Im creating a guitar effects pedal as a final year university project and I'm running it all from a single supply

I understand the concept of a virtual ground such that the input signal is biased in-between the power rails however Im having a hard time understanding the correct way to connect the input signal to the op amp

Im using OPA2350 op amps with TLE2426CLP rail splitter.

The jack lead connection from the guitar provides a mono signal with the ring being 'ground' and tip being the audio signal

  • Should I connect the ground of the jack lead to 'real' ground (supply rail) of the op amp then connect the audio signal to both virtual ground and the input of the op amp through a capacitor (biasing at \$\frac{V_{in}}2\$)
  • Or should I connect the ground from the jack lead to the virtual ground from the rail splitter then connect the audio signal straight to the op amp such that the 'ground' from the guitar would be raised to \$\frac{V_{in}}2\$?

Which way would be correct?


Door two... You connect the virtual ground to the incoming ground. That way the power floats either side of the jack ground, which is what you wanted in the first place.

  • \$\begingroup\$ That makes sense about the power floating either side of the jack ground but in that case, for circuits like this: i.stack.imgur.com/Ohebo.png. Why isn't the negative terminal of the sine input connected to the voltage divider i.e ground of the sine generator biased to Vin /2 ? \$\endgroup\$ – theguitarfreq Feb 27 '17 at 22:10
  • \$\begingroup\$ That circuit is AC coupled instead. The input to the op-amp is biased to half-rail. So the signal is effectively offset to half rail. \$\endgroup\$ – Trevor_G Feb 27 '17 at 22:22
  • \$\begingroup\$ So both achieve the same end result? Why use one method over the other? Am I right in thinking the power is therefore dissipated over half-rail with AC coupling? \$\endgroup\$ – theguitarfreq Feb 27 '17 at 22:32
  • \$\begingroup\$ The drawback of ac coupling is you are effectively adding a high pass filter to the system so you lose a little bass, and the added resistor at the output to bias it back to gnd can be problematic depending on the input resistance of the next stage. \$\endgroup\$ – Trevor_G Feb 27 '17 at 22:40
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    \$\begingroup\$ You may still get CM hum rejection problems with unbalanced impedance unless shielded with ferrite sleeve on signal pair. \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Feb 27 '17 at 23:35

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