I have been experimenting with op-amp audio amplifiers powered from a 9V battery. I am just wondering how I should go about splitting the voltage from the battery, as I have found 2 ways of doing this . . .


One method is to make the ground rail the -ve terminal of the battery and then DC offset the input signal . . .

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Or a virtual ground can be created like in the following . . .

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They both seem to do pretty much the same thing. Should I prefer one method over the other?


Based on Mark's answer, you can also have a rail splitter circuit like the one shown below. The purpose of the (optional) resistor in the feedback loop in the supply is to keep the op-amp stable in the face of heavy capacitive loads. The ground on the non-inverting input is only there to keep the CircuitLab simulator happy.

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    \$\begingroup\$ Your second design is broken in that the DC level to the opamp positive input is floating. \$\endgroup\$ Oct 31, 2012 at 13:24
  • \$\begingroup\$ Does that mean I need a pull down resistor to virtual gnd or -VCC? \$\endgroup\$
    – learnvst
    Oct 31, 2012 at 13:32
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    \$\begingroup\$ It means you can bypass C1 and put a decoupling cap from virtual ground to -Vcc to lower the series impedance to Vout load. A 3rd way is use a spare Op Amp as a unity gain follower to buffer "virtual" to reduce impedance further when necessary in future bias circuits and loads. \$\endgroup\$ Oct 31, 2012 at 15:42
  • \$\begingroup\$ So after my edit, is circuit 3 > circuit 1? \$\endgroup\$
    – learnvst
    Oct 31, 2012 at 15:48
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    \$\begingroup\$ Your third circuit has a notational problem: since you have labeled your supply as Vcc, the outputs of the rail splitter should be labeled +-Vcc/2, not +-Vcc. Some of the other schematics have this problem too. Your intent is clear when reading the question text, but that and the fact that in some circuits you show the voltages referred to the virtual ground but the ground symbol is connected to the negative rail of the supply is misleading. It would be nice if you could fix the schematics, so a casual reader won't see something technically wrong. \$\endgroup\$ Jul 4, 2014 at 15:12

1 Answer 1


You have a rather weak connection to the midpoint voltage. (50kΩ.) In the first circuit this is fine, but in the second it is not. The midpoint voltage will drift a lot, because the load current is being returned through the same 50kΩ impedance. You even get coupling from output back to input through the Rload to the bottom of the input voltage. This can cause oscillations, though probably not for a Av=1 buffer like you have here.

My suggestions:

  1. If you want to AC-couple input and output, use circuit 1. Circuit 2 gives no advantage.

  2. If you want to avoid AC-coupling, use a rail-splitter circuit (e.g. buffer the midpoint voltage with another op-amp.) Then you can dispense with the caps because you have a good +/- 4.5V supply.

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    \$\begingroup\$ Nice answer. Let me knock together a simulation of your 2nd suggestion to see if I understand it correctly. \$\endgroup\$
    – learnvst
    Oct 31, 2012 at 15:02
  • \$\begingroup\$ Would there be any advantage to using a rail splitter circuit un conjunction with my 1st circuit? \$\endgroup\$
    – learnvst
    Oct 31, 2012 at 16:17
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    \$\begingroup\$ No, the first circuit is equivalent to 50kΩ resistor to an ideal rail splitter. en.wikipedia.org/wiki/Th%C3%A9venin's_theorem \$\endgroup\$
    – markrages
    Oct 31, 2012 at 19:20

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