The discharge of a capacitor equation models the exponential decay of voltage across a capacitor when it is disconnected from a power source and allowed to release its stored energy through a resistive path. On the bench, this math dictates everything from sizing a bleeder resistor for IEC 62368-1 safety compliance to calculating the flash recycle time of a camera strobe. Below is the core formula, its algebraic rearrangements, and strict unit-tracked examples to prevent the most common calculation errors.

The Core Discharge of a Capacitor Equation and Symbol Definitions

The fundamental equation describing the voltage across a discharging capacitor at any given time is:

V(t) = V₀ · e-t / (R · C)

This formula assumes an ideal lumped-parameter circuit: the capacitance C is constant, the resistance R is purely ohmic and unchanging, and parasitic elements like equivalent series resistance (ESR) or parasitic inductance are negligible compared to the bulk discharge path. It applies strictly to transient DC decay, starting from an initial charged state at t = 0.

Symbol Standard Unit Definition & Bench Context
V(t) Volts (V) The instantaneous voltage across the capacitor at time t.
V₀ Volts (V) The initial voltage across the capacitor at the exact moment discharge begins (t = 0).
e Dimensionless Euler's number (~2.71828), the base of the natural logarithm.
t Seconds (s) The elapsed time since the discharge path was established.
R Ohms (Ω) The total resistance of the discharge path (includes load and bleeder resistors).
C Farads (F) The nominal capacitance. (Note: Bench multimeters often read in μF or nF; conversion is mandatory).
τ (tau) Seconds (s) The RC time constant (τ = R · C). Represents the time required for the voltage to drop to ~36.8% of V₀.

For deeper theoretical backing on the derivation of this exponential decay from Kirchhoff's Voltage Law, refer to the Georgia State University HyperPhysics RC circuit module.

Rearranged Forms: Solving for Time, Resistance, and Capacitance

On the jobsite or at the design bench, you rarely need to solve for V(t). Usually, you have a target safe touch voltage and need to size a bleeder resistor (R), or you have a fixed RC network and need to know the wait time (t) before probing the board. Here are the algebraic rearrangements of the core equation:

  • Solve for Time (t):
    t = -R · C · ln(V(t) / V₀)
  • Solve for Resistance (R):
    R = -t / (C · ln(V(t) / V₀))
  • Solve for Capacitance (C):
    C = -t / (R · ln(V(t) / V₀))
  • Solve for Initial Voltage (V₀):
    V₀ = V(t) · et / (R · C)

Note: ln represents the natural logarithm (base e), not the base-10 logarithm (log). Using the wrong log function on your calculator is a frequent source of error.

Worked Examples with Strict Unit Tracking

The most common reason engineers and hobbyists get wildly incorrect answers is failing to convert prefixes (micro, kilo, milli) to base SI units before calculating. The following examples enforce strict unit tracking.

Example 1: Calculating Voltage at a Specific Time

Scenario: A 50V DC rail powers a microcontroller. The bulk decoupling capacitor is 470 μF. When power is cut, it discharges through a 10 kΩ pull-down resistor. What is the voltage across the capacitor exactly 2 seconds after power-off?

  1. Identify and convert variables to base SI units:
    V₀ = 50 V
    C = 470 μF = 470 × 10⁻⁶ F = 0.00047 F
    R = 10 kΩ = 10 × 10³ Ω = 10,000 Ω
    t = 2 s
  2. Calculate the time constant (τ):
    τ = R · C = 10,000 Ω · 0.00047 F = 4.7 s
  3. Substitute into the core equation:
    V(2) = 50 · e-2 / 4.7
    V(2) = 50 · e-0.4255
  4. Evaluate the exponent and multiply:
    e-0.4255 ≈ 0.6534
    V(2) = 50 · 0.6534 = 32.67 V

Result: After 2 seconds, the capacitor retains 32.67 V.

Example 2: Sizing a Bleeder Resistor for Safety Compliance

Scenario: You are designing a 400V DC power supply using a 2200 μF bulk filter capacitor. IEC 62368-1 safety standards generally require accessible parts to discharge to a safe touch voltage (often defined as 60V DC) within a specific timeframe. If your design mandates a 45-second discharge time to 60V, what bleeder resistor value must you install?

  1. Identify and convert variables:
    V₀ = 400 V
    V(t) = 60 V
    C = 2200 μF = 0.0022 F
    t = 45 s
  2. Select the rearranged formula for Resistance:
    R = -t / (C · ln(V(t) / V₀))
  3. Calculate the natural log term:
    ln(60 / 400) = ln(0.15) ≈ -1.8971
  4. Substitute and solve for R:
    R = -45 / (0.0022 · -1.8971)
    R = -45 / -0.0041736
    R ≈ 10,782 Ω

Result: You need a bleeder resistor of 10,782 Ω or less. In practice, you would select the next standard E24 value down, such as a 10 kΩ resistor, to ensure the discharge happens slightly faster than the 45-second limit. (For further reading on practical RC timing and safety bleeders, see the All About Circuits RC Time Constants chapter).

Real-World Magnitudes and Unit Mistakes That Break the Math

Understanding what a "realistic" answer looks like is your best defense against calculator typos. In consumer electronics and power supplies, RC time constants (τ) typically range from 1 to 30 seconds for bleeder networks, and microseconds to milliseconds for snubber circuits or signal filtering. If you calculate a discharge time of 40,000 seconds (11 hours) for a standard PCB bleeder network, you have almost certainly made a unit conversion error.

⚠️ The Unit Multiplier Trap:
The most fatal mistake is plugging component values directly from the schematic into the calculator without converting to base units. If you multiply 470 (μF) by 10 (kΩ), your calculator yields 4700. You might assume this is 4700 seconds. In reality, 470 × 10⁻⁶ multiplied by 10 × 10³ equals 4.7 seconds. Always write out the scientific notation (10⁻⁶ for micro, 10⁻⁹ for nano, 10³ for kilo) during the intermediate steps.

Frequently Asked Questions

How long does it take for a capacitor to fully discharge in practice?

Mathematically, the exponential curve approaches zero asymptotically and never truly reaches 0V. In engineering practice, we use the 5τ (five time constants) rule. After 5 time constants, the capacitor has discharged to e⁻⁵, or roughly 0.67% of its initial voltage. For almost all bench and safety purposes, a capacitor is considered "fully discharged" at t = 5τ. If τ = 10s, the cap is effectively dead at 50 seconds.

Why does my measured discharge curve deviate from the theoretical equation?

If your oscilloscope trace doesn't perfectly match the calculated V(t), you are encountering real-world parasitics. The two primary culprits are Equivalent Series Resistance (ESR), which causes an immediate, instantaneous voltage drop the moment the discharge path is closed (visible as a vertical step on the scope), and dielectric absorption (soakage). Dielectric absorption causes the capacitor to "rebound" slightly after being shorted, as trapped charges in the dielectric material slowly release, making the tail end of the discharge curve decay much slower than the ideal equation predicts.

Does the discharge of a capacitor equation apply to AC or pulsed DC circuits?

No. The equation V(t) = V₀ · e-t/RC is strictly a transient DC model for a capacitor releasing stored energy into a passive resistor with no active driving source. If the capacitor is connected to an AC source or a continuous PWM signal, you must use AC steady-state analysis (calculating capacitive reactance Xc = 1 / (2πfC)) and complex impedance, or solve the differential equation with a forced function for the pulsed waveform. The simple exponential decay only applies when the driving source is removed or switched to 0V.