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RC & RL Time Constant Calculator

Time constants, charge and discharge curves, supercap hold-up time and I²C pull-ups.

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  • Free preview: the names of each table’s first rows (up to 3) and a watermarked chart, with the time constant, the hold-up time and every other figure hidden.
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Circuit

Circuit
The change

The resistance the current flows through, including the source’s own.

V
V
V
Time constant —

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Charging curve

After 1 to 5 time constants

How it was calculated

    Next steps Done. What next?

    All engineering & electronics calculators

    About the RC & RL Time Constant Calculator

    A resistor and a capacitor (or an inductor) change in the same way every time: quickly at first, then more and more slowly, with a time constant τ that sets the pace. τ = R × C for a capacitor and τ = L ÷ R for an inductor; after one τ the change is 63.2% done, after five it is 99.3% done. Enter R and C (or L), the source voltage and the starting value, and the calculator gives τ, the voltage and current at any moment you choose, the time to reach a threshold, the 1τ to 5τ table, the stored energy and a chart of the whole curve.

    Two presets cover jobs engineers do often. Supercap hold-up works out how long a charged supercapacitor keeps a load above its minimum voltage, for a constant current, a constant power (a DC-DC converter) or a resistor, including the drop across the capacitor’s ESR. I²C pull-up gives the range of pull-up resistors from the bus capacitance, the supply and the speed mode, with the limits of the NXP I²C-bus specification, and rates the standard values in that range.

    Without a pass, the free preview works on your own values and shows the names of each table’s first rows and the chart under a watermark, without its numbers; the time constant, the hold-up time, the pull-up range, every other figure, the working, Copy result and the CSV come with a Pro pass.

    How to use it

    1. Choose RC / RL transient, Supercap hold-up or I²C pull-up.
    2. For a transient, pick RC or RL and whether it is charging (energising) or discharging, then type R, C or L, the source voltage and the starting value. Values take prefixes: 10k, 4.7u, 100n, 2.2m.
    3. Optionally type a moment to read the voltage and current then, and a voltage (or current) to find when it is reached.
    4. Read τ, the curve and the table. Copy result and the CSV of the curve need a pass, or unlocking this result.

    Examples

    RC charge
    Input
    R 10 kΩ, C 100 µF, 0 → 5 V
    Result
    τ = 1 s; 3.161 V after 1 s; 4.5 V after 2.303 s; 99.3% after 5 s
    RC discharge
    Input
    R 1 kΩ, C 1 µF, from 12 V
    Result
    τ = 1 ms; 1.624 V after 2 ms; down to 1 V after 2.485 ms
    RL energising
    Input
    R 100 Ω, L 10 mH, 10 V
    Result
    τ = 100 µs; current heads for 100 mA, 63.21 mA after 100 µs
    Supercap, constant current
    Input
    1 F, ESR 0.1 Ω, 5 V → 3 V at 0.5 A
    Result
    0.05 V ESR drop; hold-up time 3.9 s
    I²C Fast-mode
    Input
    VDD 3.3 V, 200 pF
    Result
    Rp from 966.7 Ω (3 mA at 0.4 V) to 1.77 kΩ (300 ns rise time)

    The first-order step response

    Every voltage and current in a circuit with one resistor and one capacitor or inductor follows x(t) = x∞ + (x0 − x∞) × e^(−t/τ), where x0 is the value at the start and x∞ the value it heads for. Solving it for t gives the time to reach a value x: t = τ × ln((x0 − x∞) ÷ (x − x∞)).

    • After 1τ, 2τ, 3τ, 4τ and 5τ the change is 63.21%, 86.47%, 95.02%, 98.17% and 99.33% done.
    • Half of the change takes τ × ln 2 = 0.693τ; 10% to 90% takes 2.197τ.
    • The final value is approached, never reached: that is why 5τ is the usual “fully charged”.
    • The same R and C used as a filter have their −3 dB point at fc = 1 ÷ (2πτ).

    Capacitors and inductors

    • Capacitor charging from V0 toward the source Vs: v(t) = Vs + (V0 − Vs) × e^(−t/τ), and the current is (Vs − v) ÷ R. Discharging through R: v(t) = V0 × e^(−t/τ).
    • Inductor energising from a source Vs through R: the current heads for Vs ÷ R, and the voltage across the inductor is (I∞ − i) × R, Vs at the start. De-energising through R: i(t) = I0 × e^(−t/τ).
    • Energy: a capacitor stores ½CV², an inductor ½LI². Charging a capacitor from 0 V through any resistor turns as much energy into heat in the resistor as ends up in the capacitor.
    • Inductive kick: an inductor’s current cannot stop at once. When the source is removed the voltage across it jumps to I0 × R of the path it discharges through — without such a path (a flyback diode) the voltage rises until something breaks down.

    Supercapacitor hold-up time

    From i = C × dv/dt, with the capacitor’s ESR in series with the load:

    • Constant current I: the terminal voltage drops by I × ESR at once, then falls in a straight line: t = C × (V1 − V2 − I × ESR) ÷ I.
    • Constant power P (a DC-DC converter takes more current as the voltage falls): without ESR, t = C × (V1² − V2²) ÷ 2P. With ESR the calculator integrates the exact current, and warns when the load would pass the maximum-power point (the voltage at the load falling to half the capacitor’s) before the minimum voltage.
    • Resistor R: the voltage decays with τ = (R + ESR) × C: t = (R + ESR) × C × ln(V1 ÷ Vc,end), with Vc,end = V2 × (R + ESR) ÷ R.

    Only the energy between the two voltages is usable: going from 5 V down to 2.5 V uses 75% of what is stored.

    I²C pull-up resistors

    The NXP I²C-bus specification (UM10204, section 7.1) sets both limits. The resistor must be large enough for every device to pull the line down to VOL: Rp(min) = (VDD − VOL(max)) ÷ IOL, with VOL = 0.4 V at 3 mA in Standard-mode and Fast-mode and at 20 mA in Fast-mode Plus (0.2 × VDD at 2 mA when VDD is 2 V or less). It must be small enough for the line to rise from 30% to 70% of VDD within the rise time: Rp(max) = tr ÷ (0.8473 × Cb), where 0.8473 = ln(0.7 ÷ 0.3).

    • Standard-mode (100 kbit/s): tr ≤ 1000 ns, Cb ≤ 400 pF
    • Fast-mode (400 kbit/s): tr ≤ 300 ns, Cb ≤ 400 pF
    • Fast-mode Plus (1 Mbit/s): tr ≤ 120 ns, Cb ≤ 550 pF

    Sources

    Limitations

    • Parts are ideal: a real capacitor’s leakage, dielectric absorption and ESR, and an inductor’s winding resistance and core saturation, are not included (except the ESR you enter for the hold-up time).
    • R is the whole resistance in the loop: add the source’s output resistance and anything else in series.
    • Supercapacitor capacitance and ESR change with temperature and age; use the end-of-life figures from the datasheet for a hold-up time you can rely on.
    • The I²C limits assume a resistor pull-up; buses with a current-source pull-up, bus buffers or series resistors on each line need the extra terms of the specification.

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    Frequently asked questions

    What do I get without a pass?

    Without a pass, RC & RL Time Constant Calculator shows the names of each table’s first rows (up to 3) and a watermarked chart, with the time constant, the hold-up time and every other figure hidden. Until you unlock it, the result can’t be downloaded or copied. A Pro, Premium or Ultimate pass, a one-time payment that never renews, unlocks the full result. The pricing page lists the passes and their prices.

    How long does a capacitor take to charge?

    About five time constants: τ = R × C, and after 5τ it is at 99.3% of the source voltage. 10 kΩ and 100 µF give τ = 1 s, so about 5 s. To a particular voltage, use t = τ × ln((Vs − V0) ÷ (Vs − V)): to 90% of the way takes 2.3τ.

    Why is the time constant 63.2%?

    Because the change after one τ is 1 − e^(−1) = 0.632 of the way. Each further τ removes another 63.2% of what is left, so 86.5% after 2τ and 95.0% after 3τ.

    What is the time constant of an RL circuit?

    τ = L ÷ R. 10 mH and 100 Ω give 100 µs: the current reaches 63.2% of its final value Vs ÷ R in 100 µs and settles after about 500 µs.

    How do I choose an I²C pull-up resistor?

    Any value between Rp(min) = (VDD − 0.4 V) ÷ 3 mA and Rp(max) = tr ÷ (0.8473 × Cb) works. At 3.3 V in Fast-mode with 200 pF of bus capacitance that is 967 Ω to 1.77 kΩ, so 1.2 kΩ or 1.5 kΩ. Lower values rise faster; higher values waste less current while a line is low.

    How long will a supercapacitor power my circuit?

    For a constant current, t = C × (V1 − V2 − I × ESR) ÷ I: 1 F from 5 V to 3 V at 0.5 A with 0.1 Ω ESR lasts 3.9 s. A DC-DC converter draws constant power instead; choose that load and enter the power to get its time.

    Quick answers and tool search

    Type to search tools or to get a quick answer, for example 18% of 2500. Use the up and down arrow keys to move through the results, Enter to choose, and Escape to close.