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As with other ICs, it's standard practice to place bypass capacitors near the supply voltage pins of op amps. But I've seen conflicting opinions on how to properly bypass an op amp (here, for example). Some people suggest putting a single capacitor between the V+ and V- pins. Others suggest using two capacitors, one from V+ to ground and one from V- to ground. Which of these methods gives the best result? I'm using OPA827s as unity-gain voltage followers for audio signals, but I'd like to know whether the answer's the same for other situations as well.

schematic

simulate this circuit – Schematic created using CircuitLab

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    \$\begingroup\$ Your good question does not have a definative answer . \$\endgroup\$
    – Autistic
    Jan 30, 2016 at 7:35

2 Answers 2

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If the output load is primarily to ground, then two capacitors. If it's to either supply, then one capacitor will suffice.

The purpose of bypass capacitors is to provide a low-impedance close to the chip (bypassing any series inductance to the supply rails). Since most op-amps do not have a ground pin the internal circuitry does not care about the ground level, however when you apply a load to ground the current path is from the positive or negative supply, through the circuitry on the chip, out the output and through the load to ground. A capacitor from the positive and negative supplies will make sure that loop is physically small and thus low inductance.

schematic

simulate this circuit – Schematic created using CircuitLab

The left two schematics show an op-amp driving positive and negative current into a load connected to ground, and how the bypass capacitor appears in the loop. The right hand one shows a load connected to the negative rail. The capacitor in the right hand schematic is twice as effective with the load connected this way (1uF rather than 500nF) and it saves a part.

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    \$\begingroup\$ Note, though, that this only applies if the signal is referenced to V- rather than ground. \$\endgroup\$ Aug 28, 2015 at 19:36
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    \$\begingroup\$ Or V+, of course. \$\endgroup\$ Aug 28, 2015 at 19:39
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One capacitor for each supply, connected to ground.

Consider that, if you have dual supplies, and each supply produces a positive spike of equal amplitude, a capacitor between the two supplies will see a constant voltage, and will attenuate neither spike.

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  • \$\begingroup\$ The opamp between the two supplies will also see a constant voltage, so does it really matter? The inputs are too shifted together, and opamps usually are designed to have high common mode rejection, so that shouldn't be a problem either. \$\endgroup\$
    – mow
    Jun 1, 2023 at 16:19
  • \$\begingroup\$ @mow - A constant voltage difference is not nearly the same. The issue is not common-mode, it's called power supply rejection ratio (PSRR) and it's part of the spec for any op amp. The problem is that op amps, being very high gain, will respond to very small variations, especially unbalanced ones. I'm afraid you're just going to have to take my word on this that anything other than a balanced decoupling may well end in tears. And will be very difficult to diagnose. \$\endgroup\$ Jun 4, 2023 at 1:00
  • \$\begingroup\$ What is "each supply produces a positive spike of equal amplitude" if not common mode? \$\endgroup\$
    – mow
    Jun 6, 2023 at 16:51
  • \$\begingroup\$ @mow - Yes, it is common mode. However, CMRR (common mode rejection ratio) is frequency-dependent, with CMRR decreasing with increasing frequency. Spikes typically have lots of high frepuency components, so depending on the op amp not to have problems is a very bad idea. Plus, common mode at high frequencies is affected by parasitics and layout, so the input at the two inputs will typically not be as "commom" as you probably think. It's best to try to squelch this stuff at the origin. \$\endgroup\$ Jun 7, 2023 at 16:43

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