RL time constant: how fast the current rises in a coil
A capacitor resists a change of voltage; an inductor resists a change of current. Switch a coil onto a supply through a resistor and the current does not jump to V/R but climbs there along an exponential, with the time constant τ = L / R: 63% in one τ, 99% in five. Switch it off and the current falls the same way, through whatever path the coil can find. This calculator gives τ, the current and the voltages at any moment, the time to a target current, the stored energy and the filter cut-off, and draws the curve.
How to use the RL time constant calculator
- Type R and L (100, 4k7, 10m, 47uH). R is everything in the loop, including the coil's own winding resistance.
- Type the supply voltage and choose switch on (current rising from zero) or switch off (current decaying from V/R through R).
- Optional: a time to look at (blank is one τ) and a target current in mA, for the delay until a relay pulls in or a motor winding reaches its current.
- The drawing shows the circuit with the current, the voltage across each part and the curve with the τ milestones; click the resistor or the coil for its own calculation. The Milestones tab lists 1τ to 5τ.
The formula
τ = L / R I(t) = (V/R)(1 − e−t/τ) switching on I(t) = (V/R) e−t/τ switching off
A henry divided by an ohm is a second. The final current is simply V/R, because a steady current needs no voltage across an ideal inductor. Note the direction of R: more resistance makes the circuit faster (τ smaller) but the final current lower. In RC the resistance does the opposite, slowing the charge.
Switching on
At the instant the switch closes the current is zero, so the resistor drops nothing and the whole supply voltage is across the coil. The coil's voltage is L × dI/dt, so the current starts rising at V/L amps per second. As it rises the resistor takes more of the voltage, the coil less, and the rise slows: the exponential. After 5τ the current is within 1% of V/R and the coil behaves as a plain wire with its winding resistance.
Switching off and the inductive kick
The current cannot stop instantly either. Give it a path, a resistor or a flyback diode across the coil, and it decays with τ = L/R through that path. Give it none, an opening switch or a transistor turning off, and the coil raises its voltage until something conducts: hundreds of volts across a relay driver, which is why every coil driven by a transistor gets a diode across it. The energy that has to go somewhere is ½ L I²; the diode dissipates it gently, a zener or series resistor dissipates it faster and lets the relay release sooner.
The milestones
| Time | Switching on | Switching off |
|---|---|---|
| 1τ | 63% | 37% |
| 2τ | 86% | 14% |
| 3τ | 95% | 5.0% |
| 4τ | 98% | 1.8% |
| 5τ | 99% | 0.7% |
RL against RC
The two are mirror images. In RC the voltage across the capacitor follows the exponential and the current is what starts high and dies; in RL the current follows the exponential and the coil's voltage is what starts high and dies. τ = RC grows with R; τ = L/R shrinks with it. Both corner a filter at 1/(2πτ): R/(2πL) for the RL low-pass with the output across R.
Your circuit, step by step
- Time constant: 10 mH / 100 Ω = 100 µs; settled after about 500 µs.
- Final current: 12 V / 100 Ω = 120 mA, with 72 µJ stored in the field.
- At 100 µs: 63% of that, 75.85 mA; 4.41 V across the coil and 7.59 V across the resistor.
- Interrupted with no path: the coil would make about 1.2 kV across a 10 kΩ gap; fit a flyback diode.
Worked example: 10 mH and 100 Ω on 12 V
τ = 10 mH / 100 Ω = 100 µs. The final current is 12 V / 100 Ω = 120 mA. After one τ the current is 63.2% of that, 75.9 mA, with 7.59 V across the resistor and 4.41 V still across the coil; after 500 µs it has settled. The field then holds ½ × 10 mH × (120 mA)² = 72 µJ. Opened with no diode, the coil would try to push 120 mA through the gap: 1.2 kV across 10 kΩ, enough to arc a small switch or kill a transistor.
Questions
Why does more resistance make it faster?
Because τ = L/R: the resistor is where the coil's energy goes in and out, and a bigger R moves it faster. The price is a smaller final current, V/R.
Which R do I use for a relay?
The coil's own resistance (from the data sheet, or measure it) plus anything in series, such as a driver transistor's few tenths of an ohm. A 12 V relay with a 70 Ω coil draws 171 mA when settled.
How long does a relay take to release?
With a plain flyback diode the current decays with τ = L / Rcoil through the diode, so release takes a few τ, often 5–10 ms. A zener in series with the diode, or a resistor, raises the decay resistance and shortens it.
Does the curve apply to a motor?
For the first milliseconds, yes: a motor winding is an RL circuit and its current rises with τ = L/R before the back-EMF takes over as the rotor turns.