NAME
yt1z.net - BJT- Current Mirror Explained.mp3
DATE
January 12, 2025
DURATION
20m 52s
START OF TRANSCRIPT
(Transcribed by Sonix.ai - Remove this message by upgrading your Sonix account)
[00:00:12] Speaker1
Hey friends, welcome to the YouTube channel All About Electronics. So in this video we will learn about the current
mirror circuits. So first of all, let us understand what is current mirror and why it is used. And for that, let me start with
the current source. So if we take an integrated circuit then it consists of so many amplifiers. And to bias these amplifiers
we require the so many current sources. Now the advantage of biasing the amplifier with the current source is that here
the emitter current is independent of the beta as well as the RB. Moreover, by increasing the value of RB, we can
increase the input impedance. So for that, this biasing current should be very stable. That means the value of this
biasing current should not change with the temperature as well as the supply voltage. For example, if we take the case
of the smartphone, then as the battery discharges, then there is a reduction in the supply voltage. So this current
source should be able to maintain the constant current even if there is a change in the supply voltage. Well, fortunately
it is possible to design such current sources, but it involves the lot of circuitry. So for the amplifiers of the integrated
circuit we require such stable current sources. So instead of designing a stable current source for each amplifier, what
is actually done is when stable current source is fabricated in the integrated circuit itself and using the current planar
circuit, the same current source is replicated.
[00:01:50] Speaker1
So this replicated current source has the same characteristic as the reference current source. That means this current
source is as stable as the reference current source. And this current sources can be used to bias the amplifiers. So this
current mirror is the active circuit, which senses the reference current and generates the copy or the number of copies
of this reference current with the same characteristics. Now here this I copy could be same as this reference current, or
it could be either a multiple or fraction of the reference current. That means this I copy can be equal to I reference, or it
could be equal to n times I reference or I reference divided by n, but in any case it is as stable as the reference current
source. So now let us see how it can be designed using the BJT. Now we know that the BJT can be used to generate
the current source, because the collector current IC is the function of the voltage VBE. That means by controlling this
voltage VB, we can control this collector current. So to generate this current mirror somehow we need to change this
reference current. And we need to convert this current into the voltage. And we need to apply this voltage between the
base and the emitter of this transistor, so that this collector current IC is proportional to the reference current. So this is
the way by which we can convert the reference current into the voltage.
[00:03:28] Speaker1
So here as you can see the collector in the base terminal of the transistors are connected together. So basically it is the
diode connected transistor. That means here the BJT acts like a diode. And here this reference current is applied at the
collector terminal. Now here for a moment let us assume that this base current IB is negligible. That means here we
can assume that this collector current IC is approximately equal to I reference. Now we already know the relationship
between the collector current and the voltage VB, right? So in this case we can say that this I reference or the collector
current is equal to is reference times e to the power voltage VB divided by v t. So here this is reference is the reverse
saturation current of the reference transistor. And from this we can say that the voltage VB is equal to v t times natural
log of I reference divided by I is reference. So as you can see here, the voltage VP is the function of the reference
current. Now let us apply this voltage VB to the base of the second transistor. So let's call this second transistor as the
Q1. So here this voltage VB reference and the voltage VB one are same because here the base of the both transistors
are connected together and the emitter terminal of each transistor is grounded. That means here voltage VB reference
is equal to VB one.
[00:05:09] Speaker1
Let's call it as VB. And as we have seen, this collector current is the function of the voltage VB. That means IC one is
equal to ES one times E to the power voltage VB divided by v t, where ES one is the saturation current of the transistor
Q1. Now here only one thing which we need to ensure is that the collector base junction should remain in the reverse
biased, because here this transistor Q1 should remain in the active region. That means the collector voltage should be
always more than the base voltage. So by ensuring this we can replicate this reference current. So basically here what
we did. First of all we generated the voltage which is proportional to proportional to this reference current. And then we
have applied this voltage as an input to the second transistor. And this second transistor generates the collector
current based on the input voltage. So here the first transistor acts as a current to voltage converter, while the second
transistor acts as a voltage to current converter. So here this collector current IC one is proportional to the I reference.
And let us also see that mathematically. So here this I reference can be given as I is reference times e to the power
voltage vb divided by v t. So if we take the ratio of these two terms then we can write it as ic one divided by I. Reference
is equal to IS1 divided by is because these two exponential terms will get cancel out right.
[00:06:54] Speaker1
So from this we can say that this collector current IC one is equal to is one divided by is reference times I reference.
Now here, if both transistors are identical or they are the matched pairs, then the reverse saturation current of both
transistors will be same. That means in that case this current is one and the ith reference would be same. And in that
case we can say that this current IC one is equal to I reference. So in this way using this current mirror we can replicate
this reference current. Now in general this saturation current is proportional to the area of the emitter base junction. So
here basically I am talking about this highlighted area. So if the area of this transistor Q1 is five times the area of this
reference transistor, in that case this is one would be five times the ES reference, and from the expression ES one is
equal to ES one divided by ES reference times I reference. We can say that this current IC one would be five times the I
reference on the other end. If the area of this transistor Q1 is one fifth of the area of this reference transistor, or
effectively one fifth of the area of the emitter base junction of this reference transistor, then in that case, this IC one
would be one fifth of the I reference.
[00:08:37] Speaker1
So in this way it is possible to generate the multiple or the fraction of the reference current. So now let us see how we
can generate the multiple copies of this reference current. So this is the way we can generate the multiple copies of the
reference current. So as you can see over here The base of each transistor is connected to the base of the reference
transistor, or more commonly, it can be shown in this fashion. So depending on the area of the emitter base junction of
each transistor, the collector current could be either multiple or the fraction of the reference current. Now, in the
integrated circuits, the area is usually defined in terms of the area of the unit transistor. So here let's say A is the area of
the emitter base junction of the unit transistor. So here the reference transistor has the area of a. While the area of the
emitter base junction of this transistor Q1 is equal to three A. So as we have seen earlier, in this case the current ic1
would be three times the AI reference, and the same current can also be generated if the three transistors are
connected in the parallel connection. So here each transistor has an area of A and the collector terminal of each
transistor is connected together. So here the current through the collector of this reference transistor is equal to I
reference, and the same current will also flow through the collector of the each transistor.
[00:10:12] Speaker1
That means here this current IC is equal to three times the I reference. Similarly, if the three reference transistors with
the area of A are connected in the parallel connection, then the collector current of each transistor is equal to I
reference divided by three. That means we can say that here. This IC one is also equal to I reference divided by three.
So in this way we can generate the multiple or the fraction of the reference current. Now so far in our discussion, we
have assumed that the base current IP is negligible. So now let us see the effect of this base current on the replicated
current Ic1. So here both transistors are of the same area. That means here the collector current of both transistors will
be same. That means here this current will be also equal to ic1. Now we know that the base current IB is equal to IC
divided by beta right. That means the base current of this transistor Q1 will be equal to IC one divided by beta.
Similarly, the base current of this reference transistor is also equal to IC one divided by beta. That means the current
which is flowing through this branch is equal to two IC one divided by beta, right? So now to find the expression of this
IC one, let us apply the KCL at this node. So applying the KCl we can write this AI reference is equal to IC one plus two
IC one divided by beta, right? That means Ic1 is equal to I reference divided by one plus two divided by beta.
[00:12:08] Speaker1
Now usually the value of beta is very high. That means in that condition we can neglect this second term and the ic1 will
be approximately equal to I reference. That means whenever the value of this beta is very high, then the base current
IB will be negligible. And in that case, this collector current IC is approximately equal to the reference current. Now let
us see instead of the one transistor, if there are three transistors with the same area, then what would be the collector
current? So here we are. Assuming that the ic1 is equal to ic2 is equal to IC three. Let's say it is equal to IC. That
means here base current of each transistor is equal to IC divided by beta. That means here this current is equal to
three ic divided by beta, and the current at the base of this reference transistor is also equal to IC divided by beta,
right? That means current in this branch is equal to four IC divided by beta. And of course this collector current is equal
to IC. So if we apply the KCL at this node then we can write I. Reference is equal to IC plus four IC divided by beta.
That means collector current IC is equal to I reference divided by one plus four divided by beta.
[00:13:48] Speaker1
So instead of three transistors in general, if the n transistors are connected, then this collector current IC can be given
as I reference divided by one plus n plus one divided by beta, right? So as far as the value of n is moderate and the
value of beta is high, then we can assume that this collector current IC is approximately equal to I reference. Similarly,
now let us see the effect of this base current on this circuit. So in this case the collector current through each transistor
is equal to IC divided by three right. And same current will also flow through this reference transistor or to be precise
through the collector of this reference transistor. So here this base current is equal to IC divided by three beta right.
That means for the three transistors the base current would be equal to three IC divided by three beta. That means the
current through this branch is equal to four ic divided by three beta. So to find the value of this IC, let's apply the KCL at
this node. That means applying the KCl we can write I. Reference is equal to IC divided by three plus four IC divided by
three beta. Or we can say that this current current IC is equal to three times I reference divided by one plus four divided
by beta. So if we neglect the base current, then the collector current IC is equal to three times I reference, but with the
base current this is the expression of the collector current.
[00:15:43] Speaker1
Similarly, if there are n transistors are connected. In that case, this collector current IC can be given as n times I.
Reference divided by one plus n plus one divided by beta. So as you can see, when the value of n is moderate and the
value of beta is high, then IC is equal to approximately n times I reference. For example, if n is equal to five and beta is
equal to 100. In that case, this IC is approximately equal to n times I reference. Or to be precise it is equal to five times I
reference. But as the value of n increases, then that will be an error in the approximation. Similarly, in this case, also, as
we have seen, this collector current IC is equal to I reference divided by one plus n plus one divided by beta. So as the
value of n increases, then this collector current IC will be less than the I reference. And the error in the in the
approximation increases. That means here we cannot increase the value of n indefinitely. For example, if the value of
beta is 100 and the value of n is also 100. In that case, there will be a considerable amount of error in the
approximation. And in fact, the value of this collector current is almost half of this reference current. So it shows that
even though with the current mirror we can make the copies of this reference current, but we cannot increase the value
of n indefinitely.
[00:17:28] Speaker1
But there are ways by which the performance of the current mirror can be improved. So if you observe this circuit, then
one more transistor is also added over here. So let us find out how it will improve the performance of the current mirror.
So here we are assuming that the collector current of each transistor is the same. Let us call this collector current as
the I see. So for the N transistors, the base current over here would be equal to n times IC divided by beta, right? While
this current is equal to IC divided by beta. That means here the emitter current of this transistor Q is equal to n plus one
times IC divided by beta. And this current would be approximately equal to emitter current. That means we can assume
that this current is also n plus one divided by beta times IC right. So if we see the base current of this transistor q, then
it will be equal to ic q divided by beta that is equal to n plus one times IC divided by beta square. And of course this
current over here is equal to IC. So to find the value of current IC Now let's apply the KCL at this node. That means if we
apply the KCl then we can write this current I reference is equal to I bc that is the base current of this transistor q plus
ic.
[00:19:10] Speaker1
That means I reference is equal to n plus one times ic divided by beta square plus IC. That means we can say that the
skeletal current IC is equal to I reference divided by one plus n plus one divided by beta square. So as you can see
now in the expression instead of beta there is a beta square. So for example, now if the value of beta is 100 and the
value of n is also 100, then in that case still this IC is approximately equal to I reference. Or I would say that there will be
only 1% error. So with this configuration there is a significant improvement over the previous case. Similarly, in this
case also if you follow the same procedure, then this collector current IC can be given as n times I reference divided by
one plus n plus one divided by beta square. So by including this transistor, we can improve the performance of this
current mirror. And very soon we will take some examples on the current mirror on our second channel. But I hope in
this video you understood what is current mirror, why it is used and how it can be implemented using the BJT. So if you
have any question or suggestion, do let me know here in the comment section below. If you like this video, hit the like
button and subscribe to the channel for more such videos.
(Transcribed by Sonix.ai - Remove this message by upgrading your Sonix account)
END OF TRANSCRIPT
Automated transcription by Sonix
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