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PSPICE Diode Problem

Discussion in 'Electronic Design' started by siliconmike, Aug 23, 2006.

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  1. siliconmike

    siliconmike Guest

    PSpice doesn't seem to take diode recovery times into consideration.

    It is showing that 1N4007 diode is turning on in 0.5ns from a reverse
    bias of 12V to a forward 4A current condition.

    This is surprising. Any hints?

    (I'm using student version of OrCAD PSpice.)


  2. Hello Mike,

    Recovery time means the reverse recovery time.
    Reverse recovery time is the time required to switch off.

    Stimulate the diode with a forward current of 1A.
    Then switch to a negative voltage. Now look how long
    the current flows in the reverse direction.
    It's important to know the test setup if you want
    check against any number from the datasheet.

    A possible test circuit:
    I_forward=1A, I_backward=1A
    Voltage source _/-\_/-\_.... +10V/-10V, 50us on, 100us period
    Series resistor with 9 Ohm
    Diode connected to resistor and GND(0V).
    Now you should get a few microseconds of turn-off delay.
    I checked this with a 1N4002 model.

    Best regards,

    PS: I use LTspice. It's an unlimited and free of charge
    SPICE simulator with GUI(schematic entry and waveform viewer).

    This is the test circuit: "1N4002_test.asc"
    Just copy and paste it into a file named 1N4002_test.asc .
    Remove additional line feeds caused by the mail programs.
    Now you can open and run it with LTspice.

    Version 4
    SHEET 1 1352 680
    WIRE -16 64 -128 64
    WIRE 144 64 64 64
    WIRE 512 64 144 64
    WIRE -128 112 -128 64
    WIRE 512 112 512 64
    WIRE -128 240 -128 192
    WIRE 512 240 512 176
    WIRE 512 240 -128 240
    WIRE -128 272 -128 240
    FLAG -128 272 0
    FLAG 144 64 IND1
    SYMBOL diode 496 112 R0
    SYMATTR InstName D1
    SYMATTR Value DI_1N4002
    SYMBOL res -32 80 R270
    WINDOW 0 32 56 VTop 0
    WINDOW 3 0 56 VBottom 0
    SYMATTR InstName R1
    SYMATTR Value 9
    SYMBOL voltage -128 96 R0
    SYMATTR InstName V1
    SYMATTR Value PULSE(-10 11 1u 10n 10n 50u 100u)
    TEXT -144 384 Left 0 !.MODEL DI_1N4002 D ( IS=76.9p RS=42.0m BV=100 IBV=5.00u CJO=39.8p M=0.333 N=1.45 TT=4.32u )
    TEXT -136 -24 Left 0 !.tran 0 300u 0 .01u
    TEXT -136 -64 Left 0 ;Plot I(D1)
    TEXT -144 344 Left 0 ;*SRC=1N4002;DI_1N4002;Diodes;Si; 100V 1.00A 3.00us Diodes, Inc. diode
  3. colin

    colin Guest

    I think your spice diode model only uses turn off delay,
    also it doesnt use the fall time wich is sometimes
    more of an issue when using soft recovery diodes or SRD.

    Colin =^.^=
  4. Find a better set of model parameters. Some have TT (transit
    time)unspecified (0) and others use 5.76 or 5.7 usec and Colin just
    posted one with TT=4.32usec.

    Best regards,
    Spehro Pefhany
  5. Jim Thompson

    Jim Thompson Guest

    The latest PSpice diode.lib has TT=5.7us, but some time back there was
    NO TT specified, this defaults to zero.

    ...Jim Thompson
  6. siliconmike

    siliconmike Guest

    But there is forward recovery time too. I just ran LTSpice. Looks
    decent. I'll take some days to check out your circuit. Thanks for the
  7. siliconmike

    siliconmike Guest

    Use proportinal font. These are the diode params. As seen TT is 5.7uS.

    D1N4748 D1N4007 D1N4001
    IS 17.490000E-15 14.110000E-09 14.110000E-09
    N 1.984 1.984
    ISR 2.024000E-09
    IKF 94.81 94.81
    BV 22 1.500000E+03 75
    IBV .16996 10.000000E-06 10.000000E-06
    NBV 1.315
    IBVL 7.007300E-15
    NBVL 1.2735
    RS 7.976 .03389 .03389
    TT 5.700000E-06 5.700000E-06
    CJO 49.000000E-12 25.890000E-12 25.890000E-12
    VJ .75 .3245 .3245
    M .2829 .44 .44
    TBV1 840.910000E-06
  8. siliconmike

    siliconmike Guest

    I've posted the diode params in another post. Btw, I'm going to have to
    take some time to study your drawing regarding the high-side mosfet
    switch due to my inexperience in certain transistor configurations. I
    appreciate it though.

  9. Jim Thompson

    Jim Thompson Guest

    Feel free to ask questions.

    ...Jim Thompson
  10. colin

    colin Guest

    Yes this stands for the Transit Time I beleive, however it seems there are
    different conventions for TT, it is also used as Transition Time for SRDs
    wich is different again, this confused me no end when I was trying to
    simulate SRDs.

    Generaly it is the time taken for all the carriers that allow current flow
    in the PN junction to recombine wich is a random process. This cuases the
    'turn off' delay.

    However when the junction is initialy forward biased carriers are injected
    directly into the junction, the time this takes is very much smaller and so
    cunduction starts much more quickly, thus Turn on delay is usualy so short
    as to not be an issue and does not apear in the above model.

    I beleive the effect of junction capacitance usualy makes turn on delay
    unoticable anyway.

    Colin =^.^=
  11. siliconmike

    siliconmike Guest


    I understand, it does show a pronounced reverse recovery time of about
    4uS. What I'm wondering is that why is there no forward recovery period

    When the diode turns on, the voltage across diode seems to go abruptly
    from -10V to 1V and the current from 0A to 11A.

    In my case, the diode 1n4007 is used as a snubber parallel to a 40mH
    12Ohms Inductor to which 12 volts are switched using a relay.

    Therefore when the supply to the inductor is cut (assuming in a period
    of 1nS) , the diode turns on absorbing the current. But I'd like to
    focus my study on what happens "during the time" the diode turns on.

  12. colin

    colin Guest

    The turn on and turn off actions of a PN junction diode are very different

    Its like turning on a tap ... as soon as its on water flows imediatly,
    but as you turn it off it takes a while for the water to soak away.

    If you imagine the water is actualy the charge carriers that permit
    conduction (in either direction) its a bit closer to whats realy going on.

    The reason this turn off delay acts like this is that PN junction use
    minority charge carriers for conduction accros the barrier, and these stay
    in the junction untill an electron finds an empty hole or a hole finds a
    floating electron,
    they dont just simply disapear out of the terminals when you reverse the

    A schotky diode however doesnt use minority carriers, and is therefroe a lot
    faster at turning off.
    as soon as you reverse the voltage they simply move out the elctrodes.

    This is my simplified understanding of a very mathmatically complex process
    so somone correct me if im wrong please.

    Colin =^.^=
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