Table of Contents

## Current Mirror:

In this article we will learn about the Current Mirror circuits. So, first of all, let’s understand what is Current Mirror and why it is 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 R_{b}.

Moreover, by increasing the value of R_{b}, 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. 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 I_{copy} 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 I_{copy} can be equal to I_{ref}.

I_{copy}= I_{ref}

Or it could be equal to:

I_{copy}= N × I_{ref}

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.

## BJT Current Mirror:

Now, we know that the BJT can be used to generate the current source. Because the collector current I_{c} is the function of the voltage V_{BE}. That means by controlling this voltage V_{BE} 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. We need to apply this voltage between the base and the emitter of this transistor, so that, this collector current I_{C} is proportional to the reference current. So, this is the way, by which we can convert the reference current into the voltage. So 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 the BJT acts like as diode. This reference current is applied at the collector terminal. Now for a moment let’s assume that this base current I_{B} is negligible. That means here, we can assume that this collector current I_{C} is approximately equal to I_{ref}. Now, we already know the relationship between the collector current and the voltage V_{BE}.

So, in this case, we can say that this I_{ref} or the collector current is equal to:

So this Is (ref) is the reverse saturation current of the reference transistor. From this, we can say that the voltage:

So as you can see, here the voltage V_{BE} is the function of the reference current. Now, let’s apply this voltage V_{BE} to the base of the second transistor. So, let’s call this second transistor as Q1. So, here this voltage V_{BE(ref)} and the voltage V_{BE 1} are the same. Because the base of both transistors are connected together and the emitter terminal of each transistor is grounded.

That means here, voltage:

V_{(BE (ref) )}= V_{BE1}= V_{BE}

We have seen, this collector current is the function of the voltage V_{BE}.

That means

Where I_{s1} is the reverse saturation current of this transistor Q1. Now 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. Then we have applied this voltage as an input to the second transistor. The second transistor generates the collector current based on the input voltage. So, the first transistor acts as a current to voltage converter, while the second transistor acts as a voltage to current converter. So the collector current I_{c1} is proportional to I_{ref}. Let’s also see that mathematically:

Now 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 I_{s1} and I_(s (ref)) would be the same and in that case, we can say that, this current

I_{c1} = I_{ref}

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, if the area of this transistor Q1 is 5 times the area of this reference transistor, in that case:

I_{s1} = 5 ×I_{s(ref)}

I_{C} = I__{S1}/I_{(S(ref))} × I_{ref}

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 I_{C1} would be (1/5)th of I_{ref}. So, in this way, it is possible to generate the multiple or the fraction of the reference current.

## Generation of the multiple copies 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. A is the area of the emitter-base junction of the unit transistor. 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 I_{c1}1, would be:

I_{C1}= 3 × I_{ref}

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 the current through the collector of the reference transistor is equal to I_{ref}. The same current will also flow through the collector of each transistor. That means the collector current I_{C} is 3 times I_{ref}.

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 I_{ref}/3 .That means we can say that here this I_{c1} is also equal to I_{ref}/3. So, in this way, we can generate the multiple or the fraction of the reference current.

## Effect of the base current:

Now, so far in our discussion, we have assumed that the base current I_{B} is negligible. So, now let’s see the effect of this base current on the replicated current I_{C1}. So, here both transistors are of the same area. That means the collector current of both transistors will be same. That means this current will also equal to I_{C1}. Now, we know that the base current I_{B} is equal to Ic/β which 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 I_{C1}/β. That means the current which is flowing through this branch is equal to (2I_{C1})/β. So now to find the expression of the I_{C1}, let’s apply the KCL at this node. So, applying the KCL we can write:

Now, usually, the value of β is very high. In that condition, we can neglect this second term. And the I_{c1} is approximately equal to I_{ref}. That means whenever the value of this β is very high, then the base current I_{B} will be negligible. In that case, this collector current I_{c} 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 I_{c1} = I_{c2} = I_{c3}. Let’s say it is equal to Ic. That means here the base current of each transistor is equal to I_{c/β} . This current is equal to 3I_{c/β}. The current at the base of this reference transistor is also equal to I_{c/β} which means current in this branch is equal to 4I_{c/β}. this collector current is equal to Ic.

So, if we apply the KCL at this node, then we can write

So, instead of 3 transistors, in general, if N transistors are connected then this collector current Ic can be given as:

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 I_{c} is approximately equal to I_{ref}. 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 I_{c/3}. 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 I_{c/3β} which means for three transistors, the base current would be equal to 3I_{c/3β}. That means the current through this branch is equal to 4I_{C/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:

I_{ref}= I_{c/3} + (4 I_{C/3β}

we can say that, this collector current:

I_{c}=(3 I_{ref})/(1+4β)

So, if we neglect the base current, then the collector current I_{c} = 3 I_{ref}. 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:

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 NI_{ref}. For example, if N =5, and β = 100, in that case, this Ic is approximately equal to NI_{ref}. or to be precise, it is equal to 5 I_{ref}. 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 I_{C} is equal to

So, as the value of N increases, then this collector current Ic would be less than I_{ref} and the error in the approximation increases. That means here, we cannot 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. 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 cannot 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 NI_{c/β}. While this current is equal to Ic/β. That means here the emitter current of this transistor Q is equal to (N+1)I_{c/β} and this current would be approximately equal to emitter current. That means we can assume that this current is also (N+1)I_{c/β}. So, if we see the base current of this transistor Q, then it will be equal to I_{cQ/β} =(N+1)I_{c/β2} this current over here is equal to Ic.