question about multiplexer IC's

Discussion in 'General Electronics Chat' started by tuuluser, Aug 5, 2013.

  1. tuuluser

    Thread Starter New Member

    Nov 20, 2011
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    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.
     
  2. Abovethelaw

    New Member

    Aug 2, 2013
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    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.
     
  3. crutschow

    Expert

    Mar 14, 2008
    13,000
    3,229
    Maxim has a mux that apparently can handle a signal voltage larger than the supply voltage. But that type is not common.
     
  4. lightingman

    Senior Member

    Apr 19, 2007
    374
    22
    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.
     
  5. WBahn

    Moderator

    Mar 31, 2012
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    That seems like a strange part. It also looks like it requires Vpp=10V in addition to the Vdd logic supply.
     
  6. Ron H

    AAC Fanatic!

    Apr 14, 2005
    7,050
    657
    MAX14777 is a new part that might work.
    You have to use 4 of them, and add some logic, to make a 16:1 MUX.
     
  7. tshuck

    Well-Known Member

    Oct 18, 2012
    3,531
    675
    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...
     
  8. WBahn

    Moderator

    Mar 31, 2012
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    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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