The following formula is generally used for calculating balance resistor values used for overvoltage protection of capacitors in series when the individual capacitor voltage ratings are below the supply voltage.

$$R_{balance} = \frac{NV_{rate} - V_{bus}}{I_{\Delta leak}}$$

  • \$N\$ is the number of capacitors in series
  • \$V_{rate}\$ is the rated maximum voltage for any one capacitor
  • \$V_{bus}\$ is the bus voltage (the expected voltage across the whole series of capacitors)
  • \$I_{Δleak}\$ is the leakage current difference
  • \$R_{balance}\$ is the value of the balance resister between any two capacitors

After an extensive search, I could not find references to the derivation of this formula other than it is based on \$V = IR\$ .

Question 1:

How this formula is derived? Or is it just a rule of thumb?

I also found in the first reference given below that:

$$R_{balance} = \frac{NV_{rate} - V_{bus}}{I_{maxleak}}$$ This will always be the safest bet, better than any of the earlier rules of thumb. I say “safest”, by that I mean the voltage across any one capacitor will never rise above its rated voltage that way, but then you are likely burning a lot of heat in your balance resistor and you need to be careful not to be cooking your capacitor.

But it is not apparent how this ensures that the "voltage across any one capacitor will never rise above its rated voltage".

Question 2:

How this can be derived or proven? Or is that another rule of thumb with some hefty safety margins?

Note: I want to know whether these formulae always ensure that caps stay within the Voltage rating if the parameters are correct.


  1. https://va1der.ca/index.php/balance-resistors-for-series-capacitors/
  2. https://electronics.stackexchange.com/a/517165/344253

Edit: I am stuck at this point proving this:

The formula for calculating balance resistor values for capacitors in series is derived from the basic principle of voltage division.

If we assume that the initial charging stage is over, then the voltage across each capacitor is inversely proportional to its leakage resistance Rleak. Therefore, we can write:

$$V_i = \frac{R_{leak,i}}{\sum_{j=1}^N R_{leak,j}} V_{bus}$$

where Vi is the voltage across the ith capacitor, Rleak,i is its leakage resistance, and Vbus is the total bus voltage across the series of capacitors.

Now, if we add a balance resistor Rbalance between each capacitor, then the voltage across each capacitor becomes:

$$V_i = \frac{R_{leak,i} + R_{balance}}{\sum_{j=1}^N (R_{leak,j} + R_{balance})} V_{bus}$$

We want to choose Rbalance such that the voltage across each capacitor does not exceed its rated voltage Vrate. This means that we need to satisfy the following inequality for all i:

$$V_i \leq V_{rate}$$

After that,

$$ \frac{R_{leak,i} + R_{balance}}{\sum_{j=1}^N (R_{leak,j} + R_{balance})} V_{bus} \leq V_{rate}$$

Now I can remove the summation from \$R_{balance}\$ but not from \$R_{leak,j}\$ . From there I do not know how to reduce it to the original formula.

  • \$\begingroup\$ Assume your best capacitor isn’t leaking at all. What’s the worst case scenario leakage current in your worst? \$\endgroup\$
    – winny
    Commented Jul 10, 2023 at 13:37
  • \$\begingroup\$ @winny That much I understand. But I am asking about the entire formula. Can it be derived using a formal method? \$\endgroup\$ Commented Jul 10, 2023 at 16:22
  • \$\begingroup\$ That would be Ohm’s law for a resistor divider and how much voltage mismatch you can allow. \$\endgroup\$
    – winny
    Commented Jul 10, 2023 at 18:56
  • \$\begingroup\$ @winny But it doesn't guarantee. Does it? Because when leakage current "modelled as resistance" is added to the balance resister, voltage change again. Can you show this limits voltage below the rated voltage of caps? \$\endgroup\$ Commented Jul 11, 2023 at 4:31
  • 1
    \$\begingroup\$ I would just set up an array of resistors and derive it from there. In practice, I do it in excel and tune it to satisfaction. If you have the power dissipation to spare, having Rbleed << Rleak will always satisfy your equation. \$\endgroup\$
    – winny
    Commented Jul 16, 2023 at 10:41

3 Answers 3


My original answer was perhaps a bit difficult to understand so I rewrote it with schematics. Hopefully it is more clear now. The most important point is to understand what kind of model is behind the equations in question. This is by no means the most accurate model but that is a topic for different discussion.


simulate this circuit – Schematic created using CircuitLab

The model of the capacitor consists of a current source \$I_{leak}\$ and an ideal capacitor parallel to it. If we only consider DC we can omit the capacitor altogether because it has no effect to DC. Now each capacitor is modelled by just the leakage current source and there is only the associated balance resistor parallel to it.

Another important point to note is that if the leakage currents were all equal they would effectively disappear from the circuit and there would be no need for balancing resistors. Therefore we consider a situation where all current sources but one are equal. The one current source that is not equal to all the rest consists of current source that is equal to the rest and another current source that has value of \$\Delta I_{leak}\$. Now we can consider just the voltage of this one particular capacitor and the model associated with it looks like this:


simulate this circuit

Now since we defined the delta to be in the same direction as the origial leakage current, an positive delta current means higher leakage current and thus lower voltage over the capacitor under consideration. This can be formulated by the following equation: \$\Delta V=-\Delta I \cdot R_{Balance}\$ Now because the voltage of all N caps in series has to equal the bus voltage we can write:

\$ V_{bus} = (V_{c} + \Delta V)+(N-1)\cdot V_{c}\$

This can be refined into:

\$ -\Delta V=N\cdot V_{c}-V_{bus}\$

And if we now replace the definition of \$\Delta V\$ we can solve for \$R_{balance}\$:

\$R_{balance} = \frac{N\cdot V_{c}-V_{bus}}{\Delta I}\$.

Now in the original equation the \$V_c\$ of my equation is just replaced with \$V_{rated}\$. This is sensible thing to do when considering the situation where one capacitor has higher leakage current than the rest because then maximum voltage seen by any cap is \$V_c\$ and if the rated voltage is higher than that everything is fine. However, when one capacitor has smaller leakage current than the rest the situation is more difficult. Then the \$\Delta V\$ is positive and the \$V_{rated} \$ must be at least \$\Delta V\$ higher than the voltage of the other caps \$ V_c \$.

\$ V_{rated} = V_c + \Delta V => V_c = V_{rated}-\Delta V\$

Now substituting this into the equation of \$ R_{balance}\$ we get:

\$R_{balance} = \frac{N\cdot (V_{rated} - \Delta V) - V_{bus}}{\Delta I} \$

Refining and replacing \$\Delta V = -\Delta I R_{balance}\$ we get:

\$R_{balance} = \frac{N\cdot V_{rated} - V_{bus}}{\Delta I(1-N)} \$

Now because N is positive you might think the denominator would be negative which would not make any sense. The explanation is that if the voltage of the odd cap is more positive than the voltage of the rest, then the current has to be negative. This equation gives smaller values for the resistor than the previous equation making it the worst case. Therefore, the original equation does not give small enough balance resistor values for the real worst case situation where the leakage current of one capacitor is \$\Delta I\$ smaller than the leakage current of all the rest.

  • \$\begingroup\$ Many thanks for the answer, but how can it be "\$\Delta V=\Delta I \cdot R_{Balance}\$"? Isn't it "\$\Delta V=\Delta I \cdot R_{leak}\$"? And still, that's before we add balance resistors to the circuit. Then it becomes more complex. You also said my formula considers leakage current as a constant. But it isn't. I have not used leakage current directly but only leakage resistance from which leakage current is derived as proportional to the voltage across the cap. \$\endgroup\$ Commented Jul 18, 2023 at 7:49
  • 1
    \$\begingroup\$ As I tried to explain in my first paragraph, the model used to derive those equations does no include leakage resistance, but it models the leakage current as a constant current source in parallel with capacitor. You could argue if this is a good model or not but it is a different discussion. Now the \$R_{Balance}\$ is in parallel with this current source and the voltage is basically thevenin equivalent of this. The circuit is linear so we can only focus on the deltas. Even though the actual leakage current has negative sign compared to voltage we can arbitrarily select the sign for delta. \$\endgroup\$
    – Trafi
    Commented Jul 18, 2023 at 10:07
  • \$\begingroup\$ That is a well-written answer. My apologies, I seem to have misinterpreted your words. You mentioned that the original formula relies on constant leakage and my derivation does not. I got that mixed up with the opposite case. So the original formula is only good when only a few caps in a series or only one cap (or relatively few) is faulty. \$\endgroup\$ Commented Jul 19, 2023 at 7:44
  • 1
    \$\begingroup\$ I would say that the original formula was correct only for the case of 2 capacitors. For 2 capacitors, both equations give exactly the same resistance because (1-N)=1-2=-1 so the magnitude stays the same. This is expected because when you have only two capacitors the situation is always symmetrical. In all other cases the original equation does not take into account the true worst case situation that the equation derived by me considers. This is the case when one cap has \$\Delta I\$ lower leakage current than all the rest. \$\endgroup\$
    – Trafi
    Commented Jul 19, 2023 at 11:11

I don't much enjoy math and theoretical proofs of electronic design principles, especially when there are many unknown variables, and these may very well also vary with time, temperature, and various random factors. So instead of trying to validate a "rule of thumb" for capacitor balance resistors, I would rather design a simple active circuit that achieves the goal, and also can be much more efficient:


simulate this circuit – Schematic created using CircuitLab

So, here is a chart of the slow ramp voltage to 180 VDC, and the voltages across C1 and C2. You can see that although the capacitors have a 2:1 ratio of capacitance, the balance circuit stabilizes with 90 VDC on each. Thus both capacitors could have a voltage rating of 100 V.


And here is a graph of the current through R4 and R7, showing that R4 acts to stabilize the voltages by drawing a peak of 3 mA. A faster rate of rise of the applied voltage would result in a much higher current peak, but also of much shorter duration.


  • \$\begingroup\$ What is the max DC potential this circuit can withstand? \$\endgroup\$ Commented Jul 18, 2023 at 10:04
  • \$\begingroup\$ I used the default 2N3904 transistors, which are rated for just 40V and 200 mA, And while the circuit is balancing, the voltage may briefly exceed the desired capacitor voltage. Try the circuit with a simulator and see how it performs under various applied voltages and with various values of capacitance and leakage currents. \$\endgroup\$
    – PStechPaul
    Commented Jul 18, 2023 at 22:03

Theoretically perfect capacitors do not need balancing resistors. If you apply 100 V across 5 identical caps in series, each cap will have 20 V across it.

Real-world caps leak. At the concept level, this can be thought of as a high-value resistor in parallel with a perfect capacitor. If you remove the caps from the circuit, you now have a voltage source connected to five unequal resistors in series. This is why it is an Ohm's Law problem.

To "balance" the circuit (meaning re-establishing equal voltages across the elements), you add resistors in parallel with the leakage resistors, making the added resistors so much smaller than the leakage resistors that the node voltages are determined mostly by the added resistor values, not the cap internal leakage resistor values.

Making up numbers . . .

For example, if the leakage current across a batch of parts varies over a 5-to-1 range, the equivalent leakage resistances might vary from 100K to 500K. If you put a balancing resistance of 10K across each cap, the parallel equivalent resistance between each node now varies from 9.09091K to 9.80392K. This is a ratio of only 1.08:1, way less than 5:1.

Separate from this is the fact that real-world capacitors do not have zero tolerance component values. If your caps are +/-10% (common in ceramics and films, but pretty tight for electrolytics), then the worst-case node voltages will vary way more due to charge balance than that caused by leakage current.

Again, the solution is to add even lower value resistors in parallel with the caps, but because the problem is the capacitor's actual value, it is a much lower equivalent resistance. Now, power dissipation and circuit efficiency become relevant factors in the design. For +/-10% caps is might not be worth it, but for +80%/-20% caps, a much more common electrolytic spec, balancing can be very beneficial.

  • \$\begingroup\$ Thanks but I need to know whether the formula itself can be formerly derived or proved to be correct. Do you know a way to derive this? \$\endgroup\$ Commented Jul 10, 2023 at 16:25
  • 1
    \$\begingroup\$ Not sure what you mean by "formally (sp) derived". Is this school work? \$\endgroup\$
    – AnalogKid
    Commented Jul 10, 2023 at 16:42
  • \$\begingroup\$ No. This is not schoolwork. What I mean is, can this equation be derived from basic equations? I am trying to make sense of it. I understand that proportionality with voltage headroom and inverse with leakage current. But I do not understand how this ensures caps stay within the rated voltage aftermath of the charging phase when leakage currents begin to dominate voltage differences. \$\endgroup\$ Commented Jul 10, 2023 at 17:06

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