Relay coil suppression calculator

A diode across a relay coil limits the turn-off voltage to the supply plus one forward drop, so a 12 V coil leaves its transistor facing 12.7 V instead of hundreds of volts — but it also makes the coil current decay as slowly as it can, and TE's application note shows that this slow release can leave contacts welded. Putting a Zener in series with the diode fixes it: a 24 V Zener on a 12 V, 400 Ω, 0.4 H coil cuts the release time from 1.90 ms to 0.35 ms for a 36.7 V switch voltage. Enter the coil and the suppression to see what the switch sees, what the diode must be rated for, and how long the relay takes to let go.

Fig 1 — 12.0 V across 400 Ω: the switch sees 12.7 V at turn-off, and the coil current takes 1.90 ms to fall to the release level.

Across the coil

Coil supply (V)

Coil resistance (Ω)

Coil inductance (0 = unknown) (mH)

Diode forward drop (V)

Must-release voltage (% of rated)

Switching rate (0 = single event) (/s)

Coil current: 30.0 mA
Energy stored in the coil: 180 µJ
Coil voltage at turn-off · switch sees: 700 mV · 12.7 V
Switch rating, Omron ×2 rule: ≥ 25.4 V
Diode: I_F(AV) above · V_R at least: 30.0 mA · 36.0 V
Decay time constant L / R: 1.00 ms
Time to the 10 % release level: 1.90 ms
Time to zero current: 2.90 ms
Energy: suppressor · coil resistance: 17.4 µJ · 163 µJ

A 12.0 V Zener in series with the diode would cut the release time to 575 µs for a 24.7 V switch voltage.

How this is calculated

Standard: TE 13C3264; Omron relay technical information

$I_0 = \frac{V_S}{R_{coil}}, \qquad E = \tfrac{1}{2} L I_0^2$: Coil current while energised, and the energy the suppression has to absorb.
$V_{SW} = V_S + V_C, \quad V_C = \begin{cases} V_F & \text{diode} \\ V_Z + V_F & \text{diode + Zener} \\ I_0 R_p & \text{resistor} \end{cases}$: What the switch blocks at turn-off.
$L \frac{di}{dt} = -(V_C + iR) \;\Rightarrow\; i(t) = \left(I_0 + \frac{V_C}{R}\right) e^{-t/\tau} - \frac{V_C}{R}, \quad \tau = \frac{L}{R}$: Coil current after turn-off with a clamp across it. For the resistor, V_C = 0 and R becomes R + R_p.
$t_{rel} = \tau \ln \frac{V_S + V_C}{V_{rel} + V_C}, \qquad t_0 = \tau \ln \left(1 + \frac{V_S}{V_C}\right)$: Time to the must-release current, and to zero current.
$E_C = V_C \cdot Q, \qquad Q = \frac{L I_0 - V_C\, t_0}{R}$: Energy the clamp dissipates; the rest goes into the coil resistance.
$V_{CEO} \gtrsim (V_{S,max} + 0.6 + V_Z) \times 2$: Omron's transistor selection rule. The 2 is a safety factor the designer sets; V_Z is 0 for a plain diode.

Assumptions

Constant inductance: the coil is modelled as a fixed L in series with R. A real relay coil's inductance falls as the armature opens, so the tail of the decay is faster than shown; the time to the release level, which is reached with the armature still seated, is the reliable number.
The clamp is a fixed voltage: a diode at V_F, a Zener at V_Z. Both rise a little with current.
The release level is a current threshold at the datasheet's must-release voltage. Mechanical drop-out follows it by the relay's own release time, which the datasheet gives without suppression — TE's note says ratings are established that way and should be re-checked with the suppression in place.
Diode reverse-voltage minimum is Omron's two-to-three-times guidance at three times; the current minimum is the coil current, per Omron.
Without suppression the voltage is not computed. It depends on stray capacitance and the switch's breakdown, which nothing here knows.

What sets the voltage when a relay coil is switched off

A relay coil is an inductor, and an inductor's current cannot stop instantly. When the transistor driving the coil turns off, the current that was flowing — the supply over the coil resistance, 30 mA for a 12 V, 400 Ω coil — keeps flowing for a moment, and the coil raises whatever voltage it takes to find a path for it. TE's relay application note 13C3264 puts the number without a clamp at "an induced voltage transient of the order of hundreds or even thousands of volts", and that voltage, plus the supply, "appears across the coil interrupting switch". A transistor rated for 40 V does not survive it many times.

The flyback diode — freewheeling diode, snubber diode, catch diode; the names all mean the same part in the same place — gives that current a path. It sits across the coil with its cathode to the positive supply, so it is reverse-biased and idle while the coil is energised, and conducts the instant the coil end swings above the supply. The coil voltage is then pinned at one forward drop. That is why the switch sees the supply plus about 0.7 V and nothing more, and it is why every relay driver, from a single transistor to a ULN2003 with its diodes built in, has one.

What the diode costs is not obvious until the relay's own life is on the table. With only a forward drop across it the coil current decays at the slowest rate anything can arrange: the full L/R time constant, with almost all the stored energy burning off in the coil's own resistance. The armature is held in by that current, and the relay cannot start to release until it has fallen to the must-release level. TE's note is specific about what the slow decay does: the contacts of a power relay micro-weld every time they make into a fast-rising load, and breaking that weld on the next opening depends on the armature's momentum. "A slowly decaying magnetic flux (the slowest is experienced with a simple diode shunt across the coil) means the least net force integral available to accelerate the armature open", and the result of that lost momentum "can result in failure to break the stick, and a contact 'weld' is experienced". Omron says the same in fewer words: relays with a diode across the coil "tend to experience increased release times".

The fix both vendors give is a Zener in series with the diode. The coil still has its path, but now it decays against the Zener voltage instead of against 0.7 V, which pulls the current down far faster; the switch sees the supply plus the Zener plus the diode drop, which is the price. TE calls it "near optimum" and shows a 24 V Zener; Omron's rule is that the Zener "should be about the same as the supply voltage". Choose the Zener from what the transistor can block, and the calculator gives the release time that buys.

Relay coil suppression chart: the switch voltage against the release time

The worked example's coil under each suppression method, computed by the calculator above. The two columns that matter pull in opposite directions: a lower clamp voltage is kinder to the switch and slower to release the contacts. The last column is what a parallel resistor burns the whole time the relay is on, which a diode or Zener does not.

Method	Switch sees	Release time	Idle loss in the suppressor
Diode alone	12.7 V	1.90 ms	0
Diode + 12 V Zener	24.7 V	575 µs	0
Diode + 24 V Zener	36.7 V	349 µs	0
Diode + 36 V Zener	48.7 V	251 µs	0
1.2 kΩ resistor (3 × R_coil)	48.0 V	576 µs	120 mW
4 kΩ resistor (10 × R_coil)	132.0 V	209 µs	36 mW

Worked example: 12 V relay, 400 Ω, 0.4 H

A 12 V coil of 400 Ω and 0.4 H (a plausible measured value for a small power relay; the datasheet gives only the resistance), must-release at 10 % of rated, switched by an NPN transistor.

I0       = 12 / 400                       = 30 mA
E        = ½ × 0.4 × 0.03²                = 180 µJ
τ        = L / R = 0.4 / 400              = 1.0 ms

diode, V_F = 0.7 V
  switch sees   12 + 0.7                  = 12.7 V
  t_release     1.0 ms × ln(12.7 / 1.9)   = 1.90 ms
  t_zero        1.0 ms × ln(1 + 12/0.7)   = 2.90 ms
  energy        17 µJ in the diode, 163 µJ in the coil resistance

diode + 24 V Zener
  switch sees   12 + 24 + 0.7             = 36.7 V
  t_release     1.0 ms × ln(36.7 / 25.9)  = 0.35 ms
  t_zero        1.0 ms × ln(1 + 12/24.7)  = 0.40 ms
  energy        137 µJ in the Zener, 43 µJ in the coil resistance

transistor (Omron):  (12 + 0.6 + 24) × 2   = 73 V V_CEO — a 40 V part is not enough
diode:               I_F(AV) > 30 mA, V_R ≥ 36 V — any 1N4148 or 1N4001

The Zener brings the release from 1.90 ms to 0.35 ms, five times faster, and moves most of the stored energy out of the coil and into the Zener, where at any sane switching rate it is microwatts. What it asks in return is a transistor that blocks 37 V with margin; Omron's selection rule doubles that to 73 V. A 12 V Zener — Omron's "about the supply" — gives 0.57 ms at 24.7 V, which a 40 V transistor handles, and is the usual compromise. The calculator's default is the plain diode, because that is what most designs have; the note under the results says what the supply-voltage Zener would do to it.

Where the coil suppression model stops being valid

The inductance is not constant. A relay coil's L depends on the magnetic circuit, and the magnetic circuit changes as the armature opens: the air gap grows and the inductance falls. The decay to the release level happens with the armature seated, so that number holds; the tail after it is faster than the fixed-L curve, and the time to zero is an upper bound.
The must-release voltage is a static figure. It is where the armature is guaranteed to drop out with the coil voltage lowered slowly. The relay's mechanical release time then adds to the electrical delay computed here, and it is specified without suppression: TE notes that "it is normal industry practice to test relays and subsequently establish performance ratings without coil suppression", and recommends the relay "be evaluated with the suppression that will be used".
AC coils. A diode across an AC coil is a short circuit on every other half-cycle. Omron's suppressor table marks the diode and the diode + Zener as not applicable to AC and gives the varistor and the RC network for it. The calculator is DC only.
Contacts are the other half. The clamp protects the driver from the coil. The load the contacts switch has its own inductance and needs its own protection across the contacts or the load — Omron's guidance for that RC is 0.5–1 µF per ampere and 0.5–1 Ω per volt, confirmed by test — and aTVS or diode across the load is the DC equivalent.
Built-in diodes. Some relays ship with the diode inside, and the ULN2003's clamp diodes do the same job from the driver side. Both are the plain diode, with all of the above; a Zener can still be added in series with the ULN2003's COM pin, but not inside a relay that already has its diode fitted.

Common relay diode mistakes

Diode the wrong way round. Cathode (the banded end) goes to the positive supply. Reversed, it conducts the moment the coil is energised, shorts the supply through the transistor, and something burns.
No diode at all because "the transistor has never failed". Avalanche absorbs the energy each time, and the failure is statistical. The ULN2003's diodes are only in circuit when its COM pin is wired to the coil supply; left floating, there is no clamp.
Rating the diode for the supply voltage only. Omron's guidance for a diode across a DC coil in an electronic circuit is a reverse breakdown two to three times the supply; the current rating is the coil current, which the diode carries at full value the instant the switch opens.
Choosing the Zener for the coil instead of for the transistor. The Zener sets the switch voltage directly; pick the highest the transistor blocks with margin, then read off the release time — not the other way round.
Diode across the coil of an AC relay. It rectifies the drive on one half-cycle and clamps it on the other; the relay chatters or the diode fails. Varistor or RC across an AC coil.
Testing the flyback diode with a multimeter and calling the circuit good. The diode test only shows the junction exists; whether the coil voltage is actually clamped needs a scope on the switch node at turn-off, which is also where the release time is visible in the current.