# Balancing resistor values for series capacitors

I just wanted to confirm my rough calculations are correct in selecting balancing resistors for two capacitors in series.

Here are the specifications: two 10,000uF capacitors with 500V rating in series.

I found this estimation equation online: R = 10 / C where R =Mohm and C = uF.

Based on this, I got 1kohm resistors to use as balancing resistors for each capacitor.

I'm using this setup to filter out transient behavior from a power cycler supplying 900V. Are those resistor values correct (or in the ballpark?)

• Your calculation gives about 0.45 amps through the resistors. They would need to dissipate 202.5 watts each. You should probably use somewhat higher value resistors. Oct 28, 2016 at 0:35
• wrong! he said RC=10MΩ*uF= 10 Second leakage decay time constant. Critical factors are temperature vs leakage and Tolerance error between parts. 10sec seems reasonable but depends on part number, and quality and Ripple current with temp rise. Oct 28, 2016 at 1:55
• In this case, it might be worth consulting the capacitor manufacturer for advice. Oct 28, 2016 at 3:27
• For new parts that are matched, A 10 second balance RC value seems reasonable for parts with > 1000 second T but dynamic balancing may be required like batteries which tend to be balanced <1% unlike capacitors which tend to have a large tolerance "unless" from same batch. Oct 28, 2016 at 3:35

Balance resistors are intended to ensure that leakage current differences across series capacitors don't push the voltage of one or more of them over the rated voltage. There is a lot of noise online about how to calculate them. Some of them advocate resistors to draw three times the leakage current. I've seen that go as high as ten times the leakage current. The one in the question, 10/C - where does that come from? The problem is that those rules of thumb leave you blind to what is actually happening. Also, many of them give you values of resistors that burn so much electricity and produce so much waste heat that you essentially are putting little space heaters beside your capacitors, which is never a good thing. As it turns out, calculating the resistor isn't that hard. The general formula for this is:

At its core it's simple. R = V/I. In this case, it's the voltage headroom divided by the maximum difference in leakage current in your capacitors. The key is in getting these two numbers.

$$\V_{headroom}\$$ - This is the simpler of the two numbers you need. This is the amount of extra voltage you have to play with, and is simply the maximum rated voltage of the capacitors you are using times the number of capacitors you are using, then take that and subtract the actual bus voltage across all of them. In your case, 2 x 500V - 900V = 100V. This is the voltage headroom your circuit has to play with.

$$\I_{maxdeltaleak}\$$ - This is the tougher one and usually requires some estimating. This value is the maximum difference in leakage current you can expect between the capacitors in your bank. If you want to be 100% safe, then you can use the manufacturer-supplied value for the maximum leakage current in their capacitors. This is visible in the manufacturer's capacitor data sheets. For example, Nichicon is 3√(CV) to a max of 5mA. However, this assumes that one of your capacitors is leaking the maximum amount and the other one (or more) is not leaking at all, which is never the case. Most capacitors of the same value from the same manufacturer (especially if they are from the same batch) will leak about the same amount. The rule of thumb I go by is to use 20% of the overall maximum leakage current as the maximum difference in leakage currents you will see in a group of them. So:

100k is a good value to keep your voltages where they need to be. In this case, it might not be a terrible idea to use 3 x 15000uF capacitors, even lower-rated 400V ones to give some more total voltage headroom.

• Thank you very much! I was upset when I was constantly getting about 5W across the resistors. Now, max power 0.05W Dec 17, 2020 at 10:22
• Can you explain how this formula is derived or is it just a rule of thumb? Jul 9, 2023 at 6:51
• It is derrived from Ohm's Law, E=IR, or R=E/R but the final formula does involve an estimate of the maximum change in leakage current between capacitors. I wrote an article that explains the entire process in depth at: va1der.ca/index.php/balance-resistors-for-series-capacitors Jul 10, 2023 at 10:54
• Thank you, sir. I have read it. But there is no proof or derivation of the formula there. Not even on the internet. All the AI answers are gibberish. And for myself, I am stuck at step one: Trying to establish an equation for a single cap in an n caps in series. I have asked separately here: electronics.stackexchange.com/q/673317/344253 Jul 16, 2023 at 12:05
• This may be ok for just a few caps but may not behave well when the number of capacitors in series increases significantly. reference: forum.allaboutcircuits.com/threads/… & electronics.stackexchange.com/a/674349/344253 Jul 19, 2023 at 7:02

As I understand it, you want to make a 1 kV polarized capacitor from two polarized 500 V capacitors:

The bleeder resistors are intended to keep the voltages on the capacitors roughly balanced. OK so far, but this seems rather extreme.

The resistors will draw ½ A with 1 kV in! That's 500 W of power, and each resistor will dissipate 250 W. This might work if you're trying to make a small toaster, but then you wouldn't need the capacitors at all.

Another way to look at this is to see when the result looks mostly capacitive and when mostly resistive. The -3 dB point of 1 kΩ and 10 mF is 16 mHz. If all your frequencies will be substantially above 16 mHz, then you're actually OK from that point of view.

I would look carefully at the worst case this circuit will be subjected to and see if the resistors can't be made much larger.

Dissipating 500 W just to avoid more expensive high voltage caps seems like a poor tradeoff. Look at 1 kV caps, and also pop up a couple levels and re-examine why you think you need this in the first place.

interesting question.

• from these choices for 10kuF 500V price ranges from $137/ea to$341/ea (usd)

• 1st looking at the cheapest, UCC p.n. E37X501CPN103MFM9M

• summary indicates:

• Leakage current: 0.02CV(µA) or 5mA,whichever is smaller, >5 minutes @25'C
• Standard capacitance tolerance: +/-20%
• worst case is 5mA @500V = 100kΩ > 5 minutes thus RC= 1,000 seconds

The problem here is the caps like batteries have "double-layer effect" and a memory effect as well as an aging leakage that rises with age and requires burning off contaminants so that the leakage current drops below the rated limit in uA.

Pd in 100k= $V^2/R=500^2/100k=2.5W$

• THe uncertainty is the transient leakage after some unused period of time, which is why historically large e-caps were charged up with 100k series R until the voltage charged up fully then it could be put into service. thus it depends on how well matched the two caps for "Balancing".

A soft start is a safe plan or twin MOV/TVS's at rated voltage to balance fused Caps , but may be hard to find and not very precise.

simulate this circuit – Schematic created using CircuitLab

• Thanks for your informative answer Tony! I hadn't considered a lot of those factors before! Oct 28, 2016 at 16:22
• We used to buy mainframe caps with 60V 100k uF caps in the late 60's . If you didn't heed the advice of the elders about burning off leakage resistance with a high series R, the big cap had explosive results. Now they have vents and may or may not or internal fusing. FWIW Tesla version 2 LiPo cells all have fuse wires now to prevent fires from faulty cells that fail with shorts instead of normal open circuit or high ESR and Tesla puts hundreds of matched cells ( like matched cap) in parallel then many series modules to have >4k?x 16850 LiPo cells. Oct 28, 2016 at 16:58
• So, you see matching Ah and Voc or ESR, Rp & C is as important to LiPo as big HiV caps Oct 28, 2016 at 17:02

The 1965 ARRL Handbook, page 325 reads balance resistors should be "100 ohms per volt with a power rating adequate for the total resistor current." It worked in 1965. should work today?

I also match Caps in a string using ESR prior to installation. Eh. I own an ESR tester. Why not. I don't own a leakage tester with a meter indicating mA of leakage or I would likely use that in the formula above.