I am constructing a High Voltage (150 kV DC) capacitor bank. I have elected to connect 10 capacitors in series, with 10 such branches connected in parallel. My next mission is to create balancing resistor branches to make sure my capacitors don't blow up.

I have been crawling through research papers, technical documents, and here at the Stack Exchange looking for a method to select a value for my balancing resistor resistance values, but the literature I've found is either modeled for 2 capacitors in series, or produces a ludicrous number (like 3.33 TΩ).

Below is a basic model of my circuit. The purpose of the capacitor bank is to mitigate droop in the HV supply caused by our load, which acts like a pulsed current source. The current pulses drain charge from the circuit, and the power supply can't react fast enough, leading to considerable droop over time. This all occurs on the nanosecond scale.

enter image description here

A few notes on the circuit: I suspect that each individual series branch will require its own balancing branch, instead of the two that I modeled.

Additionally, the manufacturer of my capacitors couldn't have a leakage current for me, but provided a 200 GΩ resistance value, per capacitor, at its rated voltage of 20 kV. I am thinking that I should model leakage current per branch by summing the resistance values of the capacitors, and dividing the 150 kV bus voltage over that. This gives a leakage current of 0.75 µA, which seems about right. Is this a poor approach?

TL;DR: How do I choose balancing resistor values for a series string of 10 capacitors in a high voltage circuit?

Thanks in advance for any advice or insight you can offer, and I'd be happy to clarify any aspects of my problem.

  • 6
    \$\begingroup\$ You only need one resistor chain if you add 9 horizontal interconnections. This also reduces the effect of variations in the capacitor values. \$\endgroup\$
    – Jens
    Commented Nov 16, 2023 at 17:44
  • \$\begingroup\$ Isn't there meant to be a resistor across each capacitor, not one resistance across all the capacitors? Or did you mean that but the circuit diagram went wrong? \$\endgroup\$ Commented Nov 16, 2023 at 20:44

1 Answer 1


You can model this as a ladder of paralleled resistors:

RTotal = (RCap1 || RBal) + (RCap2 || RBal) + ...

Now you need to choose the balancing resistors so that no capacitor gets an overvoltage even in the worst case.

The worst case is when all capacitors are extremely leaky, except for one, which is perfect. That will put the highest possible overvoltage on the perfect capacitor.

In this case, we'll have the following string of N capacitors:

RTotalWorst = RBal + (RCapWorst || RBal) + (RCapWorst || RBal) + ...
            = RBal + (N - 1) * (RCapWorst || RBal)

Now we can treat this as a voltage divider to find the voltage across the perfect cap, relative to the total voltage applied to the bank:

RelativeVoltageWorst = RBal / RTotalWorst
                     = RBal / (RBal + (N - 1) * (RCapWorst || RBal))

To make things simpler, let's express RBal as a multiple of RCapWorst:

RBal = m * RCapWorst

RelativeVoltageWorst = m * RCapWorst / (m * RCapWorst + (N - 1) * RCapWorst * m / (m + 1))
                     = m / (m + (N - 1) * m / (m + 1))
                     = 1 / (1 + (N - 1) * 1 / (m + 1))
                     = 1 / ((m + 1) / (m + 1) + (N - 1) / (m + 1))
                     = (m + 1) / ((m + 1) + (N - 1))
                     = (m + 1) / (m + N)

In this formula, N is the number of series capacitors in the string, and m is how much larger the balancing resistor is than the worst-case leakage resistance of a cap.

Let's also solve for m:

RelativeVoltageWorst = (m + 1) / (m + N)
RelativeVoltageWorst * (m + N) = (m + 1)
m * RelativeVoltageWorst = m + 1 - N * RelativeVoltageWorst
m * (RelativeVoltageWorst - 1) = 1 - N * RelativeVoltageWorst
m = (1 - N * RelativeVoltageWorst) / (RelativeVoltageWorst - 1)

In your case, you have N=10, a total voltage of 150kV, and a conservative maximum voltage could be 18kV. That puts us at RelativeVoltageWorst = 18 / 150 = 0.12.

Plugging that in, we get m = 0.22. Or, in other words:

RCapWorst = 200 GOhms
RBal = m * RCapWorst = 0.22 * 200 GOhms = 44 GOhms

The next standard value would be 39 GOhms.

This is the value you'll need if you give every string of series capacitors its own balancing resistors. If you connect your individual capacitors in parallel, as Jens suggested in the comments under your question, you can use only one set of resistors, but each of them would have to be 3.9 GOhms instead (RCapWorst for 10 paralleled caps is 1/10th the original value, so 20 GOhms). In other words: You'll either need 100x 39 GOhm, or 10x 3.9 GOhm. The latter is probably easier.

As Neil_UK noted in the comments, you'll also need to discharge the capacitors for safety. If you repurpose the balancing resistors for this purpose, you'll get a time constant tau = 39 seconds with 39 GOhms and 1 nF. Capacitors are usually considered to be discharged after 5 tau, which would be 200 seconds in this case, or a bit more than 3 minutes. If you need them to discharge quicker, you'll have to decrease the value of the balancing resistors.

  • 2
    \$\begingroup\$ Gohm resistors are probably quite difficult to obtain. You will need capacitor discharge resistors, for safety, to discharge your bank within X minutes of removing them from power. Once you've calculated these values, you'll probably find they're less than Gohms, so the discharge resistors will double as your balance resistors. \$\endgroup\$
    – Neil_UK
    Commented Nov 16, 2023 at 17:55

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