How this instrument works
Connect a resistor and capacitor in series across a voltage source and the charge does not climb at a steady rate. Kirchhoff's voltage law says the current at any instant equals the leftover voltage — source minus whatever the capacitor already holds — divided by R, and that current is what fills the capacitor further. The result is a first-order differential equation whose solution is an exponential: charge as a fraction of full equals 1 − e^(−t⁄RC). Solving that equation for time, once a target fraction is fixed, gives the formula this instrument runs backward: t = RC·ln(1 ⁄ (1 − fraction)).
The logarithm is the honest part of the shape. Early on, charge builds almost in a straight line — for a small fraction x, ln(1 ⁄ (1 − x)) ≈ x, so the first tenth of full charge arrives in almost exactly a tenth of a time constant. Near the end the curve leans over hard: reaching 95% takes about three time constants, 99% takes about 4.6, and 99.99% takes over nine — every extra nine costs roughly as much time again. Ask for 100% and the formula wants ln of infinity; the instrument refuses that input for the same reason a real capacitor never finishes charging, only gets closer forever.
Whoever sizes R and C is choosing a stopwatch, and the people who do it professionally know the curve's lopsidedness cold. A soft-start network on a motor drive's enable pin uses a resistor and capacitor so the drive ramps up over a couple of seconds instead of slamming the DC bus on power-up; a switch-debounce network picks R and C so contact bounce, typically under 5 ms, sits well inside the flat early part of the curve before a logic threshold is crossed. The mistake worth naming is treating charge time as proportional to the target fraction, so that doubling the wait 'obviously' doubles the charge reached. It does not — past the halfway point, each further percentage point costs disproportionately more time, which is exactly the shape this instrument is built to show.
- Enter Resistance — the series resistor value, in ohms, kilohms or megohms.
- Enter Capacitance — the part's rated value, from picofarads up to millifarads.
- Set Target charge fraction, % to how close to full charge you want to measure, anywhere from 0 up to just under 100.
- Read Time to reach that charge in milliseconds, seconds or minutes.
- Try 63.2% to see the plain time constant, then 99% to see how sharply the wait stretches near full charge.
Worked example — a soft-start capacitor at 95% charge
A motor-drive engineer is checking the soft-start network on a variable-frequency drive: a 10 kΩ resistor charging a 100 µF capacitor from the drive's internal rail, feeding a comparator that releases the run relay once the cap reaches 95% of that rail. Enter R = 10 kΩ, C = 100 µF and Target charge fraction, % = 95, and the instrument returns t = 2.995732 s — call it three seconds on the bench. The time constant here, τ = RC, is a clean 1 second, so 2.995732 s is essentially three time constants, a touch under the textbook 'three-tau' mark of 95.02%.
Push the same request further and the curve's lean becomes obvious: reaching 99% of that rail takes RC·ln(100) = 4.605170 s, and 99.9% takes RC·ln(1000) = 6.907755 s. Getting from 95% to 99% — four more percentage points — costs 1.609438 s, nearly as long as the first climb to 95% itself; getting the last 0.9% costs another 2.302585 s on top of that. That lopsided cost is why the drive's designer picked a comparator threshold at 95%, not 99.9%: chasing the last sliver of charge would stall every start-up by several extra seconds for no operational benefit.
Questions
Why does time depend on a logarithm instead of a straight fraction?
Because the charging curve is exponential, not linear. Current — and therefore the rate of charging — is proportional to how far the capacitor still has to go, so the process visibly slows as it nears full. Solving 1 − e^(−t⁄RC) = fraction for t is what pulls a logarithm out of the algebra; there is no way around it once the underlying physics is an exponential closing of the voltage gap.
What happens if I set the target fraction to 100%?
The instrument rejects it, because the true answer is infinite. A capacitor charging through a resistor approaches full charge asymptotically and mathematically never quite arrives — the formula asks for ln(1 ⁄ 0), which is undefined. Enter 99.9% or 99.99% instead; either gets close enough for any real circuit, and the formula still returns a finite, usable number.
How does this relate to the time constant, τ = RC?
τ is this formula's special case at a fixed 63.2% target. Set Target charge fraction, % to 63.212 and Time to reach that charge comes out equal to R × C almost exactly, because 1 − 1/e ≈ 0.63212 by definition. Every other value on this page — 50%, 95%, 99.9% — is the same exponential law evaluated at a different point on the curve.
Why do engineers say a capacitor is 'fully charged' after 5 time constants?
Because 5τ reaches 99.326% of full charge — 1 − e⁻⁵ — which is close enough for most digital-logic and power-electronics thresholds. It is a convention, not a law: this instrument shows exactly how far 5τ actually gets you, and a design that genuinely needs 99.99% takes about 9.2 time constants instead, not 5.
Does it matter which units I enter for resistance and capacitance?
No — pick whatever the datasheet gives you. Resistance accepts ohms, kilohms or megohms, and Capacitance accepts picofarads through millifarads; the instrument converts both to base SI units before running the formula, so a 4.7 kΩ resistor and a 220 nF capacitor need no manual conversion first.
Can I use this for switch-debounce or power-on delay timing?
Yes — that is exactly the kind of RC timing this formula covers. Pick R and C so the time to reach your logic threshold, often 50% or 63% depending on what the input pin treats as 'high', sits comfortably above your worst-case bounce time or rail rise time, then read Time to reach that charge to confirm the margin before committing to component values.