# Series-Shunt Feedback Amplifier Analysis - Finding Loop Gain

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

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

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

• 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. – Alexander Villa May 3 '17 at 13:01

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.