# why nothing simulates?

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

1. ### PierreJGuest

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. ### PierreJGuest

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

TIA
P.

3. ### PierreJGuest

I tried this below question on sci.electronics.basics but got not

4. ### Helmut SennewaldGuest

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
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)

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 jardineGuest

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. ### GenomeGuest

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
No worries.

DNA

7. ### John PopelishGuest

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. ### tbellGuest

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. ### John PopelishGuest

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. ### PierreJGuest

Thankyou all for all the help.

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

P.

11. ### jasenGuest

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