Common Drain Amplifier Configuration Overview

Hey friends, welcome to the YouTube channel ALL ABOUT ELECTRONICS. So in the previous videos, we have seen the common source and the common gate amplifier configuration of the MOSFET. So in this video, let's discuss the common drain amplifier configuration of the MOSFET. And this configuration is also known as the source follower. So in this source follower, the drain terminal. is at AC ground. The input signal is applied between the gate and the ground, while the output is measured between the source and the ground terminal. And since the drain terminal is at AC ground, so we can say that the input is applied between the gate and the drain terminal, while the output is measured between the source and the drain terminal. That means the drain terminal is common between the input and the output side. And hence, it is known as the common drain amplifier configuration. So, in this configuration, the voltage gain is less than 1. Or at the most, it is very close to 1. Now, in this configuration, the output impedance of the amplifier is very low compared to other configurations. And typically, this configuration is used for driving the low impedance load. So, first of all, using the small signal analysis, let's find the input and the output impedance as well as the voltage gain of this source follower. And at the later part of the video, I will show you how this source follower is very useful for driving the low impedance load. So this is the typical source amplifier circuit along with the biasing. And here, this fixed gate voltage is applied through this gate resistor. But it can also be applied through the voltage divider biasing. Now for the DC analysis, these capacitors will act as an open circuit. And if you notice over here, then the drain terminal. is directly connected to the voltage source. And this source resistor provides a required path for the biasing current. Now if we talk about the small signal analysis, then for the AC analysis, these capacitors will act as a short circuit. While this DC voltage source VDD will act as a zero. And then after, we need to replace the MOSFET by the small signal model. So here if you observe, then this input signal is appearing between the gate and the ground terminal. And since the drain terminal is at AC ground, so these two resistors R1 and R2 will appear between the gate and the ground terminal. And here this source resistor is appearing between the source and the ground terminal. So if we see the small signal equivalent circuit, then it will look like this. So for this circuit, now let us find the input impedance, the output impedance and the voltage gain. And first of all, let us start with the input impedance. So, as I said in the earlier videos, the input impedance is the ratio of this input voltage to the input current. And in this case, since the gate acts as an open circuit at the low frequencies, so this current Iin will flow through this resistor R1 and R2. That means this Vin is equal to Iin times R1 parallel R2. Or we can say that these Vin divided by In, or the input impedance is equal to the parallel combination of R1 and R2. Now without these biasing resistors, the input impedance of this MOSFET is ideally infinite. But because of this resistor R1 and R2, the input resistance will be finite. Now to get the large input impedance, the value of this R1 and R2 should be as high as possible. And typically, the value of this R1 and R2 is in mega ohms. Alright, so that is the input impedance of this source follower. And similarly, now let us find the voltage gain. Now here, the voltage at this node is equal to Vin, right? So if we apply the KVL in this loop, then we can write voltage Vin minus Vgs minus Vo that is equal to 0. Or we can say that Vgs is equal to VIN minus V0. Similarly, now let us apply the KCL at this node. So here, this current is equal to V0 divided by Rs. So applying the KCL, we can write this V0 divided by Rs, that is equal to gm times Vgs. And this Vgs is equal to VIN minus V0. That means V0 divided by Rs, that is equal to gm times Vin minus Vo or we can say that This V0 divided by Rs plus gm times V0 is equal to gm times Vin. That means V0 divided by Vin is equal to gmRs divided by 1 plus gm times Rs. Now here if you observe, then this denominator is always greater than numerator. That means the voltage gain, that is Vout divided by Vin, will be always less than 1. And if this Gmrs is much greater than 1, then the voltage gain is approximately equal to 1. Moreover, if you observe over here, then the voltage gain is positive. That means in this source follower, the output and the input signal will be in the same phase. Now here, the same expression can also be written as this voltage gain is equal to Rs divided by 1 over gm plus rs. So this expression tells us that the voltage gain is the total resistance which is seen between the source and the ground terminal divided by that resistance plus 1 over gm. Now so far in our discussion, we have neglected the effect of the channel end modulation. That means we have assumed that the output resistance is equal to infinite. But if the output resistance is finite, then it will appear between the drain and the source terminal. And here, since the drain is at AC ground, so effectively, it will appear between the source and the ground terminal. That means now, in the voltage gain expression, instead of Rs, there will be Rs parallel R0. That means with the finite output resistance, the voltage gain will be equal to Rs parallel R0 divided by 1 over gm plus RS parallel R0. So that is the expression of the voltage gain including the effect of the output resistance. Now for a moment, let us consider that the output resistance of this MOSFET is infinite. And in that case, this voltage gain will be equal to RS divided by 1 over gm plus RS. Now from this expression, if we draw the equivalent circuit, then it can be drawn like this. So, using this equivalent circuit, it is very easy to find the output resistance. Because the output resistance is the equivalent impedance which is seen from the output side by considering all the independent sources in the circuit as zero. That means whenever this input signal is zero, then the output resistance is the parallel combination of this Rs and the 1 over gm. That means the output impedance of this source follower is equal to Rs parallel 1 over gm. So, this is the easiest way to find the output resistance. But it can also be found in a more conventional way. So, now to find the output impedance in a conventional way, let's apply a test voltage and let's find the test current by considering all the independent sources in the circuit as 0. So, the ratio of this test voltage to the test current will give us the output impedance. And for that, first of all let's consider this input voltage source as 0. Now once the input signal will become zero, then these two resistors which are in parallel with the input source will also get short circuited. That means now the gate terminal is at AC ground. Now in the earlier amplifier configurations of the MOSFET, if you have observed, then we did not apply the test voltage. Because in that case, by making the independent sources in the circuit as zero, the only thing which was left was the resistance. So intuitively, that was the output impedance. And therefore, there was no need to apply the test voltage. But in this case, after removing the independent sources, this dependent current source is still present. So to find the output impedance, it is required to apply the test voltage. Anyway, so now this voltage Vgs is the voltage between the gate and the source terminal. And since the gate terminal is at AC ground, so we can say that this voltage Vgs is equal to So, now if we apply the KCL at this node, then we can write this current Ix plus gm times Vgs that is equal to Vout divided by Rs or that is equal to Vx divided by Rs. And here this Vgs is equal to minus Vx that means Ix is equal to Vx divided by Rs plus Vx times gm That means, Ix divided by Vx or the output resistance is equal to 1 divided by Rs plus gm. That is equal to 1 divided by Rs plus 1 divided by 1 over gm. That means the output impedance is the parallel combination of this Rs and the 1 over gm. So that is the output impedance of this source follower. Now here, If the output resistance of the MOSFET is finite, then it will appear between the drain and the source terminal. And here, since the drain terminal is at AC ground, so it will appear between the source and the ground terminal. That means in this case, instead of Rs, we will have Rs parallel R0. That means with the finite output resistance of the MOSFET, the output impedance of the source follower is equal to Rs parallel 1 over gm parallel. Now, typically, this 1 over gm is in ohms. That means the overall output impedance will be less than 1 over gm. And therefore, the output impedance of this source follower is also in ohms. And this low output impedance of this source follower is particularly useful in driving the low impedance load. So let's say we have one common source amplifier and it is used to drive the 50 ohm load. That means here, RL is equal to 50 Ohm. Now here, let's say at the operating point, this 1 over gm is equal to 50 Ohm. That means at the operating point, this gm is equal to Pentimille Siemens. And let's say, this Rd is equal to 1 Kilo Ohm. So without considering the load resistor, the voltage gain of this common source amplifier is equal to gm times Rd. That is equal to 20 mSv times 1kΩ that is equal to 20. But with the load resistor, this voltage gain will be equal to gm times this Rd parallel RL. And in that case, if we find the voltage gain, then it will come out as 0.95. Because here this RL is equal to 50Ω. So as you can see, with the load resistor, The voltage gain of this common source amplifier will reduce significantly. But if we add the source follower stage or the common drain amplifier in between, then the things can be improved little bit. So let's say, a source follower stage is added between the load resistor and this common source amplifier. So this is the case of the multistage amplifier. So here, the load resistor is connected at the output of this source follower. So let's say, here this gate resistor is equal to 2 Mega Ohm. And at the operating point, this gm of the source follower is equal to 20 Millisiemens or 1 over gm is equal to 50 Ohm. And this Rx is equal to 200 Ohm. So let's say the voltage at this node is equal to Vx. That means the overall voltage gain, that is Vout divided by Vin is equal to this Vx divided by Vin times This is Vout divided by Vx. So, to find the overall voltage gain, first of all let's find the ratio of this Vx and the Vin. Now, for this common source amplifier stage, this source follower will act as a load. And the load resistance, which is seen by this common source amplifier, is the input impedance of this source follower. And in this case, it is equal to Rg. That means for the common source amplifier, this load resistance RL is equal to Zin, that is equal to 2 MΩ. That means in this case, the voltage gain of this common source amplifier or Vx divided by Vin is equal to Gm times Rd parallel RL. And here, this RL is equal to the input impedance. Or to be precise, the input impedance of this source follower. And here, since this input impedance is in MΩ. So, this Rd parallel Zin will be approximately equal to Rd. That means the voltage gain of this common source amplifier or Vx divided by Vin is approximately equal to 20. So similarly, now let us find the Vout divided by Vx. Now for the source follower, as we have seen, the voltage gain can be given as Rs divided by 1 over gm plus Rs. But here this load register will also appear in parallel with this source register. That means now the voltage gain is equal to Rx parallel RL divided by 1 over gm plus Rx parallel RL. Now here this RL is 50 ohm and the Rx is equal to 200 ohm. That means Rx parallel RL will be equal to 40 ohm. That means Vout divided by Vx will be equal to 40 ohm divided by 1 over gm that is 50 ohm plus 40 ohm that is equal to 4 by 9. So that is the voltage gain of this source follower stage. And if we see the overall voltage gain that is Vout divided by Vin, then it will be equal to Vx divided by Vin times Vout divided by Vx. And here, we are just considering the magnitude. That means the overall voltage gain will be equal to 20 times 4 divided by 9, which is roughly equal to 8.88. And as you can see, it is a significant improvement over the previous case. So here, just by adding the source follower stage, the overall voltage gain of this amplifier configuration can be improved. So in the multistage amplifier configuration, to drive a low impedance load, typically this source follower stage is used as an output stage. So, that is the application of this source follower. So, I hope in this video, you understood the small signal analysis of this source follower. And you also understood the application of this source follower through this example. So, if you have any question or suggestion, then do let me know here in the comment section below. If you like this video, hit the like button and subscribe the channel for more such videos.

And this configuration is also known as the source follower. So in this source follower, the drain terminal. is at AC ground.

The input signal is applied between the gate and the ground, while the output is measured between the source and the ground terminal. And since the drain terminal is at AC ground, so we can say that the input is applied between the gate and the drain terminal, while the output is measured between the source and the drain terminal. That means the drain terminal is common between the input and the output side.

And hence, it is known as the common drain amplifier configuration. So, in this configuration, the voltage gain is less than 1. Or at the most, it is very close to 1. Now, in this configuration, the output impedance of the amplifier is very low compared to other configurations. And typically, this configuration is used for driving the low impedance load. So, first of all, using the small signal analysis, let's find the input and the output impedance as well as the voltage gain of this source follower.

And at the later part of the video, I will show you how this source follower is very useful for driving the low impedance load. So this is the typical source amplifier circuit along with the biasing. And here, this fixed gate voltage is applied through this gate resistor. But it can also be applied through the voltage divider biasing.

Now for the DC analysis, these capacitors will act as an open circuit. And if you notice over here, then the drain terminal. is directly connected to the voltage source.

And this source resistor provides a required path for the biasing current. Now if we talk about the small signal analysis, then for the AC analysis, these capacitors will act as a short circuit. While this DC voltage source VDD will act as a zero. And then after, we need to replace the MOSFET by the small signal model.

So here if you observe, then this input signal is appearing between the gate and the ground terminal. And since the drain terminal is at AC ground, so these two resistors R1 and R2 will appear between the gate and the ground terminal. And here this source resistor is appearing between the source and the ground terminal. So if we see the small signal equivalent circuit, then it will look like this.

So for this circuit, now let us find the input impedance, the output impedance and the voltage gain. And first of all, let us start with the input impedance. So, as I said in the earlier videos, the input impedance is the ratio of this input voltage to the input current.

And in this case, since the gate acts as an open circuit at the low frequencies, so this current Iin will flow through this resistor R1 and R2. That means this Vin is equal to Iin times R1 parallel R2. Or we can say that these Vin divided by In, or the input impedance is equal to the parallel combination of R1 and R2.

Now without these biasing resistors, the input impedance of this MOSFET is ideally infinite. But because of this resistor R1 and R2, the input resistance will be finite. Now to get the large input impedance, the value of this R1 and R2 should be as high as possible. And typically, the value of this R1 and R2 is in mega ohms. Alright, so that is the input impedance of this source follower.

And similarly, now let us find the voltage gain. Now here, the voltage at this node is equal to Vin, right? So if we apply the KVL in this loop, then we can write voltage Vin minus Vgs minus Vo that is equal to 0. Or we can say that Vgs is equal to VIN minus V0.

Similarly, now let us apply the KCL at this node. So here, this current is equal to V0 divided by Rs. So applying the KCL, we can write this V0 divided by Rs, that is equal to gm times Vgs.

And this Vgs is equal to VIN minus V0. That means V0 divided by Rs, that is equal to gm times Vin minus Vo or we can say that This V0 divided by Rs plus gm times V0 is equal to gm times Vin. That means V0 divided by Vin is equal to gmRs divided by 1 plus gm times Rs. Now here if you observe, then this denominator is always greater than numerator.

That means the voltage gain, that is Vout divided by Vin, will be always less than 1. And if this Gmrs is much greater than 1, then the voltage gain is approximately equal to 1. Moreover, if you observe over here, then the voltage gain is positive. That means in this source follower, the output and the input signal will be in the same phase. Now here, the same expression can also be written as this voltage gain is equal to Rs divided by 1 over gm plus rs. So this expression tells us that the voltage gain is the total resistance which is seen between the source and the ground terminal divided by that resistance plus 1 over gm. Now so far in our discussion, we have neglected the effect of the channel end modulation.

That means we have assumed that the output resistance is equal to infinite. But if the output resistance is finite, then it will appear between the drain and the source terminal. And here, since the drain is at AC ground, so effectively, it will appear between the source and the ground terminal. That means now, in the voltage gain expression, instead of Rs, there will be Rs parallel R0. That means with the finite output resistance, the voltage gain will be equal to Rs parallel R0 divided by 1 over gm plus RS parallel R0.

So that is the expression of the voltage gain including the effect of the output resistance. Now for a moment, let us consider that the output resistance of this MOSFET is infinite. And in that case, this voltage gain will be equal to RS divided by 1 over gm plus RS. Now from this expression, if we draw the equivalent circuit, then it can be drawn like this.

So, using this equivalent circuit, it is very easy to find the output resistance. Because the output resistance is the equivalent impedance which is seen from the output side by considering all the independent sources in the circuit as zero. That means whenever this input signal is zero, then the output resistance is the parallel combination of this Rs and the 1 over gm. That means the output impedance of this source follower is equal to Rs parallel 1 over gm.

So, this is the easiest way to find the output resistance. But it can also be found in a more conventional way. So, now to find the output impedance in a conventional way, let's apply a test voltage and let's find the test current by considering all the independent sources in the circuit as 0. So, the ratio of this test voltage to the test current will give us the output impedance.

And for that, first of all let's consider this input voltage source as 0. Now once the input signal will become zero, then these two resistors which are in parallel with the input source will also get short circuited. That means now the gate terminal is at AC ground. Now in the earlier amplifier configurations of the MOSFET, if you have observed, then we did not apply the test voltage. Because in that case, by making the independent sources in the circuit as zero, the only thing which was left was the resistance.

So intuitively, that was the output impedance. And therefore, there was no need to apply the test voltage. But in this case, after removing the independent sources, this dependent current source is still present.

So to find the output impedance, it is required to apply the test voltage. Anyway, so now this voltage Vgs is the voltage between the gate and the source terminal. And since the gate terminal is at AC ground, so we can say that this voltage Vgs is equal to So, now if we apply the KCL at this node, then we can write this current Ix plus gm times Vgs that is equal to Vout divided by Rs or that is equal to Vx divided by Rs.

And here this Vgs is equal to minus Vx that means Ix is equal to Vx divided by Rs plus Vx times gm That means, Ix divided by Vx or the output resistance is equal to 1 divided by Rs plus gm. That is equal to 1 divided by Rs plus 1 divided by 1 over gm. That means the output impedance is the parallel combination of this Rs and the 1 over gm. So that is the output impedance of this source follower.

Now here, If the output resistance of the MOSFET is finite, then it will appear between the drain and the source terminal. And here, since the drain terminal is at AC ground, so it will appear between the source and the ground terminal. That means in this case, instead of Rs, we will have Rs parallel R0.

That means with the finite output resistance of the MOSFET, the output impedance of the source follower is equal to Rs parallel 1 over gm parallel. Now, typically, this 1 over gm is in ohms. That means the overall output impedance will be less than 1 over gm.

And therefore, the output impedance of this source follower is also in ohms. And this low output impedance of this source follower is particularly useful in driving the low impedance load. So let's say we have one common source amplifier and it is used to drive the 50 ohm load.

That means here, RL is equal to 50 Ohm. Now here, let's say at the operating point, this 1 over gm is equal to 50 Ohm. That means at the operating point, this gm is equal to Pentimille Siemens. And let's say, this Rd is equal to 1 Kilo Ohm.

So without considering the load resistor, the voltage gain of this common source amplifier is equal to gm times Rd. That is equal to 20 mSv times 1kΩ that is equal to 20. But with the load resistor, this voltage gain will be equal to gm times this Rd parallel RL. And in that case, if we find the voltage gain, then it will come out as 0.95.

Because here this RL is equal to 50Ω. So as you can see, with the load resistor, The voltage gain of this common source amplifier will reduce significantly. But if we add the source follower stage or the common drain amplifier in between, then the things can be improved little bit.

So let's say, a source follower stage is added between the load resistor and this common source amplifier. So this is the case of the multistage amplifier. So here, the load resistor is connected at the output of this source follower.

So let's say, here this gate resistor is equal to 2 Mega Ohm. And at the operating point, this gm of the source follower is equal to 20 Millisiemens or 1 over gm is equal to 50 Ohm. And this Rx is equal to 200 Ohm. So let's say the voltage at this node is equal to Vx. That means the overall voltage gain, that is Vout divided by Vin is equal to this Vx divided by Vin times This is Vout divided by Vx.

So, to find the overall voltage gain, first of all let's find the ratio of this Vx and the Vin. Now, for this common source amplifier stage, this source follower will act as a load. And the load resistance, which is seen by this common source amplifier, is the input impedance of this source follower. And in this case, it is equal to Rg. That means for the common source amplifier, this load resistance RL is equal to Zin, that is equal to 2 MΩ.

That means in this case, the voltage gain of this common source amplifier or Vx divided by Vin is equal to Gm times Rd parallel RL. And here, this RL is equal to the input impedance. Or to be precise, the input impedance of this source follower.

And here, since this input impedance is in MΩ. So, this Rd parallel Zin will be approximately equal to Rd. That means the voltage gain of this common source amplifier or Vx divided by Vin is approximately equal to 20. So similarly, now let us find the Vout divided by Vx. Now for the source follower, as we have seen, the voltage gain can be given as Rs divided by 1 over gm plus Rs.

But here this load register will also appear in parallel with this source register. That means now the voltage gain is equal to Rx parallel RL divided by 1 over gm plus Rx parallel RL. Now here this RL is 50 ohm and the Rx is equal to 200 ohm.

That means Rx parallel RL will be equal to 40 ohm. That means Vout divided by Vx will be equal to 40 ohm divided by 1 over gm that is 50 ohm plus 40 ohm that is equal to 4 by 9. So that is the voltage gain of this source follower stage. And if we see the overall voltage gain that is Vout divided by Vin, then it will be equal to Vx divided by Vin times Vout divided by Vx. And here, we are just considering the magnitude.

That means the overall voltage gain will be equal to 20 times 4 divided by 9, which is roughly equal to 8.88. And as you can see, it is a significant improvement over the previous case. So here, just by adding the source follower stage, the overall voltage gain of this amplifier configuration can be improved.

So in the multistage amplifier configuration, to drive a low impedance load, typically this source follower stage is used as an output stage. So, that is the application of this source follower. So, I hope in this video, you understood the small signal analysis of this source follower.

And you also understood the application of this source follower through this example. So, if you have any question or suggestion, then do let me know here in the comment section below. If you like this video, hit the like button and subscribe the channel for more such videos.

Transcript for:Common Drain Amplifier Configuration Overview

Transcript for:
Common Drain Amplifier Configuration Overview