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The following is from the book Design of Analog CMOS Integrated Circuit, Page 307.

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I think feedback is negative in my opinion.

Here is what I think, see picture below.

enter image description here

kile
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  • negative feedback and your Vtest should be on gate of M2 and the signals go in CW direction. – internet Jan 13 '24 at 21:58
  • @internet Could you show your circuit analysis diagram? – kile Jan 13 '24 at 22:39
  • @internet "$V_{test}$ should be on gate of M2 and the signals go in CW direction." What do you mean CW direction? Isn't that the same as mine? – kile Jan 14 '24 at 14:57

2 Answers2

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It's negative feedback. M1 is sensitive to the difference between its gate and source voltages. Thus this device performs the 'subtraction' of the input and feedback signal (which is in phase with the input signal).

Another way of looking at this is from the viewpoint of M1's drain -- it inverts a signal coming in from the gate, while it does not invert a signal applied to the source (the feedback signal).

jp314
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Simple analysis: Have a look at the closed loop (the loop gain with Vin=0) and find the number of signal inversions within the loop. For negative feedback, there must be an uneven number of inversions.

In your example: M1 does not cause a signal inversion between source and droain (common gate principle). However, and M2 in common source operation will cause a 180deg phase shift. T

Therefore, we have negative feedback (one single sign inversion).

Comment to your last figure: For loop gain analysis, you should try to find a node within the loop where a relatively low output resistance is connected to a much larger input resistance.

Then you can open the loop at this point and you can inject a test signal without changing too much the loading conditions at this point. Therefore, open the path at the gate of M2 and inject the test signal BETWEEN the drain node of M1 and the gate of M2. This is important, in particular, for loop gain simulation because otherwise you would destroy the DC operational point of the circuit.

However, for a rough loop gain analysis (visual inspection only) you can use the scheme as shown in the figure: Test voltage Vf at the gate of M2 and output at the drain of M1 (with Vin=0).

LvW
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  • What do you mean "output resistance is connected to a much larger input resistance"? Is $R_{in} = R_{test}$ and $R_{out} = R_F$? – kile Jan 14 '24 at 12:33
  • @kile A good breakpoint for injecting a test voltage is a point where the impedance, when looking in the forward direction, is much larger than the impedance when looking in the backward direction. You can find more details in Middlebrook's papers or lecture notes. – internet Jan 14 '24 at 13:33
  • @internet thank you for this additional explanation. That is eaxtly what I wanted to say. – LvW Jan 14 '24 at 14:11
  • @internet Can you draw your preferred test point diagram? – kile Jan 14 '24 at 14:11
  • @Lvw " open the path at the gate of M2 and inject the test signal BETWEEN the drain node of M1 and the gate of M2." Isn't that the same as my test point? – kile Jan 14 '24 at 14:54
  • There is a difference between (a) injecting the test voltage into the gate of M2 against ground or (b) against the drain node of M1. In case (a) the loop is open also for DC (no constant operational point) and in cas (b) the DC condition does not change. – LvW Jan 14 '24 at 15:34
  • @LvW Nice explanation. BTW, do you know why the book said feedback is positive $g_{21} = R_S/(R_F + R_S)$ instead of negative $g_{21} = - R_S/(R_F + R_S)$ – kile Jan 14 '24 at 20:38
  • @kile Where does it say this is positive feedback? From example 8.12 in the book you sent, it says, "The circuit consists of two common-source stages, with RF and RS sensing the output voltage and returning a fraction thereof to the source of M1. This transistor subtracts the returned voltage from Vin. The reader can prove that the feedback is indeed negative." – internet Jan 15 '24 at 04:12
  • @internet What did the book use $g_{21} = R_S /(R_F + R_S)$ if it's negative? I think it should use $g_{21} = - R_S /(R_F + R_S)$ – kile Jan 15 '24 at 09:20
  • @kile There is a difference between the feedback factor and the whole feedback loop - as far as the question positive/negative is concerned, Count the number of sign inversions within the feedback loop and you know if itis positive or negative. For example, the classical feedback block diagram contains a summing junction where the feedback signal is subtracted from the input signal (this is ONE negative sign). As a consequence, the block describing the feedback factor must ne positive (see my answer and the block diagram in another thread which you have started). – LvW Jan 15 '24 at 09:35
  • @LvW How about the feedback in previous thread? I think it's negative feedback. But the author said $h_{21} = -R_S / (R_S + R_F)$, closed-loop gain equals $A_{I,open}/(1 + h_{21} A_{I,open})$. which means $\mbox{$\Large A_{I,close} = \frac{A_{I,open}} {1 -R_S / (R_S + R_F) A_{I,open}} $}$. However, according to my analysis, $\mbox{$\Large A_{I,close} = \frac{A_{I,open}} {1 + R_S / (R_S + R_F) A_{I,open}} $}$. ▲ – kile Jan 15 '24 at 10:31
  • See my answer in https://electronics.stackexchange.com/questions/697362/where-am-i-wrong-in-understanding-feedback-network – LvW Jan 15 '24 at 11:43