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Impedance and LED on 120V.

Discussion in 'General Electronics Chat' started by AcousticBruce, Mar 25, 2014.

  1. Ratchit

    Ratchit Well-Known Member

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    What do you mean by PhD diodes, idealized diodes? Something that takes into consideration the bulk resistance? Rd is not mentioned in the link, but they do have a circuit with an idealized diode with a source voltage and a resistance they call "R". They show several methods of calculating that current. I used a source voltage of 340 volts and calculated the Vf at R=0 to be 1.53 volts and at R=100 Vf is 1.35 volts. Those figures appear reasonable, don't they? Vf doesn't change much either.

    I think you are referring to piecewise linear models. I don't think you have enough segments in your model to be viable at those high currents. I would use a graphical method instead. Don't forget that the transient pulse period is very, very small.

    http://en.wikipedia.org/wiki/Diode_modelling#Diode_with_voltage_source_and_current-limiting_resistor[/quote]

    Ratch
     
  2. ronv

    ronv Well-Known Member Most Helpful Member

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    It is called Rd. See attached.
    Also see a simulation of 2 diodes. One with a higher current rating than the other. You can see the difference that Rd makes in the slope of the curves.
     

    Attached Files:

  3. ronv

    ronv Well-Known Member Most Helpful Member

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    So Bruce, To get the impedance of your circuit use the square root of R squared Plus Z squared.
     
  4. dave

    Dave New Member

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  5. Ratchit

    Ratchit Well-Known Member

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    Two different diodes, two different curves. Schlockey's diode equation allows for two input parameters: one is "n", and the other is the reverse bias current. You can't expect to extrapolate the whole curve with just two straight lines, especially at the higher currents.

    Why don't you do a run with the values I used. R=0.1 ohms, step source voltage = 340 volts, and C = 1 uf in series and pre-energized to 170 volts with polarity to aid the source voltage. Graph the voltage across the diode and the current. That willl be a good test of Spice.

    Ratch
     
  6. Ratchit

    Ratchit Well-Known Member

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    No! The impedance is Z^2 = Xc^2 + R^2 . R is such a small value that it can be ignored.

    Ratch
     
  7. ronv

    ronv Well-Known Member Most Helpful Member

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    Yes, but not the real world. To hear you talk there is no more voltage drop across a signal diode than a 100 amp power diode at 100 amps. I've wasted all the time I need to waste with you. I'm pretty sure you know your wrong, just can't bring yourself to say it.
     
  8. ronv

    ronv Well-Known Member Most Helpful Member

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    Ratch, You just got to learn to read. Bruce doesn't care about your ideal diode.
     
  9. Ratchit

    Ratchit Well-Known Member

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    A power diode is constructed a little differently than a signal diode, but the physics are the same. I don't think I am wrong. I did not search long and hard for a 100 amp diode, but I did find a 20 amp diode which is fairly heavy duty. You can see the Vf is quite low.

    ronv.JPG

    Ratch
     
  10. ronv

    ronv Well-Known Member Most Helpful Member

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    That's what post 14 was, except with 170 volts. You won't like it any more now than you did then.

    I love your equation, just add Rd of a reasonable value for a 40 ma maximum LED.

    Ratch[/quote]
     
  11. Ratchit

    Ratchit Well-Known Member

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    The break voltage and current is way above what would be considered ideal, so the calculations will come out the same.

    Ratch
     
  12. Ratchit

    Ratchit Well-Known Member

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    Post #14 was from Northguy. There was no Spice run there. Are you referring to your Spice run where your Vf values were way up in the stratosphere?

    I posted lots of equations, can you be more specific? And show how that equation explains your assertion. [/quote]

    Ratch
     
  13. ronv

    ronv Well-Known Member Most Helpful Member

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    You didn't comment on post 82 about adding Rd (diode resistance) to make the model complete. (Don't you just hate variables?):)
    So using your math model add .53 ohms to the large diode in my simulation, in post 82, and 1.8 ohms to the small diode and see if the curves don't come very close. Without Rd the curves go almost straight up (as you have pointed out) after a small current flows. If you like that, add 17.5 ohms to the little LED and see if it doesn't match. Pretty cool.
     
  14. Ratchit

    Ratchit Well-Known Member

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    I did comment on it, see post #84. I don't know where you are coming up with those figures, but if you are happy, then so am I. The point I want to make is that you are not going to see large values of Vf on junction diodes, as the data sheet in post #88 illustrates. Another thing to keep in mind is that any resistance you include to suppress the transient will constantly leech power when the circuit enters the steady state mode. My analysis that a resistor is not needed should please the OP, because he said wanted to make the circuit efficient.

    Ratch
     
  15. ronv

    ronv Well-Known Member Most Helpful Member

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    If you agree that you need to add Rd to make the model complete then in some high current cases you will see high Vf. Don't forget the curves we are looking at for little LEDs seldom go above a 100 ma. The Vf is very different at 40 amps than at 100 ma.
    I agree the series resistor wastes power at the expense of protecting the LED. I don't think we know he doesn't need it though because there is no spec for the diodes at those currents. Only recommendations that they not be pulsed over 10 or 15 times max continuous current.
     
  16. Ratchit

    Ratchit Well-Known Member

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    In post #81, I calculated the Vf with a resistance zero ohms and 100 ohms. The two Vf's were 1.53 and 1.35.

    If that is true regardless of the shortness of the pulse, then some resistance must be included.

    Ratch
     
  17. ronv

    ronv Well-Known Member Most Helpful Member

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    The Rd is inside the diode not outside of it. Vf is measured across the diode. So higher Rd - higher Vf. - higher current higher Vf.
     
  18. Ratchit

    Ratchit Well-Known Member

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    Evidently there is not enough bulk resistance in a junction diode to make any difference. The power diode in post #88 can output 20 amps continuous with a Vf of just over 1.5 volts.

    Ratch
     
  19. ronv

    ronv Well-Known Member Most Helpful Member

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    Yes, I think so. They are built for low internal resistance.
     
  20. NorthGuy

    NorthGuy Well-Known Member

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    These all are just models.

    Simple diode model is an approximation. If you introduce Rd it gives you better approximation. If you would interoduce a second polynomial term, it would be better yet.

    These models are build over a short range of currents. It is silly to extrapolate them far away from their intended range.

    It us very hard, if at all possible, to measure a response of a small diode to 40-60A currents, and manufacturers don't post any data on that. It is impossible to predict what happens at these current either using model or with theoretical considerations.
     
  21. ronv

    ronv Well-Known Member Most Helpful Member

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    I agree.

    Unfortunately spice uses the simple Rd model which is pretty good over the standard operating range. Probably because it is the only value that can be determined from the data sheet and the sim time is short. The point of this whole exercise was to point out that at very large currents (compared to the spec) Vf is quite large, The time constant calculated is longer, but the energy the same, except now it is in the series resistor and not so much in the diode. Other than that we still don't know if the diode will survive the pulse (although a few have maximum specs at 10 usec.). The only input we have for that is failures seem to go away if the current is limited to some reasonable value.
     

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