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Question re: Calculating V drop across spark gaps, components

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LAN_403

likewhat

Sat Oct 20 2007, 05:49PM

Account deactivated by user request on 6/11/2009.
Registered Member #1071 Joined: Fri Oct 19 2007, 02:13AM
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Posts: 44

Well, if you draw a regular voltage vs current picture for a diode once you get to some point in voltage its resistance essentially goes to zero because the current blows up if you add any more voltage, so I would have to say no you wont ever be putting 25 kV across the diode. Now will the diode have some issue with arcing or something, I have no idea about that.

Marko

Sat Oct 20 2007, 05:57PM

Registered Member #89 Joined: Thu Feb 09 2006, 02:40PM
Location: Zadar, Croatia
Posts: 3145

The conducting spark gap may actually be dropping no more than few hundred volts. 10kV is needed only to break it down.

The rest will drop on the impedance of the supply. Resistive part is turned into heat in the supply; inductive part is stored as magnetic energy, a damped LC oscillation in ignition coil case, and energy is delivered to load in several cycles.

I don't see what you intend with a diode in series with a spark gap, it does nothing.
IC output is asymmetrical with high voltage only in one direction.

In ideal system it would never see voltage higher than supply voltage multiplied by turns ratio, but in reality you have stray capacitance on cathode which charges to supply voltage and needs to be held off. Also take ringing of the ignition coil output in account.

Diode really makes no purpose here.

thrival

Sat Oct 20 2007, 06:16PM

Registered Member #1019 Joined: Sat Sept 22 2007, 02:39AM
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Posts: 29

Marko:

I already said the diode was redundant in the circuit, but for computing
what the voltage drop would be across it and if it would survive. In some
circuits the diode IS necessary, but to show everything would unnecessarily
complexify the issue and elicit comments that have no bearing to the
question.

Steve Conner

Sat Oct 20 2007, 09:31PM

Registered Member #30 Joined: Fri Feb 03 2006, 10:52AM
Location: Glasgow, Scotland
Posts: 6706

What Marko says is broadly right. If you look at the graph I posted, you will see that the ignition coil output got dragged down to 1kV or so once the spark gap broke down. The diode won't get 25kV forced across it, because the output of the ignition coil itself has a high impedance and will just sag to whatever voltage the spark gap wants, plus a volt or two for the diode.

I won't dare to ask what you're planning to use the diode for.

Steve Ward

Sat Oct 20 2007, 10:17PM

Registered Member #146 Joined: Sun Feb 12 2006, 04:21AM
Location: Austin Tx
Posts: 1055

In days of early
radio, was well-known that spark gaps converted HV, low current, to low V, high current.

Impedance matching via spark gap? That above statement is either poorly stated, or you have some mis-understanding of what is actually going on with the circuit.

In my experiences with steady state arcs (from DC sources) was that the voltage could reach only a few kV from my 150kV (open circuit) CW multiplier. This was for about 6-10" of plasma, which was stable enough to use my DVM with HV divider to measure the voltage across the arc. I seem to recall it peaking near 4-5kV at about 100mA. For shorter sparks this voltage was typically just over 1kV, and i think you will find that figure to be pretty consistent with typical low power discharges under 1" in length.

But this leads to the question of: what do you intend to do? Hopefully you dont believe that there is some sort of impedance matching ability of spark gaps, as if they somehow "boost" the current output of your source, because they surely dont, unless you are after higher peak/pulsed power, in which case you need a capacitor to provide that.

thrival

Sun Oct 21 2007, 12:18AM

Registered Member #1019 Joined: Sat Sept 22 2007, 02:39AM
Location:
Posts: 29

Steve:

If I make you feel stupid by asking a question you don't know the answer to,
I'm sorry. Measure the V drop across a resistor downstream of an arc and
you'll see the V is less than the source, the arc dropped most of it, common
knowledge even today.

Again, you like some others, lead far afield from a simple question. If you don't
know the answer, I can understand, not all problems are easy, even though they
may look like it, but just say so without trying to read minds. Who said anything
about impedance matching? With a resistance? That would be stoopid. Also Jochen's
page discusses measuring approx. voltages by arc length, NOT what I'm asking.

The other Steve provided a link showing that a DC driven ignition coil's ringing core can
actually produce an AC wave at least -1.5kv, which I didn't expect, but that coil only
produced 9-10kV total, so still doesn't give me functional, mathematical tools to work
with; nothing that would allow computing V drops across other circuit elements in series
with an arc discharge(!)

I'd like to forward bias a diode in series with a DC arc discharge, without blowing
the diode, and I don't know how to determine what V the diode needs to be rated
for, end of story. Presuming there were no back EMF's or reverse bias, if I could
get away with a low-V rated diode, that would be great. I enjoy adding "useless"
components to my circuits, and there were "gaps" in my formal education, OK?
It's a real question even if you don't want to take it seriously.

Sorry if basic questions don't capture some folks' imagination. Four loaner library
books on HV Engineering don't explain. I'm not asking anyone to solve my problem,
just point me where to look. Where IS a good HV EE when you need him? It's not like
I'm not also looking elsewhere for answers; started the thread to learn something,
not consume useless bandwidth. grrr.

Marko

Sun Oct 21 2007, 12:44AM

Registered Member #89 Joined: Thu Feb 09 2006, 02:40PM
Location: Zadar, Croatia
Posts: 3145

Thrival, what are you trying to point out with your last post?

Anyone who would want to rectify an ignition coil would use diode rated for more than it's output voltage,

In your circuit it will be less than coil's output voltage but I can't tell how much because it depends a lot on parameters of your coil and gap.

There is no purpose in the diode but if I was going to build a rube goldberg machine diode would be rated for < coil's output voltage so it lasts.

What Marko says is broadly right.

Ok, gawky post.. now what was broad (wrong)? I'd be grateful if you told me...

ragnar

Sun Oct 21 2007, 01:51AM

Registered Member #63 Joined: Thu Feb 09 2006, 06:18AM
Location:
Posts: 1425

thrival wrote ...

Steve:

If I make you feel stupid by asking a question you don't know the answer to,
I'm sorry.

Stop trolling.

wrote ...
Measure the V drop across a resistor downstream of an arc and
you'll see the V is less than the source, the arc dropped most of it, common
knowledge even today.

Have you then answered your own question? Look up Kirchoff's voltage law -- you may find some webpages which discuss this in a spark gap context.

wrote ...
Again, you like some others, lead far afield from a simple question. If you don't
know the answer, I can understand, not all problems are easy, even though they
may look like it, but just say so without trying to read minds.

Stop trolling.

wrote ...
Who said anything about impedance matching?

You.

Thrival wrote ...
...converted HV, low current, to low V, high current...

That is more formally known as impedance matching.

wrote ...
I'd like to forward bias a diode in series with a DC arc discharge, without blowing
the diode, and I don't know how to determine what V the diode needs to be rated
for, end of story.

Then why didn't you ask that in your first post? The arc will have a varying resistance, consider it with respect to time:

1) Pulse goes into ignition coil, open-circuit voltage across the secondary rises to 30+kV. So long as the spark gap hasn't broken down yet, 30kV potential / "voltage drop" is present across it!
2) Corona or arc formation occurs, and as the air across the gap ionizes, current starts to flow (limited by amongst other things the secondary inductance). Now you must realise that the secondary is not open-circuit, and the voltage across it will not be 30+kV.
3) The voltage drop across the gap decreases, whilst the current flowing through the arc increases (lowering the resistance). A "higher" (than initial corona / first breakdown) current is flowing.
4) Insufficient current flows to sustain the arc, it quenches, and there may be no high voltage present in the coil (limited by amongst other things the secondary inductance) to cause breakdown again.

You may want a diode rated for ~2x the maximum reverse voltage you expect to appear, and say ~2x the peak current you expect to flow through the arc.

It's not as simple as you'd like it to be.

wrote ...
Presuming there were no back EMF's or reverse bias...

Unfortunately, there will be, and it's best to try to factor this into the design so you don't blow diodes later.

wrote ...
Sorry if basic questions don't capture some folks' imagination.

I personally find your responses and your attitudes to other people here "imaginative". You can "imagine" that you post a question on your terms, you moderating all discussion, but you will consider yourself lucky that the members here have the "imagination" to see your problem from a more complex, realistic perspective than you choose to.

We can't read your mind to say 'what you want to hear'.

wrote ...
Where IS a good HV EE when you need him?

A population of better-than-good HV EEs are here; if you stop trolling, more of them may dare to reply a second time in your threads.
(Sidenote, let us know if you ever meet a female HV EE ;)

thrival

Sun Oct 21 2007, 09:27AM

Registered Member #1019 Joined: Sat Sept 22 2007, 02:39AM
Location:
Posts: 29

Matt:

One person with reasonable answer vs. creating long threads, could suffice.

You defined 'impedance matching' on your terms as a catch-all, I simply stated
what a spark gap does. I never used the word or said they were the same thing,
you did. And doing it with R is the dumbest thing an EE might consider; more often
done with transformers/baluns, only a newbie... eh?

"You may want a diode rated for ~2x the maximum reverse voltage you expect to
appear, and say ~2x the peak current you expect to flow through the arc."

Now that bit of info is useful, basic. (Thank you.) So in other words, if there
is NO back EMF (simple DC pulses don't qualify here, but actual voltage reversals),
then I should be in the clear, at least as far as V, a low V rated diode shouldn't be
damaged, right?

How should I know how much current dumps across a spark gap? My original
question asks how to calculate just such things, including V drop across a gap vs.
other elements in a circuit. If you don't know the formula, who or where it might
have been codified, just say so. Arc gaps aren't simple R's as covered by Kirchoff,
and HV isn't first-year AC-DC principles, but I suppose I have to wait for one of
you all to say it before getting much intel. I don't know how to use the english
language any simpler, although I do realize that carbon arcs aren't so simple a
circuit element as they might appear.

I come here to learn, not waste time, solicit opinions or even as a social outlet; and
yet am not incapable of gratitude when given a reason. For example, I'm grateful for
the internet, but not when people misrepresent themselves on it.

Sulaiman

Sun Oct 21 2007, 09:30AM

Registered Member #162 Joined: Mon Feb 13 2006, 10:25AM
Location: United Kingdom
Posts: 3140

Thrival, I too find your use of language somewhat disturbing, however,

Assuming there is negligible inductance in the external circuit that will cause 'ringing'
(there's always some, but ignore it for now)
Since you are using a flyback, during the time the primary is switched ON the secondary voltage will be
Vpri x (Nsec/Npri), for 12 V supply with an ignition coil (c 1:100) that's -1200 V across the diode
IF this does not cause the spark gap to operate the current will be negligible so the diode will be safe
IF you accidentally close the spark gap then the diode needs at least 1200V rating.

When the primary is switched off and the 35 kV or whatever is generated
the diode can only be forward-biased so negligible voltage rating for the diode is required.
The forward current will be (ideally) Ipri x (Npri/Nsec) ... a few tens of miliamperes.

SO, to answer your question, the forward drop is very small. (say 1 volt)

SO, in the 'ideal' case a small-signal diode such as a 1N4148 is sufficient !

What you SHOULD have asked is "What is the worst-case REVERSE voltage across the diode"

BUT unless you ask nicely I doubt you'll get an answer !

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