The inductor voltage: seen from which terminal? V = L·dI/dt and switching supplies

Learn it paired with the capacitor

A capacitor obeys I = C·dV/dt: it can't change its voltage suddenly. An inductor is its dual: V = L·dI/dt, and it can't change its current suddenly. Cutting the current in an instant would need infinite voltage, so in practice the voltage climbs until something conducts and keeps the current flowing. Energy is ½CV² and ½LI².

V, seen from which terminal?

The sign confuses people because the direction V is measured in was never pinned down. Call the terminal where current flows in A and the terminal where it flows out B, and let V = V_A − V_B. Then V = +L·dI/dt.

While the current rises (dI/dt > 0), V > 0 and A sits higher, so the inductor absorbs power like a resistor. While it falls (dI/dt < 0), V < 0 and B sits higher than A. Current flows from A to B even though B is at the higher potential, so the inductor delivers power like a battery.

The textbook "−L·dI/dt" is the same voltage measured in the direction that opposes the change in current. The physics is identical.

What happens when current is cut off abruptly

Take an inductor (a relay coil, 100mH) fed from 12V and pulled to GND through a transistor, carrying 100mA. A sits at 12V; B sits near 0V at the transistor's saturation voltage.

Switch the transistor off in 10ns: dI/dt = −0.1A ÷ 10ns = −10⁷ A/s, so V = L·dI/dt = 0.1 × (−10⁷) = −1,000,000V. A stays at 12V, so B tries to climb to 12V + 1MV.

It never reaches 1MV. Avalanche breakdown of the transistor's collector-emitter junction (tens to hundreds of volts, sometimes destroying the part), charging of the board's stray capacitance (ceiling I × √(L/C), about 3kV for 100mA, 100mH, 100pF), or an arc through the air opens a path, and the current tapers off.

The ceiling is set not by the inductor but by whatever conducts first, so provide it yourself: a flyback (freewheeling) diode from B to A. Once B rises 0.7V above A it conducts, and current circulates inductor → diode → inductor. B stops at 12.7V, and the current decays with the L ÷ R time constant (1ms for a 100Ω coil).

Always put one across a relay, motor or solenoid. Without it, every turn-off puts a pulse of hundreds of volts on the transistor, which fails after a few to a few thousand cycles.

A diode alone makes the current decay slowly, so a relay releases slowly. For a fast turn-off, add a zener in series to raise the clamp to 12V + the zener voltage. A higher clamp means a larger dI/dt and a faster decay.

Putting that spike to use: the boost converter

If B climbs until it finds a partner, make the partner an output capacitor and you get a voltage above the input. That is the boost converter.

While the switch is closed, A = V_in, B = 0, so V = V_in > 0 and the current rises with slope V_in ÷ L. When it opens, B rises until it is clamped through a diode to the output capacitor (V_out). V = V_in − V_out < 0, so the current falls with slope (V_out − V_in) ÷ L.

In steady state the current gained equals the current lost over one cycle: V_in × T_on ÷ L equals (V_out − V_in) × T_off ÷ L, so V_in × T_on = (V_out − V_in) × T_off. With duty cycle D = T_on ÷ T this gives V_out = V_in ÷ (1 − D). D = 0.5 doubles the voltage; D = 0.75 quadruples it.

This balance of voltage × time over one cycle is volt-second balance. It only says an inductor's average voltage is zero, yet every switching supply's output voltage comes from that one line.

Stepping down with the same principle: the buck converter

In a buck converter, A is switched between V_in and open, and B is tied to the output capacitor (V_out).

While the switch is closed, A = V_in, B = V_out, so V = V_in − V_out > 0 and the current rises. When it opens, the current tries to fall, and to oppose that A's voltage drops. This time the input side moves, because the capacitor pins B. A stops 0.7V below GND, where a diode (or the low-side MOSFET in synchronous rectification) conducts. V = −0.7 − V_out ≈ −V_out, and the current falls with slope V_out ÷ L.

Volt-second balance gives (V_in − V_out) × T_on = V_out × T_off, so V_out = D × V_in. From 12V, D = 0.275 gives 3.3V.

The ripple current is ΔI = (V_in − V_out) × D ÷ (f × L), V = L·dI/dt integrated over T_on. A larger L reduces ΔI, but the inductor gets bigger and its DC resistance rises. The synchronous buck converter calculator uses this equation to work out ΔI and losses.

Which terminal moves: A or B

When the switch opens, the terminal that is not pinned to a fixed voltage is the one that moves. In relay drive and in the boost converter A is pinned to the supply, so B moves up. In the buck converter B is pinned to the output capacitor, so A moves down. Either way it stops once it finds a diode or capacitor.

On a scope both terminals can be moving, so one side against GND does not show V = L·dI/dt. Use a differential probe or the difference of two channels. The SW node (the moving terminal) shows the spike-then-clamp waveform, with ringing on its edges.

Things that catch you out on real hardware

• Saturation. Once the core's flux hits its ceiling, L collapses, dI/dt spikes and the current runs away. Rate the saturation current above the peak current (average + ΔI/2).
• A missing or backward flyback diode. Mandatory for relays, motors and solenoids. It goes from B to A (supply side is the cathode); backward, it shorts the supply.
• Diode speed. At microsecond switching, a general-purpose rectifier's slow reverse recovery can't clamp the spike. Use a Schottky or fast-recovery diode.
• Ringing on the SW node. Parasitic L and C resonate at tens of MHz on the spike's edge, radiating noise and in bad cases exceeding a device's rating. Use an RC snubber or a gate resistor.
• Continuous vs. discontinuous current. At light load the current falls to zero during the off period (discontinuous mode), and the output voltage is no longer set by duty cycle alone.
• The inductor's DC resistance. An I × R drop adds to V = L·dI/dt, hurting efficiency and output voltage at high current.