4
\$\begingroup\$

In most derivations of the output impedance of an (non-ideal) non-inverting opamp circuit, the input terminal is grounded (e.g. page 5/6 here or page 76 here). It also seems to be a common procedure for finding the output impedance of a circuit. Why is this the case?

However, I have also found a derivation that seems not to do this (Example 5.16). I am assuming that because both derivations arrive at the same result, they are both valid. But where are the assumptions similar/different and why does one ground the input pin?

enter image description here

\$\endgroup\$
1
  • 2
    \$\begingroup\$ The output impedance is defined as the change of \$I_{out}\$ as a result of a change in \$V_{out}\$. So long as \$V_{in}\$ is kept constant or is much smaller, such that it does not affect \$I_{out}\$ then either method is OK. If \$V_{in}\$ was not kept constant or not small enough, that would cause an additional current at the output and that would give the wrong result for the output impedance. Keeping \$V_{in}\$ constant (DC) is the same as grounding it from a (small) signal point of view. \$\endgroup\$ Jan 11, 2022 at 12:36

4 Answers 4

3
\$\begingroup\$

As pointed at by Bimpelrekkie, you are dealing with a small-signal analysis. The input voltage \$V_{in}\$ sets the dc operating point at which the internal \$Z_{out}\$ of the op-amp is obtained. Then, to determine the overall output impedance, you want to ac-modulate the output once the loop is closed. You inject an ac current \$I_T\$ across the output terminals and check what voltage \$V_T\$ you collect across the current source terminals. The ratio \$\frac{V_T}{I_T}\$ gives the output resistance you want.

However, when doing this analysis, the input source is not affected by the output current modulation. If you would probe this source \$V_{in}\$ with an oscilloscope, you would observe a flat dc line. Therefore, in absence of modulation, its ac or small-signal value \$\hat{v}_{in}\$ is equal to zero and does not participate to the determination of \$R_{out}\$. You replace the source by a wire in the exercise. I have drawn a quick illustration below:

enter image description here

The same philosophy is at work when you short to ground the power supplies lines \$V_{cc}\$ or \$V_{ee}\$ in the small-signal study of a circuit supplied by one or several dc rails. You do this because there is no modulation across these lines if you stimulate the input to determine the response. Same thing.

\$\endgroup\$
2
\$\begingroup\$

Why do you suppress the input when finding the output impedance of an opamp circuit?

It's purely a reason of convenience. If the input is set to a value of zero then, the natural output voltage of the op-amp is also (pretty much) zero for straightforward circuits. Then, if you inject current into the output, and the voltage rises by an amount, it's simple mathematics that tell us that the output impedance is: -

$$\dfrac{\text{voltage rise from 0 volts}}{\text{injected current}}$$

That's basically ohms law.

If alternatively we had an input voltage present, the op-amp output impedance would be: -

$$\dfrac{\text{voltage change from what it was}}{\text{injected current}}$$

So, it's slightly more convenient to force the input to be zero volts. However, if we are doing this accurately then we should note that the op-amp output may not be truly 0 volts when the input is set to zero volts. In which case we should use the voltage-change formula directly above.

\$\endgroup\$
2
\$\begingroup\$

I am not sure if I have understood your problem in full detail. Nevertheless, perhaps the following is helpful:

For any circuit, it is important that no EXTERNAL input signal (other than as injected into the output node) will contribute to the current you are going to measure through the output node.

However, in case of a feedback system that means: Injecting a signal into the output will, of course, inject a portion of this signal also into the input of the amplifier (through the feedback path).

(Added): Because this fedback signal must see the same situation as during normal operation, it is important that the terminal for the "normal" input signal is ac-grounded (and not only left open).

This effect is very important and must not be neglected - even when this portion is pretty small (it will be amplified with the large open-loop opamp gain).

This is the only method to consider the heavy influence of the feedback loop on the measurement of the output resistance.

\$\endgroup\$
1
\$\begingroup\$

There are two ways to measure output impedance: Zout=dVout/dIout

  1. Make the circuit hold its voltage constant, inject an AC current I into the output, and measure output voltage variation V. Then, Zout=V/I.

  2. Input an AC voltage at the input, measure output voltage. Connect a known load, say a resistor that will draw a current from the output. Compute the difference dV in output voltage with and without output current. Zout=dV/dI.

These two are the two methods you're talking about in the question (although you have to squint a bit to recognize the second one in the scan from the book).

The second is much more difficult because you have to do two measurements and substract. If output impedance is small, the two measurements will be very close together, which introduces error. The first method is a more practical and direct measurement, so that's usually how it's done.

You can also do it with a DC current, but it is much easier with AC, as measuring a small AC voltage is easy and ignores any DC variations due to offset, temperature, low frequency noise, etc. Also, output impedance depends on frequency, so if you're interested in that, you have to measure it using AC and do a frequency sweep.

If you have good reason to believe the output impedance will be constant over a wide range of DC output voltage, then the actual value of the output voltage doesn't matter as long as it is constant. The simplest is to just use 0V, which means to connect the input of the opamp to ground.

This is usually the case for an opamp, as long as it's not clipping: output impedance won't change much with output voltage. It will vary quite a lot with DC output current, though, because the transconductance of output devices depends on current.

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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