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Circuits, electronics and noise
Capacitor voltage ramp under constant current
Voltage across a capacitor charged by a constant current from an initial voltage V₀.
- Ref
- ELN-09
- Rev
- v1
- Field
- Circuits, electronics and noise
- Result
- V(t) [V]
A constant current I into a capacitor gives a linear voltage ramp, V(t) = V₀ + I t/C, because i = C dv/dt. The slope is I/C in volts per second. A negative current makes the voltage fall. The formula ignores leakage and series resistance, and fails when the capacitor reaches its voltage limit.
Voltage across a capacitor charged by a constant current from an initial voltage V₀. It rises linearly with time, and a negative current makes it fall.
Calculator
- Ramp rate, I/C1,000 V/s
- Charge added, I t6×10−6 C
- Energy added to the capacitor1.5×10−5 J
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.
- A 0.002 A current into 2×10−6 F ramps at 1,000 V/s, taking the voltage from 1 V to 4 V in 0.003 s.
- Doubling the time or the current doubles the voltage added, while doubling the capacitance halves it.
Chart data as tables
Ranked variables
| Variable | Typical range | Lowest result | Highest result | Swing (V) | Local step | Change in result |
|---|---|---|---|---|---|---|
| C Capacitance | 1e-9 to 0.001 F | 1.006 | 6,001 | 6,000 | +10% | −6.82 % |
| I Constant current into the positive terminal | 1e-5 to 1 A | 1.015 | 1,501 | 1,500 | +10% | +7.5 % |
| t Charging time | 1e-5 to 1 s | 1.01 | 1,001 | 1,000 | +10% | +7.5 % |
| V0 Initial voltage | −10 to 10 V | −7 | 13 | 20 | +0.4 V | +10 % |
Sweep of Charging time
| Charging time (s) | Capacitor voltage (V) |
|---|---|
| 1e-5 | 1.01 |
| 4.217e-5 | 1.042 |
| 1.778e-4 | 1.178 |
| 7.499e-4 | 1.75 |
| 0.003162 | 4.162 |
| 0.01334 | 14.34 |
| 0.05623 | 57.23 |
| 0.2371 | 238.1 |
| 1 | 1,001 |
Sweep one variable
Variables and how each one contributes
Capacitor voltage
What the equation gives you. Everything else feeds it.
Initial voltage
The initial voltage across the capacitor, in V. Adds directly to the final voltage. Raising it by 10 % (to 1.1) moves the result by +2.5 %, from 4 to 4.1 V.
Constant current into the positive terminal
The constant current into the positive terminal, in A. A larger current steepens the ramp in direct proportion. Raising it by 10 % (to 0.0022) moves the result by +7.5 %, from 4 to 4.3 V.
Charging time
The charging time, in s. A longer time raises the voltage in direct proportion. Raising it by 10 % (to 0.0033) moves the result by +7.5 %, from 4 to 4.3 V.
Capacitance
The capacitance, in F. A larger capacitor ramps more slowly, in inverse proportion. Raising it by 10 % (to 2.2×10−6) moves the result by −6.82 %, from 4 to 3.727 V.
Worked example with the values above
- Ramp rate: I/C = 0.002 / 2×10−6 = 1,000 V/s.
- Voltage added in 0.003 s: 3 V.
- V = 1 + 3 = 4 V.
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