# Does diode reverse recovery time in Cockcroft-Walton Multiplier depend on current or negative voltage?

Say there is a multiplier whose sine wave input is 1 kV, 50 kHz, output should be 10 kV, 100uA. The high frequency means a fast diode should be used (low reverse recovery time trr). However, it seems the availability of fast high-voltage diodes is limited. So I wonder if there is some room for "cheating" here by choosing a nominally slower diode.

First, if the datasheet says the trr is specified at 1 A and the current through the diodes in my application were only 10 mA, would this make the diodes faster?

Second, the multiplier is actively providing negative voltage to the diodes, does this make them stop conducting faster?

If the lower current or the negative voltage does make the diodes faster, can anyone put a figure on how great this effect could be?

Current does not have a large effect on recovery rate, basically because you'll be driving it with similar currents while forward and during reverse recovery. This can be tested even with fairly small signals, say a 3V square wave from a function generator, or with large signals as in power switching applications.

The overall effect is a matter of charge loss in the charge pump circuit (another name for a voltage multiplier circuit). Forcing you to use larger capacitors than otherwise, and causing some power loss as well. How much charge is lost, can be inferred from the recovery charge figure (which should scale proportionally to If or Ir, per the above claim). But you may also just opt to build and test one, and choose bigger capacitors if the output proves insufficient.

Note there's nothing wrong with diodes in series, and indeed the high voltage types are just that, stacks of matched dies in a single package. Hence they often have large Vf (6V or more!).

The wrongest thing about stacking diodes, is that keyword "matched": preferably they should be matched in recovery time as well, so that turn-off voltage distributes evenly between them. So, that's what they do in packaged units. With discrete diodes wired in series, one will inevitably turn off sooner, avalanching until the rest recover. So the power dissipation will be higher than expected, and in particular, intensely localized to that fastest part -- greatly reducing the current rating of the overall rectifier.

On the upside, recovery time rises with temperature, so it's not a runaway condition, at least, not to the runaway failure of a single part in the chain; the worsening recovery of an overheating chain might still compromise the whole thing, of course.

But at 100µA, that should be fine. At 50kHz, I would think types under 500ns will suffice. Whether that's strings of UF4007 or US1M, or something slower.

You may also consider getting a spare television flyback transformer, which integrates the rectifier already, and is rated for more than enough current. (A small, say 9 to 13-inch CRT should use about that voltage rating.) Depends if this is one-off or a production sort of thing, or if there's any other considerations (particular windings or supply voltage, what ground is common to, or polarity, etc.).