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?

  • \$\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

$$\frac{2 \rm{V}}{27 \Omega}=74 \rm{mA}$$

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".


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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