On studying the channel resistance of MOSFET, I found something puzzled me a lot.

The first one comes from below which related feedback with this resistance. For the most general case, let us consider an NMOS connected with a resistor R_D on its D, a resistor R_S on its S, and ignore the body effect.
Now, if we apply an signal vin to the gate of the MOSFET, and measure the response at the drain terminal of the MOSFET. Using the small signal model, we could get the gain of this circuit which should be
A_v= -g_mR_D/[1+g_mR_S+R_S/r_o+R_D/r_o]
However, we could regard r_o as a feedback and this is a voltage-voltage feedback. We could get the open-loop gain first through the first method in Chapter8 of Razavi’s book, that is, break the circuit at an arbitrary node. We get the following circuit that V_t is applied to r_o and the other node of r_o to the source of NMOS. The drain of NMOS is V_F and another r_o is connected between V_F and ground, which is parrell with R_D in small signal circuit. Now we get the loop-gain as
LG= -g_m[R_D//r_o]R_S/[R_S+r_o]
and the open-loop gain is obviously to be
A_v,open= -g_mR_D/[1+g_mR_S]
then we can get the closed-loop gain because this is actually a negtive feedback. The result, however, differs from the result we get above, unless r_o approach to infinite.
The only reason I can get is the feedback method is just an approximate way and won’t give a accurate solution. But I wondered if there is any PHYSICAL reason for this.

Secondly, if we apply the input signal vin to the sourse of the MOSFET, the circuit should be a voltage-current feedback and the open-loop gain is
A_v,open= g_mR_D/[1+g_mR_S]
just the opposite of the above, while the loop-gain should be
LG= -g_mR_DR_S/[R_S+r_o]
Therefore, we also get a closed-loop gain which is different from the one result from the small-signal model, i.e.
A_v=1/{R_S/R_D+[(1/g_m)//r_o] / [R_D//r_o]}
unless r_o is sufficiently large. I have the same question about these two things.

By the way, everything get from Miller theorem is the same as the small-signal model without any imprecision.

Finally, to the resistance r_o in the above common-gate circuit, if we do the same thing as what Allen do in Figure6.2-11 of his book, where he get the RHP zero by a simple way, i.e., using superposition, we could neither get the same result as above. What we get is
Vout={g_m/[1+g_mR_S]}Vin[R_D//(r_o+R_S)]+[R_D/(R_D+r_o+R_S)]Vin
If this is also an approximate method, how could Allen get the accurate result? I have not understood it thoroughly.

If you have any suggetion, please tell me. Thanks in advance for any help.
Please forgive me for the somewhat irregular expression and any syntactic error. Thanks a lot.

Hi, Raja,
Firstly, if we ignore the resistance ro, then this could be viewed as an amplifier, with vin at source or gate and vout at drain. Now we connect a resistor with value ro between the source and the drain, i.e. between vin and vout, then this will be a feedback, right?

However, it seemed somewhat different from the regular feedback, because there exists the resistor RS, which separate vin and the sourse(both in the common gate circuit and in the common source circuit).

Thanks.
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