low-capacitance high-voltage diode

I did a 1400 volt flyback switcher and used MMBD5004S's in the C-W multiplier. That's a dual 400-volt diode in SOT23, 300 mA rating. They ran cool, unlike some other diodes that had a lot of reverse-recovery losses.

The Diodes Inc data sheet claims 0.6 pF typ per diode at 0 volts. That's 800 volts and 0.3 pF for the diodes in series!

Isn't "low-capacitance, high-voltage" sort of like a delicious, fat-free diet cheeseburger?

Use several in series to reduce the capacitance, the increased conduction losses won't matter much at high voltage.

I have 6000 in stock if you need a few.

Hey, those are awesome! About 0.4pF each at 100V, and 1/2 0.52pF at 50V = 0.25pF at 100V, even better yet! What's-more I had saved the datasheet after seeing you use them in your T840 (but without useful identifying info). I had thought you were using a '4004 variant. They'll be perfect, thanks!

I was gonna tease with "tube" something or other, but 6AL5 is a couple pF, even 1X2A (not that you could handle the voltage drop). 2-01C maybe, but they're really rare (Eimac microwave diode -- for RF voltmeter probes, it goes in the probe end).

In my defense, you can get a very accurate peak reading -- Vf(If = 0) is a little negative in fact, so there's no "0.6V" to worry about.

How many can you series up? That's kind of along JL's idea.

I think HV diodes (>kV -- stacks of dies) are in that territory, but they often have crappy recovery, and obviously the voltage drop is high. They're also kind of scarce.

Tim

-- Seven Transistor Labs, LLC Electrical Engineering Consultation and Contract Design Website:

I used to use 1B3 HV rectifier tubes as rectifiers and variable resistors, to charge an oil cap from a neon sign transformer. I drove the filament from a D-cell and a wirewound rheostat (with a long insulated shaft!), turned the knob until I got 7KV DC on the cap. So a

Amazing I'm still alive.

A 1B3 with no filament power makes a nice 30KV capacitive probe.

TEK used some tiny wire-lead pencil tubes as the HV rectifiers in their old tube scopes.

An MMBD5004S costs 4 cents, 50 u$ per volt. Official HV diode stacks do tend to be expensive.

BAW101V is a dual, presumably matched in leakage, rated for series connection. As I read it, it's under 1 pf for series connection.

For me HV means a big gap, depletion region, distance. which looks like HV ... C = e_0*Area/distance. I know zilch about HV diodes.

George H.

That's a good start. The longest depletion regions are in PIN diodes, made for low-capacitance OFF states, but usually not any appreciable current in the forward direction. Rectifiers with HV standoff are usually limited by their surface leakages, and deep depletion regions don't help against that.

DigiKey shows under 'PIN' diodes, a 600V unit MA4PH236-1072T. Maximum Cj (10V) is 0.5 pF, but it costs several dollars, and doesn't claim to turn OFF particularly fast (carrier lifetime issue).

It sounds reminiscent of the old physics puzzle of why giant insects can't exist in the real world: essentially the compression strength of the legs is growing as the square of the bug's larger volume, but the mass is growing as the cube.

If I'm remembering the equations right the depletion region capacitance is inversely proportional to the square of the donor density, which means all things being equal a smaller capacitance is associated with a larger donor density.

But the minority carrier lifetime is nearly inversely proportional in the donor density, and the conductance is directly proportional to the donor density, but as you increase the length of the semiconductor regions to avoid breakdown the donor density is decreasing as the larger volume, so to get the same value of capacitance as the "shorter" diode you have to increase the donor density. But increasing the donor density means you decrease the minority carrier lifetime, and hence the conductance, proportional to the volume. So you increase the junction area to compensate, but you're once again increasing the junction volume, and current density is only increasing proportional to the square of the junction area.

Something like that? If I'm looking at this graph it seems like all things being equal, the depletion capacitance of a regular reversed-biased PN diode overall increases with greater path length.

Not really. The conflict would be between current rating and capacitance, not voltage and capacitance.

Imagine having a 100 volt, 100 pF diode. 10 in series is 1000 volts,