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How to compensate for voltage drop across a diode?

  1. Oct 15, 2013 #1
    Hi, I am working on a project where I am using a LCD voltage display to measure voltage at 2 points in the circuit. The first is the voltage of the batteries providing the power, and the second is to measure the output of the circuit. It is in a very small box and I don't have room for a complex circuit. Basically I want to have 2 momentary switches- one will activate the circuit and the voltmeter and display the output of the circuit. The other switch would simply activate the voltmeter and display the voltage of the batteries(they are lithium and I don't want to over-discharge them). I thought about using a diode to isolate them but I need a way around the voltage drop that it will create. Thanks in advance for any help, and let me know if I need to clarify anything.
     
  2. jcsd
  3. Oct 15, 2013 #2

    Baluncore

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    There are circuits that replace diodes with MOSFETs because they have a much lower voltage drop.
    But you could do it all with a miniature toggle switch.
    Description could be DPDT, centre off = (on-off-on). Or momentary (mom-off-mom)
    One side is used to select the input. The other is wired to power the meter when not off.
     
  4. Oct 15, 2013 #3
    Thanks for the reply. I am now thinking I can use a momentary (on)-on switch. It would need to have 2 discrete inputs to be switched and 4 outputs. Would that be dpst?
     
  5. Oct 15, 2013 #4

    meBigGuy

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    Lookup analog switch or analog mux. There are 2 input devices like the 74LVC1G3157 that are smaller than your diodes.
     
  6. Oct 16, 2013 #5

    FOIWATER

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    I am not sure if this concept is useful in your application or not, look up "superdiode" it's just an opamp feeding a diode. The op amp feedback is taken from the output of the diode and fed back to the negative pin. In this way, the op amp will output whatever is necessary (input + diode volt drop) to make the pins equal to one another. So the output of the circuit exactly resembles the input, gain of 1, but the diode prevents negative half cycle.
     
  7. Oct 16, 2013 #6

    Baluncore

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    The switch you need is probably initially described as “DPDT, centre off”.

    “DP” is double pole. One pole is used to provide power to the meter when the switch is not in the centre position. The second pole is used to select the input to be measured.

    “DT” is double throw. There are two active positions.

    “centre off”. Gives a middle position that disconnects all terminals.

    “mom = momentary”. You can select if the switch will be remain in position or must be held in position while reading the meter.

    Search for a switch that is (on-off-on), (on-off-mom) or (mom-off-mom).

    You need to better describe the states required to get better suggestions..
     
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