When a charged capacitor is disconnected from its power source and placed across a resistive load, the voltage does not drop linearly. It decays exponentially. Whether you are designing a safe bleeder network for a 400V DC bus in a motor drive, sizing a timing capacitor for a 555 monostable circuit, or analyzing a defibrillator discharge pulse, the capacitance discharge formula is the governing law. This guide strips away the abstract calculus, provides the rearranged algebraic forms you actually need at the bench, and walks through strict unit-tracked examples to prevent the most common calculation errors.

The Core Capacitance Discharge Formula & Symbol Definitions

The voltage across a discharging capacitor at any given time is modeled by the following exponential decay equation:

V(t) = V0 × e-t / RC

Similarly, the discharge current follows the exact same decay curve, starting from an initial peak of I0 = V0 / R:

I(t) = I0 × e-t / RC

Symbol Definition Standard SI Unit
V(t) Voltage across the capacitor at time t Volts (V)
V0 Initial voltage at the start of discharge (t = 0) Volts (V)
I(t) Current flowing through the resistor at time t Amperes (A)
I0 Initial peak discharge current (V0 / R) Amperes (A)
t Elapsed time since the discharge began Seconds (s)
R Resistance of the discharge path Ohms (Ω)
C Capacitance of the component Farads (F)
e Euler's number (mathematical constant ≈ 2.71828) Dimensionless

The product of R and C is known as the time constant, denoted by the Greek letter tau (τ). One time constant (τ = R × C) represents the time it takes for the voltage to drop to roughly 36.8% of its initial value. According to Georgia State University's HyperPhysics, after 5 time constants (5τ), the capacitor is considered fully discharged for all practical engineering purposes, having reached less than 1% of V0.

Rearranged Forms for Bench and Design Work

In practical design, you rarely solve for V(t). You usually know your target voltage and time, and need to find the required resistor or capacitor. Here are the algebraically rearranged forms solving for each variable:

  • Solving for time (t): t = -R × C × ln(V(t) / V0)
  • Solving for resistance (R): R = -t / (C × ln(V(t) / V0))
  • Solving for capacitance (C): C = -t / (R × ln(V(t) / V0))
  • Solving for initial voltage (V0): V0 = V(t) / e-t / RC
Bench Tip: The natural logarithm (ln) of a fraction less than 1 is always negative. This negative sign cancels out the negative sign in front of the rearranged equations for t, R, and C, yielding a positive, physically meaningful result.

When the Formula Applies (and When It Breaks)

The exponential decay model is highly accurate for standard RC networks, but it relies on strict assumptions. Understanding these boundaries prevents catastrophic design failures.

Core Assumptions

  1. Ideal Components: The formula assumes a purely resistive path and a purely capacitive load. It ignores Equivalent Series Resistance (ESR) and Equivalent Series Inductance (ESL). For high-speed snubber circuits (nanosecond discharge times), ESL dominates and the discharge becomes an underdamped LC ring, not an RC decay.
  2. Constant Resistance: R must remain static. If you are discharging through a non-linear load (like an LED or a semiconductor junction), the resistance changes with voltage, and this formula will yield incorrect times.
  3. No Dielectric Absorption: Large electrolytic and high-K ceramic capacitors suffer from dielectric absorption (soakage). If you short a 400V electrolytic capacitor to 0V and then remove the short, the voltage will "rebound" to 10V-20V minutes later as trapped charges in the dielectric relax. The formula predicts 0V, but reality requires active bleed circuits or repeated shorting for high-reliability safety.

The Unit Trap: Mistakes That Break the Math

The most common reason engineers get wildly incorrect results is failing to convert units to base SI before calculating.
Wrong: Plugging 470 μF in as "470" and 220 kΩ in as "220".
Right: Converting to Farads (470 × 10-6 F) and Ohms (220,000 Ω).
If you skip the micro-to-base conversion, your calculated time constant will be off by a factor of one million. Always write out the unit conversions explicitly in your scratchpad.

Realistic Answer Magnitudes

To sanity-check your work, compare your calculated τ against typical benchmarks:
• Signal timing / Snubbers: τ is in microseconds (μs) to milliseconds (ms).
• Power supply bulk filtering: τ is in seconds (e.g., 10s to 60s).
• Supercapacitors / Motor run caps: τ can be hours. A 1F supercap discharging through a 10kΩ resistor has a τ of 10,000 seconds (2.7 hours).

Worked Examples with Strict Unit Tracking

Let us apply the rearranged formulas to two common scenarios, tracking every unit conversion to ensure precision.

Problem 1: Calculating Time to Safe Voltage

Scenario: A 400V DC bus uses a 100 μF bulk capacitor. A 220 kΩ bleeder resistor is permanently soldered across the terminals. How long does it take for the capacitor to discharge to a safe-to-touch threshold of 50V?

  1. Identify and convert variables:
    V0 = 400 V
    V(t) = 50 V
    C = 100 μF = 100 × 10-6 F = 0.0001 F
    R = 220 kΩ = 220,000 Ω
  2. Calculate the time constant (τ):
    τ = R × C
    τ = 220,000 Ω × 0.0001 F = 22 seconds.
  3. Apply the time formula:
    t = -τ × ln(V(t) / V0)
    t = -22 s × ln(50 V / 400 V)
    t = -22 s × ln(0.125)
    t = -22 s × (-2.07944)
    t = 45.75 seconds

Result: It takes approximately 45.8 seconds for the bus to reach 50V. For a complete discharge to <1% (5τ), it would take 110 seconds.

Problem 2: Finding Unknown Capacitance from Bench Data

Scenario: You have an unmarked electrolytic capacitor. You charge it to 12V, then disconnect the supply and connect a known 10 kΩ resistor across it. You measure the voltage with a multimeter and observe it takes exactly 4.5 seconds to drop from 12V to 3V. What is the capacitance?

  1. Identify and convert variables:
    V0 = 12 V
    V(t) = 3 V
    t = 4.5 s
    R = 10 kΩ = 10,000 Ω
  2. Apply the capacitance formula:
    C = -t / (R × ln(V(t) / V0))
    C = -4.5 s / (10,000 Ω × ln(3 V / 12 V))
    C = -4.5 / (10,000 × ln(0.25))
    C = -4.5 / (10,000 × -1.38629)
    C = -4.5 / -13862.9
    C = 0.0003246 F
  3. Convert to standard engineering units:
    0.0003246 F × 106 = 324.6 μF

Result: The capacitor is approximately 330 μF (the nearest standard E6 value), likely a 330 μF electrolytic that has degraded slightly over time, which is typical for older aluminum electrolytics as noted in Electronics Tutorials.

Decision Path: Sizing a Bleeder Resistor for a 400V DC Bus

Designing a bleeder network is not just about calculating resistance; it requires balancing discharge time against continuous power dissipation. Use this decision matrix to arrive at a concrete component selection.

Design Parameter If Condition... Then Action...
Discharge Time Regulatory safety requires < 60 seconds to < 50V. Calculate max R using the rearranged formula. Round UP to nearest standard value.
Steady-State Power Resistor is permanently across the live DC bus. Calculate continuous power (P = V02 / R). Apply a 50% derating factor for enclosed PCB environments.
Surge/Pulse Energy Power supply is switched off and on rapidly. Verify the resistor's surge energy rating (Joules) exceeds the total stored energy of the cap (E = 0.5 × C × V2).
Component Topology Calculated wattage exceeds 3W or voltage exceeds 250V. Use two or more resistors in series to divide voltage stress and distribute heat.

Concrete Sizing Execution

Target Specs: 400V DC bus, 470 μF capacitance. Requirement: Discharge to < 50V in under 60 seconds.

  1. Calculate Required Resistance:
    R = -t / (C × ln(V(t) / V0))
    R = -60 / (0.00047 × ln(50 / 400))
    R = -60 / (0.00047 × -2.07944)
    R = 61,391 Ω
    Decision: Select the next highest standard E24 value to guarantee the time is under 60s. Pick R = 62 kΩ.
  2. Calculate Continuous Power Dissipation:
    P = V2 / R = (400)2 / 62,000 = 160,000 / 62,000 = 2.58 W.
  3. Apply Derating:
    For reliability inside an enclosed power supply chassis, resistors should not run above 50% of their rated power.
    Required Rating = 2.58 W × 2 = 5.16 W.
  4. Verify Voltage Rating:
    Standard 5W through-hole resistors often have a maximum working voltage limit of 350V. Since our bus is 400V, a single resistor risks internal arcing.
    Decision: Split the 62 kΩ into two 31 kΩ (nearest standard 30 kΩ or 33 kΩ) resistors in series. Let's use two 33 kΩ resistors in series (Total R = 66 kΩ). Recalculating time with 66kΩ: t = 54.2 seconds. Still passes the <60s requirement.
Final Concrete Pick: Use two 33 kΩ, 3W, 5% tolerance metal oxide film resistors in series (e.g., Vishay PR03 series or Yageo FMP300 series). Metal oxide is preferred over carbon composition for high-voltage bleeders due to superior long-term stability and pulse-withstand capabilities. Mount them with a 5mm air gap from the PCB to aid convection cooling.