question about the Rin of a simple feedback circuit

simple feedback circuit

hi, guys
i want to calculate Rin of the undermentioned circuit, when i use small-signal to calculate the Rin, the result is Rin=R1+(1/gm). but when i used the feedback theory to calculate it, the Rin(open loop)=R1//R2,which is smaller than the former result, due to shunt-shunt feedback,Rin(close loop) is even smaller than Rin(open loop), i don't know why.
hope for ur reply!
with regards
wanily1983

rines circuit 10

Why not simulate it to find the actual value and then decide which calculation is correct?

Bg,

wanily1983 said:

hi, guys
i want to calculate Rin of the undermentioned circuit, when i use small-signal to calculate the Rin, the result is Rin=R1+(1/gm). but when i used the feedback theory to calculate it, the Rin(open loop)=R1//R2,which is smaller than the former result, due to shunt-shunt feedback,Rin(close loop) is even smaller than Rin(open loop), i don't know why.
hope for ur reply!
with regards
wanily1983

Click to expand...

the result using small-signal is correct, but the point is why the method using feedback thoery cannot get the right answer.

Well, in the feedback theory, you should consider the FET as the inverting amplifier. So you must calculate the impedance looking into the gate of the FET, because that is the input of the amplifier.

The load of the FET is just R2, since the rest are current sources. The gain of the amplifier is then: A=R2*gm

Next, the feedback resistor is R2, so the impedance looking into the gate of the FET is: Zin=Zgs||R2/(1+A)≈R2/(1+A)=R2/(1+gm*R2)≈1/gm

Then the input impedance of the circuit is simply R1+Zin≈R1+1/gm

This is the same result. I think your mistake was not identifying correctly the amplifier's input and feedback resistor.

VVV said:

The load of the FET is just R2, since the rest are current sources. The gain of the amplifier is then: A=R2*gm

Click to expand...

VVV, could you explain why the load is R2? It is not connect to the ac ground, is it? Thanks.

In my opinion, seen from the input node, the effective feedback resistance is R2/(1+A), duo to the miller effect. It is so small that it could be neglected. So the transistor could be seen as diode-connected, with a resistance of 1/gm through it. Thus the total input resistance is R1+(1/gm).

R2 is not connected to the AC ground, but imagine the gate driven by a voltage source. In that case, the output voltage is Vo=Vin-gm*R2*Vin=Vin(1-gm*R2)
(All of the above should be regarded as AC).

That means the gain is actually A=Vo/Vin=1-gm*R2≈-gm*R2

With that, the rest of the explanation is the same.
The input impedance is of course that given by Miller's theorem, Zin=R2/(1+A)=1/gm.

I do not agree that the transistor is diode connected, since it has gain. It's only because of that gain that we are able to apply Miller's theorem.

Yes, you are right.

VVV,thank you for your reply.
i shouldn't regard R1 as a part of the input resistance.