The Differential Equation for Capacitor Circuits: Derivation and Worked Examples

The fundamental differential equation for capacitor behavior is i(t) = C * (dv(t) / dt). This defines the instantaneous current i(t) through a capacitor as the product of its capacitance C and the rate of change of voltage dv(t)/dt across its terminals. If the voltage across a capacitor is not changing, the current is zero; the component only passes current when the electric field between its plates is expanding or collapsing.

The Core Differential Equation and Symbol Definitions

To understand where this comes from, we start with the basic definition of capacitance: charge equals capacitance times voltage (q = C * v). Current is defined as the rate of flow of charge over time (i = dq / dt). By taking the derivative of the charge equation with respect to time, and assuming capacitance C is a constant value, we arrive at the governing differential equation for capacitor dynamics.

i(t) = C * (dv(t) / dt)

Symbol	Parameter	SI Unit	Practical Bench Unit
i(t)	Instantaneous current	Amperes (A)	Milliamps (mA)
v(t)	Instantaneous voltage	Volts (V)	Volts (V)
C	Capacitance	Farads (F)	Microfarads (µF)
t	Time	Seconds (s)	Milliseconds (ms)
dv/dt	Voltage rate of change	Volts per second (V/s)	Volts per microsecond (V/µs)

This formula assumes an ideal capacitor. As noted in standard circuit theory resources like All About Circuits, real-world components introduce parasitics that alter this behavior at high frequencies or high currents.

Real-World Component Limits and Parasitics

The differential equation applies perfectly only when we assume the dielectric is ideal, the plates have zero resistance, and the physical geometry doesn't change. In practice, every capacitor has Equivalent Series Resistance (ESR), Equivalent Series Inductance (ESL), and dielectric absorption. When dv/dt is extremely high (like in a switching power supply or a snubber circuit), the ESL term (v = L * di/dt) begins to dominate, and the simple capacitor equation no longer predicts the terminal voltage accurately.

Furthermore, physical capacitors have maximum dv/dt limits. Exceeding these causes internal heating, dielectric breakdown, or metallization vaporization. Below is a data-dense reference for common capacitor types you will encounter on the bench:

Capacitor Type	Typical Capacitance	Max dv/dt Rating	Dominant Parasitic	Common Application
MLCC (X7R Ceramic)	100 pF – 100 µF	> 100 V/µs	ESL / Piezoelectric effect	High-freq decoupling, RF filters
Aluminum Electrolytic	1 µF – 10,000 µF	~ 1 V/µs	High ESR	Bulk filtering, linear PSU smoothing
Polypropylene Film	1 nF – 100 µF	10 – 1,000 V/µs	Dielectric absorption	Snubbers, motor run, audio crossovers
EDLC Supercapacitor	0.1 F – 3,000 F	< 0.001 V/µs	High ESR / Leakage current	Energy storage, memory backup, regenerative braking

When selecting a component for a high-speed switching node, always check the manufacturer's datasheet for the maximum allowable dv/dt. As Analog Devices highlights in their capacitor selection guides, ignoring this limit in film or ceramic capacitors leads to premature failure due to internal arcing.

Rearranged Forms and Unit Pitfalls

Depending on what you are solving for, you will need to rearrange the core equation. Here are the most useful forms for bench work and circuit design:

• Solving for Capacitance: C = i(t) / (dv(t) / dt) — Use this when sizing a bulk capacitor to limit voltage droop under a known transient load.
• Solving for Voltage Slew Rate: dv(t) / dt = i(t) / C — Use this to find how fast a node will charge given a specific drive current.
• Solving for Time Increment: dt = C * dv(t) / i(t) — Use this to calculate the delay or timing interval in an RC oscillator.
• Integral Form (Solving for Voltage): v(t) = (1 / C) * ∫ i(t) dt + v(0) — Use this when the current is a known waveform (like a sine wave or a constant pulse) and you need the resulting voltage. v(0) is the initial voltage at t=0.

Worked Example 1: Current from a Linear Voltage Ramp

Scenario: You are designing a signal conditioning circuit. A function generator outputs a linear voltage ramp from 0V to 5V over a period of 2 milliseconds (ms). This signal is fed through a 10 µF coupling capacitor. What is the peak current drawn from the function generator during this ramp?

Step-by-Step Solution:

1. Identify and convert knowns to SI units:
   Capacitance C = 10 µF = 10 × 10^-6 F
   Change in voltage dv = 5V - 0V = 5 V
   Change in time dt = 2 ms = 2 × 10^-3 s
2. Calculate the voltage rate of change (dv/dt):
   dv/dt = 5 V / (2 × 10^-3 s) = 2,500 V/s
3. Apply the differential equation:
   i(t) = C * (dv/dt)
   i(t) = (10 × 10^-6 F) * (2,500 V/s)
4. Calculate final value and track units:
   i(t) = 0.025 A
   Answer: The peak current is 25 mA. Because the ramp is linear, dv/dt is constant, meaning the current is a steady 25 mA DC pulse for the duration of the 2ms ramp.

Worked Example 2: Voltage Rise from a Constant Current Pulse

Scenario: A microcontroller GPIO pin configured as a constant current sink pulls exactly 500 mA from a 470 µF bulk aluminum electrolytic capacitor for a duration of 100 ms to flash an LED. The capacitor was initially sitting at a resting voltage of 2.0V. What is the final voltage across the capacitor after the pulse?

Step-by-Step Solution:

1. Identify and convert knowns to SI units:
   Capacitance C = 470 µF = 470 × 10^-6 F
   Current i(t) = 500 mA = 0.5 A (Constant, so we can use the simple algebraic rearrangement)
   Time dt = 100 ms = 0.1 s
   Initial voltage v(0) = 2.0 V
2. Rearrange the integral form for a constant current:
   Since i is constant, the integral of i dt is simply i * t.
   v(t) = (i * t / C) + v(0)
   *Note: Since the current is being pulled OUT of the capacitor (discharging), we will treat i as negative relative to the charging convention, or simply subtract the voltage drop.*
3. Calculate the change in voltage (dv):
   dv = (i * dt) / C
   dv = (0.5 A * 0.1 s) / (470 × 10^-6 F)
   dv = 0.05 / 0.00047 = 106.38 V
4. Apply to initial voltage:
   Because it's discharging: v(final) = v(0) - dv
   v(final) = 2.0 V - 106.38 V = -104.38 V

For further reading on first-order circuit dynamics and how these differential equations scale when resistors are added to form RC networks, the MIT OpenCourseWare Circuits and Electronics lectures provide excellent foundational derivations.