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why nothing simulates?

Discussion in 'Electronic Basics' started by PierreJ, Jan 22, 2007.

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  1. PierreJ

    PierreJ Guest

    Hello friends I am Pierre living now in UK.

    I tried to use many times ltspice and get joy never.

    Even going to very basic circuit nothing ever do that it real would in
    world.

    Version 4
    SHEET 1 880 680
    WIRE -80 192 -128 192
    WIRE 160 192 -16 192
    WIRE 160 240 160 192
    WIRE -128 256 -128 192
    WIRE -128 384 -128 336
    WIRE 160 384 160 320
    FLAG -128 384 0
    FLAG 160 384 0
    SYMBOL ind2 144 224 R0
    SYMATTR InstName L1
    SYMATTR Value 0.1
    SYMATTR SpiceLine Ipk=1000 Rser=0.00000000001 Rpar=100000000000000
    Cpar=0.0000000000000000000001
    SYMBOL voltage -128 240 R0
    WINDOW 123 0 0 Left 0
    WINDOW 39 0 0 Left 0
    SYMATTR InstName V1
    SYMATTR Value PULSE(0 10 0 0.00000001 0.00000001 0.001 0.0025 4)
    SYMBOL schottky -80 208 R270
    WINDOW 0 32 32 VTop 0
    WINDOW 3 0 32 VBottom 0
    SYMATTR InstName D1
    SYMATTR Value 1N5818
    SYMATTR Description Diode
    SYMATTR Type diode
    TEXT -160 504 Left 0 !.tran 0 10ms 0

    Why v(n002) not go very minus at 1ms?

    I asked of engineer friend why but he refuse to look but say
    "simulators suck" and refuse to discuss further more.

    TIA
    P.
     
  2. PierreJ

    PierreJ Guest

    I guess answer is not know by people here. Should I try
    electronics.design group?

    TIA
    P.
     
  3. PierreJ

    PierreJ Guest

    I tried this below question on sci.electronics.basics but got not
    answer. Do anyone here help?
     
  4. Hello Pierre,

    The simulation is correct. The voltage is always one voltage drop lower
    than the input voltage. If you want negative pulses, you have to use a
    switch (transistor) instead of the diode. Please try the test circuit
    below..

    You should use the standard unit multipliers.
    Instead of 0.00000001 you should write 10n or 10e-9 for better readability.
    Example: PULSE(10 0 0 10n 10n 1m 2.5m)

    We are talking about LTspice.
    http://ltspice.linear.com/software/swcadiii.exe

    The user group is here.
    http://tech.groups.yahoo.com/group/LTspice/

    Best regards,
    Helmut


    Save the text in a file named "test.asc"

    Version 4
    SHEET 1 880 868
    WIRE -272 192 -320 192
    WIRE -80 192 -272 192
    WIRE 96 192 -16 192
    WIRE 160 192 96 192
    WIRE 160 240 160 192
    WIRE -320 256 -320 192
    WIRE 160 368 160 320
    WIRE -320 384 -320 336
    WIRE -288 496 -336 496
    WIRE -80 496 -288 496
    WIRE 144 496 16 496
    WIRE 208 496 144 496
    WIRE 272 496 208 496
    WIRE 480 496 336 496
    WIRE 560 496 480 496
    WIRE 208 544 208 496
    WIRE 480 544 480 496
    WIRE 560 544 560 496
    WIRE -144 560 -192 560
    WIRE -64 560 -64 544
    WIRE -64 560 -144 560
    WIRE -336 576 -336 496
    WIRE -192 592 -192 560
    WIRE 208 672 208 624
    WIRE 480 672 480 624
    WIRE 560 672 560 608
    WIRE -336 704 -336 656
    WIRE -192 704 -192 672
    FLAG -320 384 0
    FLAG 160 368 0
    FLAG 96 192 out
    FLAG -272 192 in
    FLAG -192 704 0
    FLAG 208 672 0
    FLAG 144 496 out1
    FLAG -144 560 ctrl1
    FLAG -336 704 0
    FLAG -288 496 in1
    FLAG 560 672 0
    FLAG 480 672 0
    SYMBOL ind2 144 224 R0
    SYMATTR InstName L1
    SYMATTR Value 0.1
    SYMATTR SpiceLine Rser=1 Cpar=100p
    SYMBOL voltage -320 240 R0
    WINDOW 123 0 0 Left 0
    WINDOW 39 0 0 Left 0
    SYMATTR InstName V1
    SYMATTR Value PULSE(0 10 0 10n 10n 1m 2.5m 4)
    SYMBOL schottky -80 208 R270
    WINDOW 0 32 32 VTop 0
    WINDOW 3 0 32 VBottom 0
    SYMATTR InstName D1
    SYMATTR Value MBRS360
    SYMATTR Description Diode
    SYMATTR Type diode
    SYMBOL ind2 192 528 R0
    SYMATTR InstName L2
    SYMATTR Value 0.1
    SYMATTR SpiceLine Rser=1 Cpar=100p
    SYMBOL voltage -192 576 R0
    WINDOW 123 0 0 Left 0
    WINDOW 39 24 132 Left 0
    SYMATTR SpiceLine Rser=10
    SYMATTR InstName V2
    SYMATTR Value PULSE(10 0 0 10n 10n 1m 2.5m)
    SYMBOL pmos 16 544 M270
    WINDOW 0 96 80 VLeft 0
    WINDOW 3 73 81 VLeft 0
    SYMATTR InstName M1
    SYMATTR Value Si7465DP
    SYMBOL voltage -336 560 R0
    WINDOW 123 0 0 Left 0
    WINDOW 39 0 0 Left 0
    SYMATTR InstName V3
    SYMATTR Value 10
    SYMBOL cap 544 544 R0
    SYMATTR InstName C1
    SYMATTR Value 100n
    SYMBOL res 464 528 R0
    SYMATTR InstName R1
    SYMATTR Value 1k
    SYMBOL schottky 336 480 R90
    WINDOW 0 0 32 VBottom 0
    WINDOW 3 32 32 VTop 0
    SYMATTR InstName D2
    SYMATTR Value MBRS360
    SYMATTR Description Diode
    SYMATTR Type diode
    TEXT -320 96 Left 0 !.tran 0 20ms 0
    TEXT 448 440 Left 0 ;external load
    RECTANGLE Normal 640 736 400 416
     
  5. john jardine

    john jardine Guest

    The sim is working perfectly and shows a true picture of the inductor
    current.
    Node 2 correctly goes as minus as it can go. I.e minus 0.29V.

    To get numbers, use the standard inductor formula ...
    Amps per second through inductor = Voltage across it / Inductance.

    At 1mS when the pulse cuts off, there's about 100ma flowing through the
    inductor and it has some potential energy stored in it's magnetic field.
    (stored joules=1/2 x L x i^2)

    The inductor's magnetic field starts to collapse and will try force a
    current flow through -any- circuit path.
    Any obstruction and the current will force a voltage across it. In this
    case there is only the diode still in circuit and this will drop 0.29V at
    the discharge rate.

    We've fixed the inductor voltage at about 0.29V (the diode) , have an
    inductor of 0.1H, so collapsing field current through the inductor flows at
    a rate of 0.3/0.1, or 3Amps per sec. Which is a drop of about 4.5ma in the
    1.5mS before the next 'charge' pulse come along. (the spice correctly shows
    this)

    Notice that the magnetic field and circulating current are seemingly taking
    ages to drop off. This is because there is little energy being burnt off by
    the diode and this is comparable with the energy stored in the inductor .
    (Noticeable in real life when a diode protected relay coil is switched off
    and the relay hesitates before opening).

    If you want node #2 to be quickly discharged then from the inductor formula
    it's simply a case of adding lots of resistance in the inductor discharge
    path. I.e a high resistance, equals high voltage, equals high power loss.
    The power can only come from that stored in the inductor's field and like a
    battery will quickly flatten. The real life case is a relay coil -not-
    protected by a diode. Or an inductive coil with say a 30V Zener across it,
    or a voltage step up power supply, etc.

    These simulator thingies are handy. Your friend is missing out.
    john
     
  6. Genome

    Genome Guest

    Welcome to the UK.
    At a philosophical level spice behaves the way things behave in the real
    world.
    They don't, you have to work out why.

    Snipped LTSpice File.
    Because you are French?

    I ran your circuit and it went negative by about 300mV, which is what I
    might expect from one of those Schottky diode things. Not only did it go
    negative, as the current increased in your inductor it went more negative,
    which is sort of what I would expect.

    I have use too many I's. LTspice was just telling you what it knew.
    Doesn't sound like a well rounded person to me. However life can do that to
    you, it helps if you are a self opinionated thick **** with problems about
    the size of your penis.
    No worries.

    DNA
     
  7. The best place to get answers related to LTspice is the
    Yahoo discussion group dedicated to this simulator:
    http://tech.groups.yahoo.com/group/LTspice/
     
  8. tbell

    tbell Guest

    Hello,

    There are a couple of answers to your question, including it is working,
    just differently than expected.

    First you might want to look for inductor time constant, L/R,
    information. It takes more time for the inductor, L1, to reach the point
    that the field is strong enough to generate the inductive spike you
    expect than your simulation actually runs. Look close at v(n002), the
    voltage starts dropping, but not much, this shows there is very little
    opposition to the current flow.

    Second have fun with the simulator, reduce the coil to 0.001h or 0.0001h
    and check the results. You can also change the time to run...

    Hope this helps.
    Tim

    To respond directly remove .snag from the email address.
     
  9. It goes as minus as it needs to, to keep the current going.
    And that is a diode drop lower than the voltage on the
    other side of the diode. What are you trying to make this
    circuit do? If you want the current through the diode, that
    is built up in the positive part of the waveform to produce
    a large negative voltage during the negative part of the
    waveform, you need something that breaks the current path
    back to the source at that time. Without getting into
    realistic devices, you could use a voltage controlled switch
    as an ideal case, as follows:

    Version 4
    SHEET 1 880 680
    WIRE -64 128 -128 128
    WIRE 16 128 -16 128
    WIRE -64 144 -64 128
    WIRE -16 144 -16 128
    WIRE 16 144 16 128
    WIRE -128 192 -128 128
    WIRE -80 192 -128 192
    WIRE 0 192 -16 192
    WIRE 80 192 0 192
    WIRE 160 192 80 192
    WIRE 80 240 80 192
    WIRE 160 240 160 192
    WIRE -128 256 -128 192
    WIRE -128 384 -128 336
    WIRE 160 384 160 320
    FLAG -128 384 0
    FLAG 160 384 0
    FLAG 16 144 0
    FLAG 80 320 0
    SYMBOL ind2 144 224 R0
    SYMATTR InstName L1
    SYMATTR Value 0.1
    SYMATTR SpiceLine Ipk=1000 Rser=0.01 Rpar=1meg
    SYMBOL voltage -128 240 R0
    WINDOW 123 0 0 Left 0
    WINDOW 39 0 0 Left 0
    SYMATTR InstName V1
    SYMATTR Value PULSE(0 10 0 0.00001 0.00001 0.001 0.0025 4)
    SYMBOL sw 16 192 R90
    SYMATTR InstName S1
    SYMATTR Value switch
    SYMBOL res 64 224 R0
    SYMATTR InstName R1
    SYMATTR Value 1k
    TEXT -160 504 Left 0 !.tran 0 10ms 0
    TEXT -136 104 Left 0 !.model switch SW( Ron=.01 Roff=1meg Vt=5)
     
  10. PierreJ

    PierreJ Guest

    Thankyou all for all the help.

    I see now my mistakes. An misunderstanding with how ltspice works not
    how world real works.

    P.
     
  11. jasen

    jasen Guest

    yes, when V1 is 0V it is ground, (not open-circuit)
    so L1 pulls current through D1
     
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