How is the RC time constant related to cutoff frequency?

For a first-order RC filter the cutoff frequency is f = 1/(2 pi RC). A 1 ms time constant gives 159 Hz. The 10 to 90 percent rise time is 2.2 RC, or about 0.35 divided by the cutoff frequency.

Why Is the RC Time Constant 63%? The Math of 1 − 1/e • The Science 360

Q: Is a capacitor fully charged after 5 time constants?

It is at 99.3 percent, which is close enough for most timing and power work. It is not enough for precision measurement: settling a 12-bit ADC input to half an LSB takes about 9 time constants, and 16 bits takes about 11.8.

Q: What percentage does a capacitor discharge to in one time constant?

It falls to 36.8 percent of its starting voltage, which is 1/e. After two time constants it is at 13.5 percent, after three at 5.0 percent and after five at 0.67 percent.

Put a 10 kΩ resistor in series with a 100 nF capacitor, connect 5 V, and start a stopwatch. After exactly 1 ms the capacitor sits at 3.16 V. Not 2.5 V, not 4 V: 3.16 V, which is 63.2% of the supply. Swap in any resistor and any capacitor and the same thing happens at t = R × C. Every textbook prints the number. Very few explain why the universe picked such an awkward one.

The short version is that nobody picked 63.2%. Engineers picked R × C because it is the only time scale the circuit has, and 63.2% is simply where the curve happens to be at that moment. Once you see why, the "5 time constants means fully charged" rule also stops looking like a law and starts looking like the rough convention it is.

Quick answerThe time constant

\tau = RC

is the natural time unit of the circuit. In each τ the remaining gap shrinks by a factor of e (2.718), so after one τ the gap is

1/e = 36.8\%

and the capacitor has covered

1 - 1/e = 63.2\%

. Equivalently, τ is how long charging would take if the starting rate never slowed.

Infographic: Why is the RC time constant 63%? A 5 V battery charges a 100 nF capacitor through a 10 kΩ resistor, and a stopwatch reads 1 ms. At t = RC = 1 ms the capacitor sits at 3.16 V, 63.2% of the supply, because 1 − 1/e = 0.632; RC is the time charging would take if its first rate never slowed. A line chart of capacitor voltage against time in time constants shows charging at 63.2% and discharging at 36.8% after one τ, a dashed initial-rate line reaching 100% at 1τ, and 99.3% after 5τ. A 12-bit ADC needs 9.01τ. Myth: each time constant does not add another 63%; it closes 63.2% of the remaining gap. — Tap the image to view it full size. View the accessible, editable SVG.

Where does the RC charging equation come from?

Two facts are enough. The resistor's current is the voltage across it divided by R. The capacitor's current is C times how fast its voltage changes. In a series loop those currents are the same current, so:

\begin{aligned}\text{Resistor: } i &= (V_s - v)/R \\ \text{Capacitor: } i &= C\,dv/dt \\ \text{Set equal: } \frac{dv}{dt} &= \frac{V_s - v}{RC}\end{aligned}

Read that last line slowly, because it contains the whole story. The capacitor charges at a rate proportional to how far it still has to go. At the start the gap is the full 5 V and the voltage climbs fast. Halfway up, the gap is half as big, so the rate is half as big. The closer it gets, the slower it moves. The only function whose rate of change is proportional to itself is the exponential, so the solution is:

\begin{aligned}v(t) &= V_s\,(1 - e^{-t/RC})\ \ (\text{charging}) \\ v(t) &= V_0\,e^{-t/RC}\ \ (\text{discharging})\end{aligned}

Notice that R and C never appear separately. Only their product matters, and it has to have units of time for the exponent to be a pure number. It does: an ohm is a volt per ampere, a farad is a coulomb per volt, so Ω × F is coulombs per ampere, which is seconds. For our parts, $10\ \mathrm{k\Omega} \times 100\ \mathrm{nF} = 10^4 \times 10^{-7} = 10^{-3}\ \mathrm{s}$ , or 1 ms.

Why is it 63.2 percent at one time constant?

Plug t = RC into the charging equation and the exponent becomes exactly −1:

\begin{aligned}v(RC)/V_s &= 1 - e^{-1} = 1 - 0.3679 \\ &= \mathbf{0.6321} \\ 5\ \mathrm{V} \times 0.6321 &= \mathbf{3.16\ V}\end{aligned}

That is the voltage after 1 ms with 10 kΩ and 100 nF.

So 63.2% is just 1 − 1/e. That answers "what", but the better question is why RC deserves to be the unit at all. Figure 1 shows the answer as geometry.

Charging curve rising from 0 to 100 percent and discharging curve falling from 100 to 0 percent over five time constants. Dashed tangent lines at the start reach 100 percent and 0 percent exactly at one tau. Dots mark 63.2 percent on the charging curve and 36.8 percent on the discharging curve at one tau. — Figure 1. Charging (blue) and discharging (orange) curves, plotted from $v = 1 - e^{-t/\tau}$ and $v = e^{-t/\tau}$ . The dashed lines continue each curve's initial slope. Both reach the finish line at exactly 1τ, which is where the real curves sit at 63.2% and 36.8%.

At t = 0 the full supply is across the resistor, so the current is V_s/R and the capacitor voltage rises at V_s/(RC) volts per second. If it kept that pace, it would cover the whole V_s in exactly RC seconds. That is what the time constant physically means: the time to finish if nothing slowed down. The curve does slow down, continuously, and by the time that imaginary straight line reaches the top the real curve has managed 63.2%.

Where does e come from? A step-by-step version

If the exponential feels like a trick, rebuild it by hand. Pretend the capacitor charges in jumps. Chop one time constant into n equal steps, and in each step let the voltage close 1/n of whatever gap remains at the start of that step (that is the "rate proportional to gap" rule, applied in chunks).

After one step, the gap is (1 − 1/n) of what it was. After n steps (one τ), the gap is (1 − 1/n)ⁿ.

\begin{aligned}n = 1&:\ \text{gap} = 0\ (\text{"fully charged"}) \\ n = 2&:\ 0.5^2 = 0.250 \to 75.0\% \\ n = 10&:\ 0.9^{10} = 0.349 \to 65.1\% \\ n = 100&:\ 0.99^{100} = 0.366 \to 63.4\% \\ n \to \infty&:\ 1/e = 0.3679 \to \mathbf{63.2\%}\end{aligned}

The percentages are the charge reached. n = 1 is the straight-line fantasy.

This is the same limit that defines e in compound interest, run backwards. The finer you slice time, the more often the circuit "notices" that the gap has shrunk and eases off, and the answer converges on 1 − 1/e. There is nothing special about capacitors here. Any system whose rate of approach is proportional to its distance from the target does it: a cup of coffee cooling toward room temperature (approximately), a thermometer settling, an inductor's current in an RL circuit with $\tau = L/R$ , a radioactive sample decaying.

The misconception: “it charges another 63% every time constant”

A common mental slip is to think the second time constant adds another 63% of the supply, which would overshoot to 126%. What actually repeats is the fraction of the remaining gap that gets closed:

\begin{aligned}1\tau&:\ \text{gap } 36.8\% \to 63.2\% \\ 2\tau&:\ 0.368 \times 36.8 = 13.5\% \to 86.5\% \\ 3\tau&:\ 0.368 \times 13.5 = 4.98\% \to 95.0\% \\ 4\tau&:\ \text{gap } 1.83\% \to 98.2\% \\ 5\tau&:\ \text{gap } 0.674\% \to 99.3\%\end{aligned}

The right-hand number is how far the capacitor has charged.

Every τ, the error shrinks by the same factor of 2.718. That is why the error plot on a logarithmic axis (Figure 2) is a straight staircase, and why the curve never formally arrives.

Is a capacitor fully charged after 5 time constants?

"Five time constants and you are done" is a rule of thumb, and it is a good one for blinking LEDs, power-on delays and filter capacitors. At 5τ the voltage is within 0.67% of final, and the RC Wikipedia article states it the same way: generally considered steady state, though in theory it never arrives. The trouble starts when "done" has to mean something precise.

Bar chart on a logarithmic scale of the remaining gap after 1 to 12 time constants, falling from 36.8 percent to below 0.001 percent. Horizontal dashed lines mark 1 percent at 4.6 tau, half an LSB for 8 bits at 6.2 tau, 12 bits at 9.0 tau and 16 bits at 11.8 tau. — Figure 2. The gap remaining after each time constant, $e^{-n}$ , on a log axis. Each dashed line is a target accuracy and the dot shows how many τ it takes to reach it, from $t/\tau = \ln(1/\text{error})$ . Half an LSB for an N-bit converter is an error of 1/2^N+1.

The general rule is a one-liner. To get within a fraction ε of the final value you need $t = \tau \ln(1/\varepsilon)$ . For an ADC, the usual target is half a least significant bit of a full-scale step, $\varepsilon = 1/2^{N+1}$ :

\begin{aligned}t/\tau &= \ln 2^{N+1} = (N + 1)\ln 2 \\ &= (N + 1) \times 0.693\end{aligned}

\begin{aligned}8\text{-bit}&:\ 9 \times 0.693 = 6.24\tau \\ 10\text{-bit}&:\ 11 \times 0.693 = 7.62\tau \\ 12\text{-bit}&:\ 13 \times 0.693 = \mathbf{9.01\tau} \\ 16\text{-bit}&:\ 17 \times 0.693 = 11.78\tau\end{aligned}

Those figures are not my invention: TI's MSP430FR267x datasheet says about 9.01τ are needed for an error under ±0.5 LSB in 12-bit mode, the Xilinx XADC application note XAPP795 uses the same 9.01, and TI's analog filter settling note tabulates 11.78 for 16 bits. A design that waits 5τ before sampling a 12-bit converter can be off by 0.674% × 4096 ≈ 28 LSB on a full-scale step.

Worked example: sampling a 12-bit ADC through a source resistor. The numbers here are illustrative, not from one specific chip. Suppose a sensor with 10 kΩ output resistance drives an ADC whose sampling switch adds 1 kΩ and whose hold capacitor is 10 pF.

\begin{aligned}\tau &= (10\ \mathrm{k\Omega} + 1\ \mathrm{k\Omega}) \times 10\ \mathrm{pF} \\ &= 11\,000 \times 10{\times}10^{-12} = 110\ \mathrm{ns} \\ t_{12\text{-bit}} &= 9.01 \times 110\ \mathrm{ns} = \mathbf{0.99\ \mu s}\end{aligned}

That is the acquisition time needed for 12 bits. If the ADC only allows 0.5 µs:

\begin{aligned}0.5/0.11 &= 4.5\tau \\ e^{-4.5} &= 1.1\%\ \text{of the step} \approx 46\ \mathrm{LSB}\end{aligned}

That last line is why datasheets publish a maximum source impedance, and why people put an op-amp buffer or a large capacitor right at the ADC pin. TI adds a caveat worth keeping: the full ln(2^N+1) only applies to a worst-case full-scale step. If successive samples are close together the real error is smaller.

Other landmarks on the same curve

Once you can write $t = \tau \ln(1/\text{gap})$ , every other figure people quote for RC circuits is the same equation at a different threshold. Figure 3 marks the common ones.

RC charging curve with markers at 50 percent at 0.69 tau, 63.2 percent at 1 tau, 90 percent at 2.30 tau and 99.3 percent at 5 tau, plus a horizontal bracket showing the 10 to 90 percent rise time spans 2.20 tau. — Figure 3. Landmark times on the charging curve. The half-life is $\tau \ln 2 = 0.693\tau$ . The 10% to 90% rise time is $\tau(\ln 10 - \ln\tfrac{10}{9}) = \tau \ln 9 = 2.197\tau$ .

Half-life, 0.693τ. If you prefer 50%, the circuit has a half-life just like a radioactive isotope. It is just less convenient to write than RC.
Rise time, 2.2τ. Oscilloscope people measure 10% to 90%, which is 2.197τ. Since a first-order filter's cutoff is f_c = 1/(2πτ), this gives the familiar t_r ≈ 0.35/f_c. Our 1 ms RC has f_c = 159 Hz and t_r = 2.2 ms.
Logic thresholds. A slow RC edge crosses a digital input's threshold at t = τ ln(V_s/(V_s − V_th)). That is how you design a reset delay or a debounce filter.

Worked example: a power-on reset delay. A 10 kΩ pull-up charges a 1 µF capacitor on a reset pin at 3.3 V. Many CMOS inputs are specified to read high above 0.7 × V_DD.

\begin{aligned}\tau &= 10\ \mathrm{k\Omega} \times 1\ \mu\mathrm{F} = 10\ \mathrm{ms} \\ V_{th} &= 0.7 \times 3.3 = 2.31\ \mathrm{V} \\ t &= 10\ \mathrm{ms} \times \ln\frac{3.3}{3.3 - 2.31} \\ &= 10\ \mathrm{ms} \times 1.204 = \mathbf{12.0\ ms}\end{aligned}

Figure 4 shows why thresholds close to the rail are dangerous to design around. The time to reach 70% is 1.2τ and barely depends on the exact threshold. The time to reach 99% is 4.6τ, and a small change in where "99%" sits moves it enormously.

Curve of time needed versus threshold level, from 0 percent at 0 tau rising gently through 50 percent at 0.69 tau, 63.2 percent at 1 tau and 70 percent at 1.20 tau, then shooting up to 4.61 tau at 99 percent and 6.91 tau at 99.9 percent. — Figure 4. Time to reach a threshold, $t/\tau = -\ln(1 - \text{threshold})$ . Mid-range thresholds give predictable timing. Thresholds near the final value make the timing hypersensitive to supply voltage, threshold spread and leakage.

Why real RC timers drift from the formula

The equation assumes a perfect resistor, a perfect capacitor and an ideal supply. Three things break that in practice, and they matter more the slower the circuit.

Capacitor tolerance. General-purpose aluminium electrolytics are commonly sold at ±20% (the "M" tolerance code), so a 100 µF part can give a τ anywhere in a 1.5 to 1 range before temperature and ageing are counted. Use a C0G/NP0 ceramic or a film capacitor when timing matters.

Leakage. A real capacitor leaks DC current, which behaves roughly like a resistor R_L in parallel with it. That turns the circuit into a divider: the final voltage drops to $V_s R_L/(R + R_L)$ and the time constant shrinks to $(R \parallel R_L)\,C$ . Nichicon's general-purpose datasheets allow leakage up to 0.01CV or 3 µA, whichever is greater, after 2 minutes at rated voltage. For a 100 µF, 16 V part that is $0.01 \times 100 \times 16 = 16\ \mu\mathrm{A}$ , which at 16 V looks like 1 MΩ. Real parts usually leak far less than that limit, but the limit is what you can design to.

Ideal: R = 1 MΩ, C = 100 µF, τ = 100 s. Crossing 3.5 V on the way down from 5 V:

t = 100 \times \ln(5/1.5) = 120\ \mathrm{s}

With R_L = 10 MΩ, V_final = 4.55 V and τ = 90.9 s:

t = 90.9 \times \ln(4.55/1.05) = 134\ \mathrm{s}

With R_L = 1 MΩ, V_final = 2.50 V, which never reaches 3.5 V.

Three charging curves over 500 seconds for 1 megohm and 100 microfarads from 5 volts. The ideal curve crosses a 3.5 volt threshold at 120 seconds, a curve with 10 megohm leakage crosses at 134 seconds, and a curve with 1 megohm leakage levels off at 2.5 volts and never crosses. — Figure 5. A slow RC timer with leakage modelled as a parallel resistance. Leakage in real electrolytics is not ohmic and varies with voltage, temperature and how long the part has been powered, so treat these as bounding cases rather than a prediction for a specific part.

The lesson is the one Figure 4 hinted at. Keep R modest (tens to hundreds of kilohms for most timers), keep the threshold in the middle of the swing, and if you need minutes, count clock cycles instead.

Source and load resistance. Whatever drives the RC has its own output resistance, and whatever reads it loads it. Both add to or divide the R you think you have. A 10 kΩ divider feeding an RC filter adds its Thevenin resistance to the filter's R, a point the companion article on using a voltage divider as a power supply works through in detail.

The short version for your bench:

\tau = RC

is "time to finish at the starting speed". The curve covers 63.2% in that time because it slows in proportion to how close it is. Count 5τ for 99.3%, use

t = \tau \ln(1/\text{error})

for anything precise, and remember 12-bit settling needs about 9τ.

FAQ

Why is the time constant 63.2 percent and not 50 percent?

Because the time constant is defined as R times C, which is the number that falls straight out of the circuit equation. At $t = RC$ the remaining gap is $e^{-1} = 36.8\%$ , so the capacitor has covered 63.2%. The 50% point exists too: it is 0.693 RC, the RC circuit's half-life.

Is a capacitor fully charged after 5 time constants?

It is at 99.3%, which is close enough for most timing and power work. It is not enough for precision measurement: settling a 12-bit ADC input to half an LSB takes about 9 time constants, and 16 bits takes about 11.8.

What percentage does a capacitor discharge to in one time constant?

It falls to 36.8% of its starting voltage, which is 1/e. After two time constants it is at 13.5%, after three at 5.0% and after five at 0.67%.

Why are ohms times farads equal to seconds?

An ohm is a volt per ampere and a farad is a coulomb per volt. Multiply them and the volts cancel, leaving coulombs per ampere. An ampere is a coulomb per second, so coulombs per ampere is seconds.

For a first-order RC filter the cutoff frequency is $f = 1/(2\pi RC)$ . A 1 ms time constant gives 159 Hz. The 10% to 90% rise time is 2.2 RC, or about 0.35 divided by the cutoff frequency.

Sources

Wikipedia, "RC time constant" (63.2% at 1τ, 99.3% at 5τ, rise time ≈ 2.2τ, f_c = 1/(2πτ)): en.wikipedia.org/wiki/RC_time_constant
Texas Instruments, MSP430FR267x datasheet, ADC sample timing (about 9.01τ for < ±0.5 LSB in 12-bit mode): ti.com document viewer, MSP430FR2672/3/5/6
Texas Instruments, SBAA535, analog filter settling for ADCs (time constants to ½ LSB = ln(2^N+1), 11.78 for 16 bits): ti.com/document-viewer/lit/html/SBAA535
Xilinx, XAPP795, "Driving the Xilinx Analog-to-Digital Converter" (12 bits requires 9.01 time constants): physics.umd.edu mirror of XAPP795
Nichicon, aluminium electrolytic capacitor datasheet, USV series (leakage current ≤ 0.01CV or 3 µA after 2 minutes at 20 °C): Nichicon USV datasheet (LCSC mirror)
Physics Forums discussion, "Time constant: why τ = RC and 63%?" (the initial-slope interpretation): physicsforums.com/threads/time-constant-why-tau-rc-and-63.188582

Method note: all curves are computed directly from v = V_s(1 − e^−t/τ). The ADC and leakage examples use round illustrative component values; check your own part's datasheet for its sampling capacitance, switch resistance and leakage limit.

Header image: NASA (image iss067e141213), regraded for this article.

Where does the RC charging equation come from?

Why is it 63.2 percent at one time constant?

Where does e come from? A step-by-step version

The misconception: “it charges another 63% every time constant”

Is a capacitor fully charged after 5 time constants?

Other landmarks on the same curve

Why real RC timers drift from the formula

FAQ

Why is the time constant 63.2 percent and not 50 percent?

Is a capacitor fully charged after 5 time constants?

What percentage does a capacitor discharge to in one time constant?

Why are ohms times farads equal to seconds?

How is the RC time constant related to cutoff frequency?

Related on The Science 360

Sources