how does this flasher stand up in the long haul!

Jamie

no it does not, I just through some values at it, I just wanted opinions about zenering the emitter. normally you would use at least 10volts or more..

I always thought avalanching the (BE) wasn't a very good idea.

Jamie

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yes, but this will unlatch and flash for you.

This circuit is very simply. I was just curious as to longevity of the transistor in such a configuration.

Jamie

It's a transistor negative-resistance effect that Spice apparently doesn't model.

In your sim, you have 23 volts base-emitter, so the Spice model is very wrong.

John

Yeah. I ran into this before when I tried to use the e-b junction as a zener. AFAIK, LTSpice does not model Veb at all.

The default LT SPice models for standard parts don't seem to be very good, especially as regards breakdown voltages. A 1N914 has a reverse leakage of 102 nA at 100,000 volts!

John

I learned about this effect years ago when I had to repair something that was oscillating because the base resistor became open on a npn circuit. I thought it was a fluke, defective transistor etc. It was later that I found out it was a known effect. Some where around 1980 or so..

And no, LtSpice wont show it. Why? I don't know.

Jamie

Can anybody explain the voltages and currents here?

Version 4 SHEET 1 880 680 WIRE -224 -32 -288 -32 WIRE -224 16 -224 -32 WIRE -96 64 -160 64 WIRE 64 64 -16 64 WIRE 112 64 64 64 WIRE 224 64 192 64 WIRE 64 112 64 64 WIRE 224 128 224 64 WIRE -224 160 -224 112 WIRE 64 240 64 192 WIRE 224 240 224 192 FLAG 64 240 0 FLAG 224 240 0 FLAG -224 160 0 SYMBOL diode 208 128 R0 SYMATTR InstName D1 SYMATTR Value 1N914 SYMBOL voltage 64 96 R0 WINDOW 123 0 0 Left 2 WINDOW 39 0 0 Left 2 SYMATTR InstName V1 SYMATTR Value -100000 SYMBOL res 208 48 R90 WINDOW 0 0 56 VBottom 2 WINDOW 3 32 56 VTop 2 SYMATTR InstName R1 SYMATTR Value 1K SYMBOL npn -160 16 M0 SYMATTR InstName Q1 SYMATTR Value 2N2222 SYMBOL res 0 48 R90 WINDOW 0 0 56 VBottom 2 WINDOW 3 32 56 VTop 2 SYMATTR InstName R2 SYMATTR Value 1K SYMBOL res -272 -48 R90 WINDOW 0 0 56 VBottom 2 WINDOW 3 32 56 VTop 2 SYMATTR InstName R3 SYMATTR Value 1K TEXT 280 80 Left 2 !.tran 0 1 0

LTSpice doesn't know how to handle the reverse voltage on the base nor the reverse voltage on the anode of the diode. That is, it doesn't handle reverse breakdown as you said in your last post in this thread.

Like our QC slogan goes at work "We Test to pass not test to fail!"

Nice how there isn't any Vbe R-break down either!

Jamie

Will Larkin even try?

No :-) ...Jim Thompson

=A0 =A0 2N2222

=A0|-

=A0 120

=3D=3D=3D

inions

s or

..

I believe you can just edit the standard.dio file to add a Vrev =3D xx theres several other parameters you can add to the different transistor/diode models

-Lasse

Any Spice is dangerous, as your recent post, with 22 volts Vbe, demonstrates. Simulation is no substitute for thinking.

Obviously it does zeners, so all the 1N914 model needs is some reverse breakdown params. It's just interesting that the model supplied doesn't have them.

The 100 kilovolts volts across a b-e junction is even more interesting. I assume that Vbr specs slow down a simulation, so weren't included.

John

But look at the currents versus time!

John

--- One would think that by making a statement like that you'd be qualified as a judge of what constitutes thinking and what doesn't.

As such, one would expect you to understand the mechanism underlying the open base's following the emitter in the circuit shown.

Unfortunately, it appears that you don't.

Since, with an open base, an NPN transistor looks like two diodes in series opposition with their anodes connected:

It's clear that we can substitute resistors for the diodes, viz.:

As such, with R1 representing the reverse biased diode and R2 representing the one that's forward biased, it's easy to see that the resistance of R1 will be very high, while the resistance of R2 will be very low, allowing the leakage current through R1 and the base region easy access to ground.

Just for grins, let's say that with 25V across the string we have 10nA of current through the string.

That means that:

E 25V R1+R2 = --- = ------ = 2.5G ohms I 10nA

Then, with the base sitting at about 3V above ground and the emitter at +25V, the values of the resistances would be:

and Vbe would be, as you found, 22V.

Here's a sim with, probably, more realistic numbers to further illustrate the concept:

Version 4 SHEET 1 880 680 WIRE 144 16 -16 16 WIRE 368 16 144 16 WIRE 512 16 368 16 WIRE 512 64 512 16 WIRE 368 80 368 16 WIRE 144 144 144 16 WIRE -16 160 -16 16 WIRE 256 192 208 192 WIRE 368 192 368 144 WIRE 416 192 368 192 WIRE 512 192 512 144 WIRE 544 192 512 192 WIRE 512 240 512 192 WIRE 368 256 368 192 WIRE -16 368 -16 240 WIRE 144 368 144 240 WIRE 144 368 -16 368 WIRE 368 368 368 320 WIRE 368 368 144 368 WIRE 512 368 512 320 WIRE 512 368 368 368 WIRE -16 448 -16 368 FLAG -16 448 0 SYMBOL npn 208 144 M0 SYMATTR InstName Q1 SYMATTR Value 2N2222 SYMBOL voltage -16 144 R0 WINDOW 123 0 0 Left 2 WINDOW 39 0 0 Left 2 WINDOW 3 24 96 Invisible 2 SYMATTR Value PULSE(0 25 0 1) SYMATTR InstName V1 SYMBOL diode 352 256 R0 WINDOW 0 -43 32 Left 2 WINDOW 3 -69 64 Left 2 SYMATTR InstName D1 SYMATTR Value 1N4148 SYMBOL diode 384 144 R180 WINDOW 0 49 31 Left 2 WINDOW 3 24 0 Left 2 SYMATTR InstName D2 SYMATTR Value 1N4148 SYMBOL res 496 48 R0 SYMATTR InstName R1 SYMATTR Value 6.8G SYMBOL res 496 224 R0 SYMATTR InstName R3 SYMATTR Value 90meg TEXT -8 392 Left 2 !;tran 1 startup uic TEXT -8 424 Left 2 !.dc V1 .01 25 .001

---

--- Rather than making ASSumptions, why not post a query to Helmut's group or email Mike for a definitive answer?

-- JF