LM393 output when both inputs same voltage

Discussion in 'General Electronics Chat' started by eblc1388, May 3, 2011.

  1. eblc1388

    Thread Starter Senior Member

    Nov 28, 2008
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    The input of the LM393 is connected to a shunt(0.01Ω) to indicate the flow of current. Effectively, its inputs are tied together via the shunt. The shunt voltage drop is well within the common mode input voltage range of the LM393.

    Another LM393 is connected to the same shunt but with (+) and (-) input swapped to detect current flow in reverse direction.

    When there is no current flow, what should we expect the output of the two LM393s be?

    [​IMG]
     
  2. Papabravo

    Expert

    Feb 24, 2006
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    It will always be about ground because you have no pull-up resistors on the outputs. You should be feeling a bit sheepish about now -- eh?
     
  3. Ron H

    AAC Fanatic!

    Apr 14, 2005
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    The outputs will be undefined. Due to input offset voltages, either could be high, low, or "rattling around" somewhere in between, due to internal and/or input noise. This assumes you have pullup resistors, which I assume you just left out for simplicity.
    Check the datasheet for input offset voltage. You might need a preamp with low input offset voltage.
     
  4. eblc1388

    Thread Starter Senior Member

    Nov 28, 2008
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    Thanks Papabravo, Ron.

    @Papabravo,

    The LM393 output state can be hi-Z, or low, even without a pullup resistor fitted. What you have replied "about ground" implies the output state is LOW.

    @Ron,

    So the output state would be arbitrary, depending on your luck.

    I was going to do a simulation but stopped right in the middle when I realized the nature of the problem so I instead asked for advice in the forum.

    This is the type of question which I know I won't be getting the correct answer via simulation.
     
  5. Papabravo

    Expert

    Feb 24, 2006
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    The output will never be Hi-Z. That would require a push-pull output stage with both transistors turned off. The LM393 has only an open-collector output stage and without a pullup resistor it can only be about ground. It can't go below ground and it can be very far above ground without something "external" to make it go there.
     
  6. tom66

    Senior Member

    May 9, 2009
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    The LM393 is open collector. When the comparator is "low", i.e. - > + the transistor is in saturation and conducts current, causing any attached pull up to be pulled down, and go low. With + > -, the comparator is high, but the transistor is in cutoff and conducts virtually zero current. So, if there is a pull up, it is pulled high. The output can then either be considered to swing from Vee to open, and with a pull-up, from Vee to Vpullup (probably Vcc; ignoring the fact that the open collector does drop about 0.2-0.3V.) With the inputs the same, it is important to note that the output is not necessarily going to be one or the other. Offset voltage tolerances are about +/-3mV, which could swing the comparator either way, so it could be either Vee or open, or somewhere in between. A comparator isn't like an op amp and isn't designed to be operated in the unsaturated mode. It is an on or off device, and it is not guaranteed to take up one of the states with equal input. Now, if you were to tweak the voltage with a potentiometer, you would find that there would be a dead band of probably a few hundred microvolts at which there is not a valid output.
     
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