Question: Guitar signal range

[wasp]

Hi,
I have a quick question that could probably be answered very simply by an expert. I plan to attempt to perform true bypass switching with a 333 quad analog SPDT switch (from Maxim, Analog, Vishay, etc.)  This IC can switch analog signals ranging from it's positive rail to it's negative rail.  Specs say up to ±20V rails, plenty of headroom.

My question is about the signal range from a guitar or a stomp box. Common sense says that voltage output would not exceed the power supply. In other words, an effect powered by a 9V battery cannot output a signal outside the range of 0 to 9V.  Unfortunately, my common sense also says that the signal will be centered around ground and operate with the range of -9V to +9V (probably a much smaller range, actually).

I'm no expert in EE (mostly self-taught), so one of the assumptions is probably the culprit.  I don't have a scope to test this with, either.

Obviously it's not the end of the world if I have to split a 18V wallwart to accommodate the signal range, but the device can also be powered with a single 9V supply which is much easier in the stompbox world.

If you have an answer, I'd love to hear from you. Thanks, all!

[Ripthorn]

I know that as long as you are using an opamp, the output signal can't exceed supply voltage peak-to-peak.  That means if you feed it +9V and have Vcc- to ground, then you will have a possible +/-4.5V swing max.  If you run at +/-18V or so, I don't think you would ever have a problem.

[amptramp]

If you have a stompbox with AC signal coupling via a capacitor (or transformer, but I have yet to see that) and there is a resistor to ground at the output, the output could go negative, below the power supply negative voltage (with a negative ground system).  Assume a simple op amp stage with the power supply at 9 VDC and an internal 4.5 VDC divider for the non-inverting input.  This forces the output to 4.5 VDC with no signal.  If this is coupled through a capacitor followed by a resistor to ground (and the resistor could be in the following amp - it doesn't have to be in the stompbox), driving the output negative from 4.5 volts will drive the output the same amount below zero volts.  If the op amp output goes to 3.0 VDC, the output of the unit will be -1.5 VDC, but due to capacitive coupling, will begin to rise asymptotically to zero with a time constant of (Ramp + Rout)*C.

[BAARON]

Another way to explain it:

You have an input cap at the start of your circuit.  On the guitar side of the input cap, the signal is indeed swinging around somewhere between -1v and +1v, centered on ground.  On the op-amp side of the input cap, there is a large resistor that connects the input to a +4.5v supply, which thus means that the input usually hovers around +4.5v.  The input cap prevents the +4.5v DC from crossing to the guitar's side, but allows the guitar's AC signal through to the op-amp side.  When the AC signal from the guitar passes through the coupling cap and combines with the +4.5v DC on the op-amp side, it'll therefore swing from about +4.5-1v to +4.5+1v, or 3.5-5.5v.

Thus, by setting the bias on the op-amp side at something such as +4.5v, the incoming AC signal will be centered around that point on that side of the input cap.

[wasp]

Thanks, everyone, for the very informative replies.  I understand now what AC coupling is and how it works and I have spotted it in several schematics.

Unfortunately, I seem to have been following in the footsteps of this article:
http://www.geofex.com/Article_Folders/cd4053/cd4053.htm

I figured out the same way to do this, right down to the D flipflop for toggling.  I later decided that a 333 switch would be easier to work with because of it's wide signal range, but it seems that the wide signal range is useless without a wide supply range.  The 4053 is much cheaper, so I'll be switching back to that (plus I ordered 10 before deciding that the 333 was better).  I further decided that a small microcontroller would be easier (and even cheaper!) than the debounce and control circuitry, while making it easier to tweak values.

I think I can take it from here by myself.

[igor12]

+/- 1V.  Stomboxes shouldn't really go over that.  Your standard overdrive never goes above .7v (usually)

[CynicalMan]

+/- 1V.  Stomboxes shouldn't really go over that.  Your standard overdrive never goes above .7v (usually)

Because the output is compressed and clipped 0.7Vp dirty sounds louder than 0.7Vp clean. With high-gain clean boosters, the output goes way above that. There are commercial effects that go above 20dB clean boost. With an input signal of 0.6Vp (from here), the output would be +/-6V.