The exact equation for capacitor discharge through a constant resistance is V(t) = V₀ · e^(-t/RC). This formula dictates how voltage decays exponentially over time when a charged capacitor is connected to a resistive load. Whether you are designing a power supply bleeder network, calculating the hold-up time for a microcontroller brownout, or simply trying to figure out how long to wait before touching a CRT flyback transformer, this single exponential relationship governs the physics of the discharge event.
The Core Equation for Capacitor Discharge
To understand the equation for capacitor discharge, we start with Kirchhoff's Current Law (KCL) at the node connecting the capacitor and resistor. The current leaving the capacitor must equal the current entering the resistor. The capacitor current is defined as I = C(dV/dt), and the resistor current is I = V/R. Because the capacitor is discharging, the current direction dictates that C(dV/dt) = -V/R. Separating variables and integrating both sides yields the natural logarithm of voltage, which exponentiates to the standard decay formula.
According to HyperPhysics, this derivation assumes a purely resistive circuit with no parasitic inductance. Below is the complete specification sheet for every symbol in the formula.
| Symbol | Parameter | Standard Unit | Description |
|---|---|---|---|
| V(t) | Instantaneous Voltage | Volts (V) | The voltage across the capacitor at a specific time t. |
| V₀ | Initial Voltage | Volts (V) | The starting voltage across the capacitor at t = 0. |
| e | Euler's Number | Dimensionless | Mathematical constant approximately equal to 2.71828. |
| t | Time | Seconds (s) | The elapsed time since the discharge path was closed. |
| R | Resistance | Ohms (Ω) | The total series resistance of the discharge path. |
| C | Capacitance | Farads (F) | The capacitance value of the component. |
| τ (tau) | Time Constant | Seconds (s) | The product of R and C (τ = RC). The time to reach 36.8% of V₀. |
Rearranged Forms: Solving for Time, Resistance, and Capacitance
In practical bench work, you rarely need to solve for V(t). Usually, you know your target safe voltage and need to size a resistor or calculate a wait time. Here are the algebraic rearrangements of the core formula. Note that solving for any variable inside the exponent requires the natural logarithm (ln), not the base-10 logarithm (log).
- Solving for Time (t):
t = -R · C · ln( V(t) / V₀ )
Use case: Calculating how long to wait before a SMPS filter cap is safe to touch. - Solving for Resistance (R):
R = -t / [ C · ln( V(t) / V₀ ) ]
Use case: Sizing a bleeder resistor to meet a specific safety discharge time limit. - Solving for Capacitance (C):
C = -t / [ R · ln( V(t) / V₀ ) ]
Use case: Sizing a hold-up capacitor to keep a microcontroller alive for 500ms during a brownout.
Worked Examples with Strict Unit Tracking
The most common reason discharge calculations fail on the bench is unit mismatch. The formula strictly requires base SI units: Farads, Ohms, and Seconds. Below are two solved problems demonstrating explicit unit tracking.
Problem 1: Finding Voltage at a Specific Time
Scenario: A 400V DC bus uses a 220 µF filter capacitor. A 100 kΩ bleeder resistor is permanently connected. What is the voltage exactly 5 seconds after unplugging the system?
- Convert to base units:
V₀ = 400 V
C = 220 µF = 220 × 10⁻⁶ F = 0.00022 F
R = 100 kΩ = 100,000 Ω
t = 5 s - Calculate the time constant (τ):
τ = R · C = 100,000 · 0.00022 = 22 seconds. - Apply the equation for capacitor discharge:
V(5) = 400 · e^(-5 / 22)
V(5) = 400 · e^(-0.22727)
V(5) = 400 · 0.7967 - Final Answer: 318.68 V. (The capacitor has barely discharged because 5 seconds is less than one-quarter of a single time constant).
Problem 2: Finding Time to Reach a Safe Voltage
Scenario: An offline AC/DC converter charges a 470 µF capacitor to 325V (peak of 230V AC mains). The bleeder resistor is 220 kΩ. How long until the voltage drops to a 'safe' 50V?
- Convert to base units:
V₀ = 325 V
V(t) = 50 V
C = 470 µF = 0.00047 F
R = 220 kΩ = 220,000 Ω - Calculate the time constant (τ):
τ = 220,000 · 0.00047 = 103.4 seconds. - Apply the rearranged time formula:
t = -103.4 · ln( 50 / 325 )
t = -103.4 · ln( 0.1538 )
t = -103.4 · (-1.872) - Final Answer: 193.5 seconds (approx. 3 minutes and 13 seconds).
Common Unit Mistakes and Realistic Magnitudes
When plugging values into a calculator, three specific errors will break your math entirely. Watch for these pitfalls:
| The Mistake | Why It Breaks | The Fix |
|---|---|---|
| Entering µF directly | Calculators do not auto-convert microfarads. Your time constant will be off by a factor of 1,000,000. | Always multiply µF by 10⁻⁶ (e.g., 1000 µF = 0.001 F). |
| Using log₁₀ instead of ln | The derivation relies on the natural logarithm. Base-10 log will yield a time or resistance value that is off by roughly 43%. | Use the 'ln' button on your scientific calculator, or multiply log₁₀ by 2.3026. |
| Ignoring ESR in high-current dumps | If discharging through a low-value resistor (e.g., 10 Ω), the capacitor's internal ESR adds to R, altering the curve. | Measure ESR and add it to your external R value for low-resistance discharge paths. |
Decision Path: Sizing a Bleeder Resistor for Safe Discharge
When designing a power supply, you must select a bleeder resistor that discharges the main filter capacitor to a safe voltage within a regulatory or practical time limit, without wasting excessive power during normal operation. Follow this decision tree to arrive at a concrete component selection.
| Condition / Constraint | Action / Calculation |
|---|---|
| Step 1: Define Limits | Set V₀ (max operating voltage), V(t) (safe target, usually 50V), and t (max discharge time, e.g., 60s). |
| Step 2: Calculate R | Use R = -t / [ C · ln( V(t) / V₀ ) ]. |
| Step 3: Calculate Steady-State Power | Use P = (V₀²) / R. This is the heat dissipated while the unit is plugged in. |
| Step 4: Check Power Rating | If P > 2W, standard 1/4W carbon film resistors will burn. You must select a wirewound or metal oxide power resistor rated for at least 2× the calculated P. |
| Step 5: Select Standard Value | Round the calculated R down to the nearest E12/E24 standard value to ensure the discharge time is slightly faster than calculated. |
Concrete Sizing Example
Requirements: Discharge a 1000 µF capacitor from 400V to <50V in under 60 seconds.
- Calculate R:
R = -60 / [ 0.001 · ln(50 / 400) ]
R = -60 / [ 0.001 · -2.0794 ]
R = 28,854 Ω - Calculate Power Dissipation:
P = 400² / 28,854 = 160,000 / 28,854 = 5.54 W - Select Standard Value & Wattage:
Rounding down to the nearest E24 value gives 27,000 Ω (27 kΩ).
Recalculating power for 27 kΩ: P = 160,000 / 27,000 = 5.92 W.
Applying a 2× safety margin for power rating requires a 10W or higher resistor.
The Final Pick: Do not use a generic cement block resistor if space is tight and reliability is critical. Select the Vishay AC10 series 27kΩ 10W wirewound resistor (or an equivalent metal oxide film like the Vishay PR03 series if pulse handling is a concern). Mount it with adequate airflow, and you have a mathematically verified, physically robust discharge path that guarantees the bus will drop below 50V in roughly 56 seconds.
