How can we evaluate the approximate output impedance of the VCCS (Voltage controlled current source) based circuit at port \$ V_0 \$ ? enter image description here

My Approach:
To calculate \$ Z_{out} \$,equivalent circuit will be :

Applying KCL at \$ V_0 \$ ,we get: $$\huge I_t + g_m v_{R_{in}}= \frac{V_0}{100k} + \frac{V_0}{R_{in}}$$ $$\implies I_t + g_m (-V_0) = \frac{V_0}{100k} + \frac{V_0}{10M}$$ $$\implies I_t = 0.01 V_0 + \frac{V_0}{100k} + \frac{V_0}{10M} \quad , \text{as} \space g_m= 0.01S$$ $$\implies \frac{10^7}{10^5 + 10^2 + 1}=\frac{V_0}{I_t}$$ $$\implies Z_0 = 99.89 \Omega \space \approx 100 \Omega$$ But answer is given as \$ 100k \Omega \$ !!
so where is my mistake? please help...


Initial quick look ...

Current source has infinite Zout, but it's in parallel with 100k, which would reduce it to 100k, but there's feedback, which will change it from 100k, so it can't be 100k!

So what's the feedback doing?

The 100k, and VCCS input 10Mohm are negligible compared to the 1/100ohm transconductance. So the circuit reduces approximately to the VCCS with the feedback path.

If we try to measure Zout by injecting a current Iout into the output, then its voltage will change, which changes the input voltage. The input voltage has to change the right amount to sink Iout, so will change by Iout/gm, or by 100ohms*Iout, giving and output impedance of approximately 100ohms. The effect of the 10M and the 100k will change that figure slightly, in the 3rd or 4th decimal places.

Has the given answer an added 'k' as a typo?

  • \$\begingroup\$ No, the given answer is stating VCCS output terminal is approximately equal to output impedance ; or \$ 100k \Omega \$ @Neil_UK \$\endgroup\$ – Suresh Mar 10 '18 at 4:50

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.