Application Note · AN-014 · Power Management
The Inrush Current That Passes Every Bench Test and Blows the Fuse in the Field
A capacitor bank that looks harmless on the bench can pull 20× its rated current in the first 100 microseconds of power-up — enough to nuisance-trip a fuse, melt a FET, or brown out a neighbouring rail. Here is why it hides during testing, and how a controlled-slew load switch like the TPS22919 fixes it for a few rupees.
1. What inrush current actually is
Every power rail in your product feeds a bank of bulk and bypass capacitors. At the instant you connect that rail, those capacitors are empty — and an empty capacitor looks, for a fraction of a millisecond, exactly like a short circuit.
The physics is one line. The current into a capacitor is set by how fast the voltage across it changes:
If a hard switch slams the rail from 0 V to 5 V in, say, 1 microsecond, then dV/dt = 5 V/µs. Across 100 µF of downstream capacitance that demands:
You never see 500 A in reality — the source impedance, trace resistance, connector resistance and the capacitor’s own ESR clamp it. But you very easily see 5 A to 30 A spikes on a board whose steady-state current is 200 mA. That ratio — inrush to running current — is commonly 5× to 100× in real equipment.
2. Why it stays hidden during testing
This is the part that catches experienced engineers, not just beginners. Inrush is a statistical, condition-dependent event, and a typical bench test samples only a tiny, friendly corner of the conditions the product will actually meet.
It is too fast for your instruments
The inrush spike lives in the 1–100 µs window. A handheld DMM averages over ~100 ms and reports a number 1000× too low. Even a bench supply’s current readout is filtered. Unless you put a current probe or a sense resistor on a scope with the timebase set to microseconds, the spike is literally invisible — the rail “just turns on.”
It depends on when in the AC cycle (or how fast) you apply power
Inrush magnitude depends on the instantaneous source voltage and on how quickly the contact closes. Flip a clean bench switch slowly and you may get a soft make; plug a connector hard, or hot-swap a board, and you get the worst case. Ten clean power-ups on the bench can all pass while the eleventh — a customer yanking a connector — fails.
It gets worse cold, and the bench is warm
Two things conspire at low temperature. Copper windings and traces have lower resistance when cold, so by Ohm’s law the peak current is higher. And many passive inrush limiters — NTC thermistors especially — start cold with high resistance but only work after they have self-heated; on a genuinely cold first power-up they have not warmed yet. Your lab at 25 °C never reproduces the −10 °C dawn start in the field.
A spike that peaks at 90% of the fuse’s melting threshold passes every test. Then a cold morning lowers the wiring resistance, a hot connector closes at the wrong instant, and the same spike now peaks at 110%. The fuse blows — or worse, it survives 50,000 cycles of near-melting and then opens months later for “no reason.” Margin you cannot see on the bench is margin you do not have.
3. Worked example: the LED beacon that died in winter
Consider a real-world style case. A 5 V industrial signalling beacon draws 0.8 A steady-state. Its input has 220 µF of bulk capacitance behind a hard P-channel MOSFET that switches in roughly 2 µs. It is protected by a 2 A fast-blow fuse. The unit passed every line of the test plan.
| Parameter | Bench (25 °C) | Field (−5 °C, hot plug) |
|---|---|---|
| Steady-state current | 0.80 A | 0.82 A |
| Effective series R (source+trace) | 0.25 Ω | 0.18 Ω |
| Peak inrush | ~1.6 A | ~2.4 A |
| Spike width | ~40 µs | ~55 µs |
| Fuse I²t margin | passes | blows intermittently |
On the bench the 1.6 A peak sat comfortably under the 2 A fuse, and because it was only 40 µs wide the fuse’s I²t (energy) rating shrugged it off. In the field, the colder, lower-resistance loop let the peak climb past 2.4 A. Fast-blow fuses respond to I²t, and the higher, slightly wider pulse pushed the delivered energy over the melting integral. Result: random “dead on arrival after a cold night” returns that no one could reproduce indoors.
4. How inrush blows fuses and overheats FETs
The same energy spike attacks two components at once.
The fuse: nuisance trips and slow degradation
A fuse element is a calibrated piece of metal that melts at a known I²t. A repeated near-threshold inrush pulse work-hardens and thins it slightly each cycle. So you get two failure modes: an immediate nuisance blow when one cold start tips over the line, and a latent failure where the fuse opens after thousands of cycles that each looked fine. Up-sizing the fuse to “fix” the nuisance trips is the dangerous wrong answer — it defeats the protection the fuse exists to provide.
The switching FET: a thermal event in microseconds
During inrush the series pass element — a discrete MOSFET or the FET inside a load switch — sees nearly the full rail across it while carrying the peak current. Instantaneous power is V×I, and it can hit tens of watts for those tens of microseconds. The die has almost no time to spread that heat, so junction temperature spikes locally. Do that thousands of times and you accumulate thermo-mechanical fatigue; do it once badly into a charged-but-shorted cap and you can exceed the FET’s Safe Operating Area (SOA) and crack the die. This is the overtemperature path the question raises: the FET can overheat from inrush alone, with no steady-state fault present at all.
A protected switch with a fixed current limit will chop the inrush peak — good. But if the downstream cap is large, the switch then sits in current-limit dropping the full rail for a long time, dissipating heat until thermal shutdown fires and the rail never comes up. The right answer is not just a current limit; it is to slow the ramp so the peak never demands the limit in the first place.
5. The fix: a controlled-slew load switch
Go back to i = C·dV/dt. The capacitance C is fixed by your design. The only free variable is dV/dt — the slew rate of the output. Slow the rise and the inrush falls proportionally.
A controlled-slew (a.k.a. controlled rise-time or soft-start) load switch is an integrated FET whose gate is driven by an internal current source so the output ramps at a defined, repeatable rate. You convert a violent step into a gentle ramp, and the inrush becomes a small, flat, predictable current you can actually design a fuse and a thermal budget around.
That is the whole strategy in one equation: pick a slew rate slow enough that C×dV/dt lands well under your fuse and FET limits across all temperatures and insertion conditions — not just the warm bench best case.
6. Designing with the TPS22919
The Texas Instruments TPS22919 is a single-channel, self-protected load switch with a fixed controlled rise time, in a tiny 6-pin SC-70 (2.1 mm × 2.0 mm) package. It is built for exactly this job — and it is the heart of the Ramanu PureSwitch 22919 load switch module, a ready-to-use breakout you can drop into a rail without spinning your own board. Its relevant specs:
| Parameter | Value |
|---|---|
| Input voltage range (VIN) | 1.6 V – 5.5 V |
| Max continuous current (IMAX) | 1.5 A |
| On-resistance (RON) @ 5 V | 89 mΩ typ |
| Controlled slew rate @ 5 V | 3.2 mV/µs → t_ON ≈ 1.95 ms |
| Controlled slew rate @ 1.8 V | 1.8 mV/µs → t_ON ≈ 1.5 ms |
| Short-circuit protection (ISC) | 3 A typ |
| Thermal shutdown | built in |
| Quick Output Discharge (QOD) | 24 Ω internal, adjustable |
| Quiescent current (on) | 8 µA typ |
Worked sizing — the TI design point
Take the datasheet’s own example. To hold inrush to 150 mA into a given output capacitance, the required slew rate works out to 3.2 mV/µs. The TPS22919 ramps at 2.3 mV/µs under that condition — slower than required — so the resulting inrush comes in comfortably below 150 mA. You go from a multi-amp uncontrolled spike to a sub-150 mA flat ramp with one part and zero external timing components.
Re-run our beacon: replace the hard P-FET with a TPS22919 driving the 220 µF bank. With a ~3 mV/µs ramp, output rises 0→5 V in roughly 1.7 ms, so:
The 2.4 A cold-field spike is gone — replaced by a 0.65 A plateau that the 2 A fuse never notices, at any temperature. As a bonus you inherit short-circuit limiting and thermal shutdown for genuine downstream faults, and QOD to discharge the rail cleanly on power-down for proper sequencing.
The TPS22919 has a fixed rise time — excellent for sub-1.5 A rails where you want a no-tuning, low-BOM solution. If you need to dial the ramp to a specific value, carry more current, or set a precise adjustable current limit, step up to a part with a C_T slew pin and adjustable I_LIM (e.g. the TPS259x / TPS22965 families). Same principle, more knobs.
Key takeaways
- Inrush current is set by
i = C·dV/dt; an empty capacitor looks like a short at switch-on. - It hides on the bench because it is microseconds wide, condition-dependent, and worse when cold than at lab temperature.
- Fuses blow on I²t energy, not just peak current — a pulse that passes warm can exceed the melting integral cold.
- The pass FET can overheat from inrush alone, with no steady-state fault, accumulating SOA/thermal-fatigue damage.
- A controlled-slew load switch like the TPS22919 slows
dV/dtso inrush becomes a small flat ramp, restoring margin at every temperature — plus free short-circuit and thermal protection.
7. Design checklist
| Step | What to verify |
|---|---|
| 1 | Sum total downstream capacitance on the rail (bulk + bypass). |
| 2 | Pick a target inrush ceiling — typically a fraction of your fuse/OCP rating. |
| 3 | Solve for required slew rate: dV/dt = I_target / C_load. |
| 4 | Choose a switch whose controlled slew is at or below that rate (TPS22919 ≈ 3.2 mV/µs @ 5 V). |
| 5 | Check the switch’s own SOA: peak current × V_rail × ramp time must stay inside its thermal limit. |
| 6 | Measure with a current probe on a scope at µs timebase — not a DMM. |
| 7 | Re-test cold and with hot-plug insertion, not just a clean warm power-up. |
8. Frequently asked questions
Can’t I just add a series resistor or an NTC thermistor?
A fixed series resistor wastes power continuously and drops voltage at full load. An NTC limits the first surge but only after it self-heats — so a genuinely cold first power-up, the exact field case that fails, gets little protection, and the NTC stays hot in steady state. A controlled-slew switch limits inrush on every start regardless of temperature and adds no steady-state drop beyond its low RON.
Why not just fit a bigger fuse to stop the nuisance trips?
Because the fuse exists to protect against real faults. Up-sizing it to tolerate inrush widens the window in which a genuine overcurrent can damage the board or start a fire before the fuse opens. Fix the inrush, keep the fuse correctly rated.
Does a current-limit-only load switch solve this?
Partially. A fixed current limit caps the peak, but into a large capacitor the switch then sits in limit dissipating heat until thermal shutdown may trip and the rail fails to come up. Controlling the slew rate so the peak never reaches the limit is the cleaner fix; the best parts do both.
How do I actually see inrush on the bench?
Put a low-value sense resistor (or a current probe) in series with the rail and trigger a scope on the rising edge with the timebase at 10–50 µs/div. A DMM or the supply’s current meter averages the spike away and will tell you everything is fine when it is not.
What package and rail does the TPS22919 suit?
6-pin SC-70, 1.6–5.5 V in, up to 1.5 A — ideal for 1.8 V / 3.3 V / 5 V sensor, MCU, module and small-rail soft-start applications. For higher current or adjustable slew, move to a larger TPS family part.
Get the PureSwitch 22919 module
Ramanu Systems designs precision power and sensor modules in India with controlled soft-start, clean rails, and field-grade margin built in — not bolted on after the returns come back. The PureSwitch 22919 puts everything in this note on a ready-to-use board.