Bias tee for audio?

[RetroSwim]

Hi there!

I'm trying to toggle an audio source on or off using a MOSFET, but when I try, this is what I get:



Blue trace is the source voltage, yellow trace is the drain voltage, measured with the gate "off".

To remedy this, and cause the MOSFET to block the negative-going part of the signal, I understand that I should bias up the audio signal by 2 volts or so, shifting the entire waveform above 0v. Is this correct?

That being the case, is a bias tee the right approach? Or is there perhaps a simpler solution to this problem? (Preferably an answer other than "Don't use a MOSFET"! :-/ )

Cheers

[c4757p]

More information, please! How are you connecting this? There are lots of ways you could switch a signal with a MOSFET, a few of which might actually work.

I'd recommend a simple analog switch IC, though:

"Don't use a MOSFET"!

[RetroSwim]

Thanks for the swift reply!

It's all on a breadboard at the moment. The MOSFET is a 2N7000, audio source is the headphone jack on my PC, and the gate is connected to a digital out from an ATMega328P (programmed to simply toggle the pin once a second), also on the breadboard. The power source for the whole setup is a USB port on my PC.

I agree on the IC comment. In fact the plan is to ultimately use analogue bus multiplexer ICs, like the Analog Devices ADG608, but the datasheet explicitly says that the analog source range is Vss to Vdd. Wouldn't that mean I'd need a negative rail for that chip to work?

Edit: A diagram, because I'm terrible at explaining myself:

- Link for larger image.

[c4757p]

No, you don't need a negative rail. What you do need is a rail underneath* your signal.

So we're back to the "bias tee", though in non-RF applications it's usually just called "AC coupling".

Use a voltage divider to get a fake "ground" halfway between the power rails, and then use a capacitor to couple the signal onto that. Note that you'll be forming a low-pass filter as well, so make sure 1/(6.28*R*C) is less than the lowest frequency you want to pass.

This can be directly switched with any analog switch IC. You can AC couple it on its way out again, to remove the bias.

You don't really have the headroom to use a single MOSFET here, even with DC bias, unless you got one with very low gate threshold. The signal would have to be less than 5V-(gate threshold) at all times. 2N7002's threshold is around 3V.



*How far underneath? Check the datasheet. ADG608 claims that the signal can range all the way from VSS to VDD, so "immediately underneath" will do.

[c4757p]

Just saw the schematic you added. See the little arrow in the MOSFET symbol? That's a diode, formed by the FET substrate.

If your signal happened to be under 0.6V peak, the diode would never conduct, and the MOSFET would appear to be symmetric.

Just food for thought

If you got a MOSFET with a separate pin for the substrate (not worth it - IIRC only one is left on the market, probably soon to be EOL'd, kind of expensive, and with crap specs), you could bias that pin negative and then this circuit would work.

Another problem, given the schematic you just added: You're not switching a signal, you're switching power, at least enough to drive headphones. You'll need large enough capacitors to carry that current, and a switch with a low enough ON-resistance as well.

[RetroSwim]

Your explanation makes perfect sense. Thanks so much for your time!