# MOSFET Common Source Amplifier - Small Signal Analysis ( Voltage Divider Bias )

https://www.youtube.com/watch?v=T-0X1N9N5V8

[00:06] Hey friends, welcome to the YouTube channel ALL ABOUT ELECTRONICS.
[00:10] So, in this video, we will learn how the MOSFET can be used as an amplifier in the common source configuration.
[00:15] And using the small-signal analysis, we will also find the expression of the voltage gain, the input impedance, and the output impedance of the amplifier.
[00:24] Now, when the MOSFET is used as an amplifier, then typically, it can be used in three configurations.
[00:30] That is a common source, Common drain, and the common gate configurations.
[00:35] And out of the three configurations, the common source is by far the most popular MOS amplifier configuration.
[00:40] So, in this configuration, the small input signal is applied between the gate and the source terminal.
[00:45] While the output is measured between the drain and the source terminal.
[00:50] That means as far as the AC signal is concerned, then the source terminal is common between the input and the output side.
[00:56] And hence, this configuration is known as the common source configuration.
[01:01] Now, for any voltage amplifier, the voltage.
[01:11] Gain, the input, and output impedance are some of the important parameters.
[01:16] And using the small-signal analysis, it is possible to find all these parameters.
[01:21] So, as we have seen in the previous video, under the small-signal approximation, the MOSFET can be replaced by the small-signal model.
[01:29] And using this model, it is possible to do a small-signal analysis of the given amplifier circuit.
[01:34] Now, in the previous video, we have briefly talked about the common source amplifier configuration.
[01:38] Where, on top of the DC biasing voltage, a small input signal was applied.
[01:45] But practically, this is how the MOSFET is biased.
[01:49] And using the coupling capacitors, the input signal is coupled to the amplifier.
[01:54] So, as we have discussed earlier, under the small-signal approximation, it is possible to perform the DC and the AC analysis separately.
[02:04] So, for the DC analysis, all these capacitors will act as an open circuit.
[02:09] And as you can see, here the MOSFET is biased.
[02:13] In the voltage divider biasing configuration, so, using the DC analysis, it is possible to set the operating point.
[02:19] Now, on top of this DC basing, a small input signal is applied through the coupling capacitors.
[02:24] So, for this AC analysis, or for this small-signal analysis, all the DC voltage sources in the circuit will act as a zero.
[02:36] While at the operating frequency, the coupling and the bypass capacitors will provide very low resistance.
[02:42] And for simplicity, they can be replaced by the short circuit.
[02:45] And then after, the MOSFET will get replaced by the small-signal model.
[02:50] That means to perform the AC analysis of any circuit, we need to follow three steps.
[02:54] That means first consider all the DC sources in the circuit as zero.
[03:00] And then replace all the coupling and the bypass capacitors by the short circuit.
[03:04] And then after, replace the MOSFET by the small-signal model.
[03:10] So, here since the bypass capacitor acts as.
[03:14] A short circuit, so this resistor Rs will also get short-circuited.
[03:20] And if we see the equivalent circuit, then it will look like this.
[03:23] So, here, this input signal is appearing between the gate and the ground terminal.
[03:29] Or effectively it will appear between the gate and the source terminal.
[03:34] Similarly, the one end of this resistor R1 and R2 is connected to the gate terminal, while the other end is connected to the ground terminal.
[03:43] That means effectively, these two resistors are connected between the gate and the source terminal.
[03:47] Similarly, this resistor Rd will appear between the drain and the source terminal.
[03:52] So, if we see the small-signal equivalent circuit, then it will look like this.
[03:56] So, for this circuit, now let's find the voltage gain, the input impedance, and the output impedance.
[04:06] And first of all, let's find the input impedance.
[04:06] So, if Vin is the input signal and this Iin is the input current, then this input impedance is the ratio of this input signal to the input.
[04:20] Current.
[04:20] Now, here since the gate and the source terminal will act as an open circuit, so this current Iin will flow through the parallel combination of this R1 and R2.
[04:29] That means we can say that, this input voltage Vin = (R1||R2)*Iin.
[04:35] Or we can say that this input impedance, that is Vin / Iin = R1 || R2.
[04:43] So, at the low frequencies, the input impedance of the MOSFET is very high.
[04:54] Or ideally, it is infinite.
[04:58] But because of these external resistors, the input impedance of this configuration is equal to R1 || R2.
[05:02] So, to achieve the high input impedance, the value of this R1 and R2 should be very high.
[05:07] And preferably, it should be in MΩ.
[05:14] Similarly, now let's find the output impedance.
[05:14] So, the output impedance is Thevenin's equivalent.
[05:20] Impedance which is seen from the output side.
[05:26] And to find that, we will consider all the independent sources in the circuit as zero.
[05:31] That means we will consider this input signal as zero.
[05:36] Now, since the resistor R1 and R2 are connected in parallel with the input signal, so they will also get short-circuited.
[05:41] And once the Vin is zero, then this voltage Vgs will also become zero.
[05:46] Because the voltage between the gate and the source is equal to the input signal.
[05:51] That means this dependent current source will also act as a zero.
[05:54] Or we can say that it will act as an open circuit.
[05:59] So, in this case, the Thevenin's equivalent impedance which is seen through the output side is equal to Rd.
[06:03] Or we can say that the output impedance is equal to Rd.
[06:08] That means for this amplifier configuration, the output impedance is equal to Rd.
[06:12] Alright so similarly, now let's find the voltage gain.
[06:19] So, here the output is measured between the.
[06:22] Drain and the source terminal.
[06:28] And since this drain current Id is flowing in this anti-clockwise direction, so we can say that this Vo is equal to -id*Rd.
[06:37] That is equal to -gm*Vgs*Rd.
[06:46] And as you can see, this voltage Vgs is equal to Vin.
[06:50] So, from this, we can say that, this voltage Vo is equal to -gm*Vin*Rd.
[07:01] That means the voltage gain Vo/Vin is equal to -gm*Rd.
[07:06] So, from this expression, we can say that, by increasing the value of Rd, we can increase the gain.
[07:11] But the value of Rd can not be increased indefinitely.
[07:17] Because as the value of Rd increases, then the voltage drop across this resistor Rd will also increase.
[07:21] And because of that, the voltage Vds will reduce.
[07:25] So, if this voltage drop is too large, then the MOSFET may come out of the saturation.
[07:31] And hence, the value of Rd, can not be increased indefinitely.
[07:36] But by cascading multiple such stages, it is possible to increase the voltage gain.
[07:42] Alright, so now so far, we have neglected the effect of the channel length modulation.
[07:48] And we have assumed that the output impedance of the MOSFET is infinite.
[07:53] But considering the finite output impedance, this is the small-signal equivalent circuit.
[07:58] So, as you can see, there is an additional output resistance between the drain and the source terminal.
[08:02] So, in this case, the input impedance will remain the same.
[08:06] That means in this case, Zin is equal to R1||R2.
[08:14] But now, this output impedance is the parallel combination of this ro and the Rd.
[08:20] That means the output impedance is equal to ro || Rd.
[08:25] And similarly, the voltage gain Av is equal to -gm*(Rd || ro).
[08:34] So, considering the finite output resistance of the MOSFET, these are the expressions of the output impedance, and the voltage gain of this amplifier.
[08:42] Alright, so far, using this small-signal analysis, we found the expression of the voltage gain, the input impedance, and the output impedance of this common source amplifier.
[08:53] And here, this amplifier is biased using this voltage divider biasing configuration right?
[08:59] So, as we have seen earlier, in this voltage divider biasing configuration, this source resistor provides negative feedback, and it improves the biasing stability right?
[09:07] But whenever, the bypass capacitor is used, then for the AC signal, it will act as a short circuit.
[09:18] And because of that, it will not have any impact on the AC analysis.
[09:23] But sometimes, to improve the frequency response, and to improve the stability of the amplifier,
[09:28] The source resistor is not bypassed.
[09:34] And in that case, this source resistor will also come into the picture during the AC analysis.
[09:41] So, because of this source resistor, the voltage gain of this amplifier will reduce.
[09:45] And through the small-signal analysis, it will get clear to you.
[09:49] So, now let's see the small-signal analysis of this common source amplifier when there is also a source resistor.
[09:53] So, the procedure for the small-signal analysis will remain the same.
[09:58] So, first of all, we will replace all the coupling capacitors by the short circuit and all the DC sources in the circuit will act as a zero.
[10:07] And then after, we will replace the MOSFET by the small-signal model.
[10:11] So, the first thing if you observe, then the source resistor is connected between the source and the ground terminal.
[10:16] And here the input signal is appearing between the gate and the ground terminal.
[10:21] Similarly, these two resistors R1 and R2 are appearing between the gate and the ground terminal.
[10:30] And this resistor Rd is appearing between the drain and the ground terminal.
[10:34] So, if we see the equivalent circuit, then the small-signal equivalent circuit will look like this.
[10:38] So, for this circuit, first of all, let's find the voltage gain.
[10:42] Now, here since the drain current Id is flowing in this fashion, so we can say that this output voltage Vo is equal to -id*Rd.
[10:55] That is equal to -gm*vgs*Rd right!
[10:55] So, now, let's find the relationship between this voltage vgs and the input signal.
[11:03] So, if we observe over here, then this drain current Id is also flowing through this resistor Rs.
[11:13] And let's say, because of that the voltage drop across this resistor is equal to Vs.
[11:20] That means this voltage Vs is equal to id*Rs.
[11:20] That is equal to gm*Vgs*Rs.
[11:32] And here, the voltage between the gate and the ground terminal is equal to Vin.
[11:36] So, if we apply the KVL then we can say that this input signal Vin is equal to Vgs + Vs.
[11:46] That is equal to vgs + gm*vgs*Rs.
[11:53] That is equal to vgs*(1 + gm*Rs).
[12:00] Or from this, we can say that this voltage vgs is equal to Vin / (1 + gm*Rs).
[12:12] So, in this way, we found the relationship between the voltage vgs and the input signal.
[12:16] So, now, let's put the value of vgs in this expression.
[12:22] That means from this we can say that, this output voltage Vo is equal to -gm*Rd*Vin / (1 + gm*Rs).
[12:33] Or we can say that the voltage gain Av, that is Vo / Vin is equal to -gm*Rd / (1 + gm*Rs).
[12:53] So, this is the expression of the voltage gain when the source resistor Rs is also present in the amplifier.
[12:58] And whenever this Rs is equal to zero, then this voltage gain is equal to -gm*Rd.
[13:05] But because of the presence of this Rs, the voltage gain will reduce.
[13:08] So, similarly, now let's find the input impedance.
[13:16] So, in this case, the input impedance will remain the same.
[13:21] Because here, the gate to source terminal acts as an open circuit.
[13:26] That means the input current Iin is flowing through the parallel combination of the R1 and R2.
[13:29] That means Vin is equal to Iin * (R1 || R2).
[13:29] Or we can say that Vin / Iin, or the input.
[13:41] Impedance is equal to R1 || R2.
[13:49] So, this is the input impedance for the common source amplifier including the source resistor.
[13:56] And similarly, now let's find the output impedance.
[14:01] So, as I said, the output impedance is Thevenin's equivalent impedance which is seen through the output side.
[14:06] And to find that, we will consider all the independent sources in the circuit as zero.
[14:11] That means this input signal will act as a short circuit.
[14:16] And once we consider this Vin as zero, then these two resistors will also get short-circuited.
[14:23] And because of that, this voltage vgs is also equal to zero.
[14:27] So, let's see how.
[14:34] So, if we apply the KVL in this loop, then we can say that voltage Vgs + Vs is equal to 0.
[14:38] Where this vs is the voltage drop across this resistor Rs.
[14:38] And as we have seen, this voltage vs is equal.
[14:44] To Id*Rs.
[14:44] That is equal to gm*Vgs*Rs.
[14:52] So, if we put the value of Vs in this expression, then we can say that, vgs + gm*vgs*Rs = 0.
[15:05] That means vgs*( 1 + gm*Rs) = 0.
[15:05] That means in this case, this voltage vgs is equal to zero.
[15:16] And since the voltage vgs is equal to zero, so this dependent current source will act as an open circuit.
[15:25] And in this case, the Thevenin's equivalent impedance which is seen through the output side or Zo is equal to Rd.
[15:29] That means in this case also, the output impedance is equal to Rd.
[15:36] So, in this way, we found the expression of the input and the output impedance as well as the voltage gain for this common source.
[15:45] Amplifier when the source resistor is also present.
[15:49] Alright, so now let's take one example so that, whatever we have discussed so far will get clear to you.
[15:54] So, here for the given MOS amplifier, we have been asked to find the input impedance, the output impedance, and the voltage gain.
[16:05] And here, we have been also given the value of threshold voltage as well as the device constant for the MOSFET.
[16:09] So, for the small-signal analysis, we need the value of the transconductance.
[16:14] And to find the value of transconductance gm, first of all, let's do the DC analysis.
[16:17] So, for the DC analysis, all the coupling and the bypass capacitors will act as an open circuit.
[16:28] And here assuming the gate current Ig is equal to zero, we can find the voltage Vg.
[16:35] So, this voltage Vg is equal to 10 MΩ * 30V / (10MΩ + 40MΩ).
[16:45] That is equal to 6V.
[16:47] So, using the voltage divider rule, we can find the gate voltage Vg.
[16:52] And here, the current through the source resistor is equal to Id.
[16:57] That means voltage Vs is equal to Rs*Id.
[17:02] That means voltage vgs is equal to 6V - Id*Rs.
[17:07] And here, assuming the MOSFET is operating in the saturation region, this drain current Id can be given as k*( Vgs - Vt)^2.
[17:23] Where this k is the device constant.
[17:27] Or we can say that, Id = K* (6 - Id*Rs- 3)^2.
[17:38] That means this drain current Id = 0.4 x 10^-3 * (3- Id*Rs)^2.
[17:49] Or we can say that, 2500*Id = ( 9 - 6*Id*Rs +Id^2*Rs^2).
[18:05] Now, here the value of Rs is equal to 1.2 kΩ.
[18:08] So, if we put the value of Rs in this expression, then we can say that, 1.44 x 10^6 * Id^2 - 9700*Id +9 =0.
[18:22] So, if we solve this quadratic equation then we will get the two roots.
[18:28] The first root is 5.62 mA, while the second root is 1.11 mA.
[18:35] Now if we take the value of Id as 5.62 mA then the MOSFET will not operate in the saturation region.
[18:46] That means the value of the Id should be equal to 1.11 mA.
[18:52] And with this value of the drain current, the voltage Vgs will be equal to Vg - Vs.
[18:59] That is equal to 6 - Id*Rs.
[18:59] That is equal to 6 - 1.2 kΩ*1.1mA.
[19:12] That is equal to 4.668V.
[19:12] And the transconductance gm can be given as 2 * K * (Vgs - Vt).
[19:20] That is equal to 0.8 * 10^-3 (4.668 - 3).
[19:33] That is equal to 1.33 mS.
[19:33] So, in this way, we got the value of the transconductance.
[19:41] And once we know the value of the transconductance, then using the small-signal analysis, we can find the value of the required parameters.
[19:46] So, for the AC analysis, we can consider all.
[19:52] These coupling capacitors as a short circuit, and we can replace all the DC sources in the circuit as zero.
[19:57] And then after, we can replace the MOSFET by the small-signal model.
[20:01] So, if we see the equivalent circuit then it will look like this.
[20:06] So, as we have seen, in this case, the input impedance is the parallel combination of this R1 and R2.
[20:16] That means in this case, it is the parallel combination of 10 MΩ resistor and the 40 MΩ resisor.
[20:22] That is equal to 8 MΩ.
[20:27] That means the input impedance of this amplifier configuration is equal to 8 MΩ.
[20:33] Similarly, the output impedance is equal to Rd.
[20:37] And in this case, it is equal to 3.3 kΩ.
[20:42] And the voltage gain is equal to -gm*Rd.
[20:47] Or in this case, it is equal to - (1.33mS)*(3.3 kΩ).
[20:57] That is equal to -4.43. So, this is the voltage gain of this amplifier.
[21:03] So, in this way, using this small-signal analysis, we can find the different amplifier parameters.
[21:10] So, in this video, we have done the small-signal analysis of this common source amplifier whenever
[21:17] it is biased using the voltage divider biasing configuration right !!
[21:21] Similarly, another popular biasing configuration for the MOSFET is the drain feedback bias.
[21:26] So, similarly, in the next video, we will see the small-signal analysis of this common
[21:31] source amplifier, when the MOSFET is biased using the drain feedback bias.
[21:37] But I hope in this video, you understood the small-signal analysis of this common source
[21:40] amplifier. So, if you have any questions or suggestions,
[21:45] then do let me know here in the comment section below.
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[21:51] videos.