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Suppose we're using a voltage multiplier to charge a high-voltage capacitor (C Big in the diagram). If you significantly exceed the capacitor's voltage you'll destroy it.

schematic

simulate this circuit – Schematic created using CircuitLab

Now suppose we're talking about something in the 4kV range, and the capacitor has very low ESR (10 milliΩ). What are good methods to protect the capacitor from overvoltage?

One idea I had was to put a Zener clipper on the AC input, and then construct the multiplier to not exceed the design voltage knowing the input voltage limit.

But of course there's all sorts of exciting "Stuff" happening on the C Big side of the circuit that could lead to feedback spikes. Because of the big capacitor's low ESR I'm assuming there's no way to use diodes to protect it by diverting voltage spikes. So, if we care, is the only other protection going to be via FETs – presumably regulated off of resistor voltage dividers?

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    \$\begingroup\$ Have you looked at "gas discharge tubes" such as littelfuse.com/products/gas-discharge-tubes.aspx \$\endgroup\$
    – scorpdaddy
    May 16 '16 at 17:22
  • \$\begingroup\$ @scorpdaddy - Yes, that's almost certainly the best answer! I had forgotten about those components. Please post as answer so I can accept it. \$\endgroup\$
    – feetwet
    May 16 '16 at 19:37
  • \$\begingroup\$ The voltage multiplier will only multiply the mains peak voltage by the number of stages. It doesn't multiply beyond that. Rate C1 for that voltage and all would be fine as far as I can see. What's the problem? \$\endgroup\$
    – Transistor
    May 16 '16 at 20:01
  • \$\begingroup\$ @transistor - I suppose the only problem on the supply side is lack of trust in the power supply. I guess it would have to be more than a transient problem to blow the main capacitor, but I never know how much to trust these cheap high-frequency DC-AC boosters. And then, like I said, there's "stuff" over on the C Big side that might create spikes on something that's already a big voltage. I'll note that on the diagram. \$\endgroup\$
    – feetwet
    May 16 '16 at 20:26
  • \$\begingroup\$ @feetwet: for a TVS diode based clamp, what does matter is the internal resistance of a spike source, not the ESR of the capacitor. That's because the voltage across the capacitor never changes instantaneously. Not so for a gas discharge tube; if the discharge is initiated, the current through the tube will be limited only by the ESR, and the capacitor will be discharged almost completely. \$\endgroup\$
    – dmitryvm
    May 16 '16 at 20:52
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A spark gap is a crude voltage-activated switch. It is open until the E field builds up to the critical point where charges jump across (a spark). That ionizes the air even more, which makes it more conductive, etc.

These things therefore exhibit hysteresis. It's not clear whether that is desirable in your case or not. In other words, a spark gap won't trip until some voltage level, but once tripped, it won't stop conducting until a much lower voltage.

You can make your own spark gaps easily enough, but they won't be very accurate since the breakdown voltage of air depends on the pressure and humidity, which you generally don't get to control. There are devices called gas discharge tubes, which are much like spark gaps in a controlled environment so that tighter specs are possible.

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  • \$\begingroup\$ In this scenario, with a GDT parallel to the Big cap, then assuming the voltage multiplier was the source of the overvoltage, presumably the GDT would conduct until it had drained the multiplier? Or rather, it effectively shorts the multiplier until the multiplier's voltage drops significantly? Beats frying a capacitor. Unless it fries something in the multiplier. If the power supply is protected from overcurrent (e.g., by fuse or breaker) is there any risk of damage to the multiplier cascade? \$\endgroup\$
    – feetwet
    May 17 '16 at 14:39
  • \$\begingroup\$ @feet: As I said, a spark gap exhibits hysteresis, which means it will trip at one level, but keep conducting until the voltage gets to a significantly lower level. \$\endgroup\$ May 17 '16 at 17:28
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This is not the complete answer to the question. I will try to clarify the role of ESR in the circuit, since there is a misconception about it in the wording of the question.

Assume zero ESR (i.e. an ideal capacitor) and try to model the circuit like this:

schematic

simulate this circuit – Schematic created using CircuitLab

\$R_{zener}\$ is not an "external" resistor. It is intrinsic series resistance of a non-ideal Zener diode (so called differential resistance). Any real Zener (TVS) diode has non-zero differential resistance, which can be modelled as ideal Zener in series with a resistor. Differential resistance is not a constant; it heavily depends on an operating point (= current through a diode).

What is the maximum voltage the capacitor will be charged to? To answer this question, assume \$V_{spike} > V_{clamp}\$, where \$V_{clamp}\$ is Zener diode clamp voltage.

Then in the steady state the current through the Zener diode will be $$I_{zener} = \frac{V_{spike} - V_{zener}}{R_{spike} + R_{zener}}$$ by the Ohm's law. This is the steady state condition, which means the capacitor has been charged to maximum voltage for the given \$V_{spike}\$ level, and no current flows through the capacitor.

The voltage across the capacitor will be $$V_{cap} = V_{zener} + R_{zener}I_{zener}$$ subsituting \$I_{zener}\$ we get $$V_{cap} = V_{zener} + \underbrace{(V_{spike} - V_{zener})}_{\text{overvoltage}}\frac{R_{zener}}{R_{spike} + R_{zener}}$$

Your can see that what does matter is the expression $$\frac{R_{zener}}{R_{spike} + R_{zener}}$$ which is the expression for voltage divider.

If we assume non-zero ESR, it will not affect the formula, since there is no current through the fully charged capacitor, which means no voltage drop across the ESR.

As I have already mentioned in the comment, the ESR does play role if gas discharge tube (GDT) will be used as a protection device. That's because I-V characteristic of a GDT dramatically differs from I-V characteristic of a Zener diode. As soon a GDS breakdown voltage is reached, the discharge begins and the voltage across the tube drops down to tens of volts (arc voltage). Take a look at https://www.bourns.com/pdfs/bourns_gdt_white_paper.pdf . Thus, ESR will limit the GDT current.

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  • \$\begingroup\$ The assumption and concern in the original question is that a voltage spike can break through the cap's dielectric, so it doesn't matter what the cap's charge is. But it looks like here you might be able to construct a voltage divider to allow a "low"-voltage Zener to shunt voltage around a capacitor? Suppose we have a 1kV supply, cap rated to 1kV, and a Zener with a 100V clamp. It appears that Zener resistance decreases with current -- e.g., 3000 to 300 ohms. Can an "R-divider" be put before the Zener to make that work? \$\endgroup\$
    – feetwet
    May 17 '16 at 14:16
  • \$\begingroup\$ The point of my answer is that a spike can not charge the capacitor to the voltage much larger than Zener clamp voltage, if internal resistance of a spike source is much larger than Zener series resistance. The ESR play no role at all (which is not completely true, but is a very good approximation). I did not mean that your can use low-voltage Zerner diode with a divider. No way! \$\endgroup\$
    – dmitryvm
    May 17 '16 at 14:23
  • \$\begingroup\$ @feetwet: to void misunderstanding: Rzener is not an external resistor. It is intrinsic series resistance of non-ideal Zener diode (so called differential resistance). Any real Zener (TVS) diode have non-zero differential resistance, which can be modelled as ideal Zener in series with a resistor. \$\endgroup\$
    – dmitryvm
    May 17 '16 at 14:31
  • \$\begingroup\$ I'm not sure how to associate a resistance with an overvoltage. Suppose we've accidentally connected the circuit to a power supply generating double the voltage limit of the cap Vc? And suppose we've managed to find a Zener with clamp voltage Vc. In that case we're fine. The problem is there are no Zeners with clamp voltages in the kV range. \$\endgroup\$
    – feetwet
    May 17 '16 at 14:31
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    \$\begingroup\$ Suppose that the 2Vc power supply is an ideal voltage source, i.e. it has zero internal resistance (Rspike = 0). Look at the formula in my answer. As your can see, the voltage across the capacitor will be 2Vc, not Vc, and the capacitor will be destroyed. That's why the spike source internal resistance plays a crucial role. True, it may be hard or impossible to find TVS (or Zener) diodes in kV range. But it is definitely possible to use diodes regardless of any ESR value, if such diodes exist. Probably your can connect several TVS diodes in series: vishay.ru/docs/88451/stacking.pdf \$\endgroup\$
    – dmitryvm
    May 17 '16 at 14:46

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