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. Try to change the voltage and a current flows, proportional to how fast you're changing it. Change the voltage in an instant and you'd need infinite current, so in practice the voltage only changes as fast as the available current allows.

An inductor is its dual: V = L·dI/dt, and it can't change its current suddenly. Try to change the current and a voltage appears across it, proportional to how fast you're changing it. Cut the current off in an instant and you'd get infinite voltage, so in practice the voltage climbs until it reaches "whatever lets something conduct and keep the current flowing."

Just as a capacitor stores charge Q = C·V to hold its voltage, an inductor stores flux Φ = L·I to hold its current. Energy is ½CV² and ½LI² respectively. "A capacitor's voltage is continuous, an inductor's current is continuous" — everything here starts from that.

V, seen from which terminal?

The sign in the equation confuses people because the direction V is measured in was never pinned down. Let's pin it down. Call the terminal where current flows in A, and the terminal where it flows out B, and let V = V_A − V_B (A's voltage relative to B, not "B as seen from A"). With this convention, V = +L·dI/dt.

While the current is increasing (dI/dt > 0), V > 0 — A sits higher than B. Current is flowing from the higher potential into the lower one, so the inductor is absorbing power, like a resistor (storing it as magnetic energy).

While the current is decreasing (dI/dt < 0), V < 0 — B sits higher than A. Current flows from A to B even though B is at the higher potential. That's the same situation as a battery: the inductor is delivering power, acting as a source (releasing the magnetic energy it stored).

The textbook "−L·dI/dt" measures this same voltage as an "induced EMF," in the direction that opposes the change in current. It's the same physics, just measured the opposite way round. When you're reading a schematic, sticking to the "in-terminal to out-terminal" convention above and using +L·dI/dt consistently removes the guesswork.

It all falls into place once you think of it as "the inductor generates whatever voltage it needs to keep the current flowing, on its own." To oppose a rise, it raises the input side; to oppose a fall, it raises the output side. Both are "opposing the change," just in different directions.

What happens when current is cut off abruptly

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

Switch the transistor off in 10ns, and the current tries to head from 100mA to 0. dI/dt = −0.1A ÷ 10ns = −10⁷ A/s. V = L·dI/dt = 0.1 × (−10⁷) = −1,000,000V. A stays at 12V, so B tries to climb to 12V + 1MV.

It obviously never reaches 1MV. Somewhere along the way, something conducts: the transistor's collector-emitter junction goes into avalanche breakdown (tens to hundreds of volts, sometimes destroying the part), the board's stray capacitance charges up (the voltage ceiling there is I × √(L/C) — about 3kV for 100mA, 100mH, 100pF), or it arcs through the air. Whichever path opens up, the inductor keeps pushing current through it at that voltage, and the current tapers off gradually.

The key point: it isn't the inductor that sets the ceiling on the voltage — it's whatever conducts first. So you arrange for that yourself. That's the flyback (freewheeling) diode: put a diode from B to A, and the instant B rises 0.7V above A, the diode conducts and current keeps circulating around the loop — inductor → diode → inductor. B's voltage stops at 12.7V, and the current decays with the L ÷ R time constant (1ms for a 100Ω coil resistance).

This is exactly why a diode always goes across a transistor driving a relay, motor, or solenoid. Skip it, and every time you switch off, a pulse of hundreds of volts hits the transistor — it fails after anywhere from a few to a few thousand cycles.

Stopping it with just a diode slows the current decay (a slow relay release), so if you need a fast turn-off, put a zener in series with the diode to raise the clamp to "12V + zener voltage." A higher clamp voltage means a larger dI/dt, so the current dies out faster — again, straight from V = L·dI/dt.

Putting that spike to use: the boost converter

If B's voltage "climbs until it finds a partner," then make that partner an output capacitor and you can generate a voltage higher than the input. That's the boost converter.

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

In steady state, the current gained in one part of the cycle equals the current lost in the other. Rise = V_in × T_on ÷ L; fall = (V_out − V_in) × T_off ÷ L. Setting these equal: V_in × T_on = (V_out − V_in) × T_off. Rewriting with duty cycle D = T_on ÷ T gives V_out = V_in ÷ (1 − D). D = 0.5 doubles it; D = 0.75 quadruples it.

This "voltage × time balances out to zero net area over one cycle" is called volt-second balance. It's just stating the obvious fact that an inductor's average voltage is zero — otherwise the current would keep climbing or keep falling forever — but every switching supply's output voltage comes out of that one line.

Stepping down with the same principle: the buck converter

A buck (step-down) converter switches the inductor's input, A, on and off to V_in, and ties the output, B, 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 current rises. Open the switch, and the current tries to fall, which means A's voltage drops to oppose that (this time it's the input side that moves, since output B is pinned by the capacitor). A stops 0.7V below GND, where a diode (or, with synchronous rectification, the low-side MOSFET) conducts. V = −0.7 − V_out ≈ −V_out, and current falls with a slope of V_out ÷ L.

From volt-second balance: (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 swing in current (ripple current) is ΔI = (V_in − V_out) × D ÷ (f × L) — again just 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.

"A moves" or "B moves"

Sorting the discussion above by which terminal's voltage actually moves gives a clean picture.

When the switch opens, whichever terminal isn't pinned to a fixed voltage is the one that moves. In the relay-drive case and in a boost converter, A is fixed to the supply, so B moves upward. In a buck converter, B is fixed to the output capacitor, so A moves downward. Either way, it moves "until it finds something that can keep the current flowing" — a diode, a capacitor — and stops there.

The same holds when you probe an inductor with a scope: both terminals can be moving. Looking at just one side referenced to GND won't show you V = L·dI/dt. Use a differential probe, or take the difference between two channels. Look at a switching supply's SW node (the moving terminal) and you'll see exactly the "spike then clamp" waveform described above, plus ringing riding on the edges (a resonance between the inductor's winding capacitance or parasitic trace inductance and the surrounding capacitance).

Things that catch you out on real hardware

• Saturation. Once the core's flux hits its ceiling, L collapses suddenly, dI/dt spikes at the same voltage, and the current runs away. Rate the "saturation current" with margin above the peak current (average + ΔI/2).
• A missing or backward diode. A flyback diode is mandatory for any transistor driving a relay, motor, solenoid, or valve. Orientation: "B to A" (the supply side is the cathode). Wire it backward and you short the supply.
• Diode speed. When switching in microseconds, as in a switching supply, a general-purpose rectifier's slow reverse recovery won't clamp the spike in time. Use a Schottky or a fast-recovery diode.
• Ringing on the SW node. A resonance between parasitic L and C rings at tens of MHz right at the spike's edge. It's a source of radiated noise, and in bad cases it can exceed a device's voltage rating. An RC snubber, or a gate resistor to soften the edge, tames it.
• Continuous vs. discontinuous current. At light load, the current can fall all the way to zero and stop during the off period (discontinuous mode). Past that point the volt-second equation changes, and the output voltage is no longer set by duty cycle alone.
• The inductor's DC resistance. On top of V = L·dI/dt there's an I × R drop. It's not negligible at high current and it eats into both efficiency and output voltage.