Series capacitance calculator

Series and parallel totals for up to six capacitors, and how an applied voltage divides across a series string.

Series total

13.03nF

In series at 12 V, with the voltage across each. Empty slots are left out.

Parallel total: 169 nF
Charge at 12 V: 156.4 nC
Stored energy: 938.3 nJ
Highest voltage, C3: 7.109 V

Voltage split across series capacitors at 12 V

Capacitor	Capacitance	Voltage	Charge
C1	100 nF	1.564 V	156.4 nC
C2	47 nF	3.327 V	156.4 nC
C3	22 nF	7.109 V	156.4 nC

Method

How this is calculated

1 / C_series = 1/C1 + 1/C2 + ... + 1/Cn
C_parallel   = C1 + C2 + ... + Cn
Q = C · V          E = C · V² / 2
V_k = Q / C_k      (series string, equal charge)

Units: Inputs in nF: 1 µF is 1000 nF, 100 pF is 0.1 nF. A slot left at 0 is ignored.
Voltage split: Ideal capacitors in series share the same charge, so the smallest takes the largest voltage. At DC, leakage current sets the split instead; high voltage strings use balancing resistors for that reason.
Assumes: Nominal values. Class II ceramics (X7R, X5R) lose capacitance with DC bias, which this does not model.

Sources

FAQ

Questions

Reciprocals add. 1 µF, 5 µF and 8 µF in series come to 0.755 µF, less than the smallest of the three.

Directly. The same 1 µF, 5 µF and 8 µF in parallel are 14 µF. This is why decoupling capacitors are placed in parallel.

Each carries the same charge, so V = Q / C. 12 µF in series with 6 µF across 12 V puts 4 V on the 12 µF part and 8 V on the 6 µF part.

Only if the voltage divides as intended. With unequal capacitance or leakage one part can see more than its share, so check the highest voltage result against each rating.

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Questions

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