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Here is a circuit I've found online for switching a guitar signal, it uses a mechanical relay in order to rout the signal through a sound processing unit or around it, bypassing the circuit.

Mechanical relays sometimes create a 'pop' noise due to coil coupling EMF when discharging or due to DC potential differences while switching (think the sound processing unit is AC coupled so when switched to a load that capacitor will discharge). Another source of noise might be the actual pulse created by the MCU to engage the relay. The square wave has a rich harmonic content that can easily couple into the audio path.

This circuit, as far as I can understand, uses a FET that switches just before the relay and creates a low impedance path to ground which enables any capacitor on the output to discharge and will mute the signal by conducting it to ground while the switch takes place. The diode is there to protect the FET (not sure about that one).

The thing is that the 2N7000, as all other MOSFETs, has a body diode which will start conducting while the signal is on the negative cycle, if larger than it's forward voltage. Since the sound processing unit is AC coupled this means that there will be negative voltage present and the diode should conductor at peaks which will square the wave shape, causing distortion. Is there an easy and quick way to solve this by replacing the FET with some other device? I imagine I could DC bias the signal so high that it will never clip but rather not do that if possible.

enter image description here

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Useful information. The thing I don't get is the diode-capacitor arrangement at the MOSFET's gate and source. It seems to me that there is no reliable way for the gate capacitance to discharge. Does that simply rely on leakage currents to keep the gate low? –  Dzarda Jun 9 at 9:54
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@Dzarda Soft turn-off, it's a "feature". LOL –  Spehro Pefhany Jun 9 at 12:26
    
@Dzarda: you could parallel a large resistor to C3, like 500K 1M or something around those numbers. It will work pretty much the same as the voltage will build up when the MCU is high (pardon). I guess the exact value will have to be determined by experimenting because too high of a value will mean slow turn-off. –  user34920 Jun 9 at 14:07
    
@user34920 Exactly. Then I don't know why such thing is missing from your schematic. It's absence is "semantically" wrong. –  Dzarda Jun 9 at 14:21
    
@user34920: I did not draw it, this is the way found online. I did not build it fully yet. The FET thing seems like a problem. –  user34920 Jun 9 at 14:23

2 Answers 2

If the signal levels are high enough to tolerate the additional voltage drop of a Schottky diode (say 0.35 Volts), then the following change should solve the issue of reverse conduction:

schematic

simulate this circuit – Schematic created using CircuitLab

The Schottky diode indicated, Rohm Semi RB531S-30, has a 350 mV forward voltage and fast recovery time. It is available in single units for under US$0.50 per unit from DigiKey and others.

Alternatively, fast recovery, small-signal Schottky diodes with even lower voltage drop, down to 200 millivolts or less, are available as well.

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But with a diode in that position, the transistor can no longer mute the negative half of the audio signal when it is supposed to be on. It also reduces its effectiveness on the positive peaks. The OP was asking about avoiding the clipping of negative peaks when the transistor is off. –  Dave Tweed Jun 9 at 11:29

Use a FET without a body diode. These are usually JFETs rather than MOSFETs. Lower power handling but commonly used in this role. The J309 or J310 is one part which has been used for this role in audio applications, but at higher supply voltages. You may have to change the gate drive voltage to turn it on and off properly.

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The problem with these is that I'll have to generate negative voltage in order to work with them? –  user34920 Jun 9 at 18:30
    
That may well be the case. My applications of them had +/-15V available. –  Brian Drummond Jun 9 at 19:23
    
I could probably get away with using a P channel jfet right? They are not as many available but there are a few... –  user34920 Jun 9 at 21:23
    
Depending on the FET, possibly. Don't know if 5V is enough to switch the ones that are available; some detailed datasheet reading required. –  Brian Drummond Jun 9 at 21:50
    
I just thought of something... my signal is both positive and negative (around 0V) so it does not matter if I use a P or N channel JFET, at some point they will close due to the polarity of the signal which will make the gate in the "wrong" potential to operate as should. –  user34920 Jun 10 at 8:41

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