Electrical
Capacitor Charging Curve
Why every supply rail needs five time constants to settle
A capacitor charges through a resistor along an exponential curve V(t) = V₀(1−e^(-t/τ)), reaching 63.2% of supply voltage in one time constant τ = RC. After five time constants the capacitor is essentially full (>99%). The curve governs every power-on transient, every flash bulb, every sample-and-hold, and every bypass cap on every IC supply pin.
- Charging lawV(t) = V₀(1 − e-t/τ)
- Time constantτ = RC
- After 1τ63.2% charged
- After 5τ>99.3% charged
- Energy storedE = ½CV²
- Charge storedQ = CV
Interactive visualization
Press play, or step through manually. The visualization is yours to drive — try it before reading on.
Watch the 60-second explainer
A condensed visual walkthrough — narrated, captioned, under a minute.
How a capacitor charges
A capacitor is a pair of conductors separated by an insulator. Push current onto one plate and an equal-and-opposite charge accumulates on the other. The accumulated charge sets up a voltage across the dielectric: V = Q/C. The instant you connect a discharged capacitor to a voltage source through a resistor, current rushes in. As charge piles up on the plates, it pushes back against the source. The harder it pushes back, the smaller the net driving voltage across the resistor, and the slower charge can keep flowing.
That feedback is what makes the curve exponential. Apply Kirchhoff's voltage law to the loop:
V₀ = i·R + V_C
= R·(dQ/dt) + Q/C
Differentiate, separate, integrate, and you get the standard charging solution:
V_C(t) = V₀ · (1 − e^(-t/τ)) where τ = RC
i(t) = (V₀ / R) · e^(-t/τ)
Voltage rises from zero asymptotically toward V₀; current starts at V₀/R and decays toward zero. Both share the same time constant τ. The product RC has units of seconds: ohms × farads = (V/A)·(A·s/V) = s.
Worked example: 1 µF through 1 kΩ
Take the canonical lab circuit. R = 1 kΩ. C = 1 µF. Source V₀ = 5 V. The time constant is
τ = R·C = 1000 Ω × 1×10⁻⁶ F = 1×10⁻³ s = 1 ms
Plug t = 1 ms into the charging law:
V(1 ms) = 5 · (1 − e⁻¹) = 5 · (1 − 0.368) = 5 · 0.632 = 3.16 V
That 63.2% number is the universal signature of the exponential — it's where every RC charging curve sits at one time constant, regardless of values. Continuing:
t = 2τ: V = 5·(1 − e⁻²) = 5·0.865 = 4.32 V (86.5%)
t = 3τ: V = 5·(1 − e⁻³) = 5·0.950 = 4.75 V (95.0%)
t = 5τ: V = 5·(1 − e⁻⁵) = 5·0.993 = 4.97 V (99.3%)
t = 7τ: V = 5·(1 − e⁻⁷) = 5·0.9991 = 4.995 V (99.9%)
The energy finally stored on the capacitor is
E = ½ · C · V² = 0.5 · 1×10⁻⁶ · 5² = 12.5 µJ
An equal amount of energy is dissipated as heat in R during charging. That 50% efficiency cap is why charging a discharged capacitor from a voltage source through a resistor is intrinsically lossy — power supplies recover this with switching converters that don't dump energy into a series resistor.
Charging curve, drawn
The shape is universal — the only knobs are τ (which stretches the time axis) and V₀ (which scales the voltage axis):
V/V₀
1.0 ┤ ╶─────── 100%
0.99┤ ╶───── ← 5τ
│ ╶─────
0.86┤ ╶───── ← 2τ
│ ╶─
0.63┤ ╶─ ← 1τ
│ ╱
0.0 ┤─╱
└──────────────────────────────── t/τ
0 1 2 3 4 5 6
Discharging is the mirror — replace the source with a short and you get V(t) = V₀·e^(-t/τ), falling from full to 36.8% in one τ.
Capacitor families compared
| Family | Cap density | ESR | Voltage | Tolerance | Typical role |
|---|---|---|---|---|---|
| Aluminum electrolytic | High (1 µF–47 mF) | 0.05–1 Ω | 10–500 V | ±20% | Bulk supply, ripple smoothing |
| Tantalum electrolytic | High (1–680 µF) | 0.1–0.5 Ω | 2.5–50 V | ±10% | Compact bulk, low-ripple analog |
| Ceramic class 1 (C0G/NP0) | Low (pF–nF) | <0.01 Ω | 10–500 V | ±1–5% | Timing, RF, precision filters |
| Ceramic class 2 (X7R/X5R MLCC) | Medium (nF–100 µF) | 0.005–0.05 Ω | 4–100 V | ±10–20% | Bypass on every IC pin |
| Film (polypropylene/polyester) | Low (nF–10 µF) | <0.05 Ω | 50 V–10 kV | ±1–10% | Audio, snubbers, AC line |
| Supercapacitor (EDLC) | Very high (0.1 F–5000 F) | 0.5 mΩ–1 Ω | 2.5–3.0 V/cell | ±20% | Memory back-up, regen brakes |
The choice is driven by the τ you need, the ripple current you must handle, and the frequency band where ESR/ESL matter. Electrolytics dominate where you need millifarads in cubic centimeters; MLCCs dominate everywhere noise must be flat above 1 MHz; supercaps dominate where you need to dump or absorb tens of joules in a second.
Real-world numbers
- 0.1 µF MLCC bypass cap placed within 3 mm of every IC supply pin. With a 0.5 Ω parasitic loop, τ ≈ 50 ns — fast enough to source the current spike of a CMOS gate switching in 2 ns.
- 10 mF aluminum electrolytic on a 12 V regulator output. Charged through a 0.1 Ω source, τ = 1 ms — power-on settles in 5 ms.
- Maxwell BCAP3000 supercapacitor: 3000 F at 2.7 V, holding ½·3000·2.7² = 10,935 J ≈ 11 kJ per cell. Bus regen-brake banks chain dozens of these to reclaim 30–40% of braking energy.
- Camera flash capacitor: 200 µF at 330 V, charged through a 5 V boost converter. Energy ½·200µ·330² = 10.9 J. Dumped through a xenon tube in ~1 ms — that's 10 kW peak.
- Tantalum on a smartphone PMIC rail: 47 µF, ESR 0.1 Ω, volume of a grain of rice. Replaces three aluminum electrolytics that would not fit.
Variants
- Charging vs discharging. Same τ, mirrored curve. V_charge = V₀(1−e^(-t/τ)); V_discharge = V₀·e^(-t/τ). The 36.8% number for discharge is the natural counterpart of the 63.2% charge number.
- Tantalum vs aluminum electrolytic. Tantalum is smaller, has flatter ESR vs frequency, and tolerates higher temperature — but fails as a short circuit when overstressed (sometimes catastrophically, with audible pops). Aluminum is bigger, dries out over decades, and fails open or as ESR creep.
- Constant-current charging. Dump a fixed current i into the cap and the voltage rises linearly: V(t) = (i/C)·t. Used in flash strobes and photodiode readouts; the integrating cap on a CMOS image sensor pixel works exactly this way.
- Two-stage RC. Cascading two RC sections gives an S-shaped step response — the second section can't begin to charge until the first develops voltage. A useful trick to soft-start motors and avoid inrush.
- Constant-current discharge. Replace the resistor with a current source and the cap voltage falls linearly. This is how supercapacitor energy is metered: linear droop means E = ½C(V₁² − V₂²) is easy to bookkeep.
Failure modes
- ESR-driven heating. Ripple current i flowing through ESR dissipates i²·ESR. For a 1 A RMS ripple in a 0.5 Ω ESR electrolytic that's 0.5 W — enough to dry out the electrolyte over years and accelerate failure. Switching-supply caps are the #1 reason consumer electronics die at the 5–10 year mark.
- Voltage derating ignored. Class 2 ceramics lose 50–80% of their nominal capacitance at rated voltage due to dielectric saturation. A 10 µF X7R rated at 10 V might be only 2 µF at 9 V. Designers who don't read the DC bias curves blow stability margins.
- Inrush current. An empty large cap looks like a short circuit at t=0. A 4700 µF cap on a 24 V rail through 0.05 Ω draws 480 A peak — enough to weld switch contacts and trip breakers. Mitigated with NTC thermistors, soft-start FETs, or pre-charge resistors.
- Reverse polarity on electrolytics. Aluminum and tantalum electrolytics rely on a chemically formed oxide layer that only insulates one direction. Reverse them and the layer breaks down, the electrolyte boils, and the can vents (or, for tantalum, ignites). Always check the silkscreen polarity stripe.
- Mechanical cracking on ceramics. MLCCs are brittle and small; PCB flex during depanelization or in-field shock cracks them silently. Cracks short or open intermittently. Place chip caps perpendicular to PCB stress lines and avoid edge placement.
- Dielectric absorption. A discharged cap can spontaneously redevelop voltage as molecules in the dielectric relax — high in electrolytics, near-zero in C0G ceramics. Hazardous in high-voltage systems where shorted-out caps still bite hours later.
Where the curve shows up
- Power-on settling on every regulator output.
- RC oscillators (555 timer, neon-lamp relaxation oscillators).
- Sample-and-hold amplifiers in ADCs — the hold cap must charge to within ½ LSB before the converter samples.
- Camera flashes, defibrillators, rail-gun pulsers.
- Touchscreen capacitance sensing — a finger changes C, which shifts τ, which an MCU reads as a count.
- Soft-start circuits that ramp gate voltage on a power FET to avoid inrush.
Frequently asked questions
Why is the charging curve exponential rather than linear?
Charging current is set by the voltage across the resistor, which is V₀ minus the voltage already on the capacitor. As the capacitor fills, the driving voltage shrinks, so current — the rate of charge accumulation — shrinks proportionally. That self-limiting feedback is what produces dV/dt ∝ (V₀ − V), whose solution is V(t) = V₀(1 − e^(-t/RC)).
Why do engineers say five time constants is fully charged?
After 5τ the capacitor reaches 1 − e⁻⁵ ≈ 99.33%. The remaining 0.67% is below the noise floor of most measurements, so designs treat 5τ as the practical settling time. For 0.1% precision you need about 7τ; for 12-bit ADC accuracy (≈0.024%) you need 8.3τ.
Does ESR change the time constant?
Yes — equivalent series resistance adds to the source resistance, so the effective τ is (R_source + ESR) × C. For a 100 µF aluminum electrolytic with 0.5 Ω ESR charging through a 1 Ω source, ESR raises τ by 50%. ESR also dissipates I²·ESR as heat, which is why bulk caps in switching supplies run warm.
Why does every IC supply pin need a 0.1 µF bypass cap?
When a CMOS gate switches it draws a fast current spike. PCB trace inductance prevents the bulk supply from delivering that spike quickly, so a small ceramic capacitor placed within millimeters of the pin acts as a local reservoir. The 0.1 µF MLCC has low ESR/ESL up to ~100 MHz, exactly the band where digital edges live.
Can supercapacitors replace batteries?
Only for short-term or high-power roles. A 5000 F Maxwell supercap stores about 70 Wh/kg energy density vs 250 Wh/kg for lithium-ion, but delivers 5–10× more power density and survives a million cycles. Buses and cranes use them for regenerative braking; cameras use them for memory back-up. They lose voltage linearly during discharge, which simplifies energy accounting but complicates load regulation.
Why does the first cycle of a flash circuit take seconds but the second seems instant?
First charge starts from zero so it takes ~5τ. After firing, the capacitor still holds residual voltage (the trigger only dumps part of the stored charge through the lamp), so the next charge has less voltage to make up. Modern flashes also use boost converters with low-impedance output, shrinking τ. Recovery time is roughly proportional to the energy delivered in the previous flash.