How this instrument works
Energy stored in an inductor grows with the square of current, not with current itself. Push twice the amps through the same coil and you have not doubled the stored energy, you have quadrupled it, because building that field means fighting the coil's own back-EMF the entire way up: the first milliamp costs almost nothing to establish, the last milliamp is shoved in against nearly the full induced voltage the winding can produce. Integrate that rising cost over the whole charging ramp and the halves cancel neatly into ½LI² — the same bookkeeping that gives a spring ½kx² and a capacitor ½CV².
The formula falls straight out of Faraday's law once you track power rather than force. Instantaneous power flowing into a coil is p = v·i, and since v = L·(dI/dt), that is p = L·i·(dI/dt). Integrate p with respect to time from zero current to I and the dt cancels against dI/dt, leaving ∫L·i di = ½LI². Nothing about frequency, waveform shape or how quickly the current got there survives that integral — only the final current and the inductance matter, which is why a coil ramped up over one second and one ramped up over one microsecond end up holding identical energy at identical current.
The number is only honest while L stays constant, and real cores do not cooperate forever. Iron and ferrite saturate as current climbs, permeability collapses, and the henry value printed on the part quietly shrinks well before its current rating is reached — so a coil driven past its saturation current stores less than ½LI² predicts, not more, and the missing joules show up as extra heating rather than extra field. Below saturation, though, this is exact physics, not an approximation, and it is the number a protection circuit has to absorb in full the instant the current path is broken.
- Type the part's rating into Inductance, H — in henries, so a 100 mH choke is entered as 0.1.
- Enter the operating current into Current, choosing mA or A from its unit menu to match your measurement.
- Read Stored energy in joules, or switch its unit to kJ for large industrial coils and magnet windings.
- Rerun at your circuit's peak current, not its average, since a switch typically opens somewhere near the peak of a switching ripple.
Worked example — a 0.1 H relay coil at 2 A
A control relay's coil measures 0.1 H and draws 2 A while its contacts are pulled in. Set Inductance, H to 0.1 and Current to 2 A, and the sheet returns E = ½ × 0.1 × 2² = ½ × 0.1 × 4 = 0.2 joule — an exact, round answer, since 0.1 and 4 are both kind numbers to multiply. That fifth of a joule is sitting in the coil's magnetic field the entire time the relay is energised, and it does not vanish when the driving transistor turns off; it has to be absorbed somewhere within microseconds.
Without a flyback diode across the coil, that stored 0.2 J collapses through whatever path offers itself, usually an arc across the opening switch contacts or a destructive avalanche breakdown inside the driving transistor, because the coil resists the sudden drop in current by forcing voltage to spike as high as it takes to keep current flowing somewhere. A diode wired to conduct only during turn-off gives the field a lazy path to decay through instead, dissipating that same 0.2 J gently in the diode's forward drop and the coil's own resistance rather than in a spark. Push the same coil to 4 A instead of 2 A and stored energy does not double to 0.4 J, it quadruples to 0.8 J, which is precisely why flyback protection is sized for peak current, not for typical current.
Questions
Why does stored energy scale with current squared rather than current itself?
Because charging a coil means fighting its own induced voltage the whole way, and that voltage grows with the rate of change of current. Power flowing in is p = L·i·(dI/dt); integrating that from zero to I produces ½LI², a squared term, the same way integrating a linearly rising force over a linearly growing displacement gives ½kx² for a spring. Doubling current quadruples the energy because both the current and the effective resistance to increasing it have gone up together.
Where does that stored energy go when I switch the circuit off?
It has to go somewhere within a few microseconds, because an inductor opposes sudden changes in current far more strongly than in voltage. Cut the current path abruptly and the coil forces voltage upward, arbitrarily high if nothing limits it, until some path — an arc, a transistor's avalanche breakdown, or a deliberately added flyback diode or snubber — lets the current keep flowing long enough to bleed the field down to zero. Unprotected inductive switching is the single most common cause of destroyed relay drivers and motor-control transistors.
Does this energy figure depend on how fast the coil was charged?
No. Only the final current and the inductance value determine ½LI² — the charging time drops out of the integral entirely. A coil ramped to 2 A over one second and the same coil ramped to 2 A over one microsecond both end up holding exactly the same 0.2 J at 0.1 H, though the microsecond ramp demands a far higher instantaneous voltage to get there, since v = L·(dI/dt) scales inversely with the time allowed.
How does an inductor's stored energy compare with a capacitor's?
They are complementary rather than identical. A capacitor's ½CV² builds because pushing charge onto a plate gets harder as voltage climbs; an inductor's ½LI² builds because pushing current higher gets harder as the coil's back-EMF climbs. One field forms across an insulating gap under a steady voltage, the other forms around a conductor under a steady current, and swapping current for voltage and inductance for capacitance in the formula, and in the physical behaviour, carries you cleanly from one component to the other.
Why does my henry value seem to drop off at high current?
Core saturation. Iron and ferrite cores lose permeability once their internal magnetic domains run out of room to align further, and inductance falls as a direct consequence, often to a fraction of its low-current value well before the winding's thermal current rating is reached. Below that saturation point ½LI² is exact; above it, the true stored energy is less than the formula predicts using the nameplate inductance, and the shortfall shows up as extra core heating instead.
Is inductance entered in henries even for a small radio coil?
Yes — this instrument always works in henries internally, so a 220 microhenry RF choke is typed as 0.00022. Millihenry and microhenry parts are common enough that it is worth writing the exponent out on paper first, since a slipped decimal point here changes the reported energy by orders of magnitude rather than by a small error.