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The following is from the book Design of Analog CMOS Integrated Circuit, page 319.

enter image description here

Let's break the loop at the gate of M1.

enter image description here

$$V_{out} = - V_T g_m \biggr[ (R_S + R_F) \parallel R_D\biggr] $$

$$V_F = V_{out} \frac{R_S} {R_F + R_S} $$

$$\Rightarrow \frac{V_F} {V_T} = - \frac{g_m R_S R_D} {R_D + R_S + R_F}$$

$$\Rightarrow L(s) = \frac{V_F} {V_T} = -\frac{g_m R_S R_D} {R_D + R_S + R_F}$$

\$\beta\$ negative since it's negative feedback.

$$V_G = \frac{V_{in} R_F + V_{out} R_S} {R_F + R_S} $$

$$\Rightarrow \beta = -\frac{R_S} {R_S + R_F}$$

$$A_{v, close} = \frac{L} {1 + L} \frac{1} {\beta} = \frac{g_m R_D (R_F + R_S)} {R_D + R_S + R_F - g_m R_D R_S} $$

However, the exact close loop gain is

$$A_{v,close} = \frac{(1 - g_m R_F) R_D} {R_D + R_F + R_S + R_D R_S g_m}$$

You can see how I derive this exact close loop gain from my previous post previous post.

the closed-loop frequency response of a feedback circuit can be expressed without reference to A(s)

$$A_{CL} = \frac{1} {\beta} \frac{L(s)} {1 + L(s)} $$

In conclusion, these two gains are not the same. Is there anything wrong here?

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  • \$\begingroup\$ For a negatively feedbacked system, the voltage gain with feedback is given by: Av=A/(1+Aβ), where A is the voltage gain of the amplifier without feedback. \$\endgroup\$ Commented Mar 6 at 9:14
  • \$\begingroup\$ @Franc Can you explain the difference Gain with different method? \$\endgroup\$
    – kile
    Commented Mar 6 at 10:01
  • \$\begingroup\$ Beta is wrong, this is an inverting amplifier, you have the term for a non-inverting one. \$\endgroup\$
    – Designalog
    Commented Mar 6 at 10:29
  • \$\begingroup\$ In the book "Microelectronic circuits" by -Sedra-Smith-, on page 601, there is a BJT polarized in the same way, the study of the reaction should be very similar. However, it is an example found in many student texts. \$\endgroup\$ Commented Mar 6 at 10:30
  • \$\begingroup\$ @kile - you have two different expressions for the closed-loop gain. Both are wrong. The denominator in the 1st one must not show a minus sign - but for Rs=0 it gives the correct gain without negative feedback. And the 2nd one must not show a minus sign in the numerator. More than that, for Rs=0 it does not give the correct gain without feedback. \$\endgroup\$
    – LvW
    Commented Mar 6 at 14:47

2 Answers 2

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When you write the formula in this format, you can easily see that \$ \frac{1}{\beta} \$ should be the ideal closed-loop gain when the loop gain approaches infinity.

$$\frac{1}{\beta} = \lim_{{L \to \infty}} \frac{1}{\beta} \cdot \frac{L(s)}{1 + L(s)}$$

When the loop gain approaches infinity, you can see that \$V_{GS}\$ should be 0, and from that, you can easily derive \$V_{out}/V_{in}\$ and \$\beta\$: $$ \frac{V_{\text{out}}}{V_{\text{in}}} = -\frac{R_{\text{F}}}{R_{\text{S}}} \quad \text{and} \quad \beta = -\frac{R_{\text{S}}}{R_{\text{F}}} $$

You may want to check Middlebrook's feedback theory lectures if you want to learn more about this.

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  • \$\begingroup\$ But you mentioned this equation before. $$V_G = \frac{V_{in} R_F + V_{out} R_S} {R_F + R_S} $$ electronics.stackexchange.com/questions/701003/… \$\endgroup\$
    – kile
    Commented Mar 6 at 17:16
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    \$\begingroup\$ Yes, that is a simple result obtained by superposition. What you referred to as \$\beta\$ above is the ideal transfer function when the loop gain is infinity, not the feedback factor. \$\endgroup\$
    – internet
    Commented Mar 6 at 17:35
  • \$\begingroup\$ "loop gain approaches infinity, you can see that \$V_{GS}\$ is 0" I don't know why you say \$V_{GS}\$ is 0. Can you explain in more details? \$\endgroup\$
    – kile
    Commented Mar 6 at 17:45
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    \$\begingroup\$ Imagine the transistor as an op-amp. If the open-loop gain of the op-amp is infinite, the two inputs become virtually shorted. \$\endgroup\$
    – internet
    Commented Mar 6 at 17:47
  • \$\begingroup\$ Do you mean Gate and Source of this transistor can be seen as two inputs of an op-amp? \$\endgroup\$
    – kile
    Commented Mar 6 at 18:09
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@kile: I think, the loop gain calculation L(s) is correct.

However, can you please explain how you arrived at the term Av,close=[L/(1+L)]/beta ?

I rather think, that we always must use the form:

Av,close=af*Ao/(1+L) with loop gain L=Ao * beta and the forward damping factor af.

Added (edit): This forward damping factor (af) gives the input signal at the amplifiers input (gate) when there is no signal coming back from the amplifier output: af=(Rf+Rd)/(Rs+Rf+Rd)

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  • \$\begingroup\$ It come from the book Analog Integrated Circuit Design by TONY CHAN CARUSONE (2nd version), Page 221 \$\endgroup\$
    – kile
    Commented Mar 6 at 10:28
  • \$\begingroup\$ The formula is simply the closed-loop gain with an error term duebto finite loop gain. It's most likely part of most loop gain models; I know that part of the asymptotic gain model it is. \$\endgroup\$
    – Designalog
    Commented Mar 6 at 10:32
  • \$\begingroup\$ However, I think his beta term is wrong. \$\endgroup\$
    – Designalog
    Commented Mar 6 at 10:33
  • \$\begingroup\$ @Designalog What's \$\beta\$ according to you? \$\endgroup\$
    – kile
    Commented Mar 6 at 10:41
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    \$\begingroup\$ @Designalog $$\beta = -\frac{R_S} {R_F}$$? \$\endgroup\$
    – kile
    Commented Mar 6 at 10:56

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