Why is DC voltage biasing applied to a capacitor in distortion pedals

Question

I've been looking at some distortion pedal schematics and many of them are a variation of the same circuit, such as these two:

and

I know that most of these pedals are used with 9V batteries or power supplies, so a 4.5V bias is needed in the circuits, however whats the purpose of appliying the bias voltage to the C3 capacitor in the first image or C2 in the second image?, why not connect it directly to ground?, the capacitor would block any DC applied to it. I also know that the capacitor is probably there to achieve unity gain at DC, and a different gain for AC, but I still cant figure out why the DC voltage is applied there aswell.

Answer 1

You are correct that in the steady state C3 will have a fixed average potential across it and the actual value, 0 V or 5 V, won't matter.

The answer may be that bringing up both the inverting and non-inverting inputs together on switch-on - typically by plugging in the jack - minimises switch-on thumps on the output.

• As the 5 V supply rises the pin 3 will rise due to R4.
• Meanwhile the left side of C3 will be lifted which will cause the right side to follow, at least for a moment.
• The result will be to minimise the difference between the inputs and prevent the output driving hard and fast into saturation.

Answer 2

Theroetically, you are right, you could connect C2 to ground.

However, you have to ask, what is creating that bias voltage. That part of the circuit is not shown.

Ask yourself how does the current going through C2 make it back around through R1 to keep the balance at the inputs to the op-amp. Or if you prefer, what happens when the two Vr points in your schematic are disturbed by a different amount.

If one side has to go through a resistor divider and maybe even a voltage regulator, you can see there is an imbalance.

Whether it matters in this type of circuit though is debateable.

There are also the power on issues Transistor mentions.