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In this lecture, we will discuss what application of the operation amplifier. In the previous
class, I had shown you how to make a voltage regulator using transistor. If you use transistor
for voltage regulation, we will have problem due to temperature drift, where as that problem
will not be there if you use operation amplifier. Now, first we see how to make a voltage regulator
using operation amplifier, then I will explain some more characteristics of the operation
amplifier.
For example, we have a input that is coming from 230 voltage that is AC 230 volt, that
I feed to a transformer primary and then I have the transformer secondary here. So, this
voltage I rectify then I will get a rectified DC voltage here. So, what we are done is,
you have taken the 230 volt AC and then using a step down transformer, we have reducing
the voltage and then reduced voltage is rectified and then we will get the rectified DC voltage,
we will see how much voltage we have to select and so on little later.
So, you got the rectified DC voltage, this rectified; you know the voltage the whatever
voltage that we are getting between whatever voltage we are getting between this point
and this point. That actually depends upon what is the mains voltage as well, because
if this increases this voltages also increase, if this decreases this also will decrease.
But then we want this voltage across this to be constant irrespective of change in the
input voltage that is what we going to do it here. What I do is, I will take transistor
then then I also since you want high current I put here and then connect the transistor
to this and this I connect to ground. Then what I do is, this I connect the operation
amplifier to this, this connect one serialization to connect this then the output of the transistor.
If put the voltage divider connect this, this is the minus and then plus input I connect
an zener diode. Then for biasing the zener diode I put on resistance and connect it here
and you will get the output voltage V zero, in fact, we also have put one capacitor here
to the output to ground. Now, this is the voltage regulator circuit using operation
amplifier operation amplifier. Now, let us see how this works and what is the advantage
of the operation amplifier, we have got over and above that what we got from the transistor.
Now, first if you see that we got a DC voltage here and that DC voltage this, DC voltage
through this resistance we are applying to this zener diode.
So, if you see only this part now, this is only used biased zener diode so, assume that
you know the zener voltage is 5 volt here. And say the zener can be a very good zener
like for example, we have zener like L M 3 3 6 is there, which is the band gap zener
and this is very stable against temperature. For example, with temperature the voltage
changes only 30 parts per million so, we called as about 30 P P M zener, because per zener
it is very important spec, you know how much voltage will change with temperature. Now,
I have to bias this for that these resistance R 1 that is through this R 1 only, current
is flowing to the zener, this zener needs about one milli ampere current, unless one
milli ampere current goes through that you will not get five volt correctly here.
So, I have to get five volts I such that one milli ampere current goes all the time. Now,
assume that, you know I have some voltage at this point, now how much voltage I need
that depends upon, what is the output voltage that I require. For example, if I take this
is 20 volt that means this is five, that means voltage across this, you know if I take this
you know this is the if I take it as 20 and if this take as five then you have 15 volt
across this. Then if I need to keep one milliampere current flowing through this then I have to
take 15 volt divided by one milliampere that will give me 15 k resistance. Because I have
to maintain one milliampere current, anyway the current flowing through the input of the
op-amp is small, so, I have to keep 15 k. So, depending upon this voltage and the current
requirement, I have to select this resistance, in this example by taking this voltages 20
volt, this comes at 15 k, now this voltage is five volt. Now, you have output voltage
here and assumed that I have put the resistance here ten k, assumed that I have put ten k,
we will talk about this values little later more in detail. Assumed that now this is also
this this you know; this and this are ten k then, it means that whatever voltage that
we are getting here V zero this will be V 0 by 2, because these main acting as a voltage
divider. And there is no current flowing through this; so, if this is say ten volt then this
will be five volt that means whatever voltage is here half of that it is coming here.
We will see what is happening in that op-amp that this point is kept at five volts fix
and this point depends upon the output voltage. Suppose if this voltage is higher than five
, because this is you know this point is at five, if this goes higher than five then the
operation amplifier output will be the different between these two multiply by the open-loop
gain. The open-loop gain of op-amp is that without any feedback, what is the gain that
this op-amp is providing. Even for the simple general purpose operation amplifier the open-loop
gain will be around 20000 so, we say the open-loop gain around 20000.
That means if this is 5 volt and if this comes to 4.9 9 9 it says 1 millivolt difference,
assumed that this is at 4.9 9 9 and this is 5 then you can expect this difference of 1
millivolt multiplied by this 20000 that what suppose to appear at this point. Now, what
is happening is the difference 1 millivolt multiply by 20000 will give you 20 volt at
the output. Now, this operation amplifier also needs for power supply so, what do is
normally, we will apply power to op-amp and that invariably the same as what actually
given here so, we connect the op-amp here. So, this 1 millivolt difference 20 volt at
this point and assuming that this current is very small, base current is very small
then 20 volt, here you know 20 volt at this point will appear as the 20 volt here.
And then this 20 volt actually will become 19.4 volt at this point, because you need
base emitter voltage difference of 0.6. The 19.4 will appear as 18.8, again because you
need only 0.6 volt difference between base and emitter, so, 19.4 volt here will come
as 18.8 here. So, then 18.8 will be appear at this point 18.8 appears then this will
go 9.4 volt, the 9.4 volt is much higher than 5 volt. So, now this is higher this is lower,
the difference is 9.4 minus 5, that is this point is higher by 4.4 volt. That means output
supposed to go minus side that is 4.4 volt into 20000 that much minus voltage come, anyway
we have no minus supply so, output of the op-amp drops to zero, because it cannot go
to minus eight. So, otherwise, what did it can go is zero
so, it is goes to the zero volts. Once it goes to zero this also goes to zero then output
will also ten to zero. If the output goes to zero then this also will become zero, once
if this is zero and if this is sitting at five you know; if this is zero and if this
is sitting at five then again output will go. If the difference is extremely high five
volt, five volt into 20000 have to go to some 10000 volt and so on, but that much is not
possible. So, whatever supply voltage is there 20 volt, the 20 also will appear here, then
again this will 20 and 19.4 18.8 will come, again this voltage will increase.
Now, like that if you see this has to go on increase and decrease increase and decrease
what will happen. But reality what happens is that once you know this is set at five,
this start raising and then we know this voltage across this has to be five then voltage across
this also will be five and if this has come to ten volt. If this has to come to ten volt,
if this has come to for example, if this this point as come to ten volt I know that this
voltages come to 10.6, this has come to 11.2 and if assume the drop across is negligible
then this is 11.2. We know that to get 11.2, I need some difference so, what is the difference
that we need to get the 11.2, because if I need output 11.2 then 11.2 divided by open-loop
gain that is 20000 is the V difference require. There is the difference between these two
point should be this much so, if I calculate this that will be roughly 11 volt divided
by 20 into 10 power minus 3 millivolt; that roughly comes 0.5 millivolt, nearly 0.5 millivolt
difference. That means if this is sitting at five then obviously this has to sit at
4.9 9 9 5 volt, that is only 0.5 volt difference, that this is less, this is more. And then
you got 0.5 millivolt multiplied by 20000 will gives you around 11.2 volt and 11.2 comes
10.6 and this come to ten and that roughly gives you 4.9 9 9 5. Eventually what is happening,
if I keep this at five this comes very close to five, all pattern purpose I can take this
also coming to five, that means the voltage across this is five and voltage across this
also five now output goes to ten. Now for some reason, if this voltage trying
to go more than ten you know; if this voltage if this voltage goes to more than ten, then
this will go more than five then if it is trying to go more than five this will start
decreasing then output will decrease. If this decreases, then this will also try to go lower
than five in case if this tries to go lower than five then this voltage will increase
and the output will also increase. And that make sure that, this point is always kept
at five and this point is always kept at; where this point is always kept at ten and
this point always kept at five, this is how the operation amplifier regulates.
So, if carefully watch that what the operation amplifier doing is whatever voltage here is
given to plus input, the same voltage force to come to the minus input because of this
feedback, because of the closed-loop feedback that is taken place here. And also because
there is i gain in the operation amplifier. And so, what is happening is, that if I keep
here five then automatically this is; this have to stay at five, actually this is not
exactly five, if this little less than five that difference whatever that you are getting
that multiple open-loop gain is what is put here and that what is coming here. So, very
small difference is maintain to generate this ten volt, because open-loop gain is very high.
Now, this is very important, because all that we have to worry is that I keep this five
then I know you will get five here. And if I want say for example, 15 volt, what I do
is? I will remove this ten k, I will put here 20 k for example. In that case then this ten
volt will disappear now assume that you have some higher voltage here, then what happen
is, I know across ten k it is five volt then since the same current is flowing through
this across 20 k it must be ten volt. Because you know across ten k if I get five volt across
20 k I have to get ten volt, then total voltage will be 10 plus 5 then I will get 15 volt
here so, in this case I will just get 15 volt.
So, that means all that need is if I want to change the output voltage I will change
either this or either this or either change the zener voltage. So, the essentially the
we can say that the output voltage, if I see the output voltage V zero output voltage can
be written as V zero is actually. The V reference what we had divided by assume that this is
your R 3 and this is R 4 then V reference by R 3 in to R 3 plus R 4 is the relation
between preference R 3 and R 4. Because V R by R 3 gives you the current through this
then the same current is flowing through this so, total voltage will be total resistance
into the current that is why we have to put this current in to total resistance R 3 plus
R 4 so that gives you the relation for the output voltage.
Of course, you can adjust this or adjust this or adjust this to get any voltage you would
like provided input voltage is more than the output, for example, if you want 15 volt then
I need to have minimum of 15 plus three volt drop is require for transistor work so, I
had get at least 18 volt here. So, if 20 volt I need get 23 volt here so, as long as, I
have the sufficient voltage at the input I can get whatever voltage that I need by varying
either this or this or this. Now, this is how the regulator is working in fact, the
working principle of transistor regulator also same that we had seen in the previous
class. But then the difference here is that, if temperature of this operation amplifier
is changes you will find that is very little change in output.
Because you know we had shown in the previous class that the operational amplifier the temperature
drift introduced by the transistors that basically will be changed due to the temperature is
compensated and they have very little effect. So, if this 5 and this always come to 5 and
with temperature whatever the change that it demand is very small that all in microvolts.
So whereas, in transistor the base emitter voltage changes 2.2 millivolt every one degree
c, whereas, here it changes only few microvolt that is the difference in between the two
base emitter voltage change that alone will be coming here.
That is one of the great advantage using op-amp and then now we have op-amp with very high
gain that also give us you know makes us assume that if I am sitting this is at five and this
also coming at five, that we can assume because open-loop gain operational amplifier is very
high. So, this gives us good insight into how the op-amp operation amplifier is useful
and then how it is very easy compare to the transistor in using the electronic circuits.
Now, this is how the basic regulator is working, now we will let us see how to make practical
voltage regulator that is how to decided on this transformer then we have to also decide
on this decide on this capacitor value. And how to select the diodes, then how to
select this transistors and then transistor also displace heat and we need to put heat
scene for this and then why you need this capacitor here. All these things have to be
decided then only you will get a practically working circuit, because in real life if we
make a circuit that must be practically usable and only then it has any value, simply text
book circuit really has no value and it has only academic interest.
So, we have to make a circuit, which is really useful and someone wires that that must work
without giving any problem. So, from that angle now we see how to design a circuit,
assume I need 15 volt output, I have to use in some circuit. And then at the maximum I
also to draw some current here, assume that this my load resistance and my load resistance
my load resistance never going to be less than 15 in the sense I am not going to draw
more than one ampere current more than one ampere current here. So, that means I am fixing first, what
is my output requirement? That is my output requirement is 15 volt 15 volt 15 volt and
one ampere current and one ampere current means, I am not going draw more than one ampere,
I may draw less than that. So, this is my output requirement, I fix my
output requirement then if I need 15 volt here, I should fix what is the require voltage
at the input here. If I need 15 volt then I need to have at least 18 volt at this point,
because the transistor needs minimum 3 volt to be effectively working. Because we have
to reverse biased the base collector junction that I needs actually minimum 3 volt so, we
need 18 volt at this point and also I have to see that 15 volt is here. And even when
the mains voltage, for example, the mains voltage is 170 then also I want you know this
170 also I want output to be 15, even at 270 the output to be 15.
That means if the output voltage at to be 15 then I know that input has to be 18, that
means I need to get a 18 volt here when the input here is 170. Because at 170 I should
get 80 then only at 170 I would able to get 15. Of course, when the input goes to 270
this will go much higher than 18 and they does not matter that anyway that 15 volt will
be here and the extra voltage dropped on the transistor that work is done by the operation
amplifier automatically. Next that what we have to fixe is, what is the input voltage
range? So, input voltage range is input voltage range is 170 to 270 volt, there is a from
170 to 270 input, I should get constant 15 volt.
If the voltage goes below 170 volt output is not going to be 15 volt that is what we
are accepting it by telling that input voltage range is 170 volt to 270 volt. So, having
fixed the input voltage and then the output voltage and current, now we can decide what
is rating required for the transformers. Now, if we take this now the input voltage that
the we need at this point is minimum 18 volt, so, 18 volt is what is required minimum and
that is at 170 volt. Then we know that when it have AC voltage here if I am rectifying
and then connected to the capacitor, the capacitor charges goes to the peak value because if
I look at the voltage at this point that you have, I will shown you in the next diagram.
See the input voltage is, if I take, if I draw the rectifier schematic then the primary
you have, then the secondary, now the diodes we are putting it here. If I see the waveform,
this is A and this is B and this my ground point C and this D then you will get the voltage
at A and B will be like this then if I look at C and D. If I see the voltage actually
if we have some load here then if you know the first cycle it will climb up like this
then it will come like this it will come like this, this is time versus voltage; this is
voltage across C and D. This is the voltage that we are getting if I start from zero time, what is happening
is, when this first time then current comes here and then at that time the capacitor voltage
was not there it represents was empty so, it charges up to the peak.
This is nothing but the peak voltage, peak voltage of the AC then once it peak is charged
up then what is happening at the output C if I see here that is I will take this point
as if take it as C if we take E then at between C and E you will have the rectified voltage
like this. This is between C and E C and E you will have the voltage coming like this,
it is time versus voltage. So, essentially the first cycle goes like this, it charge
up to peak then AC voltage drop very fast, but then the DC comes slowly because the capacitor
discharges slowly and the next cycle it charges only during this time. The next cycle again
if drops like this and it charges only during this time.
So, you see that the diodes are conducting only near the peak for a small fraction of
the time this time only, that is the actual conduction takes place only here, that is
will start somewhere here and stops at the peak, sometimes it will goes little below
peak also. So, you have you find that the diodes are conducting fraction of every half
cycle and charges of the capacitor close to the peak value. In practically case for example,
one half cycle last for ten milliseconds and if the 50 years, then one cycle is 20 milliseconds,
one half cycle is ten milliseconds. You will find diodes are conducting only near the peak
about three millisecond time, remaining seven millisecond diodes are not conducting.
That means the diodes are actually seven millisecond it is half where disturb positive half cycle
the current actually goes like this and then it comes like this and then goes through this.
Even during that time only three milliseconds they are conducting and remaining seven milliseconds
it is off, anyway next half cycle the totally off, then totally ten millisecond off. That
means out of 20 millisecond each diode as conducting only three millisecond and the
17 millisecond they are off. But nevertheless they charge very close to the peak value so,
the voltage that you are getting here is this secondary voltage that normally given in R
M S. For example, if I have 20 volt AC then I will
get 20 into 1.4 1 4 that is roughly 28 volt DCI will get across this, then this point
is 28 volt then depending upon the load will decreases little bit again comes up again
goes up. So, normally, we forget about the first half cycle so, we assume that the voltage
is like this that is what we see and that is the ripple this what is the bottom, you
know where it is discharging and stopping and then from here is next cycle start conducting
and loading.
So, if I see this carefully one one half cycle I put and show you that how the diode conducts,
because this is very important to select the rating for the capacitor and so on so. If
I take time and voltage you know the it take one half cycle and then I put the next half
cycle. If the first cycle we leave it then at this point the diodes would stop the conduction
then the voltage decreases, then the decreasing the capacitor voltage is taking place.
At the same time the AC voltage is decreasing and reached zero because reached zero then it started rising up then when’s this
and this will started dropping and when at some point they meet. Now, again connect so,
during this time and this time diodes are not conducting, so, only during this time
diode is conducting. So, the story is that you know if you have 20 volt AC you will get
28 volt DC with small ripple, the ripple depends upon the load resistance. That means if I
want 28 volt, I have to specify 20 volt AC here.
Now, in our discussion if we go back to our earlier discussion that you know we want 15
volt output for that 18 volt here, at 18 volt is needed when the input is 230 volt. So,
to get 18 volt we need some AC voltage, the required AC voltage for 18 volt is 18 volt
DC, the required AC is 18 divided by 1.4 1 4 so, that comes nearly 180 divided by 14
that will call for about 13 volt AC is require. That mean to have 13 volt AC here that will
give me 18 volt DC here after rectification, but then I know that I need 18 volt when the
input is 170 volt. That means I have to find out what is the voltage that I need at this
point at 230 volt, because if I am giving specifications for a transformer I should
give at 230 volt what is the voltage require. So, you need 13 voltage AC at 170 volt, because
at 170 I have to get 13 that will give me 18 here and that will make sure I get the
15 voltage at the output, that is at 170 volt at 170 volt AC input secondary voltage is
the 18. So, at 230 volt at 230 volt input secondary voltage actually is equal to you
will have 13 into 230 divided by 170 that is the voltage required. That if you work
out that tense out to be, you know if I can calculate and show you, so the next actually
see this thing. At 230 volt AC input and secondary voltage is actually that 13
volt into 230 divided by 170. So, if I calculate this that actually comes, if we multiply this
you will get divided by 17 that will come get 12 and 19 so, that will be 8 times roughly
18 volt. So, you need you need secondary voltage of 18 volt is required at 230 volt that means
higher transformers take as primary zero to 230 volt, secondary zero to 18 volt, because
at 230 you get 18 and that will make at 170 to get 13 volt at 170 volt.
At 170 volt here then automatically I will get 13 and that will give me 18 and I was
assured of 15 volt here. Of course, at 230 I need 18 volt and that will give more voltage
here 18 into 1.4 1 1 4, does not matter this you will still get 18. That means we have
decide on the transformers first the transformers can be rating can be decided like this. So,
now what the voltage is required is, I need to have 18 volt that is at 230 I will get
18 volt here and then this voltage go 18 into 1.4 1 4 1 so, that is how the transformer
has to be decided. Then the question of deciding on the capacitor
here, because this capacitor also we have to decided that the value of the capacitor
will be decided depending upon the load. Now, how we calculate this capacitor value? Now
I can calculate the value of this capacitor by considering how much time that the energy,
how much time this discharging to the load, how much time it is charging up. Now, I will
show you soon that in the most of the cases, we can see that the conduction angle is always
around three millisecond in one half cycle. Then of course, the capacitor have to discharge
remaining time that is during three millisecond the diodes are supplying energy to the capacitor
as well as energy to the output. And then remaining seven millisecond the capacitor
have to discharge, that means I can take the charging time of capacitor is three millisecond
and then discharging time as seven millisecond. So, if the capacitor discharges for seven
milliseconds we know that the voltage will be decrease and if I put large value capacitor
then you find that the voltage decreases small that means ripple will be small.
The larger value capacitor will get up smaller the ripple, but then if you keep too much
larger capacitor again is not that going to ripple I will show you little later that if
keep on increase on the capacitor we are not going to get much benefit. Of course, the
ripple will be come down, but the conduction angle is not going to change much and the
output voltage also will decrease. So, depending upon the ripple requirement I had to select
the capacitor. There is a small complication is there between the conduction angle and
capacitor size, right now we will not get into that argument h7owever, if I put large
value capacitor you will find smaller ripple at that point.
So, the ripple voltage the ripple voltage is the one which decides the size of the capacitor.
The ripple voltage is to be determined so, assume that I want ripple voltage of say three
volt. If I need a ripple of the three volt then I have to select what is the capacitor
require. We know that basic formula i into t is equal to C into V this is the basic equation
governing the charge and the voltage. Now, in this case, I want ripple of the capacitor
that is how much this capacitor discharging so, I take C into delta V is this. Then i
is the load current that is in the case one ampere and actually the discharge time is
seven millisecond 7 into 10 power minus 3, because capacitor discharging for seven millisecond.
And we will have discharging seven millisecond, the voltage will be decrease that voltage
is taken as delta V. So that C actually is given by now the C actually is given by 7
into 10 power minus 3 divided by delta V. So, in this case, I wanted three volt ripple
that means 7 by 3 into 10 power minus 3 that is actually 2.3 into 10 power minus 3 farad
R that is equal to 2300 microfarad. The standard value is 2200 microfarad very well suits here,
so, I can put 2200 microfarad that will give me ripple voltage of three volt when the load
current is one ampere. So, that is how the capacitor value is selected now we are selected
the transformer voltage and then we have selected the capacitor value.
Now, one thing that you know the diodes are conducting only for three millisecond and
during that time only all the energy is coming and remaining seven millisecond diodes are
not conducting and the transformer also not giving any power. But we are getting continuously
one ampere power at this point, that obviously obviously puts some questions that if I am
drawing one ampere continuously and if I am drawing power only three millisecond and out
of ten millisecond ten millisecond half cycle. Then obviously during that, three millisecond
more current suppose to come from the transformer, more than one ampere current suppose to come
from the transformer. That is true because if this giving one ampere current continuously
and if this charging only three millisecond. And during this three millisecond definitely
more than one ampere current have to come and this is very important, because very often
we make this mistake in selecting the current rating for the transformer. Because there
are two additional problems that we will get in this.
For example if this transformer continuously supplies one ampere the heating will be less,
but if it supplies say two ampere half the time and other half the time if it slips without
giving any power you will find the heating is more. Even though, it not given the same
amount of power during average time; during you know long average period of time, but
if are not taking power continuously then heating will be more. That is what happens
in this diode capacitor arrangement, because the diodes are not conducting, continuously
we are drawing large power for three millisecond and rest of time it is quit.
But that makes the transformer to heat up more than the normal amount that means, if
I am designing the transformer then if I have to rate the current rating at the transformer
I have to give that actually calls for over rating of this transformer. If I need one
ampere here definitely I need more than one ampere rating for this transformer. Now, for
many students it may be a worry that finding out why the transformers have to heat up more,
if we are not drawing the power continuously. This is because the power loss is actually
given by I square R is the power loss. Whereas, where I is the current nor is the
resistance of the equivalent resistance of the transformer take in the case only the
secondary resistance alone. Assume that I have resistance as one ohm R is equal to one
ohm and then I need load current load current I need one ampere. Assume I am taking only
three milliseconds current then the peak current is much more than one ampere. It is assume
during three millisecond I am trying average current of three ampere. That means in this
case the same I can take heat loss in two cases that is assuming I am taking continuously
one ampere current from here. You know if I am taking one ampere current
from the transformer continuously then what is the power loss I calculate. Then next case
I assume the transformer is conducting only one-third of time, but it is taking three
ampere current, it is one ampere current it delivers three ampere current for one-third
of at time. That means total power plan is same in both cases so, we take two cases and
find out in which case the heating is more, so, what I do is I will take this two examples.
See case 1, continuously power is taken, that is continuous power one ampere continuously
taking and case 2, three ampere for one-third of a time one-third of a time taking three
ampere, that means average power is same in both cases. Now, here I square R, if I find
I square R in this case will be one square and R is one ohm that I will have one watt
power distribution. In this case I will have I square that is three ampere that is nine,
that is if I take I square that is 3 square into R then of course, we have only one-third
of time is conducting, so I have to divide this by 3. So, here in this case 9 into R
is 1, we get 3 that is equal to 3 watt. That means in this case we had a heat dissipation
of one watt and if I take three ampere per one-third of a time then heat dissipation
is three watt. So, it is better to take the power continuously in the pulsed form. When
we are taking current in a pulsed form more heating is taking place, but then if you go
back and see we have no choice because once you put the regulator, once you put the; once
you put the diode and capacitor the current comes only in the pulsed form. That means
you will have more heating on the transformer that automatically happens. Now, only choice
that we have is that we design a transformer such that it takes, it accepts that much heat
other around telling this would be over heat the transformer that is if I want one ampere
current then I put more than one ampere rating for this transformer.
I can show you in some other calculation later that for transformer of I need to rate 1.8
times actually. That is you know if I want one ampere current for one ampere load secondary current current
requirement requirement will be one ampere into 1.8 times. Now, you ask what is this 1.8 that is for example, in these cases
I will put 1.8 ampere rating. This is one ampere I will put give 1.8 that means you
going to use the thick wire and the heating going to come down. Of course how many times
we have to over rate depends upon what is the conduction angle.
The conduction angle is smaller the conduction angle is smaller then overwriting have to
be more, the conduction angle is more than overwriting may not be more, because the conduction
angle is more means is more or less it is conducting continuously. So, depending upon
the conduction angle I have over rate the transformer, now you will be surprised later
that even if I change this you know 2200 M F to 4700 M F you will find the conduction
angle almost remains same. Because in our classical text books there is one misconception
that if will increase the capacitor value the conduction angle will come down.
That is not really true, that is true only for ideal transformer for our practical transformer,
by increasing this you cannot actually increase the, reduce the conduction angle. This happens
is because for example, if I put one farad capacitor here, then the ripple is almost
zero, that means the conduction angle will be very very narrow, that you may get few
microsecond only conduction angle. For example, if I put one farad capacitor the conduction
angle comes to few microseconds, obviously the transformer will not able to charge this
capacitor of one farad in few microseconds. So, you will find the conduction angle still
remains close to three milliseconds only. And this needs a big explanation, at this
point of time we will not look at it, you will find for practical values of 2200 microfarad
or 4700 microfarad or 10000 microfarad you will find the conduction angle is almost close
to three milliseconds only. So, the classical misconception that if we increase the capacitor
value that conduction angle will come down and current will suit up that is not correct.
You can practically see increase this capacitor and you will find conduction angle remains
three millisecond more or less and current also only about the three times the load current.
So, considering this this factor of 1.8 was selected and we can mathematics how will this
1.8 ampere are come and that is not require at this point of time, we will see that at
later time. So, all that for most of the practical cases
are we need to have overwriting of 1.8 times, so that decides what is the rating require
for the transformer. That is in this case for one ampere and 15 volt output I need 18
volt and 1.8 ampere current rating for this transformer and I need 2200 microfarad capacitor
to work this. So, that decides what is the capacitor size and the transformer so on.
Now, we have go on now we have go and decide and various others comments, particularly
the heat sink and then what type of transformer transistor you have to use, where you need
Darlington pair. And then if I use different op-amps 7 4 1 or 7 1 4 like that if I use
different operation amplifier, what happens; that remains to be seen.
Similarly, we have to that worry why I need this capacitor, now it looks that designing
at simple regulator seems to be very complicated. But actually it is not that way we will show
you that some more complications that you know in case if I ground the output accidently.
For example, if the you have a load here and if I accidently ground this then I know that
this goes to zero and then this goes to zero then this is sitting anyway at five and that
make the output voltage go very high. Because the difference is now five 5 volt, the five
into open-loop gain V i and this will go to 20 volt, that will come to 20 and that will
come to 19.4 and this is zero and the 19.4 then first distance will goes bad then once
that goes bad this also goes bad so, this and the transistor go bad.
That means we also had provided short circuit production, particularly it happens you know
the short automatically at beginning. At the beginning this capacitor is empty, you know
this not charge so, large current runs like a short and most of the cases as soon as switch
on this regulator goes bad. Even though the text book circuit it is ok, but then practically
if I provide short circuit protection it goes bad, you will never able to use this regulator.
So, we also has provided as short circuit protection, also we have to decide on this
resister and then this resister. And then also I may ask why did you select here 20
k and 10 k and I can also have one k and two k, like that there so many questions come
in and all this things have to be answer. So, we need not get fainted with simple regulator
like this so many design issues there. It is so, natural that design involves you know
putting very comments and that comments must be perfect in all respect and that is actually
the design. So, you know learning electronic circuit, without getting into design is very
simple, but that is useless. So, your focus would be always how to design a circuit, how
to select the different values and then what is the extreme conditions. You know when it
will failed and so, what happens the main voltage changes and then what is the really
happening and that should be looked into that very carefully then only the design will be
complete. In fact the focus of the entire course will be like this only that everything
will be getting into details and then we will select the values considering very many factors.
Now, for example, in this case we are selected the transformer and then the specification
related to that. Now, let us see how will design this transformer itself, there is another
question, because many times you may say I miss give specification for the transformer
manufacturing and give this one. But then that is only of this worry because if you
are power supply manufacture, you will find cost of this transistor, then cost of this
operation amplifier and all this electronic component if we take you know the cost is
very small. Most of the cost goes only to the transformer
and if you are buying the transformer, you order the transformer and getting it from
supplier; you will find that all the money actually goes to the transformer manufacture
and I say the voltage manufacture you get nothing at all. So, if you are in practical
world, if you are making a regulator, voltage regulator then you must also make the transformer.
And as a analog circuit designer you should know, how to design the transformer also.
If you do not know how to design the transformer then really the no use in studying the voltage
regulator. Because you are studying the voltage regulator
only with an intention to produce them, because all our circuit that what we going to discuss
that I assume that you are running a industry and for the industry we are developing that
product. That is the approach that I am going take to throughout this course I am not going
to teach you from the academic point of you, how the circuit works and write the equation
and then move on from there. And as far as I am concern that is useless and I advised
you not to do, not to take that kind of approach, that will do in over a period of time. That
is what happens is you know many times when we study in our college that is what we have
done, we are actually taken approach that you want understand the circuit, that is not
correct approach. We get assume that I am going to manufacture
this circuit and for that how I going to, what are things I have to learn and how I
have to optimize the cost and so on so, that is what we going to do. So, next we will take
up how to design the value of R 1 and this resister, then how will select the value of
this, this and then which transfer you will select and how will select the heat sink for
this then also we discuss how will you design this transformer itself. Because that transformer
design learning will help you at many places, because in analog circuit we will be designing
transformer for very many purpose. Like for example, when I want to make a accelerator
I will have to design a transformer and then we also design a filter that time I need to
design a inductor. For the inductor designer I have to anyway
learn how to find the coil and so on so, that way learning how to design a transformer also
will be very great help to you in future. So, the next class I will talk about how to
design various other components that is associated with this and then also tell you how to design
the transformer in a simpler way. To design a transformer you do not need any computer
or any calculator you know, we can; I will show you how design it very easily. Then you
can look back and see how the text books are shown how to design the transformer, so, this
I will stop this lecture will continue in the next class. Thank you.