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Capacitor Charging Voltage
Describes the exponential rise of voltage across a charging capacitor in an RC circuit.
Interpretation
Capacitor charging voltage follows an exponential rise: V(t) = V₀(1 − e^(−t/RC)).
The voltage asymptotically approaches the source voltage V₀ but never reaches it in finite time.
Example: V₀=10V, RC=1s, t=2s → V = 10(1 − e^(−2)) ≈ 8.65V.
Variables
| Symbol | Quantity | Unit |
|---|---|---|
| τ | Time constant (RC) | s |
| R | Resistance | Ω |
| C | Capacitance | F |
| V(t) | Voltage across capacitor at time t | V |
| V₀ | Source voltage (final voltage) | V |
| t | Time | s |
| e | Euler's number (≈2.71828) | — |
What it means
The voltage across a charging capacitor in an RC circuit as a function of time is V(t) = V₀ * (1 − e^(−t/τ)), where V₀ is the final voltage (supply voltage) and τ = RC is the time constant. The voltage rises exponentially from 0 towards V₀, never quite reaching it in finite time. This equation is derived from solving the differential equation for the circuit. It is fundamental for understanding timing circuits, such as those in 555 timers, integration circuits, and pulse delay generators. The exponential rise is also used in ramp generators and in analog computing. The time to reach a certain percentage of V₀ can be calculated: for 50%, t = τ * ln2; for 90%, t = τ * ln10. Example: V₀=10V, R=1kΩ, C=100µF → τ=0.1s. At t=0.1s (one τ), V = 10*(1 − e^(−1)) = 10*(1−0.368) = 6.32V. At t=0.2s (2τ), V = 10*(1−e^(−2)) = 10*(1−0.135) = 8.65V.
Worked example
Capacitor Charging – Practical Example
Real‑World| Parameter | Value |
|---|---|
| V₀ | 10 V |
| τ = RC | 0.1 s (from previous example) |
| t | 0.1 s |
| Formula | V(t) = V₀(1 − e−t/τ) |
Common mistakes
- V₀: The final voltage (source voltage) – not the initial voltage.
- Exponent: The term is −t/(RC) – ensure the time constant is correct.
- Initial condition: At t=0, V(0) = 0 (assuming capacitor initially uncharged).
- Asymptotic: The voltage never exactly reaches V₀; it asymptotically approaches.
- AC signals: This is for DC step input; for AC, use phasor analysis.
Applications
The capacitor charging voltage follows an exponential curve V(t) = V₀(1 − e^(−t/RC)), asymptotically approaching the source voltage. This formula is used to design timing circuits, to ensure proper operation of switched‑capacitor circuits, and to control the startup of power supplies. Engineers use it to calculate the charge time for a capacitor, to determine the time to reach a threshold voltage, and to design analog‑to‑digital converters. It also applies in oscillators and pulse generators. By understanding the charging curve, professionals can predict the voltage at any time and ensure that circuits operate reliably. This exponential response is a classic example of first‑order system behaviour.
- Timing and delay circuit design
- Switched‑capacitor filter and converter design
- Soft‑start circuits for power supplies
- ADC sample‑and‑hold circuit analysis
- Educational demonstration of exponential charging
Frequently Asked Questions
The voltage across a charging capacitor is V(t) = V₀(1 − e^(−t/RC)), where V₀ is the source voltage.
As t → ∞, V(t) → V₀ (the source voltage).
V(τ) = V₀ (1 − e⁻¹) ≈ 0.632 V₀.
After 5τ, the capacitor voltage reaches about 99.3% of V₀.
I(t) = (V₀/R) e^(−t/τ). It decreases exponentially from V₀/R to zero.
If the capacitor has initial voltage V_i, the formula becomes V(t) = V₀ + (V_i − V₀) e^(−t/τ).
Larger R or C increases τ, slowing the charging process.
Timing circuits, switched-mode power supplies, and flash circuits in cameras.
t = −RC ln(1 − V/V₀).
Common errors include: 1) using the discharge formula instead, 2) forgetting the initial voltage, 3) using the wrong sign, 4) mixing up V and V₀, and 5) applying the formula to constant current charging.