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.

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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.

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

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

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 $v_x$, 

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

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

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

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

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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.

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