What is the function of the capacitors in this virtual ground circuit?

enter image description here

I've read on the fly that C75 is to maintain a linear impedance for higher frequencies, but under which conditions I don't know.

And the purpose of C40? What is it for? On what depends its capacitance value?

This is in the context of an audio circuit. In particular the splitter should feed 2 variable state filters in series consuming around 50mA.

Reference: https://sound-au.com/articles/st-var-f2.gif

I ask because the circuit works fine without any both caps and I don't know if I'll be able to live without them in the long term, but I would like to skip them for space reasons if safe/possible.

  • \$\begingroup\$ Actually, C75 messes up the virtual ground when VCC is not stable. As Vcc varies, the virtual ground is not held exactly between the two rails, Instead, its voltage with respect to the negative rail remains relatively constant (with a time constant or 5 ms), and with respect to the positive rail it varies. That may not be the desirable behavior. \$\endgroup\$ Jan 15 at 16:07

1 Answer 1


Both capacitors are noise filters. C75 works with the Thevenin equivalent of R124 and R178 to form a single-pole lowpass filter. You want this corner frequency to be significantly below the lowest signal frequency of interest.

C40 both filters noise on the power supply rail, and lowers its impedance caused by wiring and pc board trace resistances and inductances. Typically you want one such large capacitance per xx square inches of circuit area, plus smaller capacitors right at the power pins of each IC. The standard part is a 0.1 uF ceramic cap with the shortest possible leads/traces to the chip pins, one cap per power pin.

Check your numbers - 50 mA is way too high for the linked schematic. Also, most of the power supply current does not go into/out of the virtual GND. The GND circuit output sees only the algebraic sum of all circuit ground currents.


C40 filters noise on the power rail, but more importantly it filters noise caused by the signal going through the opamp. A portion of the current at the power pin is the static current needed to operate the internal circuits, but another portion is the amplified signal current going into and out of the output pin. If the supply rail impedance at the IC pin at the signal frequencies is not zero, the signal current will appear as a varying voltage at the power pin. This varying voltage can now appear across the opamp's input circuit as an unwanted form of feedback, sometimes causing the circuit to break into oscillation. This is especially true with linear audio power amplifier chips. The LM386 makes a fair AM radio transmitter.

The non-zero impedance of the power connection comes from both the output impedance of the source (at signal frequencies, not just at DC), and wire and pc board resistance and inductance. Because electrolytic capacitors usually do not have good high-frequency performance, a compromise is to have large caps in the area for low frequency decoupling, and smaller caps with better high-frequency behavior right at the IC pins to get the impedance at the signal frequencies down as low as possible. These caps also supply some signal current during transients, so the size does matter. 0.1 uF is the standard value for general-purpose work at low signal currents. At 10 mA and above I jump to a 1 uF ceramic at the pins. In many Analog Devices application circuits, they show a 0.1 uF ceramic and 10 uF electrolytic in parallel right at the device pins.


Besides acting as an energy store for signal transient currents, C40 is the shunt leg of a lowpass filter. The series leg is the combination of resistances and inductances in the wires and pc board traces between the power source and C40, plus the output impedance of the power source itself. Note that these are complex impedances, not simple resistances; their values vary with frequency. You can expect them to be higher at a signal current frequency of 10 kHz compared to a signal current of 1 kHz.

These are very small impedance values, difficult to measure and messy to calculate; for normal circuits and wiring, less than 1 ohm. Because an electrolytic capacitor has a relatively high equivalent series inductance, its impedance increases at high frequencies. This is why good design practice is to add a second lowpass filter right at the IC power pin, this time with a ceramic capacitor that has a much higher self-resonant frequency. Some Analog Devices app notes (love me ADI docs!) show a 10 ohm resistor in series with the power source, to give the decoupling capacitor(s) a higher series resistance to work with.

A search for power supply decoupling (no quotation marks) returned many documents and videos in exactly 0.5 seconds. Like this:


And, of course, this:


At the end are links to eleven (11) other documents on this topic.

  • \$\begingroup\$ Provided it is directly powered by a stack of batteries, would C75 still be necessary? Impedance at that point should be quite low. The AD8397 has a max. output current of 250mA. Wouldn't it be enough to provide 50mA even with half amp (it's a dual opamp)? \$\endgroup\$
    – Domingo
    Jan 11 at 16:12
  • 1
    \$\begingroup\$ C75 is necessary since your power supply line will have noise on it. The power supply line has resistance, inductance, and devices drawing power. Battery impedance increases as battery life is used up. You also have thermal noise from the resistors (about \$ 30nv/\sqrt{Hz}\$) and opamp input current noise multiplied by input resistance (about \$75nV/\sqrt{Hz}\$ giving about \$80nv/\sqrt{Hz}\$ of noise. This noise will be injected in your whole system. Then there is noise picked up by the + input since the input is relatively high impedance. C75 is a pretty important capacitor. \$\endgroup\$
    – qrk
    Jan 11 at 16:51
  • \$\begingroup\$ I can trust that. So the capacitance value of C75 cannot be easily calculated by the circuit's power consumption, or any other general parameters (apart from PCB size, which is quite relative)? And what about the mentioned circuit not being able to supply 50mA? Did you not notice the opamp I was using or you were talking about a limitation of the circuit as such? 50mA is the overall consumption of the circuit to supply. You are right that the virtual ground will see less than that, but I brought in my worst case scenario. \$\endgroup\$
    – Domingo
    Jan 11 at 17:06
  • \$\begingroup\$ I didn't say anything about a limitation. I pointed out that the circuit linked to in post #1 almost certainly will not draw 50 mA from the system power supplies; it will draw much less. \$\endgroup\$
    – AnalogKid
    Jan 11 at 20:20
  • \$\begingroup\$ Sorry, I misread and understood the virtual-ground wouldn’t be enough for the linked circuit. Thanks a lot for the answers. C40 is a filter just because it stores current before releasing it, or it also forms a low pass filter? If the second is true, what is the resistance it works with to form a filter? \$\endgroup\$
    – Domingo
    Jan 11 at 21:13

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