The exact formula for capacitor discharge is V(t) = V₀ · e^(-t/RC). This exponential decay equation dictates how voltage drops across a capacitor when it releases its stored energy through a resistive path. Whether you are sizing a bleeder resistor for a high-voltage power supply or calculating the hold-up time for a microcontroller brownout circuit, this formula is the mathematical foundation of your design.
Below, we break down the derivation, map out every variable, solve bench-ready problems with strict unit tracking, and examine a real-world failure where relying solely on this formula without considering thermal limits destroyed a component.
The Core Formula for Capacitor Discharge
When a charged capacitor is connected across a resistor, the current flow is highest at the very first instant and tapers off as the voltage drops. The voltage at any given time t is modeled by the following equation:
V(t) = V₀ · e^(-t / RC)
To use this formula correctly on the bench, you must understand the physical meaning and expected magnitude of every symbol. Here is the definitive spec sheet for the variables:
| Symbol | Definition | Standard Unit | Realistic Bench Magnitude |
|---|---|---|---|
| V(t) | Voltage across the capacitor at time t | Volts (V) | 0V to 400V (depending on supply) |
| V₀ | Initial voltage at t = 0 (fully charged state) | Volts (V) | 3.3V, 5V, 12V, 24V, 170V (peak AC) |
| t | Time elapsed since discharge began | Seconds (s) | Milliseconds to tens of seconds |
| R | Resistance of the discharge path (load or bleeder) | Ohms (Ω) | 10Ω to 1MΩ |
| C | Capacitance of the component | Farads (F) | 100pF to 10,000µF (0.01F) |
| e | Euler’s number (mathematical constant) | Dimensionless | ≈ 2.71828 |
| τ (tau) | Time constant, defined as R × C | Seconds (s) | Microseconds to seconds |
A critical rule of thumb for realistic answer magnitudes: after one time constant (1τ), the voltage drops to roughly 36.8% of V₀. After five time constants (5τ), the voltage drops to 0.67% of V₀, which is universally considered "fully discharged" in practical electrical engineering.
Rearranged Forms: Solving for Time, Resistance, and Capacitance
On the bench, you rarely just want to find V(t). Usually, you have a target safe voltage and need to size a bleeder resistor or calculate the required wait time before touching a circuit. By applying the natural logarithm (ln) to both sides, we can rearrange the formula for capacitor discharge to isolate any variable.
- Solving for time (t):
t = -R · C · ln(V(t) / V₀)
Use case: Calculating how long a high-voltage bus takes to drop to a safe 50V touch threshold. - Solving for resistance (R):
R = -t / (C · ln(V(t) / V₀))
Use case: Sizing a bleeder resistor to ensure a power supply discharges within 2 seconds of unplugging. - Solving for capacitance (C):
C = -t / (R · ln(V(t) / V₀))
Use case: Selecting a hold-up capacitor to keep a 3.3V microcontroller rail above 2.7V for 50ms during a brownout. - Solving for initial voltage (V₀):
V₀ = V(t) / e^(-t / RC)
Use case: Back-calculating the peak charge voltage based on a measured voltage at a known time delay.
Solved Problems with Strict Unit Tracking
The most common reason hobbyists and students get garbage results from this formula is failing to convert microfarads (µF) into base SI units (Farads). Let's walk through two bench scenarios with explicit unit tracking.
Problem 1: Finding Voltage at a Specific Time
Setup: You have a 10,000 µF filter capacitor charged to 24V DC. You connect a 470 Ω bleeder resistor across it. What is the voltage after 2 seconds?
- Convert units to SI base: C = 10,000 µF = 10,000 × 10⁻⁶ F = 0.01 F.
- Calculate the time constant (τ): τ = R × C = 470 Ω × 0.01 F = 4.7 seconds.
- Set up the formula: V(2) = 24 · e^(-2 / 4.7)
- Calculate the exponent: -2 / 4.7 = -0.4255
- Apply Euler's number: e^(-0.4255) ≈ 0.6534
- Final multiplication: 24V × 0.6534 = 15.68V
Result: After 2 seconds, the capacitor still holds a potentially hazardous 15.68V.
Problem 2: Finding Time to Reach a Safe Voltage
Setup: A motor controller bus uses a 4,700 µF capacitor charged to 50V. The discharge circuit has a 1 kΩ resistor. How long does it take to drop to a safe 5V?
- Convert units: C = 4,700 µF = 0.0047 F. R = 1 kΩ = 1,000 Ω.
- Select the rearranged formula for time: t = -R · C · ln(V(t) / V₀)
- Plug in the values: t = -(1,000) · (0.0047) · ln(5 / 50)
- Simplify the ratio: 5 / 50 = 0.1
- Calculate the natural log: ln(0.1) ≈ -2.3026 (Note: Ensure your calculator is set to 'ln', not 'log₁₀').
- Final multiplication: t = -4.7 · (-2.3026) = 10.82 seconds
Result: You must wait at least 10.82 seconds after power-off before the bus is safe to touch.
Real-World Scenario: The VFD Bleeder Resistor That Vaporized
The formula for capacitor discharge perfectly predicts voltage over time, but it is completely blind to instantaneous power dissipation. Ignoring this distinction is a classic bench mistake.
The Setup: While repairing a 400V DC bus on a small Variable Frequency Drive (VFD), I needed to replace the bleeder network. The bus capacitance was 470 µF, charged to 400V DC. To ensure a fast discharge, I installed a 10 kΩ, 1/2W carbon film resistor.
The Numbers: According to our formula, the time constant τ = 10,000 Ω × 0.00047 F = 4.7 seconds. The voltage would drop to a safe level in about 25 seconds (5τ). The math looked perfect.
The Outcome: The first time I cycled the VFD power off, a loud 'crack' echoed from the bench. The 1/2W carbon film resistor had literally split its ceramic casing open and failed open-circuit, leaving the 400V bus fully charged and highly lethal.
What Went Wrong: The discharge formula doesn't calculate energy. The total stored energy in the capacitor was E = ½CV² = 0.5 × 0.00047 × 400² = 37.6 Joules. Furthermore, the instantaneous peak power at t=0 was P = V²/R = 400² / 10,000 = 16 Watts. Dumping 37.6 Joules into a tiny carbon film element designed for a continuous 0.5W steady-state dissipation vastly exceeded its pulse thermal mass. The element overheated and fractured in milliseconds.
The Fix: I replaced the carbon film part with a Vishay CW005 series 5W wirewound ceramic resistor. Wirewound resistors have a much higher thermal mass and specific pulse-energy ratings (often capable of absorbing hundreds of joules for short durations). Alternatively, I could have increased the resistance to 100 kΩ to drop the peak power to 1.6W, though that would have increased the discharge time to over 4 minutes.
Assumptions, Limits, and Unit Mistakes That Break the Math
To use the formula for capacitor discharge reliably, you must understand the boundaries of the model and the traps that invalidate your calculations.
When the Formula Applies (and Its Assumptions)
- Purely Resistive Loads: The formula assumes R is constant. If you are discharging through a diode, a transistor, or a non-linear load (like an LED or a switching regulator), the resistance changes with voltage, and this simple exponential model fails. You must use differential equations or SPICE simulation.
- Ideal Capacitors: The formula ignores Equivalent Series Resistance (ESR). In reality, a high-ESR capacitor will discharge slightly faster in the first few microseconds due to internal I²R heating. It also ignores dielectric absorption, where a capacitor seemingly "recharges" itself slightly after being shorted, due to trapped charges in the dielectric relaxing back to the plates.
- Constant Temperature: Resistance values drift with heat. As a bleeder resistor heats up during discharge, its resistance changes (usually increasing for carbon/metal film, altering the decay curve).
Unit Mistakes That Guarantee Failure
- The Microfarad Trap: Plugging '4700' directly into C while R is in Ohms and t is in seconds will yield a time constant millions of times too large. You must convert µF to Farads (multiply by 10⁻⁶) or use matching sub-units (e.g., Megohms and Microfarads yield seconds directly).
- Log Base 10 vs. Natural Log: The inverse of the exponential function e^x is the natural logarithm ln(x) (base e). If you use the standard log(x) button on your calculator (base 10), your calculated time or resistance will be off by a factor of roughly 2.302.
- Ignoring Peak Power: As proven in the VFD scenario, calculating a valid time constant does not mean your chosen resistor will survive the physical reality of the energy transfer. Always verify the resistor's pulse-energy datasheet specifications alongside the RC time constant.
For deeper theoretical proofs regarding RC time constants and the derivation of Euler's number in circuit analysis, refer to the foundational tutorials at Electronics Tutorials and the DC circuit textbooks at All About Circuits. Mastering this formula—and knowing its physical limits—is what separates a textbook student from a reliable bench engineer.
