The capacitor decay formula for voltage in a discharging RC (resistor-capacitor) circuit is V(t) = V0 × e-t / (RC). This exponential equation dictates exactly how quickly a capacitor sheds its stored electrical 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 bedrock of transient DC analysis.
Below, we break down every variable, algebraically rearrange the equation for bench troubleshooting, and walk through two rigorously unit-tracked solved problems.
The Capacitor Decay Formula and Symbol Definitions
When a charged capacitor is connected across a resistor, the voltage across its terminals does not drop linearly. It decays exponentially. The rate of this decay is governed by the interplay between the capacitance (energy storage) and the resistance (energy dissipation). The primary voltage decay equation is:
V(t) = V0 × e-t / (RC)
Similarly, the discharge current follows the exact same exponential curve, defined as I(t) = I0 × e-t / (RC), where the initial current I0 = V0 / R.
Every symbol in this formula must map to a strict SI unit for the math to hold. Here is the definitive spec sheet for the variables:
| Symbol | Name | Standard SI Unit | Definition & Bench Context |
|---|---|---|---|
| V(t) | Instantaneous Voltage | Volts (V) | The voltage across the capacitor at a specific time t during discharge. |
| V0 | Initial Voltage | Volts (V) | The voltage across the capacitor at t = 0, right before the discharge path is closed. |
| t | Time | Seconds (s) | The elapsed time since the discharge began. |
| R | Resistance | Ohms (Ω) | The total resistance in the discharge path, including the explicit resistor and any parasitic trace/wire resistance. |
| C | Capacitance | Farads (F) | The nominal capacitance value. (Note: Real-world capacitance drops under DC bias in Class II ceramics). |
| e | Euler's Number | Dimensionless | The base of the natural logarithm, approximately 2.71828. It defines the continuous decay rate. |
| τ (tau) | Time Constant | Seconds (s) | Defined as R × C. The time required for the voltage to decay to ~36.8% of V0. |
Rearranged Forms: Solving for Time, Resistance, and Capacitance
On the bench, you rarely just solve for V(t). More often, you know your target voltage and need to size a bleeder resistor (R), or you are characterizing an unknown capacitor (C) by measuring its discharge time (t). By applying the natural logarithm (ln) to both sides, we can isolate any variable.
Here are the algebraically rearranged forms of the capacitor decay formula:
- Solving for Time (t):
t = -R × C × ln( V(t) / V0 )
Use when: Calculating how long a power supply takes to reach a safe-to-touch voltage after unplugging. - Solving for Resistance (R):
R = -t / [ C × ln( V(t) / V0 ) ]
Use when: Sizing a bleeder resistor to ensure a capacitor discharges within a specific safety window. - Solving for Capacitance (C):
C = -t / [ R × ln( V(t) / V0 ) ]
Use when: Verifying the actual capacitance of an aging electrolytic capacitor using a known resistor and a stopwatch. - Solving for Initial Voltage (V0):
V0 = V(t) / e-t / (RC)
Use when: Forensic analysis to determine what voltage a circuit was operating at before a power loss event.
Worked Examples with Strict Unit Tracking
The most common reason hobbyists and students get wildly incorrect answers is failing to convert microfarads and milliseconds into base SI units. Let's walk through two problems with rigorous unit tracking.
Problem 1: Calculating Discharge Time for a Safe Touch Voltage
Scenario: A 470 μF electrolytic capacitor in a linear power supply is charged to 24V DC. It discharges through a 2.2 kΩ bleeder resistor when the mains power is cut. How long does it take for the voltage to drop to 5V (a generally safe threshold)?
- Convert to base SI units:
C = 470 μF = 470 × 10-6 F
R = 2.2 kΩ = 2,200 Ω
V0 = 24 V
V(t) = 5 V - Calculate the time constant (τ):
τ = R × C = 2,200 Ω × (470 × 10-6 F) = 1.034 seconds. - Calculate the natural log ratio:
ln( V(t) / V0 ) = ln( 5 / 24 ) = ln( 0.20833 ) ≈ -1.5686 - Apply the rearranged time formula:
t = -τ × ln( V(t) / V0 )
t = -(1.034 s) × (-1.5686)
t = 1.622 seconds
Reality Check: 5V is roughly 20% of 24V. We know that at 1τ (1.03s) voltage is at 36.8%, and at 2τ (2.06s) it is at 13.5%. An answer of 1.62 seconds sits perfectly between 1τ and 2τ, confirming our magnitude is realistic.
Problem 2: Finding Unknown Capacitance via Discharge Testing
Scenario: You are testing a salvaged capacitor of unknown value. You charge it to 12V and discharge it through a precision 10 kΩ resistor. Your oscilloscope shows the voltage dropping to 1.5V in exactly 450 milliseconds. What is the capacitance?
- Convert to base SI units:
R = 10 kΩ = 10,000 Ω
t = 450 ms = 0.450 seconds
V0 = 12 V
V(t) = 1.5 V - Calculate the natural log ratio:
ln( V(t) / V0 ) = ln( 1.5 / 12 ) = ln( 0.125 ) ≈ -2.0794 - Apply the rearranged capacitance formula:
C = -t / [ R × ln( V(t) / V0 ) ]
C = -0.450 / [ 10,000 × (-2.0794) ]
C = -0.450 / -20,794
C = 0.00002164 Farads - Convert back to standard component units:
C ≈ 21.6 μF
Reality Check: 1.5V is exactly 12.5% of 12V. This represents exactly two time constants (since 0.3682 ≈ 0.135, close to 12.5% - actually, 1.5V is exactly 1/8th, which is 3 half-lives, or roughly 2.08τ). If 2.08τ = 0.45s, then 1τ ≈ 0.216s. Since τ = RC, C = 0.216 / 10,000 = 21.6 μF. The math aligns perfectly.
Assumptions, Edge Cases, and Unit Traps
The capacitor decay formula is an idealized mathematical model. According to standard circuit theory outlined by resources like Georgia State University's HyperPhysics, the formula relies on several assumptions that break down in extreme real-world scenarios.
When the Formula Applies (and When it Doesn't)
- Ideal Components: The formula assumes the capacitor has zero Equivalent Series Resistance (ESR) and zero leakage current. In reality, large electrolytic capacitors have ESR (often 10mΩ to 100mΩ), which slightly alters the initial rapid discharge spike.
- Constant Parameters: It assumes R and C remain constant. However, Class II ceramic capacitors (like X7R or Y5V) suffer from severe DC bias effects, where their actual capacitance can drop by 50% or more at rated voltage. As the voltage decays, the capacitance actually increases, making the real-world decay curve slightly non-exponential.
- Purely Resistive Path: If your discharge path includes parasitic inductance (like long test leads or wirewound resistors), the circuit becomes an RLC circuit, and the decay will ring or oscillate rather than following a smooth exponential curve.
Unit Mistakes That Break the Math
The absolute most common error on the workbench is plugging component values directly from the schematic into the formula without converting them. If you type "470" for a 470 μF capacitor, your calculated time will be off by a factor of one million. Always convert microfarads (μF) to Farads (multiply by 10-6), milliseconds to seconds (multiply by 10-3), and kilo-ohms to Ohms (multiply by 103) before touching the natural log button on your calculator.
Realistic Answer Magnitudes
Use the "Rule of 5 Tau" to sanity-check your answers. In RC circuits, 5τ is universally considered the time it takes for a capacitor to fully discharge (dropping to 0.67% of V0). If you are calculating the discharge time for a macro-scale power supply bleeder circuit and your answer comes out to 0.0004 seconds, you likely forgot to convert microfarads to Farads. Conversely, if a tiny 100nF decoupling capacitor takes 4 hours to discharge on your spreadsheet, you misplaced a decimal point in your resistance conversion.
Frequently Asked Questions
How do you calculate the capacitor decay time constant?
The time constant, denoted by the Greek letter tau (τ), is calculated by simply multiplying the resistance in Ohms by the capacitance in Farads (τ = R × C). The resulting unit is seconds. One time constant represents the time it takes for the capacitor's voltage to decay to exactly 1/e (approximately 36.8%) of its initial starting voltage. It is a quick benchmark used by engineers to estimate circuit timing without running the full exponential equation.
Why does the capacitor decay formula use Euler's number (e)?
Euler's number (e ≈ 2.71828) is the mathematical constant that describes continuous, proportional decay. In a discharging capacitor, the rate of voltage drop (the current) is directly proportional to the voltage currently present across the capacitor (Ohm's Law: I = V/R). Because the voltage drops, the current drops, which in turn slows down the rate of further voltage drop. The only mathematical function whose rate of change is proportional to its own value is the exponential function based on e. For a deeper dive into the calculus behind this, Electronics Tutorials provides an excellent breakdown of the differential equations involved.
Does the capacitor decay formula apply to AC circuits?
No, the standard V(t) = V0e-t/RC formula applies strictly to DC transient events (like a switch opening or closing). In continuous AC circuits, capacitors introduce capacitive reactance (XC = 1 / 2πfC), which causes a phase shift between voltage and current. While the fundamental physics of charge and discharge still occur during every AC half-cycle, analyzing AC RC circuits requires complex numbers, phasor diagrams, and impedance (Z) calculations rather than the simple DC exponential decay formula.
What happens to the decay formula if the capacitor has high ESR?
If a capacitor has high Equivalent Series Resistance (ESR)—common in aged or cheap electrolytic capacitors—the total resistance in the discharge loop is actually R_total = R_external + ESR. For a slow discharge through a large bleeder resistor (e.g., 10 kΩ), a 0.5Ω ESR is mathematically negligible. However, if you are analyzing a rapid, low-resistance discharge (like a short-circuit fault or a camera flash dump where R_external is near zero), the ESR dominates. In that edge case, the effective R in the formula becomes the ESR itself, and the capacitor will dissipate significant internal heat (I²R losses), potentially causing venting or catastrophic failure.
