I am building a preamp for a sinusoidal current source. I have a two stage amplification process, with a bandpass RC filter in the middle. The second stage amplification is has two outputs from an inverting, and a non inverting OP-AMP. I want the output from both inverting and non-inverting amps to be as close as possible, but inverted. I also require V_out/V_in for the second amplifiers to be about 20 and -20 for the inverting and non-inverting respectively. This would normally mean $$-R7/R6 = -20$$, and $$1+ R4/R5 = 21$$

enter image description here I have two questions.

Firstly, for the inverting amp, does the resistor in the bandpass RC filter (R2) affect the amplification of the inverting OP-AMP? Making the gain $$-20 = -R7/(R2+R6)$$ Would R3 have any effect too?

Secondly, for the Non inverting OP-AMP, as the input inpedance is infinite on the +ve input anyway, does R2 make any difference?

  • \$\begingroup\$ Your bandpass looks wrong, specifically C1. \$\endgroup\$
    – JimmyB
    Jun 29, 2016 at 14:20
  • \$\begingroup\$ "I want the output from both inverting and non-inverting amps to be as close as possible" - You could of course let one amp do the amplification and have the other amp just invert it's output with unity gain. \$\endgroup\$
    – JimmyB
    Jun 29, 2016 at 14:30
  • \$\begingroup\$ @Jimmy, thanks, I'll make sure to correct that. And yes, that looks like that would work. Would there be any phase delay because of that? My signals are only about 20KHz \$\endgroup\$ Jun 29, 2016 at 15:00
  • \$\begingroup\$ What is the purpose of C1? Why didn`t you use a parallel LC bandpass? \$\endgroup\$
    – LvW
    Jun 29, 2016 at 15:00
  • \$\begingroup\$ @LvW, the signals I'm using are quite low frequency. About 20KHz. I think for that case RC filters are fine? \$\endgroup\$ Jun 29, 2016 at 15:04

1 Answer 1


@JimmyB is correct - C1 is attached to the first stage preamp op amp output and sees the output impedance of the amplifier (basically zero), so you would be better placing R2 between the op amp and C1. The op amp will not behave well trying to drive a capacitive load for all but the smallest capacitor values.

Answering your questions: First one: The inverting op amp gain you have calculated (assuming no saturation) would apply only to the signal appearing at the junction of C2 and R3, which is the output of your network. Since the negative input of the op amp is always zero, treat R6 as a resistor to ground when calculating the output of your network. R6 is parallel to R3.

Second Question: Leaving aside the capacitor on the output of the preamp op amp and looking at the circuit as you have drawn it, R2 does affect the gain from start to finish, because the signal is divided down by the ratio of R2 and the parallel combination of R3 and R6 at very high frequencies (where the impedance of C2 is negligible). At low frequencies, you will have to account for additional voltage drop because C2 is in series with R2.

  • \$\begingroup\$ ...or much better: leave out R3 and use R6 as only "load" to C2 in the highpass cell \$\endgroup\$
    – carloc
    Nov 28, 2016 at 7:50

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