Mastering Inductor Charging and Discharging Equations: Formulas & Worked Examples

The inductor charging and discharging equations dictate exactly how current ramps up and decays in an RL (resistor-inductor) circuit over time. When a DC voltage is applied, the charging current follows I(t) = (V/R) × (1 - e^-(R/L)t). When the source is removed and the circuit is shorted or clamped, the discharging current follows I(t) = I₀ × e^-(R/L)t. The speed of these transients is governed entirely by the time constant τ = L/R.

The Core Inductor Charging and Discharging Equations

Inductors oppose changes in current. When you close a switch on a DC RL circuit, the inductor generates a back-EMF that limits the initial current to zero, forcing it to ramp up exponentially. The governing equations for these transient states are derived from Kirchhoff’s Voltage Law (KVL) applied to the loop.

Charging Equation (Step Response)

I(t) = (V / R) × (1 - e^-(R/L)t)

Discharging Equation (Source-Free Response)

I(t) = I₀ × e^-(R/L)t

Symbol Definition Table

When the Formula Applies and Its Assumptions

These equations assume an ideal step-input (voltage goes from 0 to V instantly), a constant inductance L (the core does not saturate as current rises), and a linear resistor R. In real-world power electronics, if your inductor core saturates, L drops sharply, τ shrinks, and the current spikes much faster than the equation predicts. Furthermore, the formulas assume the temperature remains constant; in high-current solenoids, copper winding resistance increases with heat, altering R dynamically.

Real-World RL Circuit Parameters & Magnitudes

To build intuition, you need to know what a realistic answer magnitude looks like. A time constant of 5 seconds is normal for a massive contactor coil, but would indicate a catastrophic design flaw in a switching power supply. Below is a data-dense reference of common RL applications you will encounter on the bench.

Unit Mistakes That Break the Math

The most common way to brick your calculations is mixing prefixes. The formula τ = L / R strictly requires base SI units: Henries and Ohms.

• The Millihenry Trap: If L = 10 mH and R = 50 Ω, τ is NOT 10/50 = 0.2 seconds. You must convert 10 mH to 0.01 H. τ = 0.01 / 50 = 0.0002 s (or 0.2 ms).
• The Microsecond Blindspot: In switching regulators, L is in microhenries (µH). 4.7 µH / 0.5 Ω = 9.4 µs. If you forget the µ prefix and treat the answer as seconds, your PWM frequency calculations will be off by a factor of one million.
• Ignoring DCR: Using only the external resistor value and ignoring the inductor’s internal DC Resistance (DCR) will result in an artificially low R, predicting a slower charge time than what your oscilloscope will show.

Rearranged Forms for Circuit Design

On the bench, you rarely just solve for I(t). Usually, you have a target current and need to size the inductor, or you have a fixed inductor and need to know exactly when a threshold will be crossed. Here are the algebraic rearrangements for the charging equation, solving for each critical variable.

• Solve for time (t):
  t = -(L / R) × ln(1 - (I(t) × R / V))
  Use case: Finding the exact delay before a solenoid pulls in.
• Solve for Inductance (L):
  L = (-R × t) / ln(1 - (I(t) × R / V))
  Use case: Sizing a filter inductor to limit inrush current to a specific value at time t.
• Solve for Target Current (I(t)):
  This is the base equation: I(t) = (V / R) × (1 - e-(R/L)t)
• Solve for Discharge Resistance (R) to achieve a specific decay time:
  R = -L / (t × ln(I(t) / I0))
  Use case: Sizing a discharge bleeder resistor to ensure a relay drops out within 5ms.

Worked Example 1: Calculating Inductor Charge Time

Scenario: You are driving a 12V DC solenoid valve using a MOSFET switch. The solenoid has an inductance of 50 mH and a total circuit resistance (coil DCR + wiring + MOSFET R_DS(on)) of 24 Ω. The solenoid requires 95% of its maximum steady-state current to generate enough magnetic force to pull the valve open. How many milliseconds after the MOSFET turns on will the valve open?

Step 1: Identify Knowns and Convert to Base Units

• V = 12 V
• R = 24 Ω
• L = 50 mH = 0.05 H
• Target I(t) = 95% of I_max

Step 2: Calculate Maximum Steady-State Current and Target Current

Once the inductor is fully charged, it acts as a short circuit (ignoring DCR, which is already bundled into R).

I_max = V / R = 12 V / 24 Ω = 0.5 A (or 500 mA)

Target I(t) = 0.95 × 0.5 A = 0.475 A

Step 3: Apply the Rearranged Time Equation

t = -(L / R) × ln(1 - (I(t) × R / V))

Notice that (I(t) × R / V) is simply the ratio of the target current to the max current (0.475 / 0.5 = 0.95). This simplifies the math beautifully:

t = -(0.05 H / 24 Ω) × ln(1 - 0.95)

Step 4: Execute the Math with Unit Tracking

• Calculate τ: 0.05 [H] / 24 [Ω] = 0.002083 [s] (or 2.083 ms)
• Calculate the natural log term: ln(1 - 0.95) = ln(0.05) ≈ -2.9957
• Multiply: t = -0.002083 [s] × -2.9957 = 0.00624 [s]

Final Answer: The valve will open 6.24 milliseconds after the MOSFET is triggered. As a rule of thumb, reaching 95% always takes approximately 3τ (3 × 2.083ms = 6.249ms), which confirms our exact calculation.

Worked Example 2: Discharging a Relay Coil via Flyback Diode

Scenario: You are switching off a 5V PCB relay. The relay coil has an inductance of 100 mH and a DC resistance of 125 Ω. To protect your driving transistor from the inductive voltage spike, you placed a 1N4148 flyback diode in reverse parallel across the coil. When the transistor turns off, the coil discharges through the diode. The diode has a dynamic forward resistance of roughly 0.5 Ω during this decay phase. How long does it take for the coil current to drop to 10% of its initial value?

Step 1: Identify Knowns and Total Discharge Resistance

During discharge, the 5V source is gone. The current circulates through the coil's internal resistance AND the flyback diode. According to Electronics Tutorials, the total decay resistance is the sum of all components in the freewheeling loop.

• L = 100 mH = 0.1 H
• R_coil = 125 Ω
• R_diode = 0.5 Ω
• Total R = 125 + 0.5 = 125.5 Ω
• I₀ = 5V / 125 Ω = 0.04 A (40 mA)
• Target I(t) = 10% of I₀ = 0.004 A

Step 2: Apply the Discharging Equation

I(t) = I0 × e-(R/L)t

We need to solve for t. Rearranging for the discharge case:

t = -(L / R) × ln(I(t) / I0)

Step 3: Execute the Math with Unit Tracking

• Calculate the ratio: I(t) / I₀ = 0.004 A / 0.04 A = 0.10
• Calculate τ (Discharge): 0.1 [H] / 125.5 [Ω] = 0.0007968 [s] (or 0.797 ms)
• Calculate the natural log: ln(0.10) ≈ -2.3026
• Multiply: t = -0.0007968 [s] × -2.3026 = 0.001834 [s]

Final Answer: It takes 1.83 milliseconds for the relay current to decay to 10%.

Mastering these equations means moving beyond abstract textbook problems. By rigorously tracking your units, accounting for parasitic resistances like DCR and diode dynamic resistance, and understanding the physical limits of core saturation, you can predict exactly how your magnetic components will behave when the switch flips.