because this is...V z is 7 volts and R 2 is taken as equal to

R 1. 10 K, 10 K. So, 1 plus R 2 over R 1 is 2. So, 2 times 7 volts is the output voltage.

So, regulated output voltage is 14 volts; 2 times V z - that is gone. And let us say

R L...we would like to now know, how much current I can draw before this circuit starts

failing, stops functioning as a voltage regulator.

First of all, is there a minimum current, is there a maximum current, etcetera? So,

the current drawn here is V naught by R L. So, it keeps changing depending upon R L.

The first question that we would like to pose is what will be the limitation on R L during

which this will continue to function for an input voltage which is changing from 20 to

30 volts.

Let us now see other bias currents that are going to flow through this; whether our design

is properly drawn.7 volts is here and the voltage across this baizing arrangement is

7 volts minus the point 6 volts drop here for the diode connected transistor. This is

a current mirror; point 6 volts drops here. So, 7 minus point 6 is 6 point 4; divided

by 5 point 4. We will change this resistance suitably so that our current is some round

number. Otherwise, we have to use calculator. 7 volts minus point 6 is 6 point 4 volts;

and across 6 point 4 K we have 1 milliampere drawn here. This 1 milliampere is a constant

current that is going to be drawn and therefore this is going to be 1 milliampere. That is made independent of the input voltage

variation.

If we have connected this here it would have varied. We do not want that to vary. So, this

current is made constant, bias current is made constant by connecting it here. So, this

current, of course, keeps varying here. This is the bias current needed for the Zener.

Please remember, this has to be greater than 1 milliampere - here this current - so that

some current is left for the Zener. This current is going to supply the Zener as well as this

1 milliampere.

So meanwhile, we are ignoring the current drawn by the transistor, of course. So, this

current is going to be V i which is varying from 20 to 30, divided...minus 7 volts divided

by 10 K. So basically, this current is varying from, let us say 20 minus 7 is 13. 30 minus

7 is 23 by 10 K. So, it is varying from 1 point 3 milliamperes to 2 point 3 milliamperes.

So, this current is varying; corresponding to this is 1 point 3 milliamperes; corresponding

to this is 2 point 3 milliamperes.

Let us keep a note of this because when 1 point 3 milliamperes is flowing, this 1 milliampere

is taken away. The Zener is left with point 3 milliamperes. So, this current is going

to vary from point 3 milliamperes to 1 point 3 milliamperes.

So, that information is very useful for us. When it is changing from point 3 milliamperes

to 1 point 3 milliamperes, obviously, the Zener voltage will change. It will not be

constant at 7 volts. That change in Zener voltage is what is going to appear at the

output as 2 times that. So let us say, at point 3 milliampere, it is some value - 7

volts; and at 1 point 3 milliamperes, it is 7 point 1 volt; higher. Then correspondingly,

output will be changing from 14 to 14 point 2 volts. So, that is what is called as line

regulation.

So, as V i is changing, output voltage is changing. So, we can therefore find out the

line regulation factor as 14 point 2 minus 14 divided by 14 expressed as a percentage.

So, that is the line regulation factor for this circuit.

So, line regulation can be found out from the information about Zener voltage change here. So, let us say

this...at point 3, it is 7 volts Zener and this at 1 point 3, is 7 point 1 volt. So,

output voltage is going to change from 14 volts to 14 point 2 corresponding to this.

So, line regulation factor is going to be 14 point 2 minus 14 by 14. So, that is equal to point 2 by 14, expressed

as a percentage. That means that into 100.

So, it is really 20 by 14 or 10 by 7 which is 1 point...how much is it? 1 point 4 percent.

So, that is what is called as line regulation factor for this circuit. So, please understand

this is one way. Because of line voltage variation, output voltage can vary, direct. Other way

is also through the link here because this is not a truly very high impedance point.

This...there is a reverse bias junction here. Then it is connecting.

So, if the reverse bias impedance is not too high, then there will be transmission on this

side also. So, all these factors will come into picture; but the primary factor in this

circuit, I told you...because compared to this impedance, this is very low and therefore

most of the transmission occurs through this path.

Next, we will also know what is the variation of current that is occurring here. That is,

this we had mentioned about earlier. Because V i is changing from 20 to 30 volts, the current

here is changing. What is the voltage here? This is 14 volts and this is 14 point 6; let

us say, V Gamma more. So, this voltage is going to change. This is going to change from

20 to 30. That means this voltage across this is going to change from, let us say 5 point

4 to 15 point 4 volts. So, the current is correspondingly going to change from 5 point

4 by 15 to 15 point 4 by 15. What is the value?

Current is going to change. That is most important for us. 5 point 4 by 15 to 15 point 4 by 15

milliamperes. So one is, this is point 36 milliamperes and the other one is 1 point

zero 3,1 point zero 3. So, the point 36 to 1 point zero 4 something... So, this is the

variation of current in this.

Now, what is the current demanded by this is what is important, so that we can find

out how this works here. This load is going to change. That we said. To what extent we

can change the load so that this circuit still functions? This is an interesting point so

that you see exactly how this circuit functions, trying to maintain 14 volts here. So, the

current in this is 14 by R L.

So, 14 by R L plus, of course, 14 by 20 K. This current, 14 by R L plus 14 by 20 K - 1

point naught 3. That does not matter; not much of a change. So, 14 by R L plus 14 by

20 K, which is how much? This is point 7 milliamperes. So, what should be the range of variation

of R L so that this transistor, this circuit still functions satisfactorily? This divided

by Beta. Beta is given as 200. So, this divided by Beta; this becomes a very small part of

the current, you can see. Now we have to concentrate on this 14 by Beta R L.

So, this current is being delivered here; base current of this. Now, let us consider.

It should keep functioning at all times, let us say. This current is changing from point

36 to 1 point zero 3; the current required for biasing this is totally equal to 1 milliampere.

So, when this is point 36, what can happen? Let us see. When this current is point 36,

what can happen? This can demand most of it. So, and therefore, this is left with almost

nothing. So that means, point 36 is going to be one limit of current where this is left

with almost nothing and this point 36 is straightaway going into this. That means point 36 into

Beta, that is 200 is the maximum current this can deliver. So, or this current can be equal

to point 36, limiting situation, R into Beta. That is 36 into 2 - 72 milliamperes. That

is the maximum current it can deliver under that situation. If you demand more, there

is nothing to give that, under the situation that this is point 36 that corresponds to

input voltage of 20 volts.

So, you can now see that R L should be greater than some value because this you can now find

out. 14 by R L should be equal to 72 minus point 7. That is, 71 point 3. So, R L should

be greater than 14 by 71 point 3. The...this is in terms of Kilo ohms, K. So roughly, this

will be equal to about 200 ohms. How much is it? 196. So, this is equal to 196 ohms.

That means R L has to be greater than this. If it is less than this, this circuit will

stop functioning for input voltage of 20 volts.

Obviously, it will happily work if R L is 196 ohms with 2 point 3 milliamperes. That

is 30 volts because 30 volts corresponds to 1 point zero 3; and this is demanding only

point 36. Rest of it can go into this. That is 1 point zero 3 minus point 36, whatever

it is. That is, about point 64 can go into this very happily because it still has some

current to be given to this; that is point 36. So, this circuit will function up to this

kind of load; lower than this, it will stop functioning.

Now, we can also determine the maximum limit; that is R L max. The...that will deliver the

lowest current. Obviously, can we...can we see whether it can be open circuited. We will

see. Suppose R L is open circuited. Nothing is demanded from it. Then what happens is

this current is zero. The only current that is to be supplied is point 7 milliamperes

for this biasing circuit; point 7 milliamperes divided by 200, extremely small.

So, most of the current in this will go into this. That is alright as long as it is point

36, because if this point 36 going...goes here, point 64 will go into this. No problem.

But, if 1 point zero 3 goes into this, there is a problem because only 1 milliampere should

have gone here and there is nothing that will take the rest of the current. So, that means

it has to deliver the rest of the current to this. That means it is not possible to

have open circuit here. So, there is a maximum value for R L also. So, we will find out that.

So, we have 14 by R L divided by Beta plus point 7 becoming equal to, let us say it is

1 point zero 3 here because it is of no consequence to discuss it for point 361. Point zero 3,

when it is highest here, 1 milliampere is taken from this. Point zero 3 will go to this.

If point zero 3 goes 200 times, that will appear there. That means 6 milliamperes will

be appearing there. So, this current can be at the limit equal to...lowest value is 6

milliamperes.

Point zero 3 comes here, into 200 is 6 milliamperes. So, this 14 by R L... This Beta is already

taken. 14 by R L plus point 7 milliamperes is equal to 6 milliamperes; or this is 5 point

3 milliamperes; or R L should be less than this 14 divided by 5 point 3 K - 2 point 6

K.

That means we could have done this more easily. A better design would have been to select

this R 1 R 2 combination itself as 2 point 6 K. So, there is no question of our...not

2 point 6 K, 2 point 6 K parallel 20 K. Whatever resistance there is. That R 1 plus R 2 itself

could be that so that we do not have to worry about this lower limit at all. They can keep

it open. That is one way of design; or, you restrict the load resistance to change from

2 point 6 K to 1...196 ohms.

So, load resistance can vary for this particular problem from 196 ohms to 2 point 6 K. Therefore,

it starts malfunctioning. So, let us consider that the load under consideration normally

is a constant, let us say. Let us discuss other parameters for a constant load. This

is the variation of load and we can also plot load regulation by varying only this kind

of resistance, because load regulation becomes meaningless beyond this range. Vary the load

from 196 ohms to 2 point 6 K and find out how output voltage varies with respect to

this variation of load. So, you can plot that characteristic which will be looking almost

as 14 volts constant. This load current is changed from... let us say, corresponding

to R L equal to 196 to...; this is the maximum current, this is the minimum current.

So this variation, you have to...this may be very small variation that is going to occur

primarily. Once again, as the load...if this input voltage is kept constant, this variation

is going to be occurring because of the non-idealities of this differential amplifier that you have.

That is going to be extremely small. That is, we are assuming that the voltage difference

is zero; it is not strictly zero.

So, the error voltage is zero is what it says. So, the error voltage keeps changing as the

load changes because the operating point keeps shifting. So, the error voltage keeps changing

and therefore that error voltage variation will appear at the output as the variation

in the output voltage. So that, you have to determine experimentally basically because

this is a non-linear phenomenon here. Voltage is...the load current is varied over a large

range and the operating point keeps shifting and we have to find out how much the error

voltage changes.

So based on this, we can obtain the load regulation characteristic. And as far as the other parameters

are concerned, like output impedance, this can be evaluated only for a specific load.

Let us take the load as, let us say for making matters simple, 14 K. Is it possible? No.

We cannot take because we can take it as, let us say 2 K - between 2 point 6 and 196

- 2 K. So, the current here is 7 milliamperes and this is again point 7 milliamperes. So,

the total current here is 7 point 7 milliamperes. This is 7 point 7 and therefore the base current

is 7 point 7 divided by 200, which is going to be point zero 38. Point zero 38 milliamperes,

correct? Point zero 7... point zero 38 milliamperes.

Let us assume that this is nominally fixed at, let us say 25 volts, 25 volts. So, this

is fixed at 25 volts. This is 14 point 6. So, this is 25. 25 minus 14 point 6 is the

voltage which is 5, 10 point 4. 10 point 4 by 15 K. How much is that? Point 69 milliamperes.

Now for the 25 volts here. This is 14 point 6 volts. The current is point 69, out of which,

only point zero 4 volts...let us say, this is point zero 4. So, what is remaining here

is point 65. So, this will have point 35 milliamperes.

So, these will give us an idea about the operating currents of each of these. This is going to

be how much? 25 minus 7 - that is 18 by 10 K - 1 point 8 milliamperes. So, these are

the operating currents. Now, how do we fix the output impedance? First, let us find out

the open loop gain of this structure; open loop gain before feedback.

Feedback factor, Beta in this case is half; 10 K, 10 K. So, R 1 by R 1 plus R 2 is half;

and open loop gain is going to be g m. Now, g m is to be evaluated. r e 1 plus r e 2,

1 over that is g m. So, v t divided by i e; i e 1 is point 35, i e 2 is point 65. So,

25 by point 35. How much is that? 25 by point 35 is 71 point 4; plus 25 by point 65. 25

by point 65 - 38 point 46. So, this is the g m of the transistor stage. That into r c

by 2, because single...this is the...this thing and this need not be divided by 2 because

this is the total resistance.

So, this is the g m, into R c. R c is point 15 K. Of course, please remember that this

2 K is shunted by 20 K. So, 20 K effect can be ignored. So, this load is 2 K and this

2 K will appear here as 2 K into 200. So, 400 K. But compared to 15 K, 400 K also can

be ignored. So essentially, it is 15 K. Otherwise, you have to take the loading effect of this

into account in evaluating the open loop gain.

So, this is 15 K - R c. So, that is the open loop gain. This is 15 K by this total, which

is point 8 6 9 1 naught 9 -- 110, about... So, no... What is the...this thing? Total?

15,000 divided by this thing? That means about 150; 2...136 point 3 6; 136. That is the open

loop gain. That into Beta. This is, let us say A naught.

So, A naught Beta is going to be half of 136 -- 68.

So, the original output impedance is going to be decreased by a factor of 1 plus loop

gain. That is what we have learned in our earlier lecture. So, original output impedance

is this r e plus 15 K divided by Beta. Without any feedback, it is simply r e plus 15 K by

Beta, apart from the shunting effect of this. So essentially, original output impedance

without feedback is going to be r e. r e is 25 by 7 point 7 ohms plus 15 K divided by 200. So, you can notice that it

is essentially 75 ohms plus about 3 ohms; 25 by 8 - about 3 ohms. That means 78 ohms.

So, this has to be divided by with feedback, 78 divided by 69 ohms. How much is this? Very

nearly 1 ohm; and that is the output. 1 point 1. So, this is the kind of output impedance

of the structure.

I think that you have to remember is this is a feedback structure; and that is why it

is giving pretty low output impedance. If you use an op-amp, it can become fractions

of ohms; but that is valid only for low frequencies. At high frequencies, we cannot assume this

kind of gain. Loop gain reduces and output impedance increases. This has to be borne

in mind because this regulator may be used for high frequency application.

So, output impedance is important only for high frequencies. Low frequency output impedance

is of no significance. So, how much the circuit that is being supplied...this power is...loop

seeing. That impedance becomes important only at the frequency at which the circuit is functioning.

This is functioning at low frequency. This is a D C regulator and therefore output impedance

on the other hand is important only looking at the circuit from the circuit that is using

this as a power supply.

So, that circuit may be work...working as an oscillator at, let us say 1 megahertz.

Then, you have to evaluate the output impedance at 1 megahertz for this circuit. So, that

may not be purely resistive. So, that output impedance may become pretty high and it may

cause problem, unless you purposely reduce it using special technique.

Like now, this, if it does not use an op-amp, the output impedance at low frequency will

be almost same as output impedance at high frequency. The moment you use an op-amp, that

gets drastically changed. So, output impedance is evaluated. You can also evaluate the repel

rejection by replacing the whole by its equivalent circuit and varying this voltage by a small

extent and finding out what this voltage is; or, most of the repel rejection occurs because

of this. The...when there is a repel, there is a transfer here which is going to be R

z divided by R z plus 10 K; that will be simply multiplied by a factor of 2. So, the repel

rejection is going to be due to this fact that R z by R z plus 10 K into 2 is the repel

rejection factor expressed as a percentage is really what is normally indicated as repel

rejection.

So, these are the important parameters associated with this. Apart from that, we can also find

out the efficiency. Efficiency here is essentially 14 by 25. I have told you that it is just

ratio of output voltage by input voltage; 14 by 25 is the efficiency of this.