I've been looking deeper into the idea of performing feedback analysis for MOSFET differential pair configurations. This particular amplifier circuit has been giving me some trouble, where I am looking to find the closed-loop gain \$(v_{o}/v_{s}) \$:

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The circuit states that the transconductances of the MOSFETs in the pair are equal but the transconductance of the 3rd MOSFET is not \$(g_{M1}=g_{M2}\ne\ g_{M3}) \$. Output resistances of the transistors are also taken to be infinity.

If we continue under the assumption that \$R_1=R_2\$ and perform our standard \$A\beta\$ feedback analysis, I arrive at the following equations:

$$A=-g_{M3}R_{eq}g_{M1}R_{1},\space\ \beta=\frac{R_5}{R_5+R_4}$$

where \$R_{eq}=(R_{22}||R_3)=((R_4+R_5)||R_3)\$ with the final closed-loop gain expression being of the form:

$$\left( \frac{v_o}{v_s} \right)=\frac{A}{1+A\beta}$$

Here is my issue. I do not see how the gain can be solved for unless \$R_1\$ is taken to be equal to \$R_2\$. This assumption allows us to convert the differential pair into a half circuit and makes solving the \$A\$ circuit relatively straightforward.

Is this a legitimate assumption though, or can the circuit be solved assuming \$R_1\ne R_2\$?

  • \$\begingroup\$ Typing error? Req=(R4+R5)||R3. More than that, the gain of the 1st stage is A1=(1/2)gm1R1. \$\endgroup\$
    – LvW
    Nov 26, 2020 at 9:54
  • \$\begingroup\$ @LvW Yes, thanks for the catch, I've edited to fix it. Ah, I believe I understand your reasoning for including the (1/2) factor in the first stage now. Is it because we wish to solve for gain with respect to the single-ended output at the drain of M2, as opposed to the difference in drain voltages of M2 and M1? \$\endgroup\$
    – JTaft121
    Nov 26, 2020 at 15:09
  • \$\begingroup\$ Simulate the circuit and ommit R2, is there a big difference? measure THD at the output with and without R2. \$\endgroup\$
    – S.s.
    Nov 26, 2020 at 15:20

2 Answers 2


Of course, the gain at the drain node of M1 will NOT be (totally) independent on the value of R2.

HOWEVER, the influence of R2 will nearly be neglectable. This can be verified as follows:

  • The transistor M1 is operated in common source configuration with signal feedback caused by the (rather low) input resistance r_in2 at the source of M2. This input resistance is (nearly) independent on the resistor R2 and we are allowed to set r_in2=1/gm2.

  • Of course, such a simplification causes a small error which, however, will be smaller than other uncertainties: Resistor tolerances, non-ideal current source, differences between gm1 and gm2, neglection of transistor output impedances,...)


Apologies for having neglected this question. So I've worked through the problem without the assumption that \$R_1= R_2\$ and I believe my original answer still holds.

My reasoning: When we perform a small signal analysis, we ground DC voltage sources and short any current sources. Thus, \$I_{bias}\$ would be shorted, causing the source voltages of \$M_1\$ and \$M_2\$ to be zero. Because of this, there is no small signal influence of \$M_2\$ on \$M_1\$, hence it does not matter really if \$R_1= R_2\$.

I also found an alternative version of this question, which replaces \$M_3\$ with a PNP BJT in the same layout. Part of that question asks why the base of the PNP BJT is connected to the drain of \$M_1\$ and not the drain of \$M_2\$. I believe my above reasoning explains why this is.

I believe my reasoning to be correct, but if there is an error that I am not seeing, please do let me know.


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