I'm testing a voltage controlled amplifier circuit in a simulator. The circuit is powered by a single 9V supply, so a "virtual ground" equal to 1/2 VCC is used. I have a question about the output of this circuit.

Due to the virtual ground, the output is offset by 4.5 V. I can only remove this offset using a filter from a capacitor and a resistor to ground. But some other single-supply circuits that I'm looking into have just a capacitor at the output, no resistor to ground. I tried this connection, but one capacitor does not remove the offset. Is this really true, or is this a simulator bug? Maybe it's not necessary at all?

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  • \$\begingroup\$ The resistor to ground is assumed to be at the input of the destination where it is not at the output. \$\endgroup\$ Feb 23, 2023 at 23:30
  • \$\begingroup\$ Interesting, I didn't think about that. I have often seen a resistor to ground at the input of various audio circuits. \$\endgroup\$
    – Alex_G
    Feb 23, 2023 at 23:38

2 Answers 2


In the total absence of any reference to ground, the end of a capacitor will take on any voltage it "wants to". You can't get this in real life -- but a simulator will give it to you if you ask. To simulate a real circuit, you need a resistor to ground or some other voltage source.

Note that simulation (of anything, not just electronic circuits) is full of little pitfalls like this. There's no way you can simulate a full model of what you're actually doing, so you need to know the shortcomings of the model you're presenting to the simulator, and know what valid stuff you can learn from it and what is simulation oddities.

  • \$\begingroup\$ So it's a problem in the simulator. Or more precisely, in my understanding, how the simulator works. I should build real circuits on a breadboard more often. \$\endgroup\$
    – Alex_G
    Feb 23, 2023 at 23:41
  • \$\begingroup\$ Well, it was a problem in your understanding and use of the simulator. Now you know :). \$\endgroup\$
    – TimWescott
    Feb 23, 2023 at 23:51

It is not a bug in the simulator. It is the way capacitors work. WHen power is applied, The mid-point bias ia like a step voltage being applied to the capacitor.

If the other lead of the capacitor is not connected, then no current can flow through the capacitor. The capacitor cannot charge so the voltage drop is zero. The dc voltage will appear on both sides of the capacitor.

If a 1000uF capacitor is loaded with a 10Mohm DMM, then the time constant is 10,000 sec (6.9 days). The capacitor is discharging, but it will take a long time.

A simulator is can simulate an infinite resistance by using some enormous value for the load resistance. This allows us to explore "ideal" circuit behaviour that we cannot do with real components.

This is why a smaller value of resistance is used to charge/discharge series connected input and output capacitors.


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