Switching in a bipolar power buffer/mixer/whatever

[G. Hoffman]

So, I'm having trouble figuring this out.

I'm doing a thing, not really a name for it, but the audio section is on a bipolar +-9V power supply, and the switching is driven by a 5V microcontroller.  Basically, the audio is just a bunch of buffers and mixers, and I was going to use JFETs, but I can't seem to figure a good way to get the μC to drive the JFETs. 

Do I just need to go with MOSFETS?  Some CMOS switch?  I definitely don't want to do relays.


Gabriel

[R.G.]

You can use either N-channel or P-channel JFETs, but you have to observe their necessary gate-source voltages.

The simplest way to do it conceptually is to power your uC from 0V and +5V, and use P-channel JFETs. For a P-channel, with the gate open (that is, Vgs~0) the JFET conducts. To turn it off, you pull the gate UP. The uC can do this to 5V directly. The P-channels should have both the drain and source at 0V, either through resistors for series switching audio, or with the drain pulled to ground by a big resistor and the source grounded for shunt switching.

If you use N-channels, it would be simpler to run the uC from 0V and -5V, with the uC thinking that  0V is Vdd and -5V is Vss. Now the uC can drive N-channel JFETs the same way.

In either case, you're limited to audio signals less than about 1-2V peak, because the JFET gate must be at least Vgsoff plus the signal voltage in the "off" direction to keep the signal voltage from pulling the JFET out of shutoff. The uC voltage limits you to a total swing of 5V on the gates if you drive them directly, so you need a low-Vgsoff JFET to have some Vgsoff left over after the signal peaks.

If you can live with doing some level shifting and buffering, you can run the uC from 0V and +5V, and have its outputs drive other transistors to make quite-large gate drive voltages. If your circuit runs from, say, +/-12V as an example, you can level shift you uC output from 0-5V to 0 - +12V and drive P-channels very well and use larger signals. Likewise, you could use N-channels and have the 0-5V uC signals converted to -12 to 0V signals for the gates of N-channels.

[G. Hoffman]

You can use either N-channel or P-channel JFETs, but you have to observe their necessary gate-source voltages.

The simplest way to do it conceptually is to power your uC from 0V and +5V, and use P-channel JFETs. For a P-channel, with the gate open (that is, Vgs~0) the JFET conducts. To turn it off, you pull the gate UP. The uC can do this to 5V directly. The P-channels should have both the drain and source at 0V, either through resistors for series switching audio, or with the drain pulled to ground by a big resistor and the source grounded for shunt switching.

If you use N-channels, it would be simpler to run the uC from 0V and -5V, with the uC thinking that  0V is Vdd and -5V is Vss. Now the uC can drive N-channel JFETs the same way.

In either case, you're limited to audio signals less than about 1-2V peak, because the JFET gate must be at least Vgsoff plus the signal voltage in the "off" direction to keep the signal voltage from pulling the JFET out of shutoff. The uC voltage limits you to a total swing of 5V on the gates if you drive them directly, so you need a low-Vgsoff JFET to have some Vgsoff left over after the signal peaks.

If you can live with doing some level shifting and buffering, you can run the uC from 0V and +5V, and have its outputs drive other transistors to make quite-large gate drive voltages. If your circuit runs from, say, +/-12V as an example, you can level shift you uC output from 0-5V to 0 - +12V and drive P-channels very well and use larger signals. Likewise, you could use N-channels and have the 0-5V uC signals converted to -12 to 0V signals for the gates of N-channels.

Sounds like P-Channels and a NPN for the level shifting are in order!


Gabriel

[PRR]

> +-9V power supply, and the switching is driven by a 5V microcontroller.

CD4051/CD4052/CD4053

[G. Hoffman]

Turns out, the answer was a 4016 between -9v & ground to glue things together.

[R.G.]

CD4051/CD4052/CD4053

Hmm. They've updated those. The earlier versions (and perhaps other maker's versions still) would only go 15V total, so that the bipolar power supply range was +/- 7.5V. Good to know.