Protection

Intent

Estimate the peak current and the joule integral when a supply is switched onto an empty input capacitor.
Choose a voltage ramp for a given inrush current, and know where its energy goes.
Know what warm air does to a fuse, and what a series diode against a reversed battery costs.

Theory

Inrush from a step

At the instant a supply is switched onto an empty capacitor, the capacitor has no voltage across it. When the wiring’s inductance is small, its own time constant (the inductance over the series resistance) far shorter than $R_\mathrm{ser} C_\mathrm{in}$ , the whole step falls across the resistance in series with the capacitor: the source’s, the wiring’s and the capacitor’s own ESR. That resistance then sets the peak [Alexander & Sadiku 2017, Ch. 7 (First-Order Circuits)]:

I_\mathrm{inrush} = \frac{V_g}{R_\mathrm{ser}}

$I_\mathrm{inrush}$: peak inrush current
$R_\mathrm{ser}$: series resistance of the charging path
$V_g$: input voltage

Peak inrush current into an empty capacitor✓ verified by 7 tests

Assumptions, convention and source

Assumptions

an empty capacitor charged from a voltage step through a constant series resistance; the wiring's inductance negligible (its time constant, L over the series resistance, far shorter than the R-C one)

Notes

at the step the empty capacitor has no voltage across it, so the whole step falls across the series resistance, and the current then decays with the time constant

R_\mathrm{ser} C_\mathrm{in}

. In this R-C model the capacitance does not limit the peak; inductance in the leads slows the rise, lowers the peak (which then depends on the capacitance too) and, with little resistance, makes the capacitor's voltage overshoot the supply

Source

Alexander & Sadiku 2017, Ch. 7 (First-Order Circuits); derived here

Derivation

step-by-step sympy derivation

Equation id

inrush.I_pk · equations.yaml entry

The current then decays with the time constant $R_\mathrm{ser} C_\mathrm{in}$ . The integral of its square over the pulse, its joule integral, is

I^2 t = \frac{V_g^{2} C_\mathrm{in}}{2 R_\mathrm{ser}}

$I^2 t$: joule integral of the pulse
$C_\mathrm{in}$: input capacitance
$R_\mathrm{ser}$: series resistance of the charging path
$V_g$: input voltage

Joule integral of the inrush pulse✓ verified by 7 tests

Assumptions, convention and source

Assumptions

an empty capacitor charged from a voltage step through a constant series resistance; the wiring's inductance negligible (its time constant, L over the series resistance, far shorter than the R-C one)

Notes

the integral of the decaying current squared; times

R_\mathrm{ser}

it is the energy the resistance dissipates while the capacitor charges,

C_\mathrm{in} V_g^2/2

, as much as the capacitor stores whatever the resistance. Inductance in the leads does not change it: the source still delivers

C_\mathrm{in} V_g^2

and the capacitor keeps half

Source

Alexander & Sadiku 2017, Ch. 7 (First-Order Circuits); derived here

Derivation

step-by-step sympy derivation

Equation id

inrush.I2t · equations.yaml entry

A resistance in the path times the joule integral is the heat the pulse leaves in it. The series resistance as a whole takes as much as the capacitor stores, whatever its value (energy stored in a capacitor), and each part of it (a fuse’s element, a switch, the wiring) takes its own share.

More inductance in the leads slows the current’s rise and lowers its peak, which then depends on the capacitance too; with little resistance, the capacitor’s voltage rings above the supply. The joule integral does not change: the source still delivers twice what the capacitor stores, and the resistance dissipates the rest.

A controlled ramp

Making the voltage across the capacitor rise more slowly reduces the inrush [Texas Instruments 2015]. A capacitor’s current is its capacitance times the rate of change of its voltage, so a linear ramp from zero draws a constant current:

I_\mathrm{ramp} = \frac{C_\mathrm{in} V_g}{t_\mathrm{ramp}}

$I_\mathrm{ramp}$: current of a controlled ramp
$C_\mathrm{in}$: input capacitance
$V_g$: input voltage
$t_\mathrm{ramp}$: ramp time

Inrush current with a controlled voltage ramp✓ verified by 7 tests

Assumptions, convention and source

Assumptions

an empty capacitor whose voltage is ramped linearly from zero to its final value
the ramp is set by an element in series with a fixed supply V_g (a load switch or a hot-swap switch)

Notes

a capacitor's current is

C \, dv/dt

, so a constant slope draws a constant current, inversely proportional to the ramp time. With no other resistance in the path, the element that sets the ramp in series with the supply drops

V_g - v_C

while carrying it and dissipates

C_\mathrm{in} V_g^2/2

over the ramp, as much as a series resistance would from a step: the ramp lowers the current and spreads the energy over time, but does not remove it. A source that ramps its own output (a converter's soft start) has no such element: its path loses only the resistance's share,

2 R_\mathrm{ser} C_\mathrm{in}/t_\mathrm{ramp}

of the stored energy

Source

Alexander & Sadiku 2017, Ch. 6 (Capacitors and Inductors); derived here

Derivation

step-by-step sympy derivation

Equation id

inrush.I_ramp · equations.yaml entry

When an element in series with a fixed supply (a load switch or a hot-swap switch) sets the ramp, it carries that current while it drops the rest of the supply’s voltage. Over the ramp, the path still dissipates as much as the capacitor stores, and nearly all of it in that element when the ramp is long against $R_\mathrm{ser} C_\mathrm{in}$ . The ramp lowers the current and spreads the energy over time; it does not remove it. A source that ramps its own output, such as a converter’s soft start, has no such element to heat.

A fuse in warm air

The higher the ambient temperature, the hotter a fuse operates, and the shorter its life [Littelfuse, Fuseology]. A fuse beside a hot heat sink or inside a warm enclosure runs in a higher ambient than the same fuse on an open bench.

A reversed battery

A diode in series with the battery protects against a battery connected backwards [Texas Instruments 2003]: reversed, the diode blocks. The right way round, it carries the whole load current for as long as the circuit runs, and costs its forward drop times the average current, plus its resistance’s share (diode conduction loss). The drop is a larger fraction of a lower battery voltage, and so is the loss.

Worked example

A supply switched onto an empty input capacitor, and the same capacitor brought up by a controlled ramp:

Inrush example

Symbol	Value	How it is obtained
$V_g$ input voltage	48 V	given
$R_\mathrm{ser}$ series resistance of the charging path	100 mΩ	given
$C_\mathrm{in}$ input capacitance	100 µF	given
$t_\mathrm{ramp}$ ramp time	4.8 ms	given
$I_\mathrm{inrush}$ peak inrush current	480 A	Peak inrush current into an empty capacitor
$I^2 t$ joule integral of the pulse	1.15 A²s	Joule integral of the inrush pulse
$I_\mathrm{ramp}$ current of a controlled ramp	1 A	Inrush current with a controlled voltage ramp

Through 100 mΩ in all, the empty 100 µF capacitor draws 480 A at the step, and the pulse’s joule integral is 1.15 A²s. Ramped to 48 V over 4.8 ms, the same capacitor draws a constant 1 A. Nearly all the energy the series resistance took at the step now goes to the ramp’s element.

Try it

▶ Open in the equation explorer: Peak inrush current into an empty capacitor
Preset from the example bench-inrush

Sweep the series resistance: the peak follows its inverse. A capacitor of lower ESR draws a higher peak.

▶ Open in the equation explorer: Inrush current with a controlled voltage ramp
Preset from the example bench-inrush

Sweep the ramp time: doubling it halves the current.

Bench exercise

Start from a low voltage and a current-limited source. With a current probe on the input lead, switch the source onto a converter’s empty input capacitor and capture the current’s peak and its decay; a current limit in the source caps the peak, so note it. Estimate the path’s resistance: the wiring’s and the switch’s with a four-wire milliohm meter, the capacitor’s ESR from its data sheet or an impedance meter at a high frequency, and the source’s from its data sheet or its voltage drop under a known load step. Compare the peak with the step’s equation: a lower peak and a capacitor voltage ringing above the supply point to the leads’ inductance. Square the captured current in the oscilloscope’s math, integrate it, and compare with the joule integral. Then slow the switch’s turn-on and repeat: the peak falls. Capture the switch’s voltage as well; its product with the current, integrated over the ramp, is the energy the switch now absorbs.

Gotchas

In the R-C model the capacitance does not limit the peak from a step; only the series resistance does (peak inrush current). Inductance in the leads lowers the peak and can make the capacitor’s voltage ring above the supply.
A ramp set by a series element does not remove the energy. The path still dissipates as much as the capacitor stores, now mostly in that element (ramped inrush).
A warm fuse ages faster [Littelfuse, Fuseology].
A series diode costs power for as long as the circuit runs, not only when something goes wrong (diode conduction loss).

Go deeper

Application note Managing Inrush Current (SLVA670A) — Texas Instruments · retrieved 2026-09-24
Where inrush current comes from when a load's capacitance is switched on, and how a slower voltage rise reduces it.
Application note Fuseology: Fuse Facts and Fuse Selection Guide — Littelfuse · retrieved 2026-09-24
How a fuse's ratings are defined, and what its current, its pulses and its ambient temperature do to it.
Application note Reverse Current/Battery Protection Circuits (SLVA139) — Texas Instruments · retrieved 2026-09-24
Circuits that protect a load against a battery connected backwards, starting from a diode in series with the battery.

Quiz

A 24 V supply is switched onto an empty 470 uF capacitor through 0.05 ohm in all, with negligible wiring inductance. What is the peak current?
1. 24 A
2. 11.3 A
3. 480 A
4. 0.2 A
Answer and explanation
Correct answer: C The whole step falls across the series resistance: 24 V over 0.05 ohm.
The capacitor is replaced by one of half the capacitance with the same ESR. What happens to the peak?
1. It stays the same
2. It halves
3. It doubles
4. It falls to zero
Answer and explanation
Correct answer: A In the R-C model, at the step the capacitor has no voltage across it, whatever its size: only the series resistance limits the peak. The pulse is shorter, since the time constant halves.
What is the joule integral of the first pulse (24 V, 470 uF, 0.05 ohm)?
1. 5.41 A^2 s
2. 1.35 A^2 s
3. 2.71 A^2 s
4. 0.14 A^2 s
Answer and explanation
Correct answer: C V^2 C / (2 R) = 576 times 470e-6 over 0.1, which is 2.71 A^2 s.
The same 470 uF capacitor is brought up to 24 V by a linear ramp. How long must the ramp be for a 2 A inrush?
1. 2.82 ms
2. 11.3 ms
3. 0.564 ms
4. 5.64 ms
Answer and explanation
Correct answer: D The current is C V over the ramp time, so the ramp time is C V over the current: 470e-6 times 24 over 2.
A load switch in series with the fixed 24 V supply sets that ramp. Compared with the step, how much energy does the path dissipate while the capacitor charges?
1. None: a slow ramp is lossless
2. The same, as much as the capacitor stores, now mostly in the load switch
3. Half as much
4. Twice as much
Answer and explanation
Correct answer: B The supply delivers C V^2 and the capacitor keeps C V^2/2 either way; the ramp lowers the current and spreads the loss over time.
What does a higher ambient temperature do to a fuse?
1. Nothing, a fuse depends only on its current
2. It opens at a higher current
3. It lasts longer
4. It operates hotter and its life is shorter
Answer and explanation
Correct answer: D The higher the ambient temperature, the hotter the fuse operates and the shorter its life.
A diode in series with a 12 V battery protects its load against reversal. The load draws 2 A on average and the diode drops 0.7 V. Ignoring its resistance, what does the diode cost?
1. Nothing: it acts only when the battery is reversed
2. 1.4 W, for as long as the circuit runs
3. 24 W
4. 0.35 W
Answer and explanation
Correct answer: B It carries the whole load current whenever the battery is the right way round: 0.7 V times 2 A.

The numbers in the examples, presets and exercises on this site are synthetic: round teaching values, not data from any real design.