Mastering the Voltage Equation of a Capacitor: Derivation, Worked Examples, and Sizing

The voltage equation of a capacitor defines the exact relationship between accumulated electrical charge, the physical capacitance value, and the resulting electrical potential across its dielectric. Whether you are sizing a supercapacitor for a 3-second RTC hold-up or calculating the ripple on a 500kHz buck converter, the math starts with one fundamental integral. Below is the direct derivation, the rearranged forms you actually need at the bench, and a decision matrix to move from abstract math to a concrete DigiKey part number.

The Core Voltage Equation of a Capacitor and Its Assumptions

The most general time-domain expression for the voltage across a capacitor is the integral of the current flowing through it over time, plus any initial voltage:

V(t) = V₀ + (1/C) ∫₀^t i(τ) dτ

For the vast majority of DC power supply and timing applications, we assume a constant current (I) over a specific time window (Δt). This simplifies the integral into the algebraic form used for 90% of bench calculations:

ΔV = (I · Δt) / C

Rearranged Forms for Circuit Design

You rarely solve for V(t) in isolation. Usually, you have a voltage constraint (like a maximum allowable ripple) and need to find the component value or the time limit. Here are the rearranged forms for the constant-current linear model:

• Solving for Capacitance (Sizing): C = (I · Δt) / ΔV
• Solving for Time (Hold-up / Timing): Δt = (C · ΔV) / I
• Solving for Current (Inrush / Charge Rate): I = (C · ΔV) / Δt
• Solving for Voltage Drop (Ripple / Sag): ΔV = (I · Δt) / C

For RC timing circuits where current is not constant (exponential decay/charge), the voltage equation shifts to: V(t) = V_s(1 - e^-t/RC), where V_s is the source voltage and R is the series resistance.

Worked Examples with Strict Unit Tracking

The most common point of failure in capacitor math is unit misalignment. Below are two solved problems tracking every unit conversion.

Problem 1: Constant Current Supercapacitor Sizing

Scenario: You are designing a backup power circuit for an ESP32 RTC. A constant current charging source supplies 50 mA to a supercapacitor. You need the capacitor to charge from 2.0V to 4.5V in exactly 120 seconds. What is the minimum capacitance required?

Step-by-Step Solution:

1. Identify knowns and convert to SI base units:
   I = 50 mA = 0.050 A
   Δt = 120 s
   V₀ = 2.0 V, V_final = 4.5 V → ΔV = 4.5 V - 2.0 V = 2.5 V
2. Select the rearranged formula:
   C = (I · Δt) / ΔV
3. Substitute and track units:
   C = (0.050 [A] · 120 [s]) / 2.5 [V]
   C = 6.0 [A·s] / 2.5 [V]
4. Calculate final value:
   C = 2.4 Farads (F)

Bench Reality Check: A 2.4F supercapacitor is physically large (roughly the size of a D-cell battery). You would select a standard 5.0V, 4.7F radial supercapacitor (like the Eaton/Vishay series) to provide a safety margin for dielectric absorption and aging.

Problem 2: Exponential RC Timing Circuit

Scenario: A 555 timer monostable circuit uses a 100 kΩ resistor and a 47 μF capacitor. The supply is 9V. What is the voltage across the capacitor exactly 3 seconds after the trigger pulse begins?

Step-by-Step Solution:

1. Convert to SI base units:
   R = 100 kΩ = 100,000 Ω
   C = 47 μF = 47 × 10^-6 F
   V_s = 9 V, t = 3 s
2. Calculate the RC time constant (τ):
   τ = R · C = 100,000 [Ω] · 47 × 10^-6 [F] = 4.7 seconds
3. Apply the exponential voltage equation:
   V(t) = V_s(1 - e^-t/τ)
   V(3) = 9 · (1 - e^{-3 / 4.7})
4. Compute the exponent and final voltage:
   -3 / 4.7 ≈ -0.6383
   e^-0.6383 ≈ 0.5282
   V(3) = 9 · (1 - 0.5282) = 9 · 0.4718 = 4.246 V

At 3 seconds, the capacitor has charged to approximately 4.25V. For deeper theory on RC transient responses, refer to the foundational texts at All About Circuits.

Common Unit Traps and Realistic Magnitudes

If your calculator spits out a wildly incorrect number, you likely fell into one of these unit traps:

• The Microfarad Trap: Forgetting to multiply μF by 10^-6. If you type '47' instead of '0.000047' into the C variable, your calculated time constant will be off by a factor of one million.
• The Milliamp Trap: Mixing mA and A. 100 mA is 0.1 A. If you use 100 in the numerator, your required capacitance will be calculated as 1000x larger than reality.
• The Millisecond Trap: Switching power supplies operate in μs or ns. If your switching frequency is 500kHz, the period is 2 μs (0.000002 s), not 2 ms.

What does a realistic answer magnitude look like?

• Logic Rail Decoupling (Ripple ΔV): Should be in the millivolt (mV) range (e.g., 10mV - 50mV). If your equation yields 2V of ripple on a 3.3V rail, your capacitor is drastically undersized or your ESR is too high.
• Audio Coupling / DC Blocking: Signal swings are typically 0.5V to 2V RMS.
• Mains Snubbers / Motor Run: Steady-state voltages are 120V to 480V AC. (Always use the peak voltage, √2 × V_RMS, when checking dielectric breakdown limits).

Decision Path: Sizing and Selecting Your Capacitor

Math gives you the ideal value; physics dictates the actual component. Use this decision tree to translate your calculated ΔV, C, and frequency requirements into a specific capacitor chemistry and part number. For comprehensive component behavior across different dielectrics, consult the Electronics Tutorials capacitor guide.