why nothing simulates?

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

TIA P.

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.

The user group is here.

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

"PierreJ" schrieb im Newsbeitrag news: snipped-for-privacy@11g2000cwr.googlegroups.com...

[...]

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

--
Posted via a free Usenet account from http://www.teranews.com

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 cunt with problems about the size of your penis.

No worries.

DNA

The best place to get answers related to LTspice is the Yahoo discussion group dedicated to this simulator:

"PierreJ" wrote in news: snipped-for-privacy@a75g2000cwd.googlegroups.com:

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.

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)

Thankyou all for all the help.

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

P.

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

--

Bye.
   Jasen