Library / Circuits, electronics and noise / ELN-06
Circuits, electronics and noise
Series capacitance of two capacitors
Equivalent capacitance of two linear capacitors in series with an isolated middle node.
- Ref
- ELN-06
- Rev
- v1
- Field
- Circuits, electronics and noise
- Result
- Ce [F]
For two capacitors in series with an isolated middle node, the charge on each is the same and the voltages add, so 1/Ce = 1/C₁ + 1/C₂. This is the product over the sum, C₁C₂/(C₁ + C₂). The equivalent is always smaller than the smaller capacitor, and the smaller one carries the larger share of the voltage.
Equivalent capacitance of two linear capacitors in series with an isolated middle node. It is always smaller than the smaller of the two.
Calculator
- Parallel capacitance, C₁ + C₂9×10−6 F
- Share of the voltage across C₁, C₂/(C₁ + C₂)0.6667
- Share of the voltage across C₂, C₁/(C₁ + C₂)0.3333
What drives the result
Each variable is moved across its typical range while the others stay at the values above. The longest bar is the key driver.
- 3×10−6 F in series with 6×10−6 F gives 2×10−6 F, less than the smaller capacitor.
- Across a supply, the smaller capacitor takes 66.7 % of the voltage, which is why series capacitors need voltage-sharing resistors.
Chart data as tables
Ranked variables
| Variable | Typical range | Lowest result | Highest result | Swing (F) | Local step | Change in result |
|---|---|---|---|---|---|---|
| C1 Capacitance 1 | 1e-9 to 0.001 F | 9.998e-10 | 5.964e-6 | 5.963e-6 | +10% | +6.45 % |
| C2 Capacitance 2 | 1e-9 to 0.001 F | 9.997e-10 | 2.991e-6 | 2.99e-6 | +10% | +3.13 % |
Sweep of Capacitance 1
| Capacitance 1 (F) | Equivalent capacitance (F) |
|---|---|
| 1e-9 | 9.998e-10 |
| 5.623e-9 | 5.618e-9 |
| 3.162e-8 | 3.146e-8 |
| 1.778e-7 | 1.727e-7 |
| 1e-6 | 8.571e-7 |
| 5.623e-6 | 2.903e-6 |
| 3.162e-5 | 5.043e-6 |
| 1.778e-4 | 5.804e-6 |
| 0.001 | 5.964e-6 |
Sweep one variable
Variables and how each one contributes
Equivalent capacitance
What the equation gives you. Everything else feeds it.
Capacitance 1
The capacitance of capacitor 1, in F. A larger C₁ raises the series capacitance, but never above C₂. Raising it by 10 % (to 3.3×10−6) moves the result by +6.45 %, from 2×10−6 to 2.129×10−6 F.
Capacitance 2
The capacitance of capacitor 2, in F. A larger C₂ raises the series capacitance, but never above C₁. Raising it by 10 % (to 6.6×10−6) moves the result by +3.13 %, from 2×10−6 to 2.063×10−6 F.
Worked example with the values above
- Product: C₁C₂ = 3×10−6 × 6×10−6 = 1.8×10−11 F².
- Sum: C₁ + C₂ = 9×10−6 F.
- Cₑ = 2×10−6 F.
Assumptions and limits
- Ideal linear capacitors and inductors with constant values.
- Leakage current, series resistance and saturation are neglected.
Common mistakes
- Adding capacitances in series as if they were in parallel.
- Forgetting the sign of a voltage when two capacitors are connected with opposite polarity.
- Assuming no energy is lost when two charged capacitors at different voltages are joined.
Statistics and uncertainty
Real capacitors have tolerances of 5 % to 20 % and change with voltage, temperature and age, so results scale with those errors. Inductor values fall when the core saturates.
Where it is used
- Timing, smoothing and switched-capacitor circuits.
- Energy storage in inductors and capacitors for converters and pulse circuits.
Maths you need
- Capacitance: C = Q/V
- Charge conservation: charge on isolated nodes does not change
- Energy in a capacitor or inductor: ½CV² and ½LI²