question about multiplexer IC's

Thread Starter

tuuluser

Joined Nov 20, 2011
23
I have a circuit where I need to multiplex 16 analog channels to a single ADC pin. I set it up with a CD4067B. It works great. I just select the channel I want using the 4 address pins, and it serves the channel on the common pin.

I have Vdd at 5V. It seems as though the chip can only handle input signals less than about Vdd+0.4V. Above that, regardless of the input voltage, output is fixed at Vdd+0.4V.

Problem is, my analog channels range from 3.3 - to 15V. Is there a multiplexer chip than can handle this switching with only a 5V rail?

Or do I have to add a 15V rail to the circuit?

This is a great forum.
 
Any voltage > Vdd+.4V on that chip is likely turning on ESD/parasitic diodes, which is why the part is spec'd that way. I'd definitely recommend adding a 15V rail if your analog channels include that range. Someone might be familiar with a mux that doesn't have the Vdd + .4V restriction, but I'm not aware of one off the top. It's a common restriction.
 

lightingman

Joined Apr 19, 2007
374
Drive the address inputs of the CD4067 through a CD40109. It is a quad level shifter. You can then supply the CD4067 with a higher voltage (18v max). You can then switch higher signal levels through it. I use this method to drive 0 to 10 volt dimmer signals with a CD4067.

Dan.
 

tshuck

Joined Oct 18, 2012
3,534
As an alternative, you could scale the voltage to be within the range of the mux.

Remember, your Vref on your ADC must also be >=15V in order to digitize the signal, so if your Vref is 5V, you can kill two birds with one scaling circuit, which, may even be as simple as a voltage divider, depending on application...
 

WBahn

Joined Mar 31, 2012
30,066
As an alternative, you could scale the voltage to be within the range of the mux.

Remember, your Vref on your ADC must also be >=15V in order to digitize the signal, so if your Vref is 5V, you can kill two birds with one scaling circuit, which, may even be as simple as a voltage divider, depending on application...
If you do the scaling before the MUX, you have to scale all 16 lines individually. If you scale them after the MUX, then you can do it with just one scaling circuit.

You could split the difference, assuming that a simple resistive voltage divider is adequate, and put the lower resistor on the output (just one resistor) and put the top resistor on the input (one resistor per input). This would allow you to put trim pots there and calibrate the channels individually, which may or may not be useful in this app.
 
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