[sdiy] Best way to limit current through an analog switch IC?

[Roman Sowa]

An elegant, but complex solution is to measure voltage across the mux 
with differential amplifier. When it exceeds 3V the diff-amp shuts the 
mux with Enable input and locks. Possibly reconnects after 1s so you 
don't have to manually reset it.
You could use one amp and switch its input with another mux addressed 
from the same address lines. Having 2 muxes in one chip as in MUX507 
simplifies things a bit.

But frankly, if you worry about pitch CV erros in MUX, you should rather 
use part with Ron in range of 5 ohms or less and not 125-250 ohms.

I'd put 1k in series with the mux and be happy it works. Many modules 
have 1k at outputs anyway. And for offset-free pitch CV switching it's 
so much easier to make one-direction mux.

Roman

W dniu 2024-07-02 o 17:05, Paul Glass-Steel via Synth-diy pisze:
> Hello all,
> 
> I'm working up a bi-directional switch module around the TI MUX507 
> analog switch IC. The datasheet lists the maximum IO current as 30mA. 
> I'd like to proof the switch against the fault condition of having the 
> IO spanning my +/- 12V rails, however remote that mis-connection might be.
> 
> By Ohm's law, the bulk series resistor needed would be 820R 1W, rounding 
> to common values. I imagine that resistance would be high enough to 
> audibly affect pitch CV. Is there a better or more elegant solution to 
> limit the current?
> 
> Thanks,
> -Paul
> 
> 
> 
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