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I have the following circuit which I'm asked to calculate the output impedance Zout, assuming the op-amp is ideal.

enter image description here

I use the following equivalent circuit:

schematic

simulate this circuit – Schematic created using CircuitLab

I get: $$i_1 = \frac{V_x}{R_{i}+R_1}$$ $$V_i = -\frac{V_x R_{i}}{R_{i} + R_1}$$ $$i_2 = \frac{V_x - A_{vo} V_i}{R_2 + R_o} = \frac{V_x}{R_2 + R_o} + \frac{A_{vo} V_x R_{i}}{(R_{i} + R_1)(R_2 + R_o)}$$ $$I_x = I_1 + I_2$$ $$Z_{out} = V_x/I_x = \left( \frac{1}{R_{i} + R_1} + \frac{1}{R_2 + R_o} + \frac{A_{vo} R_{i}}{(R_{i} + R_1)(R_2 + R_o)} \right)^{-1} \approx \frac{R_2}{1 + A_{vo}} \approx 0$$ $$\implies Z_{out} = 0$$

Using the fact \$A_{vo} = \infty\$, \$R_{i} = \infty\$ and \$R_o = 0\$ for an ideal op-amp.

The solution says \$Z_{out} = R_2\$. My question is : what's wrong in my approach?

Thanks

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    \$\begingroup\$ The solution is wrong. Zout = 0. \$\endgroup\$
    – τεκ
    Mar 10, 2018 at 1:34
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    \$\begingroup\$ You're both correct. \$\endgroup\$ Mar 10, 2018 at 1:48

3 Answers 3

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  • For Vin that does not saturate the output of A1 when loaded, Zout=0

  • When A1-out saturates Zout= R2.

    • U1 is no longer linear with linear feedback so the gain is zero when saturated
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  • \$\begingroup\$ Thanks alot for the answer. Could I ask you why Avo is considered zero when Vi saturates the op amp? Is it because Avo=Vo/Vi will be approximately zero when Vi >> Vo? Would that mean if Vi=Vsat, then Zout=R2/2? \$\endgroup\$
    – Yannick
    Mar 10, 2018 at 2:01
  • \$\begingroup\$ No An ideal OA-out has 0 Ohms so Zout becomes R2 \$\endgroup\$ Mar 10, 2018 at 2:37
  • \$\begingroup\$ What you need to know is that an OA runs out of gain with rising f , the Zout rises from near zero towards the internal current limiting and/or fixed R which defines the short circuit current. Just like this circuit. except during linear operation. and gain=0 when saturated same as above. Zout=R/Aol(f) for linear and Zout=R when saturated. \$\endgroup\$ Mar 10, 2018 at 2:42
  • \$\begingroup\$ Oh I got it. Avo is zero because the Vi-Vo curve hits a ceiling, which makes just plain sense. That means either way my approach was working. Thanks a lot \$\endgroup\$
    – Yannick
    Mar 10, 2018 at 5:38
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When saturated, the output of the opamp no longer responds to changes in output ( Flat transfer curve). Means, \$ A_{vo} = \frac {\delta V_o}{\delta V_i} = 0 \$. So now you can substitute it in your expression to get the output impedance Zo = R2.

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I was taught an "ideal Op-Amp" (theoretical) has an output impedance of zero Ohms. Add that to R2 and I get Zout = R2. (as others have said)

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