So I'm trying to solve a problem I found in a textbook (won't say which one so people can't use it to cheat on their hw) and I haven't been able to arrive at the answer found in the appendix. Please note this is not a homework problem, I graduated from college last year and just reviewing for my own benefit and so I can finish an analog MPPT controller I'm designing.

The problem:


simulate this circuit – Schematic created using CircuitLab

Assume each mosfet in the above figure is biased so gm = 4mA/V and that the ro of each mosfet can be ignored. Find the loop gain AB assuming the value for RF is 900 Ohms (the value I calculated for making the closed-loop gain ideally (1/B) 10V/V in the first part of the problem, which is correct)

My attempt at an answer
I used the test-voltage method, where a break is made in the feedback loop and a test voltage is applied and then measured at the other end with t-model for Mosfets:
$$I_{o} = g_{m}V_{T}$$ $$I_{D1} = I_{F}\frac{R_{S1}}{R_{S1} + 1/g_{m}}$$ $$I_{F} = I_{o}\frac{R_{S2}}{R_{S2} + R_{F} + R_{S1} || 1/g_{m}}$$ $$I_{D1} = \frac{g_{m}V_{T}R_{S2}}{R_{S2} + R_{F} + R_{S1} || 1/g_{m}} \cdot \frac{R_{S1}}{R_{S1} + 1/g_{m}} $$

Output voltage of M1: $$V_{o1} = I_{D1}R_{D1}$$ Voltage at opposite end where feedback loop broken: $$V_{r} = -g_{m}V_{o1}R_{D2} = -g_{m}I_{D1}R_{D1}R_{D2}$$

Loop Gain (Where Vt is test voltage): $$A\beta = -\frac{V_{r}}{V_{T}}$$ $$A\beta = \frac{g_{m}^{2}R_{D1}R_{D2}R_{S1}R_{S2}}{R_{S2} + R_{F} + R_{S1} || 1/g_{m} \cdot (R_{S1} + 1/g_{m})} $$

However, this gives me the wrong answer, the value for loop gain should be 31.33 according to the appendix.

I must be doing something wrong in calculating the voltage gain of each stage but I haven't been able to wrap my head around it. Any insight in how to approach the problem and some background I might be lacking would be highly appreciated. Guess I'm a little rusty on my circuit theory...


3 Answers 3


Your circuit has voltage-series feedback (Series-Shunt). So, we can draw this equivalent circuit:


simulate this circuit – Schematic created using CircuitLab

The open-loop voltage gain \$A_{OL}\$ can be found out by inspection:

\$M_1\$ voltage gain is:

$$\frac{R_{D1}}{R_{S1}||R_F + \frac{1}{g_{m1}}} \approx 29.41 $$

\$M_2\$ voltage gain:

$$\frac{R_{D1}}{\frac{1}{g_{m2}}}=\frac{10k\Omega}{250 \Omega} = 40 $$

And \$M_3\$ voltage gain:

$$\frac{\left ( R_{F}+R_{S1} \right )||R_{S2}}{\frac{1}{g_{m3}}+\left ( R_{F}+R_{S1} \right )||R_{S2}} \approx 0.267 $$

Hence the open-loop voltage gain is :

$$A_{OL} \approx 314 $$

The feedback factor

$$\beta = \frac{V_{S1}}{V_O} = \frac{R_{S1}}{R_{S1}+R_F} = 0.1 $$

And finally the loop gain:

$$A_{OL}\,\beta = 314*0.1 = 31.4$$

And the Closed loop gain

$$A_{CL} = \frac{A_{OL}}{1+A_{OL}\,\beta} \approx 9.69$$

  • \$\begingroup\$ Thank you! I wasn't familiar with finding the equivalent circuit like that, but looking through the textbook I see the author did a similar thing for a different problem. \$\endgroup\$ Commented May 3, 2017 at 13:01
  • \$\begingroup\$ @G36....I must admit that I am not too happy with your approach for the loop gain, however, at the moment I have no time to present another solution. My question is: For finding the loop gain we must open the loop and restore the original loading conditions at the opening. Therefore, what about the small input resistance rs1 at the source of M1? Don`t you think that the feedback factor should contain the term Rs1||rs1 instead of Rs1 alone? \$\endgroup\$
    – LvW
    Commented Dec 10, 2019 at 9:17
  • \$\begingroup\$ Completing my comment above, I have calculated the loop gain based on opening the loop at the drain of M2 (as indicated in the original diagram). To me, this should give a more correct result because only in this case the simplified "Middlebrook method" with voltage injection only is allowed (assuming no loading caused by the gate of M3). And the result is: Loop gain Aloop=-31.28. Of course, this is very close to the result as given by G36. \$\endgroup\$
    – LvW
    Commented Dec 10, 2019 at 11:06
  • \$\begingroup\$ @LvW Interesting comment. When I use "your" method I've got: loop gain = -31.311. And -31.3727 using "my" method Also, I disagree that the Rs1||rs1 term should be include in the feedback factor. At least not in the method that I used to find the loop gain. And this method that I used is taught for example in Sedra and Smith's book. \$\endgroup\$
    – G36
    Commented Dec 10, 2019 at 18:25
  • \$\begingroup\$ And the loop gain using your method $$ A_{OL}\:\beta = \frac{\left( \frac{1}{g_{m1}}||R_{S1}+R_F\right)||R_{S2}}{\frac{1}{g_{m3}}+ \left( \frac{1}{g_{m1}}||R_{S1}+R_F\right)||R_{S2}} \times \frac{\frac{1}{g_{m1}}||R_{S1}}{\frac{1}{g_{m1}}||R_{S1} +R_F} \times g_{m1}R_{D1} \times g_{m2}R_{D2} \approx -31.3112$$ Any mistake? \$\endgroup\$
    – G36
    Commented Dec 10, 2019 at 19:53

I've no time to write a more complete answer right now. Anyway, the last equation must be wrong, since the dimensions don't match.

Aβ must be a pure number, since it is a ratio between two voltages, whereas you have inconsistent units in the ratio in the right-hand side.

In fact in the numerator you have \$ \Omega^2 \$, whereas in the denominator you have a mismatch: you add \$ \Omega \$ to \$ \Omega^2 \$. Therefore there is an error in the math leading to that denominator (assuming the numerator is correct, you should have \$ \Omega^2 \$ as unit, in the denominator).

Time to check your math between your algebraic passages.


Your forward voltage gain (provided by M1 and M2) is over 1,000. The rightmost Resistor is simply a voltage divider that reduces the 1.000 gain of M3 to something < 1.000. Again, the forward gain is over 1,000.

The feedback is 10:1, so the gain is 20dB.

By the way, I assume these 3 are NFETs. I prefer the bubble nomenclature on FET symbols.


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