simulate this circuit – Schematic created using CircuitLab

This is the virtual ground circuit from a battery-powered headphone amplifier(direct link to schematic PDF). What is the behavior of the pseudo-ground as the opamp attempts to sink/source current to/from it?

A suggestion for improving performance of the amplifier is to switch from this passive circuit to an active one (using opamps, rail splitter ICs, etc.) to provide a "lower-impedance" virtual ground. Before going down that road I'm curious what my starting point so I can know what sort of improvements I might see.

EDIT TO ADD: Another way of asking the question is: does this circuit have any benefits over grounding the load to V- (or V+) and decoupling it from the opamp with a 470uF capacitor?

  • \$\begingroup\$ Impedance to what? Note that you have not drawn a ground (symbol) in the circuit so it appears to be floating. Look up what the AC impedance means. Then have a guess what it is in this circuit. \$\endgroup\$ Commented Sep 22, 2017 at 21:13
  • \$\begingroup\$ There's still no ground even after the edit so again: impedance to what? You circuit is still floating. \$\endgroup\$ Commented Sep 22, 2017 at 21:33
  • \$\begingroup\$ Can you post a link to the original schematic? I can't find it on the site you linked. \$\endgroup\$
    – Transistor
    Commented Sep 22, 2017 at 21:39
  • 2
    \$\begingroup\$ @Frosty I'll fix it up for you. Hopefully, you'll like the result. \$\endgroup\$
    – jonk
    Commented Sep 22, 2017 at 21:42
  • 1
    \$\begingroup\$ Ahem, your amplifier has positive feedback and will not work as a linear amplifier. \$\endgroup\$
    – Andy aka
    Commented Sep 22, 2017 at 21:46

1 Answer 1


It's amazing what passes for circuit design in some of the DIY forums.

Your link goes to this schematic: enter image description here

The center point created is limited by the charge on the 220 uF capacitors, so particularly at high amplitude/low frequencies (as the amp draws longer current and perhaps asymmetrical pulses) the center point will wander.

This is just the same as if the amplifier output had been capacitor coupled which negates what was probably thought of an advantage (DC coupled output).

To fix up this amplifier, I'd simply use a 2 pole supply switch and ground the center point of the batteries ...Voila ....dual supplies and proper DC coupling.

enter image description here

  • \$\begingroup\$ Good point if you have two batteries to work with, but I'm working with a single supply. Also I'm curious what the differences in performance would be between using this virtual ground vs. Just capacitively coupling the outputs to the speakers (and using a resistive divider to bias the opamp. \$\endgroup\$
    – Frosty
    Commented Sep 23, 2017 at 1:46
  • \$\begingroup\$ There is no difference between a capacitively coupled amp and the original circuit shown. For either positive or negative swings the output is capacitively coupled. \$\endgroup\$ Commented Sep 23, 2017 at 3:35
  • \$\begingroup\$ Oh good grief, they really did this from what was a split supply in the first place??? \$\endgroup\$
    – user16324
    Commented Sep 23, 2017 at 11:33
  • \$\begingroup\$ Note that +-4.5v (for a true rail-to-rail op amp) or less (don't even think of using a 741!) is not much "headroom" for a headphone amp and is probably why they used two 9v batteries in the first place. That and 9v's have ridiculously low energy capacity. Also note, headphones are different than earphones - the former require more power at a lower impedance to drive them properly. \$\endgroup\$
    – rdtsc
    Commented Sep 26, 2017 at 1:03

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.