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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:

schematic

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...

Alexander Villa
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3 Answers3

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

schematic

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

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

G36
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  • 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 03 '17 at 13:01
  • @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? – LvW Dec 10 '19 at 09:17
  • 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. – LvW Dec 10 '19 at 11:06
  • @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. – G36 Dec 10 '19 at 18:25
  • 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? – G36 Dec 10 '19 at 19:53
  • @G36,,,yes, I arrive at the same expression - consisting of 4 factors. Here are the factor values acording to my calculation: 0.266*0,0735*40*40=31.28 – LvW Dec 11 '19 at 08:26
  • @G36,,,Independent on these values I like to ask you: WHY do you "disagree that the Rs1||rs1 term shouls be included" ? I rather think, for a correct image of the load we must, of course, consider the dynamic input resistance rs1 at the source node of M1. We must consider everything that is connevted to the source node of M3: 3 resistors and the source node of M1. – LvW Dec 11 '19 at 08:35
  • @LvW If I include Rs1||rs1 term in feedback factor then the Acl = 13 V/V which is obviously wrong. – G36 Dec 11 '19 at 16:40
  • @G36...are you really satisfied with this explanation? I do not know how you arrive at 13V/V for the closed-loop gain - however, that plays no role at all! The only question is "what is the correct loop gain"....independent on the consequences on other parameters. By the way - the influence of rs1 is certainly not very large because 100 ohms are in parallel. Note that I spoke only about rs1 as far as the most right part of your equivalent circuit is concerned (image of the load). It must not appear, of course, in the gain expression for M1. – LvW Dec 11 '19 at 17:24
  • @LvW No, I'm not satisfied but the method I use just works and gives enough accuracy in the real world. But, how would you calculate the Aol and Acl then? – G36 Dec 11 '19 at 18:05
  • @G36....I have simulated the circuit using an ideal FET model with 4mA/V. The result for the loop gain using "my method" is Aloop=-31.311. The result using "your" opening again is -31.311, however, only with considering rs1||Rs1=71,428Ohms. Without rs1=250 ohms the result is -31.77 (which is not correct). One further remark: In your drawing the source Vs1 is not placed correctly.. – LvW Dec 11 '19 at 18:11
  • Yes - I agree, in the "real world" (and in our example !!) the difference is pretty small. However, that is not the point. Here, we are discussing the question: Correct yes/no ? In some cases, we perhaps can neglect the influence of a certain part of a circuit, but it is important to know WHY we are allowed to do this. Imagine, the transconductance would be 40mA/V (instad of 4mA/V). In this case, the resistance would be rs1=25 ohm (instead of 250 ohms). We would make a fatal error not to consider the parallel connection Rs1rs1. Don``t you agree? – LvW Dec 11 '19 at 18:18
  • @LvW Hmm...for gm1 = 40mA/V and gm2 = gm3 = 4mA/V .. the loop gain using "your" method and opening is -92.219. And using "my method" and opening we have -92.753 (0.6% of a drift from "correct answer") and Acl = 9.89334 ( any error ?). Also, can you explain to me how to find Acl using "your method"? And how to tell which method is correct? – G36 Dec 11 '19 at 18:45
  • What do you mean with "my method" for finding Acl? I have calculated the loop gain with opening the loop at a suitable node - thats all. This has nothing to do with the calculation for finding the closed-loop gain. This is a complete other task - of course without opening the loop! This task is, certainly, more complicated due to the overall feedback loop. We cannot use fromulas readily available (like Blacks feedback expression). – LvW Dec 11 '19 at 19:31
  • @LvW You wrote that " We cannot use fromulas readily available (like Blacks feedback expression)." Why? As you can see I used the Black's formula in "my method" with success. – G36 Dec 12 '19 at 15:11
  • Blacks formula can be (errorless) applied only if the various blocks (Aol, beta, summing junction) are well separated and do not influence (load) each other. This is not possible in the circuit under discussion. For example: Your definition of the feedback factor involves only Rs and Rf. Why? Rs1 is not connected to a high-resistive amplifier input (like an opamp) and the voltage divider Rs-Rf is not driven by a low-resistive voltage source (as provided by an opamp). I do not know why your result (Acl) is - more or less - correct. As you know - sometimes two errors can cancel each other.... – LvW Dec 13 '19 at 08:35
  • More than that - what the calculation of the closed-loop gain complicates is the following: The circuit is bi-directional (not allowed for Blacks formula). The voltage at the source of M1 determines the gain of M1 and, hence, the ouput voltage Vo - but there is also a direct way (parallel to the 3 transistors) between the source of M1 and the output node (carying Vo). And - vice versa - Vo acts back (feedback effect) to the source of M1. – LvW Dec 13 '19 at 15:41
  • @LvW is there a name for this type of breaking loop (at an internal node inside the amplifier) like that? – emnha May 29 '21 at 16:52
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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.

LorenzoDonati4Ukraine-OnStrike
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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.

analogsystemsrf
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