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's understand what is
current mirror and why it used. And for that, let me start with the current
source. So, if we take any integrated circuit, then
it consists of so many amplifiers. And to bias these amplifiers, we required
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 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 a lot of circuitry. So, for the amplifiers of the integrated circuits,
we required such stable current sources. So, instead of designing a stable current
source for each amplifier, what is actually done is, one stable current source is fabricated
in the integrated circuit itself and using the current mirror circuit, the same current
source is replicated. 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 these 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 Icopy could be the same as
the reference current or it could be either a multiple or the fraction of the reference
current. That means this Icopy can be equal to Iref. Or it could be equal to NIref or Iref / N. But in any case, it is as stable as the reference
current source. So, now let's 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 Vbe,
we can control this collector current. So, to generate this current mirror, somehow,
we need to sense 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. So, here as you can see, the collector and
the base terminal of the transistor are connected together. So, basically, it is the diode-connected transistor. That means here the BJT acts like as diode. And here this reference current is applied
at the collector terminal. Now here for a moment let's assume that this
base current Ib is negligible. That means here, we can assume that this collector
current Ic is approximately equal to Iref. Now, we already know the relationship between
the collector current and the voltage Vbe right !!
So, in this case, we can say that this Iref or the collector current is equal to Is (ref)
exp (Vbe / Vt). So, here this Is (ref) is the reverse saturation
current of the reference transistor. And from this, we can say that the voltage
Vbe (ref) = Vt ln (Iref / IS (ref)) So, as you can see, here the voltage Vbe is
the function of the reference current. Now, let's apply this voltage Vbe to the base
of the second transistor. So, let's call this second transistor as Q1. So, here this voltage Vbe (ref) and the voltage
Vbe1 are the same. Because here, the base of both transistors
are connected together and the emitter terminal of each transistor is grounded. That means here, voltage Vbe (ref) = Vbe1. Let's call it as Vbe. And as we have seen, this collector current
is the function of the voltage Vbe. That means Ic1 = Is1* exp (Vbe/ Vt)
where Is1 is the reverse saturation current of this 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 the 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 Ic1 is proportional
to Iref. And let's also see that mathematically. So, here this Iref can be given as Is (ref)*
exp (Vbe / Vt) So, if we take the ratio of these two terms
then we can write it as Ic1/ Iref = Is1 / Is(ref) Because these two exponential terms will get canceled out right !!
So, from this, we can say that this collector current Ic1 =
Is1*Iref/ Is(ref) . Now, here if both transistors are identical,
or they are matched pairs then the reverse saturation current of both transistors will
be the same. That means in that case, this current Is1
and Is(ref) would be the same. And in that case, we can say that, this current
Ic1 = Iref. 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 5
times the area of this reference transistor, in that case, this Is1 would be 5 times Is
(ref). And from the expression, Ic1 = Is1/ Is (ref)
- Iref, we can say that, this current Ic1 = 5 * Iref. On the other end, if the area of this transistor
Q1 is (1/5)th of the area of this reference transistor, or effectively (1/5) th of the
area of the emitter-base junction of this reference transistor, then it that case, this
Ic1 would be (1/5)th of Iref. So, in this way, it is possible to generate
the multiple or the fraction of the reference current. So, now let's 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 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 the transistor Q1 is equal to
3A. So, as we have seen earlier, in this case,
the current Ic1, would be 3Iref. And the same current can also be generated
if three transistors are connected in the parallel connection. So, here each transistor has the 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 Iref. And the same current will also flow through
the collector of each transistor. That means here this current Ic is 3 times
Iref. Similarly, If 3 reference transistors with
the area of A are connected in the parallel connection then the collector current of each
transistor is equal to Iref/3. That means we can say that here this Ic1 is
also equal to Iref/3. 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 Ib is negligible. So, now let's 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 also equal
to Ic1. Now, we know that the base current Ib is equal
to Ic/β right !! That means the base current of this transistor
Q1 will be equal to Ic1/β. Similarly, the base current of this reference
transistor is also equal to Ic1/β. That means the current which is flowing through
this branch is equal to 2Ic1/ β right !! So, now to find the expression of the Ic1,
let's apply the KCL at this node. So, applying the KCL we can write, Iref = Ic1
- (2Ic1/β) right !! That means Ic1 = Iref / [ 1 + (2 /β) ]
Now, usually, the value of β is very high. That means in that condition, we can neglect
this second term. And the Ic1 is approximately equal to Iref. That means whenever the value of this β 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's see, instead of 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 = Ic2
= Ic3. Let's say it is equal to Ic. That means here the base current of each transistor
is equal to Ic/β. That means here, this current is equal to
3Ic/β. And the current at the base of this reference
transistor is also equal to Ic/β right !! That means current in this branch is equal
to 4Ic/β. And of course, this collector current is equal
to Ic. So, if we apply the KCL at this node, then
we can write, Iref = Ic + (4Ic/β) That means collector current Ic = Iref / (1
- (4/β)) So, instead of 3 transistors, in general,
if N transistors are connected then this collector current Ic can be given as Iref / [ 1 + (N+1)/β]
right !! So, as far as the value of N is moderate and
the value of β is high, then we can assume that this collector current Ic is approximately
equal to Iref. Similarly, now let's see the effect of this
base current on this circuit. So, in this case, the collector current through
each transistor is equal to Ic/ 3 right !! And the 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/3β
right !! That means for three transistors, the base
current would be equal to 3Ic/3β. That means the current through this branch
is equal to 4Ic / 3β. So, to find the value of this Ic, let's apply
the KCL at this node. That means the applying the KCL we can write,
Iref = Ic/3 + (4Ic/3β) Or we can say that, this collector current
Ic = 3Iref / (1 + 4/β) So, if we neglect the base current, then the
collector current Ic = 3Iref. But with the base current, this is the expression
of the collector current. Similarly, if there are N-transistors are
connected in that case, this collector current Ic can be given as
Ic = NIref/ (1 + (N+1)/β) So, as you can see, when the value of N is
moderate and the value of β is high, then Ic is approximately equal to NIref. For example, if N =5, and β = 100, in that
case, this Ic is approximately equal to NIref. Or to be precise, it is equal to 5Iref. But as the value of N increases, then there
will be an error in the approximation. Similarly, in this case also, as we have seen,
this collector current Ic is equal to Iref/ [1 + (N+1)/β]
So, as the value of N increases, then this collector current Ic would be less than Iref. And the error in the approximation increases. That means here, we can not increase the value
of N indefinitely. For example, if the value of β 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 we can make
copies of this reference current, but we can not increase the value of N indefinitely. 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's 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's call this collector current as Ic. So, for N transistors, the base current over
here would be equal to NIc/β right !! While this current is equal to Ic/β. That means here the emitter current of this
transistor Q is equal to (N+1)*Ic/β And this current would be approximately equal
to emitter current. That means we can assume that this current
is also (N+1)*Ic/β right !! So, if we see the base current of this transistor
Q, then it will be equal to ICQ/β. That is equal to (N+1)*Ic/ β^2
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 Iref = IBQ, that is the base current of this transistor Q plus Ic. That means Iref = (N+1)Ic/β^2 +Ic. That means we can say that, this collector
current Ic = Iref/ [ 1 + (N+1)/β^2] So, as you can see, now in the expression,
instead of β, there is a β^2. So, for example, now if the value of β is
100, and the value of N is also 100, then in that case, still this Ic is approximately
equal to Iref. Or I would say that there will be only 1 percent
error. So, with this configuration, there will be
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
NIref/ [ 1 + (N+1)/β^2] 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 is used? and how it can be implemented using the BJT. So, if you have any questions or suggestions
do let me know here in the comment section below. If you like this video, hit the like button
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