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Op amp with zener in feedback loop

  1. Sep 21, 2014 #1
    Getting confused on understanding how opamp works with zener in Feedback loop.
    Referring to the ckt here (attached.)
    (replace diode with 5V zener)

    Using virtual short, when the input voltage is zero, opamp output is 5V?
    Basically the opamp sources current into the zener.
    When the input is -1V, the opamp output is still 5v?
    Opamp has to pump 5V to keep the zener conducting.
    The current flows from output of opamp thru the zener into the input resistor and to the -1v souce?
    The current flowing in the resistor will be (-1/R)?
     

    Attached Files:

  2. jcsd
  3. Sep 21, 2014 #2

    jim hardy

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    I think you already know the answer.

    Let's take that one last....


    Back to opamp basics.
    What is the most basic assumption?
    The designer has surrounded the opamp with a circuit that allows it to hold its inputs equal. Else it isn't "operating".

    The designer nailed your opamp's + input right down to 0 volts.
    Question is then, can it hold its - input node at zero?
    Sure, so long as it can deliver current to the node that's equal and opposite the current from Vin, Vin/R.
    If there's ANY current of either polarity into the -input node from Vin, Vout will be very near either +5 or -0.6 so as to push equal and opposite current through the zener. So the opamp can hold its -input at zero, equal to + input.

    Not sure what you mean by 'virtual short'...
    Now we have to depart from practical thinking to near-idealistic thinking.
    That opamp has open loop gain of around 100,000 to maybe a few million. Let's just pick an even million to demonstrate.
    So,,,, if voltage at its - input node were one microvolt and gain is a million, output would be 1 volt and Zener would not conduct.
    You'd need five microvolts at -input node to get 5 volts output and make Zener conduct..
    AHA ! There's the secret ! Five microvolts is so close to zero that we assume the -input node stays at zero , which is only a a mild exaggeration.

    To make it more precise, we could re-state your question as a statement:
    To be even a little more precise we'd say
    Avol being open loop gain of the opamp....
    But - then we should add in factors for the amplifier's offset voltage...

    The possible refinements are endless.
    For practical work It's best to just remember the opamp wants to hold its inputs equal, and be aware you can refine ad nauseum if you want or need to.
    Zero needn't go out to six decimal places. We just take things to extremes sometimes to keep our thinking straight.

    old jim
     
  4. Sep 22, 2014 #3

    Baluncore

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    A zener diode in the feedback loop converts the op-amp into an input sign comparator. The circuit shown is a cheap and dirty way to generate logic level voltages from op-amps running on higher voltage split supplies. For positive input currents, output is –Vpn = logic low. For negative input currents, output is +Vz = logic high.
     
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