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Usually it is recommended to put resistors in parallel to each diode like this:

enter image description here

It is supposed that the resistors will equalize leackage currents and make the voltage drop more even.

However, I doubt that. And that's why: diode I-V curve looks like this:

enter image description here

All diodes will share the same leakage current. And if the voltage is low it CAN be distributed unevenly. But it is not a problem until any diode will aproach to the breaking point. At this point the curent will go higher for all of the diodes. And all diodes will have a voltage drop near to its breaking point (V1, V2 and V3 on my ugly drawing).

So even different diodes (like 200V plus 1000V to have about 1200V of reverse voltage) should work well in reverse without any additional resistors. The only thing should be considered is the reverse leakage current of the most leaky diode multiplied to the highest breakdown voltage which in turn can make the last one to melt.

So is there anything I missed?

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  • \$\begingroup\$ Actually in the HV diode strings that I've seen (in projects in The ARRL Handbook, mostly) the diodes have parallel caps as well as resistors, to keep them balanced when they're switching. This may not be a good idea if they're being switched rapidly, though. \$\endgroup\$
    – TimWescott
    Aug 16 '19 at 20:05
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All diodes will share same leakage current.

That is true.

But the diode with the least leakage current (assume, the best diode) will control the overall current. Assume the leakage currents of the other two diodes (assume poor diodes) are higher. Correct me if I am wrong, this is equivalent to having a high value resistor (best diode) and a lower value resistor (worse diode).So, the resistor with higher resistance will have higher voltage across it. Hence, the ideal diode will be at more risk than the other poor diode.

At a given temperature and reverse leakage current, there will be a difference in the VI characteristics of the diode, though they are from the same manufacturer.

It is an analogy I tried to make. Once it fails (assume fail to short), the other diodes will be then under stress of higher voltage than before.

Below is a simple graph spec showing the possibility of variation of reverse leakage current of a random diode.

And below is another graph showing the current and voltage in the same image. What I want to point is that, if we draw a horizontal line in the reverse voltage reverse current quadrant, then we can see that, for a given fixed reverse current, there are two different voltages possible and the one which has higher voltage across it is at higher risk.enter image description hereenter image description here

Unless there is too much buffer (if not using 5 200V diode for a 400V application), Using forced voltage sharing via resistors avoids the above huge voltage difference developing across diodes. The wattage of the resistors also needs to be considered.

The energy dissipated by the diode with lower leakage current spec will be having higher voltage across it which equivalents to higher heat dissipation.

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  • \$\begingroup\$ If diodes behaves like resistors - I'd agree with you. However diodes have the breakdown at some certain (well specified) voltage. It should make all diodes I-V condition more or less equal. \$\endgroup\$ Aug 16 '19 at 20:58
  • \$\begingroup\$ @RomanMatveev Since when do diodes have a breakdown at some well specified voltage? Zener diodes are among the most accurately specified and those are still only 5%. TVS diodes have a wild range and diodes not meant for use in breakdown only really list a minimum guaranteed breakdown with no accuracy at all \$\endgroup\$
    – DKNguyen
    Aug 17 '19 at 6:07
  • \$\begingroup\$ @DKNguyen. That is true. But in any case each diode will work about its breaking point. The only thing I wanted to mention is that there will be no situation like one diode will take most of the voltage as the leakage current will make the drops at their breaking point or so. \$\endgroup\$ Aug 17 '19 at 17:32
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The main thing that you missed is that you may not know which diode is the "most leaky". The datasheet for each diode will tell you the maximum leakage current for some specific reverse voltage. Any given diode out of the bin may have a leakage current much lower than the specified maximum, or it could be right at the specified maximum. So you can't assume that you know the actual value of the leakage current.

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  • \$\begingroup\$ And if you have one or more diode that's constantly riding at breakdown, it may be damaged. Eventually it fails, then other diodes fail, then the magic smoke comes out. \$\endgroup\$
    – TimWescott
    Aug 16 '19 at 20:04
  • \$\begingroup\$ @TimWescott what makes diode fail? Heat produced by Amps*Volts? \$\endgroup\$ Aug 16 '19 at 20:56
  • \$\begingroup\$ There are several failure mechanisms, and heat tends to accelerate all of them. Excess current can destroy bond wires, excess voltage can cause breakdown, this list goes on. Just stay within the specified maximum conditions. \$\endgroup\$ Aug 16 '19 at 20:59
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Some diode junctions are designed to avalanche safely at the reverse breakdown. You often see a specification on MOSFETs for instance, saying how many mJ they can absorb with repetitive overvoltage breakdown.

On the other hand, most diode junctions fail with overvoltage, and usually to short circuit. This is the case with many rectifier diodes.

Whether they are safe or fail with overvoltage is an expensive behaviour that has to be designed into the junction, the uniformity has to be well controlled to make the whole junction area share the heat, rather than a hotspot developing.

One technique for putting low cost low voltage diodes in series I have seen is to make a series string with a theoretical PIV of at least twice the intend voltage, without employing balancing resistors or capacitors, but using devices from the same batch to get a good first stab at them being reasonably matched. On first use, the weakest diodes fail short, leaving you with the self-selected better balanced survivors in the string. With luck, more than half will survive.

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