Diode with floating terminal

Discussion in 'General Electronics Chat' started by vinodquilon, Oct 22, 2013.

  1. vinodquilon

    Thread Starter Member

    Dec 24, 2009
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    Consider the circuit arrangement shown in the attachment. (Ideal diode)
    If I connect a MM in voltmeter mode at floating terminal A, What would be the reading on the voltmeter?
    If I terminate the floating A terminal with a 10K resistor to GND, What would be the change in reading shown?
     
  2. MrChips

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    Oct 2, 2009
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    Sounds like Homework question.

    What is an "ideal" diode?

    To answer the second question you need to know the I-V characteristics of the diode.
     
  3. vinodquilon

    Thread Starter Member

    Dec 24, 2009
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    Ideal diode means diode cut in volt is at 0V instead of 0.7V
     
  4. wayneh

    Expert

    Sep 9, 2010
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    No change. The ideal diode has an essentially zero impedance - it's like a wire that conducts in one direction. Your meter might have a 10MΩ impedance versus the 10kΩ for the resistor, so adding the 10k resistor in parallel has no impact on the meter reading.

    A real "ideal" diode needs a few millivolts of drop, I believe. For instance if your ideal diode is made using a MOSFET, it will have a few mΩ of Rdson resistance. Still insignificant compared to 10kΩ.
     
  5. wayneh

    Expert

    Sep 9, 2010
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    We don't know what the OP means, but this term comes up as a synonym for "active rectifier", for instance using a comparator to turn on a MOSFET when a small voltage is seen across the body diode, thereby eliminating the diode drop.
     
  6. WBahn

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    Mar 31, 2012
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    Though my guess is that the OP's question implies that they are not dealing with sophisticated active circuits, but rather idealized paper compoments (which probably includes an idealized ∞-resistance voltmeter, as well) in order to gain a feel for the zero-order behavior of these devices.
     
  7. #12

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    Nov 30, 2010
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    I'm going with the, "wire that only conducts in one direction" theory. The ideal diode does not have a V-I curve. It's only on or off. Superconductor quality.
     
  8. vinodquilon

    Thread Starter Member

    Dec 24, 2009
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    Here is my interpretation considering the case of non-ideal diode,

    CASE 1(without 10K)-- Depending on the output impedance of DC voltage source and the input impedance of the MM, the drop across the diode is not enough to turn it ON. And the MM will reads electric filed at floating terminal. And it will be high depending on the neutral to earth voltage of ac supply distribution to the voltage source (if above 1V).

    CASE 2(with 10K)--The effect of high input impedance of MM get reduced due to parallel 10K. In this case diode conducts and the output at non-floating terminal will be [5V-(forward diode drop)].
     
  9. wayneh

    Expert

    Sep 9, 2010
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    I can buy that, but Case 1 might have two sub-cases; the one you mentioned plus it could also be already "on" (conducting) if the active components haven't turned it off. It could, in theory, be programmed to not turn off until it sees -3mV, for example. Or you could say it hasn't seen a zero crossing since the voltage hasn't crossed zero yet.

    I guess it come down to how you want to define the "ideal diode". I'm thinking of real devices. If this is homework, your professor is unlikely to share my bias.
     
  10. WBahn

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    Mar 31, 2012
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    Oh, it has a very well defined V-I curve. It is identically zero current for Vd<=0V and identically 0V for Id>0A. You can use that curve in load-line analyses just fine.
     
    #12 likes this.
  11. MrChips

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    Oct 2, 2009
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    Now we have to declare, is the MM "ideal" or not?
     
  12. WBahn

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    Mar 31, 2012
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    I'm guessing it is. This seems like a very entry-level treatment.
     
  13. MrChips

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    Oct 2, 2009
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    Well then we have a dilemma.

    If there is zero current through the diode and there is 0V across the diode, what is the potential at the cathode with respect to ground?

    This is not a trivial question.

    If we consider the diode as a device with zero current and zero voltage then the potential at the cathode wrt ground is indeterminate, similar to a small capacitor instead of a diode.

    If we consider the diode to be related to a real diode, then the voltage at the cathode wrt ground is Vs - Vd.
     
    Last edited: Oct 23, 2013
  14. WBahn

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    If there is zero voltage across the diode, then the voltage on the right side of the diode is the same as the voltage on the left side of the diode.

    There IS a problem, which is what I think you are trying to get at, in that the voltage across the diode doesn't have to be 0V. If could, in theory, be reverse biased by some arbitrary amount which creates the situation you are talking about with the capacitor if you are using an ideal DMM. But that is probably at a level a bit beyond what this exercise is intended for since it requires the notion of charge storage and hence junction capacitance (or at least lead wire capacitance).

    How is that? That means that it is forward biased and conducting. But if we are still using the ideal DMM that is causing the indeterminate issue, we would still have the same problem. If we use a DMM with a realistic resistance with an ideal diode, we would get Vs at the right side. If we use a diode with a real V-I characteristic and a DMM with a realistic resistance then we get Vs-Vd, but the Vd would be much less than the 0.7V we usually bat around because the current would be so small.

    The simplest way to remove the indeterminacy would be to introduce the notion of a reverse leakage current. No matter how small, this would eventually remove any reverse voltage.

    But we can't just If we use a real diode, then
     
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