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INTRINSIC GAIN METHOD EX-1

FINDING GAIN OF DEGENERATED & CASCODE AMPLIFIER

In this post, we shall discuss two basic amplifiers.

Q1) Find the gain.

Solution:

From the signal flow, it is evident that the amplifier is in CS mode.

Let's directly write the gain expression. What is the Intrinsic Gain Method?

$GAIN=V_{IN} \times \frac{R_{LOAD}}{R_{LOAD}+R_{OUT}}$

Now, ${R_{LOAD}=R_{D}}$ ; ${R_{OUT}=r_{0}+M \times R_{S}}$, where $M= 1+g_{m}r_{0}$

How did I write the above expression? Go check.

$GAIN= -g_{m}r_{0} \times \frac {R_{D}}{R_{D}+(r_{0}+M \times R_{S})} \approx -g_{m}r_{0}\times \frac {R_{D}}{M \times R_{S}} \approx -\frac{R_{D}}{R_{S}}$

Let's try to solve it by another method called, $G_{m}r_{out}$ method and match the results with the above result.

Another Method:

We know the equivalent transconductance of the source-degenerated CS is $G_{M}= - \frac {g_{m}}{1+g_{m}R_{S}}$

$R_{eq}=R_{LOAD} || R_{OUT} \approx R_{D}$

$GAIN=G_{M}R_{eq}=- \frac {g_{m}}{1+g_{m}R_{S}} \times R_{S}$

Considering, $g_{m}R_{S}>>1$

$GAIN \approx -\frac{R_D}{R_S}$, which matches with the previous result.

Q2) Find the gain.

Solution: Let's draw the schematic for a small signal.

The above diagram shows that M1 and M2 are in CS and CG mode respectively.

$GAIN= -g_{m1}r_{01} \times M_{2} \times \frac {R_{LOAD}}{R_{LOAD}+R_{OUT}}$

where, $R_{LOAD}= r_{03}+M_3 \times r_{04} \approx M_3 \times r_{04}$

$R_{OUT}= r_{02}+M_2 \times r_{01} \approx M_2 \times r_{01}$

Now,

The $GAIN= -g_{m1} \times \frac {M_{2} \times r_{01} \times R_{LOAD}}{R_{LOAD}+R_{OUT}}$ =$-g_{m1}(R_{LOAD}|| R_{OUT})$

Let's also try with $G_{m}r_{out}$ method.

Another Method:

$\frac{I_{SC}}{V_{IN}}= G_{M}=-g_{m1} \frac {r_{01}}{\frac{1}{g_{m2}}+r_{01}} \approx -g_{m1}$

$GAIN=G_MR_{eq}=-g_m \left(R_{LOAD}||R_{OUT} \right)$, which matches with above result.

We will solve more complex circuits in the upcoming posts.

Check previous posts of the series we started "INTRINSIC GAIN METHOD".

PART1: INTRINSIC GAIN METHOD

PART2: INTRINSIC GAIN METHOD PARAMETERS

PART3: FINDING EQUIVALENT IMPEDANCE

For part 5 follow our blog and don't forget to comment

Fundamental Concept Of Differential Amplifier

This post is going to deal with a very basic concept of differential amplifier. What I personally feel is that, not many students are clear about the virtual ground/ half circuit concept in diff amp. Textbooks will make the half circuit and simply find the gain, which is the root of the problem. But remember, always go with the basics. So, personally I would prefer to write 5 extra lines instead of the half circuit approach and you will find its worth when you come across the 2nd problem of this post.

Before starting let me tell you the convention. I have used Capital Letters for DC and Small letters for AC. Suppose $V_{out}=V_{OUT}+v_{out}$ where $V_{OUT}$ is the common mode value and $v_{out}$ is the small signal value.

I shall start this post with a very simple and common question. Let's consider the following circuit of Fig:1(a). Let's try to find the differential gain $A_{dm}$ and common mode gain $A_{cm}$. Ignore $g_{ds}$. Take $g_m$ of M1= $g_m$

Let's start with DC analysis, by applying a common voltage of $V_{CM}$ at the gate of M1 and M2. As the gates have same potential and the sources are shorted, hence $V_{GS1}=V_{GS2}$. Also $(W/L)_1=(W/L)_2$. So, the bias current $I_B$ will divide in 1:1 ratio, hence M1 and M2 will carry a current of $\frac{I_B}{2}$.

$V_{OUT1}=V_{OUT2}=V_{DD}-\frac{I_BR}{2}$

Hence, $A_{cm}=\frac{V_{OUT1}-V_{OUT2}}{VCM}=0$

Now, $g_m=\mu_nC_{ox}\frac{W}{L}\left(V_{GS}-V_{TH}\right)$

So, $g_{m1}=g_{m2}$

Now, let's jump to the ac analysis by applying $+\frac{vin}{2}$ at the gate of M1 and $-\frac{vin}{2}$ at the gate of M2. The $V_{DD}$ is ac grounded and also the $I_B$ is opened as these sources won't change small signal wise.

Let's try to find out what will happen to the $v_x$ small signal wise. Applying KCL at $v_x$,

$g_{m1}v_{gs1}+g_{m2}v_{gs2}=0$

$g_m\left(\frac{v_{in}}{2}-v_x\right)+g_m\left(-\frac{v_{in}}{2}-v_x\right)=0$

$v_x=0$

So, the $v_x$ node won't move small signal wise. Hence, $v_x$ will be at ac ground.

So, $i_1=g_m\frac{v_{in}}{2} => v_{out1}=-i_1R=-g_mR\frac{v_{in}}{2}$

$i_2=-g_m\frac{v_{in}}{2} => v_{out2}=-i_2R=g_mR\frac{v_{in}}{2}$

Hence, $A_{dm}=\frac{v_{out1}-v_{out2}}{v_{in}}=-g_mR$

Now, let me make a small change in this circuit with the same question. Let's try to find the differential gain $A_{dm}$ and common mode gain $A_{cm}$ of Fig:2(a). Ignore $g_{ds}$. Take $g_m$ of M1= $g_m$. (First give it an honest try without seeing the solution)

Let's start with DC analysis, by applying a common voltage of $V_{CM}$ at the gate of M1 and M2. As the gates have same potential and the sources are shorted, hence $V_{GS1}=V_{GS2}$. But, $(W/L)_2=5(W/L)_1$. So, the bias current $I_B$ will not divide in 1:1 ratio in this case.

Assuming square law is valid,

$I_1=\frac{\mu_nC_{ox}}{2}\frac{W}{L}\left(V_{GS}-V_{TH}\right)^2$

$I_2=\frac{\mu_nC_{ox}}{2}\frac{5W}{L}\left(V_{GS}-V_{TH}\right)^2$

So, $I_1=\frac{I_B}{6}$ and $I_2=\frac{5I_B}{6}$

$V_{OUT1}=V_{DD}-I_1\left(5R\right)=V_{DD}-\frac{5I_BR}{6}$

$V_{OUT2}=V_{DD}-I_2R=V_{DD}-\frac{5I_BR}{6}$

Hence, $A_{cm}=\frac{V_{OUT1}-V_{OUT2}}{VCM}=0$

Now $g_m=\mu_nC_{ox}\frac{W}{L}\left(V_{GS}-V_{TH}\right)$

So, $g_{m2}=5g_{m1}$

Let's try to find out what will happen to the $v_x$ small signal wise. Applying KCL at $v_x$,

$g_{m1}v_{gs1}+g_{m2}v_{gs2}=0$

$g_m\left(\frac{v_{in}}{2}-v_x\right)+5g_m\left(-\frac{v_{in}}{2}-v_x\right)=0$

$v_x=-\frac{v_{in}}{3}$

So, the $v_x$ node will move small signal wise. Hence, $v_x$ will not be at ac ground.

So, $i_1=g_{m1}v_{gs1}=g_m(\frac{v_{in}}{2}+\frac{v_{in}}{3})=\frac{5g_mv_{in}}{6}$

$\Rightarrow v_{out1}=-i_1\left(5R\right)=-\frac{25g_mRv_{in}}{6}$

$i_2=g_{m2}v_{gs2}=5g_m(-\frac{v_{in}}{2}+\frac{v_{in}}{3})=-\frac{5g_mv_{in}}{6}$

$\Rightarrow v_{out2}=-i_2R=-\frac{5g_mRv_{in}}{6}$

Hence, $A_{dm}=\frac{v_{out1}-v_{out2}}{v_{in}}=-5g_mR$

Now, you may get a small doubt here. Suppose, you are applying the virtual ground (which is indeed wrong) and you have 2 half circuits.

For the left half circuit, $v_{out1}=-g_m\left(5R\right)\frac{v_{in}}{2}$

For the right half circuit, $v_{out2}=-5g_m\left(R\right)\frac{-v_{in}}{2}$

So, $A_{dm}=\frac{v_{out1}-v_{out2}}{v_{in}}=-5g_mR$ which is matching with the answer. But I would like to say, it's not the right way as your assumption of virtual ground is completely wrong. But it's also important to know, then why is it giving the right answer? Let's find out. For that, we have to know, whether the differential gain depends on the value of $v_x$ or not.

$i_1=g_{m1}v_{gs1}=g_m\left(\frac{v_{in}}{2}-v_x\right)$

$\Rightarrow v_{out1}=-i_1\left(5R\right)=-\left(5R\right)g_m\left(\frac{v_{in}}{2}-v_x\right)$

$i_2=g_{m2}v_{gs2}=5g_m\left(-\frac{v_{in}}{2}-v_x\right)$

$\Rightarrow v_{out2}=-i_2\left(R\right)=-\left(5R\right)g_m\left(-\frac{v_{in}}{2}-v_x\right)$

$v_{out1}-v_{out2}=-\left(5R\right)g_m\left(\frac{v_{in}}{2}-v_x\right)+\left(5R\right)g_m\left(-\frac{v_{in}}{2}-v_x\right)$

$\Rightarrow v_{out1}-v_{out2}=-5g_mRv_{in}+5Rv_x-5Rv_x=-5g_mRv_{in}$

So, basically the terms containing $v_x$ are cancelling each other, that's why the differential gain is not depending on the value of $v_x$. But for dc wise imbalanced circuit, it won't happen and the answer will be wrong if you consider $v_x$ as virtual ground.

I hope, next time you won't face any confusion related to half circuit analysis or virtual ground. Happy learning.

MOSFET INTRINSIC GAIN

This post will be dedicated to find out the intrinsic gain for 3 different MOSFET amplifier Configurations.

First Consider the small signal model of MOSFET.

Here Body effect on incremental MOSFET model is neglected.

Common Source:

Here biasing is not shown and only MOSFET in common source is analyzed.

$V_{out}= -g_{m}V_{gs}r_{o}$

=> $V_{out}= -g_{m}V_{in}r_{o}$

=> $\frac{V_{out}}{V_{in}}= -g_{m}r_{o}$

Common Drain:

$V_{out} = g_{m}V_{gs}r_{o}$

=> $V_{out} = g_{m}(V_{in}-V_{out}r_{o})$

=> $V_{o}(1+g_{m}r_{o}) = V_{in}g_{m}r_{o}$

=> $\frac{V_{out}}{V_{in}} = \frac{g_{m}r_{o}}{M}$

Common Gate:

Vin Voltage source does not draw any current due to unavailability of path to ground. All $g_{m}V_{gs}$ current flows through $r_{o}$

$V_{gs}=V_{in}$

=>$V_{out}= V_{in}- g_{m}(-V_{in})r_{o}$

=>$\frac {V_{out}}{V_{in}}=(1+g_{m}r_{o})$

=>$\frac {V_{out}}{V_{in}} = M$ >

Extra: If anywhere you want to consider body effect on intrinsic gain expression (Body grounded, where source has some incremental signal nature).

just Put $M=1+(g_{m}+g_{mb})r_{o}$

This is 2nd post of the series "Finding MOSFET gain". In 3rd post we will learn the intuitive way to find impedances of any MOSFET circuit. From the 4th post onwards we will take different amplifier circuits and try find out their gains and impedances.

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Post 2

FINDING MOSFET CIRCUIT GAIN USING INTRINSIC GAIN METHOD

Here in this post, I'll help build your basics of the MOSFET incremental model and try to find some parameter values, which will be used extensively to find out the gain of any MOSFET circuit.

There are three method to find out the gain of the amplifier

Conventional Method (using the small-signal method)
Intrinsic gain method approach
$G_{m}, R_{out}$ method

Here in this tutorial, We will only be discussing the second method in this post. For this you only need to remember three expressions along with a voltage divider equation and some basic methods of finding equivalent impedance (from inspection) at any node.

That's all!

(There will be a separate tutorial dedicated to the last method.)

First, we shall define the intrinsic gain of different configurations (CS, CD, CG) and find their value once.

Let's start-------

Configuration Intrinsic Gain

Common Source $-g_{m}r_{0}$

Common Drain $\frac{g_{m}r_{0}}{M}$

Common Gate $ M $

But what is $M$ in the expressions?
Well, I call it Mirroring Factor. For the time being just remember its expression $M= (1+g_{m}r_{0})$. ( It will be explained in the subsequent blog.)

Other than above 3 expressions you have to remember the Thevenin's equation.

$Actual \, Gain = Intrinsic \, Gain \times \frac{R_{L}}{R_{L}+R_{out}}$

$R_{L}$= Actual load seen by the circuit

$R_{out}$= Output impedance seen into the circuit >

Without justifying the above equation, let's just jump into one of the simplest MOSFET amplifier circuit.

Observation: Common Source Configuration

So, $\frac{v_{out}}{v_{in}}= A_{Intrinsic}\times \frac{R_{L}}{R_{L}+R_{out}}$

$ = -g_{m}r_{0} \times \frac{R_{D}}{R_{D}+r_{0}}$

$ = -g_{m} \times \frac{r_{0}R_{D}}{r_{0}+R_{D}}$

$ = -g_{m}(r_{0}||R_{D})$

This is a very well known result, doesn't seem surprising to us.

You will appreciate this, as we are going to solve quite a complex amplifier circuit. I promise, you will never have any fear while dealing with MOSFET Circuits ever again.

In the upcoming lectures, We will discuss more about the basics of intrinsic Gain. So get ready!

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Analog Intuition (GATE & VLSI)

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