The direct answer for the instantaneous voltage across an inductor is defined by the voltage of inductor equation: v(t) = L (di/dt). This fundamental relationship dictates that the voltage v is strictly proportional to the inductance L and the rate of change of current di/dt. If the current is constant (DC steady state), di/dt is zero, and the voltage drops to zero. If the current changes rapidly, the voltage spikes proportionally. Below, we break down the derivation, rearranged forms, unit pitfalls, and worked examples to bridge textbook theory and bench-level reality.
The Core Formula and Symbol Definitions
The differential form of the voltage of inductor equation is expressed as:
v(t) = L · (di(t) / dt)
This equation assumes a linear, time-invariant inductor. Every symbol must be tracked with strict SI units to prevent calculation errors.
| Symbol | Parameter | SI Unit | Practical Bench Units |
|---|---|---|---|
| v(t) | Instantaneous voltage across the inductor terminals | Volts (V) | mV, kV (for flyback spikes) |
| L | Inductance (ability to store energy in a magnetic field) | Henrys (H) | μH, mH |
| i(t) | Instantaneous current flowing through the inductor | Amperes (A) | mA |
| t | Time | Seconds (s) | ms, μs, ns |
| di/dt | First derivative of current with respect to time (slew rate) | Amperes per second (A/s) | A/μs |
For deeper foundational reading on how magnetic flux linkage creates this relationship, refer to the core principles outlined by Electronics Tutorials.
Rearranged Forms for Circuit Analysis
Depending on what variable you are solving for in a circuit simulation or design phase, you will need to rearrange the voltage of inductor equation. Here are the standard algebraic and integral forms:
- Solving for Inductance (L):
L = v(t) / (di/dt)
Use this when sizing an inductor for a switching regulator (e.g., buck converter) where you know your target voltage, switching frequency, and allowable current ripple. - Solving for Current Slew Rate (di/dt):
di/dt = v(t) / L
Use this to determine how fast current will ramp up when a specific DC voltage is applied across a known inductor. - Integral Form (Solving for Current i(t)):
i(t) = (1 / L) ∫ v(τ) dτ + i(t0)
This form is critical for transient analysis and SPICE simulations. It states that the current at any time t is the integral of the applied voltage over time, scaled by 1/L, plus the initial current i(t0) already flowing through the component. Inductors resist changes in current; the integral form mathematically proves that current cannot change instantaneously without an infinite voltage.
Assumptions, Applicability, and Unit Pitfalls
When the Formula Applies (and When It Fails)
The standard voltage of inductor equation assumes an ideal, linear inductor. This means:
- No Core Saturation: The permeability of the core material remains constant. If the magnetic flux density exceeds the core's saturation limit (e.g., ~0.3T for standard ferrite), L drops precipitously. The equation then requires the product rule: v = d(Li)/dt = L(di/dt) + i(dL/dt).
- Below Self-Resonant Frequency (SRF): Real inductors have parasitic parallel capacitance between windings. At the SRF, the inductor acts as a parallel LC tank. Above the SRF, it behaves capacitively, and the v = L(di/dt) model is invalid.
- Negligible Winding Resistance: The equation models the purely reactive voltage. In reality, the total measured voltage is vtotal = L(di/dt) + iRDC, where RDC is the wire's parasitic resistance.
Unit Mistakes That Break the Math
The most common bench and exam error is failing to convert practical units to base SI units. The dimensional analysis requires Volts = Henrys × (Amperes / Seconds).
If you have a 10 μH inductor and a current ramping at 5 mA/μs, plugging in "10" and "5" yields 50V. This is wrong.
Correct conversion: 10 μH = 10 × 10-6 H. 5 mA/μs = (5 × 10-3 A) / (1 × 10-6 s) = 5,000 A/s.
Correct math: v = (10 × 10-6) × (5,000) = 0.05V (50 mV). Always convert to H, A, and s before multiplying.
Realistic Answer Magnitudes
What should your answer look like? In low-power RF circuits, inductors are in the nH range, yielding mV-level reactive drops. In power electronics (like a 48V solar charge controller), inductors are in the mH range. When a MOSFET switches off in 50 ns, the dt is so small that the voltage of inductor equation routinely generates 100V to 500V flyback spikes, which is exactly why snubber networks and flyback diodes are mandatory.
Worked Examples with Unit Tracking
Let’s apply the voltage of inductor equation to two distinct scenarios: a DC transient switching event and an AC sinusoidal steady state.
Problem 1: DC Transient Flyback Spike
Scenario: A 10 mH relay coil is carrying a steady-state current of 2.0 A. A mechanical switch opens, forcing the current to drop to 0 A in 5 μs. Calculate the induced voltage spike.
- Identify and convert variables to SI base units:
L = 10 mH = 10 × 10-3 H
di = ifinal - iinitial = 0 A - 2.0 A = -2.0 A
dt = 5 μs = 5 × 10-6 s - Calculate the slew rate (di/dt):
di/dt = -2.0 A / (5 × 10-6 s) = -400,000 A/s - Apply the voltage of inductor equation:
v(t) = (10 × 10-3 H) × (-400,000 A/s)
v(t) = -4,000 V
Bench Reality Check: The negative sign indicates the voltage polarity opposes the change in current (Lenz's Law). A 4,000V spike will instantly arc across the mechanical switch contacts, causing pitting and EMI. In a solid-state circuit, it would avalanche a standard MOSFET. This magnitude proves why a reverse-biased flyback diode is required to clamp the voltage to ~0.7V.
Problem 2: AC Sinusoidal Steady State
Scenario: A current defined by i(t) = 3 sin(120πt) Amperes flows through a 50 mH inductor. Find the instantaneous voltage at t = 5 ms.
- Convert Inductance:
L = 50 mH = 0.05 H - Take the derivative of i(t) with respect to t:
di/dt = d/dt [3 sin(120πt)]
Using the chain rule: di/dt = 3 × 120π × cos(120πt) = 360π cos(120πt) A/s - Multiply by L to find v(t):
v(t) = 0.05 × 360π cos(120πt)
v(t) = 18π cos(120πt) V - Evaluate at t = 5 ms (0.005 s):
v(0.005) = 18π cos(120π × 0.005)
v(0.005) = 18π cos(0.6π)
Since 0.6π radians is 108°, cos(108°) ≈ -0.309
v(0.005) = 18π × (-0.309) ≈ -17.5 V
Bench Reality Check: For AC steady-state RMS calculations, engineers rarely use the differential form directly. Instead, they use Vrms = Irms × XL, where XL = 2πfL. See All About Circuits for a deep dive on inductive reactance.
Frequently Asked Questions
Why does the voltage of an inductor equation result in a massive spike when a switch opens?
When a switch opens, the transition time (dt) approaches zero. Because di/dt is in the numerator, dividing a finite current change by an infinitesimally small time yields a mathematically infinite voltage. Physically, the inductor's collapsing magnetic field forces electrons across the air gap of the opening switch (creating an arc) to maintain current continuity. The energy stored in the magnetic field (E = 0.5 L i2) must go somewhere, and it dissipates as heat, light, and electromagnetic interference in that arc.
Can I use the voltage of inductor equation for AC RMS calculations?
Technically yes, but it is highly inefficient. The equation v(t) = L(di/dt) calculates instantaneous voltage at a specific microsecond. To find RMS voltage, you would have to integrate the square of that instantaneous function over a full period and take the square root. In practice, for linear AC circuits, we use the phasor domain: V = I × jωL, where ω is the angular frequency (2πf). This algebraic shortcut bypasses calculus entirely while yielding the exact same magnitude.
What happens to the equation if the inductor core saturates?
Core saturation occurs when all magnetic domains in the ferrite or iron core are aligned. At this point, the core's relative permeability drops toward 1 (the permeability of air), causing the inductance L to plummet. If L is no longer a constant, you cannot pull it out of the derivative. The true equation becomes v(t) = d(L(t)i(t))/dt. On the bench, this looks like a sudden, massive spike in current ripple (di/dt increases) for a given applied voltage, which frequently leads to overcurrent faults and destroyed switching transistors in power supplies.
How does parasitic capacitance affect the voltage of inductor equation at high frequencies?
Every physical inductor has stray capacitance between its adjacent wire windings. This creates a parallel resonant tank circuit. At low frequencies, the inductive reactance dominates, and v = L(di/dt) holds true. However, as frequency increases, the capacitive reactance drops. At the Self-Resonant Frequency (SRF), the inductor acts purely resistive. Above the SRF, the parasitic capacitance dominates, and the component behaves like a capacitor. If you attempt to use the inductor equation above the SRF, your phase and magnitude calculations will be completely inverted.
