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Switching a voltage divider with a FET

Discussion in 'Electronic Design' started by Michael, Jan 2, 2008.

  1. Michael

    Michael Guest

    Hi - I'd like to measure the voltage of a ~55V DC power source with
    the ADC of a 3.3V MCU. Whenever the MCU is on, I would like to be
    measuring the voltage of this power source. When the MCU is off, all
    of its pins go high impedance. I don't want to hang a normal resistive
    divider to ground across the power source, as that'd be a constant
    load (even if that load was in the uA range). Instead, I'm thinking
    about putting a resistive divider on it, with a FET switched ground.
    The gate of the FET would be driven by the microcontroller with a 10K
    pull down to ground. I'm thinking I'd use a 2M (1%), a 105K (1%), and
    a BSS138 (RDSon =~ 1.5). Nothing special about the BSS138 - I just
    have a bunch of 'em hanging around. Looks like it'll let half a uA
    through when turned off. More than I'd like - but I can probably
    survive that.

    Is this a bad idea? The ADC will be using a VREF of 3.0V. The MCU will
    shut off when the power source goes down to 36V or below. How do FETs
    handle low current situations like this? Will I see a large VDS? The
    BSS138 datasheet is more interested in higher current situations.
    Maybe there are FETs better suited for this?

    Thanks!

    -Michael
     
  2. John Larkin

    John Larkin Guest


    When the low-side fet is off, the full 55 volts will be applied,
    through the 2M resistor, to a pin of the ADC. That will turn on esd
    diodes internal to the adc and drag the voltage down, so you're no
    better off for adding the fet. "High impedance" doesn't mean you can
    pull a pin up to 55 volts.

    Consider a high-side switch, like a solid-state relay.

    John
     
  3. Michael

    Michael Guest

    Argh! Foiled by logic! Hmmmmm... I'll actually have 55V+10V available
    to me (though very, very low current), so it would be conceivable for
    me to drive a high side N-FET. I am incredibly space constrained
    though - so this is definitely pushing it for me. How about this: a
    BSS138 between the 55V supply and the 2M, then with the 105K on the
    bottom of the voltage divider. Drive the BSS138's gate with the
    emitter of a FOD617 optoisolator that has its collector tied to 55V +
    10V. Have a 1M resistor between the gate and the source of the BSS138.
    Total cost to me is one more part - the unfortunately fairly large
    FOD617. Current drain on the +10V supply should be fairly minimal, in
    the nano or pico amps I think?

    -Michael
     
  4. John Larkin

    John Larkin Guest

    You could just use the optoisolator itself as the only series switch.
    Lots of optos will handle 55 volts (some will handle 400) and the "on"
    voltage drop will be small.

    Or an ssr!

    John
     
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