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I have a difference amplifier built with an op-amp (specifically, I'm using the LT6237) with 10dB gain:

schematic

simulate this circuit – Schematic created using CircuitLab

As it is, without external compensation, the frequency response has a bump of more than +12dB at approx. 15MHz (as per an LTSpice simulation). This is undesirable, so I add a compensation capacitor in parallel with RF1 (approx. 2pF as per LTSpice simulations).

My question: do I need to exactly match this modified feedback network by placing an identical capacitor in parallel with RF2? The rationale in favor is that this difference amplifier requires ZF2 / RG2 to exactly match ZF1 / RG1 (where ZFn is the impedance of RFn || Cc --- Cc being the compensation capacitor). The rationale against could be that the mismatch occurs only at high frequencies that are in principle of no interest (where the frequency response starts to fall anyway). LTSpice simulations show that the bump in the frequency response goes away with just the one compensation capacitor; placing the same Cc in parallel with RF2 changes the frequency response making the bandwidth a little bit lower (this is consistent with the intuition --- Cc in parallel with RF2 introduces a low-pass filter in that branch).

Notice that this relates to a modification I need to make to a circuit where the layout is already done, and I'm struggling to fit additional components in it (that's why I'm looking for "excuses" ( :-) ) to avoid having to place the additional capacitor with RF2)

Any suggestions? Should I just place the one Cc in parallel with RF1 and be done with it?

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  • \$\begingroup\$ The quick-and-easy answer is "simulate it, and see if your results look acceptable". But I'd imagine you want more analytical results. \$\endgroup\$ – Hearth Apr 29 '18 at 13:03
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If you a have high-frequency common-mode voltage (including noise) that you wish to differentially eliminate, then don’t compromise and add the capacitor across RF2. If this gives too much attenuation at the top end of your desired frequency response, then reduce the value of capacitance but keep them identical in value.

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