biggest rookie question in the league. / Tesla Coils / Forums

Forums

4hv.org :: Forums :: Tesla Coils

« Previous topic | Next topic »

biggest rookie question in the league.

previous 1 2

Move Thread

LAN_403

Ash Small

Thu Jun 09 2011, 12:12AM

Registered Member #3414 Joined: Sun Nov 14 2010, 05:05PM
Location: UK
Posts: 4245

Andyman wrote ...

Not sure if this is right, but lets see if i can give it a shot.
Since the diodes have some reverse standoff voltage, that means there's a resistance R to the reverse current flow. V=I*R but in our case, the standoff voltage V is not really dependent on amount of current, so we could pretend I=1 and say that V is roughly proportional to R.
Since resistors in series add up, the standoff voltage would add up in the same way. V1+V2+V3â‰ˆR1+R2+R3

Does that seem right? Or am i just pulling stuff outta thin air?

.

I don't think that is quite right. if V=I*R then, in an 'ideal' diode, reverse current is zero.(this would give V=0) (or zero X infinity, whatever that is, my calculator says 'Math ERROR')

but, as nothing is ideal, there is 'some' current flow (a very small amount). R is very large, but not infinite.

The resistance of each diode is ~equal, so the voltage is spread over the diodes. The same voltage drop over each diode.

EDIT: I think I=V/R is more appropriate here.)

Forty

Thu Jun 09 2011, 03:15AM

Registered Member #3888 Joined: Sun May 15 2011, 09:50PM
Location: Erie, PA
Posts: 649

it's not really a resistance thing either. it's a hill of potential energy that unless the voltage exceeds the height of the hill, the material doesn't let the charge flow. If you stack up a bunch of small hills, then you get a larger one. the hill analogy works the other way as well, in the forward biased direction, as charge can easily 'roll down the hill' to a lower potential at the bottom.

Ash Small

Thu Jun 09 2011, 11:05AM

Registered Member #3414 Joined: Sun Nov 14 2010, 05:05PM
Location: UK
Posts: 4245

Forty wrote ...

it's not really a resistance thing either. it's a hill of potential energy that unless the voltage exceeds the height of the hill, the material doesn't let the charge flow. If you stack up a bunch of small hills, then you get a larger one. the hill analogy works the other way as well, in the forward biased direction, as charge can easily 'roll down the hill' to a lower potential at the bottom.

I agree that there is a breakdown voltage, but data sheets generally have a figure for the reverse leakage current, which indicates that there must be a resistance.

Dr. Slack

Thu Jun 09 2011, 04:46PM

Registered Member #72 Joined: Thu Feb 09 2006, 08:29AM
Location: UK St. Albans
Posts: 1659

Yes, the diode forward voltage drop is 0.7V, but that isn't what i am misunderstanding. I guess I'm trying to figure out how stacking 25 in series I will get effectively a maximum rating of 25kV reverse voltage before breakdown occurs (each diode rated 1000V reverse voltage). As I see it, the first diode will "see" 15kV, and as it has no drop, its rating is surpassed by 15x.

You are right and you are wrong, and you've really got to get into the nitty-gritty detail of all the parasitics to figure out what happens.

With boggo IN4007 type diodes, the general concensus is that you halve the theoretical reverse working voltage for a safe working voltage, so 25 diodes in series = 12kV.

Assume the string has been conducting a forward current. Each junction will have large charge stored. Now the voltage reverses.

Current continues to flow, depleteing the stored charge. Eventually one diode depletes fully and its voltage begins to rise, at a rate limited by its intrinsic capacitance and the current flowing - so it's going to be pretty fast. The other diodes still have some stored charge, and so still have a low voltage across them. This first diode will probably die, and they tend to go short circuit. The same thing will happen to the next few early depleters too. Now we are starting to get the bulk of the diodes depleteing at the same time, enough such that their intrinsic capacitance is enough to balance the voltage between them as it rises, and the bulk of them survive.

That's why you rate a series string like that at much less than a simple sum of all the reverse voltages.

That's the dynamic behaviour, controlled by storage times and capacitance, what about the static behaviour? As the reverse biassed string continues to sit there, differences in reverse leakage across each junction will stress some diodes more heavily than others. The more heavily stressed ones may die.

After a few cycles of switch-off and high voltage soak, the low charge storage and low leakage diodes will all have turned into bits of wire, and you will be left with a fairly well-matched string, but with a reverse voltage withstanding of rather less than 25kV.

Shunting each diode with perhaps 10Mohm attempts to balance the leakage currents. Shunting each diode with perhaps 100pF attempts to swamp the charge storage differences. Either will allow you to use fewer series diodes safely, but will cost money and electrical efficiency. Do you buy more diodes, or other components?

Using "avalanche rated" diodes mean that the early depleters don't die when their reverse breakdown is exceeded. They are robust enough that they can just sit there and eat it until their slow co-workers join in and share the voltage. But they're more expensive than 4007s.

Whether you use cheap or avalanche rated diodes, it's absolutely essential to use the same type, and a really really good idea to get them all from the same batch, to maximise the number that turn out to be well matched.

previous 1 2

Moderator(s): Chris Russell, Noelle, Alex, Tesladownunder, Dave Marshall, Dave Billington, Bjørn, Steve Conner, Wolfram, Kizmo, Mads Barnkob