Relay coil and its flyback diode
The coil keeps going after switch-off
A relay uses a magnetic coil to move switch contacts. A flyback diode provides a path for coil current after the driving switch opens. Watch the stored magnetic energy drain instead of disappearing instantly.
EXPERIMENT BENCH
Change the circuit settings, predict the readings, then check a challenge. These are learning models, with the assumptions explained in each lesson.
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Why it works
A settled coil is represented by resistance R and inductance L. Release samples its decay through an always-present ideal 0.7 V diode; the laboratory stays powered. No unprotected-voltage mode, contact movement, bounce or real relay release time is simulated.
Vs is coil supply, R winding resistance, L inductance, I coil current and t time after opening the switch. Vsw is switch-node voltage measured from 0 V; D marks the diode. mH is one thousandth of a henry, ms one thousandth of a second and µJ one millionth of a joule of energy. Before release I0 = Vs/R. After release the equation uses exp, exponential decay, and max, which prevents negative current. While current flows Vsw = Vs + 0.7 V; at zero current Vsw = Vs.
Difficulty
Build · Build it
Power on at supply 5 V, winding resistance 100 Ω and inductance 100 mH. Keep the coil energized: read 50 mA and 125 µJ.
Need a hint?
Vs = 5 V; R = 100 Ω; L = 100 mH → I = 50 mA; E = 125 µJ
Explore · Explore the boundary
With the same 5 V, 100 Ω and 100 mH, release the coil and sample 1 ms after switch-off. Read about 13.969 mA and switch-node voltage 5.7 V.
Need a hint?
Vs = 5 V; R = 100 Ω; L = 100 mH; t = 1 ms → I ≈ 13.969 mA; Vsw = 5.7 V
Solve it · Solve the challenge
Keep 5 V, 100 Ω and 100 mH; sample the released coil at 3 ms. Reach zero current and zero stored energy.
Need a hint?
Vs = 5 V; R = 100 Ω; L = 100 mH; t = 3 ms → I = 0 mA; E = 0 µJ
Extra reading and circuit recipes
The coil keeps going after switch-off
The driving switch opens, yet the current cannot stop instantly. The coil’s magnetic field gives up its stored energy and pushes current through a new loop: coil winding and diode. The diode’s orientation makes it block normal supply current and conduct during this decay.
I0 = Vs/R; I(t) = max((I0 + 0.7/R) × exp(−Rt/L) − 0.7/R, 0); E = ½LI²
Initially stored energy is E = ½LI². The winding and diode dissipate it during release. The fixed diode drop adds a voltage opposing the current, and the model stops the decay when the current reaches zero.
A useful pattern
With these 5 V, 100 Ω and 100 mH settings, the current reaches zero at about 2.097 ms. A fixed 0.7 V diode drop makes it reach zero in finite time; a pure exponential alone would approach zero forever.
Beyond the model
A low-voltage flyback path limits voltage stress but keeps current circulating. Real relay contacts may therefore release later than with a higher-voltage clamp; this electrical model does not predict their timing.
A useful trap to avoid
Release samples a previously settled coil; it does not simulate a coil that was energized for only a short time. Zero calculated coil current is not a measurement of when real contacts moved.
Only inductance doubles; supply, resistance and the 1 ms sampling time stay fixed. The remaining coil current…
- is larger
- is unchanged
- immediately becomes zero
Read more
Initial current Vs/R is unchanged, but larger inductance slows its decay, so more current remains at the same elapsed time.
Build it
Use 5 V, 100 Ω and 100 mH with the coil energized. The steady 50 mA stores 125 µJ, and the closed ideal switch sits at 0 V.
I = 50 mA; Vsw = 0 V; E = 125 µJ
Explore the boundary
Keep those values, release the coil and choose 1 ms. Read approximately 13.969 mA, 5.7 V at the switch node and 9.757 µJ still stored.
I ≈ 13.969 mA; Vsw = 5.7 V; E ≈ 9.757 µJ
Change one thing
At 5 V and 100 Ω, compare 100 mH with 200 mH at 1 ms after release. The initial current stays the same, but the larger inductance stores more energy and retains more current.