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I was studying negative feedback in voltage amplifiers. I found this on Wikipedia for calculating input resistance as shown below.

enter image description here

Following the similar method, I tried to calculate the output resistance with feedback \$Z_{of}\$.

Connected a voltage source \$V_o\$ which supplies a current \$I_o\$. Applying Kirchoff's voltage law to the output loop, we have

\$V_o = I_oZ_o + A_v.V_{in}\$

For negative feedback, \$V_{in} = V_x - V_f\$ and \$V_f = \beta V_o\$. Therefore

\$V_o = I_oZ_o + A_v.(V_x - \beta V_o)\$

\$V_o = I_oZ_o + A_v.V_x - A_v.\beta V_o\$

\$V_o + A_v.\beta V_o = I_oZ_o + A_v.V_x\$

\$V_o(1 + A_v.\beta) = I_oZ_o + A_v.V_x\$

Dividing both the sides by \$I_o\$, we get

\$Z_{of} = \frac{Z_o}{(1 + A_v.\beta)} + \frac{(A_v.V_x)}{(I_o)(1 + A_v.\beta)}\$

The above expression is what I am getting. But the answer is \$Z_{of} = \frac{Z_o}{(1 + A_v.\beta)}\$. I am getting an extra term \$\frac{(A_v.V_x)}{(I_o)(1 + A_v.\beta)}\$.

What is the mistake in my calculation? Please help.

Essar
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  • [Related question](https://electronics.stackexchange.com/q/565339). This may answer your question. You need to assume that input voltage is zero. – AJN May 17 '21 at 15:43
  • In V=IR, R is the slope of the voltage vs current line. Therefore R is the *change* in output voltage with *change* in output current. Since you are only looking at changes, you can ignore the extra constant term (Av.Vx)/(1 + Av.B). – user4574 May 17 '21 at 15:50
  • @AJN I looked up the internet and found out the same, input signal is assumed zero. But what may the reason of doing so? To eliminate the extra term? – Essar May 17 '21 at 16:16
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    https://electronics.stackexchange.com/questions/242397/why-should-we-set-input-source-equal-zero-for-calculating-output-resistance-of-c/242453#242453 – G36 May 17 '21 at 16:37

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