Electronic – measuring output impedance of op-amp

impedanceoperational-amplifier

I've just took for granted the fact that the input impedance of op-amp is high and output impedance is low.

I was reading "the art of electronics" lab book and there is an exercise that asks you to measure the output impedance of an op-amp by having the negative input connected to the output (1) and after the 1k resistor (2) while having and removing the 1k load resistor.

It isn't clear to me how it allows them to measure the output impedance of the op-amp.

The only equation I could write down was with the load resistor in place and the negative input connected to the output of the op amp, I have the following equation

\$\displaystyle V_{out} = V_{in} \frac{R_{load}}{R_{load}+1k \Omega + Z_{out}} \$

I thought this would give you the output impedance of the op-amp.

I wasn't sure what the rest of the steps are for.

From(Learning the Art of Electronics: A Hands-On Lab Course)

enter image description here

Best Answer

The exercise is as follows.

In configuration feedback #1

  1. You measure V_out without R_load, with some high impedance voltmeter. Say you measure 5V
  2. Then you attach R_load. You get an output of 2.5V. You deduce the output impedance is 1k (which is expected because you have a 1k resistor in series with the load)

Then you switch to feedback #2

  1. You measure V_out without R_load, with some high impedance voltmeter. Say you measure 5V again
  2. Then you attach R_load. What do you get? Well, you get 5V. You are forced to conclude that the output impedance is 0
  3. Moreover, for any other load, 10K, 500R, 2K, you'll get 5V. Agree?

So, what happened? Feedback is producing the low output impedance. You can think the opamp is compensating the drop on the 1k resistor (raising the opamps output voltage).

Related Topic