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While trying to measure the output impedance of a follower circuit involving a 411 op amp, I connected a small resistor (27ohm) to the output (at Vout). enter image description here

Using an oscilloscope to measure the voltage drop across this resistor, I saw this (blue is the drop across the resistor, yellow is the input at Vin): enter image description here

Is this a quirk of op amps? Or is the current so large for the small resistor that it's essentially shorting as the voltage increases?

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  • \$\begingroup\$ You are right about the current being so large it is essentially shorting. In order to measure the output impedance you need to load it with a larger resistor, which will cause a small change in the output. You can then treat the load and output impedance as a voltage divider to find the output impedance. \$\endgroup\$ – Austin Oct 25 '16 at 11:18
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$$\frac{2 \rm{V}}{27 \Omega}=74 \rm{mA}$$

Based on the datasheet, the LF411 is not able to supply more than 25 mA (at 25 C temperature).

Here's the relevant figure from the datasheet:

enter image description here

You can see they don't even specify the capability below 100 ohms, but it's rolling off fast enough to tell you "don't go there".

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See Figure 28 "Detailed Schematic" in this Texas Instruments datasheet for the LF411 op amp. Mentally connect your \$27\Omega\$ load resistor \$R_{LOAD}\$ between the op amp's output pin \$V_{O}\$ and GROUND. Assume transistors Q9 and Q10 are shut off (not conducting) and transistors Q8 and Q11 are ON. The output path is now a resistor voltage divider consisting of R5 (\$22 \Omega\$) and \$R_{LOAD}\$ (\$27 \Omega\$). This being the case, the maximum positive voltage you can develop across \$R_{LOAD}\$ is approximately \$V_{CC}/2\$. For example, given \$V_{CC}=5V\$,

$$ V_{R_{LOAD},MAX} \approx (V_{CC}-V_{CE,Q8}) \frac{R_{LOAD}}{R5 + R_{LOAD}} \approx 5V \frac{27\Omega}{22\Omega+27\Omega}\approx 2.76V $$

Likewise, when transistors Q8 and Q11 are shut off, and transistors Q9 and Q10 are ON, the maximum negative voltage across the load resistor \$R_{LOAD}\$ will be about \$V_{EE}/2\$.

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