# How does this single supply opamp circuit work, and what does the output look like?

When building audio circuits, I have often used a single-supply non-inverting configuration in which the feedback resistors are connected to via 10uF electrolytic capacitor to ground instead of directly to ground. I found this circuit online once, but don't generally see it when looking up single-supply circuits.

How does this circuit work? And how does the output end up more or less in the middle of the output range of the opamp (0v -> 9v in my particular case).

simulate this circuit – Schematic created using CircuitLab

EDIT: The reply that states that this doesn't work isn't quite correct - since it does :). However I did omit to mention that the input is not referenced to ground but instead somewhere around 5v or so. It's not half-supply though, since it came out of a class-A FET input stage and ultimately from an electric guitar input.

Anyway, the answer that you get a DC gain of 1, and an AC gain of 10 is the one I was looking for. I'd like to understand more about this though. Is it that at DC the capacitor is effectively an open circuit, and so the op amp becomes a voltage follower? And at AC it is a resistor of some frequency-dependant size, and thus it becomes an amplifier of some gain?

• It will also work much better now that you edited it to have the supply pin connected to 9V rather than floating. ;-) Yes, your understanding is correct. In fact, at frequencies much above 1/(2*piRC) a capacitor in series with R can be ignored (so you just have R). Commented Jun 23, 2014 at 5:18
• Wouldn't the AC gain be a function of the input frequency since the reactance of C1 changes? Commented Jun 23, 2014 at 5:55
• It is a function of input frequency, but above a few tens of Hz it hardly changes at all. Think of sqrt(R^2 + Xc^2).. If Xc << R then the exact value hardly matters. Commented Jun 23, 2014 at 11:46

It doesn't work.. at least it needs a power supply on the op-amp.

1. The TL081 has a common-mode input range that goes down about 3V above the negative rail, so it cannot deal with voltages at the negative rail (in this case, ground).

2. The reason to have the 10uF there is to have a DC gain of 1 and an AC gain that is higher (in this case the AC gain is 11 and it rolls off at the 16Hz -3dB cutoff frequency).

3. With the capacitor there, the DC gain is still one, so you would need to bias the input to +4.5V or so and capacitively couple the input to get it to work (say two resistors across the supply, with a capacitor to the input- but that has no power supply noise rejection).

• Fair points. 1) My input happens to be referenced around half supply. 2) This is what I'm asking about. 3) As above. Commented Jun 23, 2014 at 1:14

When this circuit starts up, C1 is discharged and at 0 volts. This will cause the junction of R1 and R2 to be close to 0 volts. This junction goes to the inverting ( - ) input of the op amp. Assuming the input signal is not at ground (you have to bias this up somewhat) the op amp will swing fully to the positive rail. In that condition, current will flow from the op amp's output, through the two resistors, and charge C1. Once the capacitor has charged to the point where the inverting input voltage rises above the (non inverting) input voltage, the output will swing towards 0 volts. That will stop charging the capacitor, and the circuit will come into balance.

If C1 is leaky (as electrolytics can be), it will throw the balance off.

As mentioned, the input (your audio source) has to be biased positive, because if it swings below zero volts, it's below the supply rail and the signal will be clipped.