Silly question: PIV for diodes in series

[jmaf]

Alright, here's a silly doubt I just had.

If I don't add large shunt resistors or capacitors across series connected diodes, the voltage drop between every two will be just 0.7 volts(right?). That is, they're not dividing the potential like caps or resistors do.

So I have this 900V power supply and for safety I'm using two 1n5408's rated for 1000V on each rectifier leg. But I just realized that adding two in series just adds a bit of redundance without actually making my PIV 2000 V, correct? Or will they add up?

Silly question, I know, but this just ocurred to me.

[JoeM]

They add, 2 1kv diodes in series will handle 2kv. (Or use three etc)


Some really good info here:

The Valve Wizard

[exclamationmark]

PIV is the maximum voltage it can withstand (safely) in reverse before breakdown occurs (key word is reverse - not forward!). So the 0.7 volts forward drop really has nothing to do with PIV, and indeed series diodes will sum their PIV.

[TransLucid]

As is pointed out by the Valve Wizard, you should add caps in parallel with the series diodes to ensure equal voltage across each...

[jmaf]

Thanks everyone who responded.

I always took for granted that they did add up, all of a sudden it downed on me that maybe they didn't. Brainfart of the day

[R.G.]

For sudden applications of voltage, the voltage across two series reverse biased diodes is controlled by the diode and wiring capacitance acting like a capacitive voltage divider. It is common in applications where this needs to divide equally to swamp the internal and wiring capacitances with a bigger capacitor across each diode to force the transient voltage to divide nearly equally.

As the capacitors charge up, the longer term DC voltage across each diode is controlled not by its PIV but its DC leakage current charging/discharging the capacitances. If you have two 1000V PIV (actual breakover, not rated; diodes rated at 1000V break over above 1000V) diodes, one which leaks twice as much as the other, and put 1900V across them, they will *both* break over, because the leakage lets one have 1/3 of the applied voltage across it, and 2/3 across the other one. This breaks over the 2/3 one, it dies shorted, and then this breaks over the 1/3 voltage one.

It is common to use resistors in parallel with diodes to allow "leakage" currents to be more nearly equal and not let one take a higher % of the applied reverse voltage.

The closer you get to the applied voltage being equal to the sum of the actual PIVs, the more you need to use capacitive and resistive swamping parts to force equal division of the reverse voltage.

[trobbins]

jmaf - a 900V supply is pretty damn high! Is that a 900VDC output designed supply, or 900VAC secondary, or 450-0-450VAC secondary or ?

[jmaf]

Thanks RG, appreciate that.

What about the opposite phase when the diodes are forward conducting? During the initial transient the reservoir caps have near 0V, the first diode drops only 0.7 V so a diode ahead gets nearly the full +B voltage across it. Is PIV an issue at all when they're forward biased or is only the current rating relevant in that case?

[jmaf]

jmaf - a 900V supply is pretty damn high! Is that a 900VDC output designed supply, or 900VAC secondary, or 450-0-450VAC secondary or ?

Yeah, you bet that's high! It's just a theoretical question at this point, I don't have such a supply here but I was thinking about the 1000V rating for the hundreds of diodes I have and whether I could safely string them up and use them for a higher voltage supply.

Edit: My typical English-as-second-language miscommunication: when I said "so I have this 900V supply", etc, I meant "assume I had this 900V supply here and was wiring it up like so".

[trobbins]

It's just the peak current rating for forward bias - if one of the diodes is forward biased, then per se both are )if ignoring transient effects of added C & R) - as they are both conducting the same current. The peak current is a complex function of resistances and effective voltage - it's not easy to 'design' for a certain diode rating.