Hey friends, welcome to the YouTube channel ALL ABOUT ELECTRONICS. So, in this video, we will understand, how to design the differential amplifier using the BJT. The differential amplifier amplifies the difference between the two input signals. And it is widely used in integrated circuits. If you see the internal structure of the op-amp, then the first stage of the op-amp is the differential amplifier. So, let's understand, what are the advantages of the differential amplifier and how it can be designed using the BJT. But before that, first of all, let's understand what is a differential input and what is the difference between the single-ended and the differential input. And to understand that, let's take the example of the BJT amplifier. So, all the BJT amplifiers which we have discussed so far are the single-ended. where the input is applied between the input and the ground terminal, while the output is measured between the output and the ground terminal. So, in such a single-ended amplifier, when some noise or the interference gets coupled to the input, then it will also get amplified and will appear at the output side. But in the case of the differential input, such common-mode noise on the input side can be eliminated at the output. So, let's say, these are the two input terminals of the differential amplifier. And this input Vin1 is the input to the first terminal w.r.t ground. Similarly, this Vin2 is the input to the second terminal w.r.t ground. And here, this Vn is the noise, which is common to both input terminals. So, with the noise, if we see the input at the first terminal, that is Vin1' then it is equal to Vin1 +Vn. Similarly, this Vin2' will be equal to Vin2 + Vn. And the differential input between the two terminals, that is Vd is equal to Vin1' - Vin2'. That is equal to Vin1- Vin2. That means at the output, the noise won't appear. So, that is the advantage of this differential amplifier. which amplifies the differential input or the difference between the two input terminals and eliminates the common-mode noise. Or the common-mode signal, which is common to both input terminals. Moreover, by taking the output in the differential configuration, it is possible to eliminate the supply hum or any noise which is common to the amplifier circuit. And when such differential amplifier stages are cascaded together, then it also eliminates the need of the coupling capacitors which is required in the case of the single-ended amplifiers. So, these are some of the advantages of the differential amplifier. And this is the typical circuit of the differential amplifier. So, as you can see, the differential amplifier consist of the pair of BJTs And for the differential amplifier to work properly, the transistor pair should be matched perfectly. That means all the transistor properties like the current gain beta, voltage Vbe should match perfectly. Moreover, these collector resistors should also be the same. And if there is any mismatch, then this differential amplifier won't be able to eliminate the common-mode noise perfectly. Anyway, so as you can see, the emitter of both transistors is tied together to the constant current source. So, in the integrated circuits, it is possible to match these two transistors as well as the collector resistor. And in that case, this constant current source is designed using the current mirror. But when this circuit is designed using the discrete components, then there will be some mismatch between the transistor pair. And in that case, this emitter can be biased in this fashion. But here during our discussion, we will assume that both transistors are matched perfectly. So, in this differential amplifier, the input is applied between the base of the two transistors. But for DC analysis, both inputs will act as a ground. So, let's say, here the current through the emitter resistor is equal to Iee. And since both transistors are matched perfectly, so this current Iee will get equally divided between the two transistors. That means the current through the emitter of both transistors will be equal to Iee/2. And applying the KVL in this particular loop, we can find the emitter current. So, if we apply the KVL, then we can write, -Vbe - Iee*Re +Vee = 0 Or from this, we can say that current Iee = (Vee - Vbe) / Re. And the emitter current Ie1 and Ie2 will be equal to Iee/2. And the collector currents Ic will be approximately equal to the emitter current. But if we see the exact value, then this Ic1 and Ic2 will be equal to α*Iee/2. That means the collector voltage, at both transistors, that is Vc will be equal to Vcc - Ic*Rc. That is equal to Vcc - α*Iee*Rc/2. So, this collector voltage will appear at these two terminals. And in the case of the constant current source also, the same expression is true. So, now let's understand qualitatively, how the differential amplifier responds to the common-mode as well as the differential input. And in the next video, we will go through the large and the small-signal analysis. So, first of all, let's see the response of this differential amplifier to the common-mode signal. Now, this common-mode signal could be some DC voltage or it could be some form of noise that is common to both input terminals. Now, here since both transistors are matched, that means this current Iee will get equally divided between the two emitters. That means this current Iee will be equal to Iee/2. And the voltage at this node will be equal to voltage VCM - Vbe. And since the transistors are perfectly matched, the voltage drop across the collectors will also be the same. That is equal to Vcc - α*Iee*Rc/2. So, if we see the differential voltage, that is Vo2 - Vo1, or Vc2 - Vc1, then it will be equal to 0. That means for the common-mode signal or the signal which is common to both input terminals, the differential output will be zero. And even if there is a change in the voltage VCM, then because of this constant current, the current through each emitter will remain the same. That is Iee/2. And hence, the differential voltage will remain zero. Now, when this differential pair is biased with the emitter resistor, then in that case also, the differential output will remain zero. So, in that case, let's say, the voltage at this node is equal to Vp. And this voltage Vp will be equal to Vcm - Vbe right !! And the current Iee will be equal to (Vee - Vp) / Re. Now, depending on the common-mode voltage, this current change a little bit. But since both transistors are identical, the current will get equally divided between the two emitters. That means the current through the emitter, that is Ie1 and Ie2 will be equal to Iee/2. And because of the identical transistors, the drop across the collector will also be the same. That means in that case also, the differential output voltage will be equal to zero. So, this is how the differential amplifier rejects the common-mode input signal. Now, here, to ensure the operation of BJT in the active region, the value of the VCM should be in such a way that, the collector-base junction remains in the reverse biased condition. That means the value of the VCM can not be arbitrarily very high. Alright so now, let's see, how the differential amplifier behaves for the differential input signal. So, let' say, 1V of common-mode signal, is already present at both inputs. And on top of it, +1V is applied at the Vin1. That means Vin1 is equal to 2V, while Vin2 is equal to 1V. That means we can say that 1V of the differential input voltage is present between the two input terminals. And because of this large differential input voltage, this transistor Q1 will draw the entire current Iee. And in this case, this transistor Q2 will remain OFF. So, let's see how. Now, since the base voltage of the Q1 is at 2V, that means this node Vp should be at 1.3 V right !! But the base of the second transistor is at 1V. And according to that, this voltage Vp should be equal to 0.3V. And that is not possible right !! Because if that is the case, then the voltage Vbe1 will be equal to 1.7V right !! And with such a large value of Vbe, the collector current and eventually, the emitter current will be very high. Because if you are aware, this collector current Ic can be given as Is* exp (Vbe/Vt) So, for such a very large value of the Vbe, the collector current will be very high. And in this case, it will get restricted by the bias current. And the entire current Iee will flow through the transistor Q1. And in that case, this voltage Vp will be equal to 1.3V. And because of that, this transistor Q2 will remain in the OFF condition. That means for the large value of this differential input voltage, the entire bias current is drawn by the particular transistor. And the reverse is also true. That means whenever the transistor Q2 is at 2V and the transistor Q1 is at 1V, at that time, the entire current Iee will be drawn by this Q2. And in that case, the collector voltage Vc2 will be equal to Vcc - α*Iee*Rc. And since the transistor Q1 is in OFF condition, that means voltage Vc1 will be equal to Vcc. So, as you can see, even with 1V of differential input voltage, it is possible to steer the entire bais current from one side of the pair to the other side. And because of this current steering property of this differential amplifier, it is widely used in the logic circuits. But to use this differential pair as an amplifier, this differential input voltage should be very small. Typically, it should be less than 2Vt. That means whenever this differential input voltage is in mV, then it is possible to use it as a differential amplifier. So, let's understand qualitatively how the circuit works for the small differential input. So, let's say, on top of this common-mode voltage, a small differential input ΔV is applied at the Vin1. Similarly, the voltage at the Vin2 is equal to Vcm - ΔV. That means the differential input voltage Vin1 - Vin2, that is Vid is equal to 2ΔV. Now, let's say, because of this small increase in this input voltage, this emitter current has increased slightly. And now, it is equal to Iee/2 + ΔI. Similarly, on this side, because of the reduction in the input voltage, now let's say, the emitter current is equal to Iee/2 - ΔI. And if you see the summation of these two currents, then still it is equal to Iee. So, now let's find the relationship between the change in the collector current and the change in the input voltage. And qualitatively, let's find the voltage gain. Now, we know that this current ΔIc can be given as gm*ΔVbe, right !! So, now because of the increase in this input voltage, the voltage at this node P has increased by ΔVp. That means we can say that this collector current ΔIc1 is equal to gm*ΔVbe that is gm*(ΔV -ΔVp). So, this indicates, the change in the voltage Vbe. Similarly, the reduction in this collector current Ic2 will be equal to gm*(-ΔV - ΔVp) That is equal to -gm*(ΔV +ΔVp) Now, here if we apply the KCL at this node P, then the outgoing current should be equal to Iee. That means the increase in the current Ic1 and the reduction in current Ic2 should be same. Or in other words, we can say that this ΔIc1 should be equal to -ΔIc2. And from this, we can say that, gm*( ΔV - ΔVp) = gm*(ΔV +ΔVp) Or from this, we can say that this Δvp is equal to 0. which indicates that, for small differential input voltage, the voltage at this node P will remain the same. Or we can say that the voltage at this node P will not change. Or we can say that, for small signal, this node will act as a virtual ground. And we will talk more about it during the small-signal analysis. But since ΔVp is zero, so we can say that this ΔIc1 = gm*ΔV and this ΔIc2 = -gm*ΔV. Or from this, we can say that this voltage Vo1 or ΔVc1 is equal to -gm*ΔV*Rc. Because as we have seen previously, this collector voltage Vc1 is equal to Vcc - Ic*Rc. Now, with the change in the input voltage, if the change in the collector current is ΔIc, then we can say that the change in the voltage Vc is equal to -gm*ΔV*Rc. Similarly, this voltage ΔVc2 will be equal to gm*ΔV*Rc. That means if we see the differential output voltage, that is Vo2 - Vo1, then it will be equal to 2*gm*ΔV*Rc. And here this differential input voltage, that is Vin1 - Vin2 = 2ΔV. That means the differential voltage gain, that is ΔVo / ΔVin will be equal to 2*gm*ΔV*Rc/ 2ΔV. That is equal to gm*Rc. And if we consider the voltage Vo1- Vo2, then it is equal to -gm*Rc. And if you notice, this gain is the same as the common emitter amplifier gain. But here, if both transistors are matched perfectly, then this common-mode signal or the common mode noise will get eliminated perfectly. So, that is the advantage of this differential amplifier. So, I hope in this video, you got the general overview of this differential amplifier. In the next video, we will go through the large and the small-signal analysis and we will learn more about this differential amplifier. 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 and subscribe to the channel for more such videos.