How to measure open loop freq response

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Measure open loop etc

Generally it is done by modifying temporarily the feedback loop, so that the DC feedback stays intact to keep the amp in DC balance, but making the AC feedback zero. For instance, you can split the resistor going from the output back to the input stage, and put a cap from the split point to ground. To make the measurements easier, you should also attenuate the input signal 100 times or more, otherwise you overdrive the amp with just a tiny little input signal. Details depend on your particular amp of course.

Cheers, Jan Didden
Remaining signal on the (almost) virtual earth.

Maybe you could hook up the opamp as an inverting gain of 1amplifier and drive it to full output swing with a swept frequency. The higher the internal gain at any particular frequency the less residual signal on the inverting input. So if the opamp has an open loop gain of 100,000 and swings 10 volts then the change in voltage between the + and - inputs will be 10/100,000 volts = 100uV. Just guessing here, I don't really know for sure.

Measure open loop etc


You're absolutely right. I have seen it done that way. The output level was kept constant, and the input level plotted vs frequency. But this will involve measuring reliably very low input levels. The method described above avoids that. But there is no reason it shouldn't work.

Jan Didden
The one and only
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If you have an Audio Precision, you can typically tap
across the diff pair going to the balanced input of
the AP and sweep. This avoids having to modify
the circuit.

If you don't have an AP, build a high impedance
instrumentation amp out of op amps, and measure
across the input diff pair with it.
Your method still involves a very small signal, which is
ok if the signal is not too small.

Let's assume R1 is resistor from opamp output
to diffpair inverting input and R2 is shunt resistor
which is connected from the inverting input to
ground thru a large value capacitor so that
the low freq 3db point is well below the
first open loop pole.
If I understand your concept correctly, you split
R1 into 2, say, R1 from output to R11 to inverting
input. You connect a large cap from R1/R11 junction
to ground.
If the open loop gain is say, 100db, then it still means
the signal has to be attenuated to very small levels.
I was trying to avoid this as the noise can be
not insignificant, no?

I don't have an AP set, does your method have
the feedback network intact? If so, how does
one get the open loop response - Vout / Vdiff_at_input?


The one and only
Joined 2001
Paid Member
So maybe it loads the inputs with a couple pF.
If your circuit is typical, it has a couple K impedance
to ground on the feedback loop, so no problem.

Open loop is Output voltage / diff input voltage. If you
need a bandpass filter to take out any noise, DIY :)
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