100nF

Inrush current limiter calculator

An NTC inrush current limiter is chosen by three numbers: a cold resistance of at least the peak voltage over the current the fuse or bridge allows, an energy rating above ½CV² of the capacitance it charges, and a steady-state current rating above what the equipment draws. For a 120 V supply with 4700 µF and a 20 A limit that is 8.5 Ω, 67.7 J and the running current — Ametherm's own example, which rounds up to a 10 Ω disc. Enter the line, the capacitance and the limits to get the three ratings, the peak with and without the part, and the charging time; or switch to the MOSFET mode for the ramp current, the FET's dissipation and the timing resistor of TI's soft-start circuit.

025.0 Awithout NTC: 339 A peak, off the topallowed 20.0 A18.9 A with 8.49 Ω cold42.2 ms84.5 ms127 ms169 mscapacitor charging current, NTC held cold (worst case)
Fig 1 — Charging 4.70 mF from 170 V peak: 8.49 Ω of cold NTC holds the first peak to 18.9 A instead of 339 A. The real current falls faster than this once the part heats.
Peak voltage the capacitor charges to
170 V
Peak with only 500 mΩ in the loop
339 A
Cold resistance R25, at least
8.49 Ω → peak 18.9 A
Energy per turn-on, ½CV²
67.7 J · TDK C_test ≈ 963 µF at 375 V
Steady-state I_max needed
3.00 A
Charging time constant · 95 %, part cold
42.2 ms · 127 ms
A fixed 8.49 Ω resistor would burn
76.4 W continuously

Round up: pick the next stock R25 above 8.49 Ω, an I_max above 3.00 A and an energy or C_test rating above 67.7 J. If no single part meets the energy, put two in series — never in parallel (TDK).

How this is calculated

Standard: Vishay Ametherm 24002; TDK/EPCOS NTC ICL application notes; TI SLVA156

Vpeak=2 VRMS(AC),Ipeak=VpeakRloopV_{peak} = \sqrt{2} \, V_{RMS} \quad (\text{AC}), \qquad I_{peak} = \frac{V_{peak}}{R_{loop}}
The first-cycle peak into a discharged capacitor.
R25≥VpeakIallowedR_{25} \geq \frac{V_{peak}}{I_{allowed}}
Ametherm's zero-power resistance: peak voltage over the fuse or breaker rating.
E=12CVpeak2,Ctest=2E3752E = \tfrac{1}{2} C V_{peak}^2, \qquad C_{test} = \frac{2E}{375^2}
Energy per turn-on. The second form is TDK's rating expressed as a capacitance discharged from 375 V; equal energy, derived here.
Imax,required=Issk(TA),k=1 to 65 °C,  0.9 at 75 °CI_{max,required} = \frac{I_{ss}}{k(T_A)}, \quad k = 1 \text{ to } 65\,°\text{C}, \; 0.9 \text{ at } 75\,°\text{C}
Steady-state rating with Ametherm's derating point; interpolated between.
τ=(R25+Rloop) C,t95%=3τ\tau = (R_{25} + R_{loop})\, C, \qquad t_{95\%} = 3\tau
Charging with the part held cold — the slow bound.
Iramp=CVtSS,EFET=12CV2,RT≈(VO−VTH) tSSCGD VOI_{ramp} = \frac{C V}{t_{SS}}, \qquad E_{FET} = \tfrac{1}{2} C V^2, \qquad R_T \approx \frac{(V_O - V_{TH})\, t_{SS}}{C_{GD}\, V_O}
MOSFET soft start: the constant ramp current, the energy the FET dissipates, and SLVA156 eq 1 for the timing resistor.

Assumptions

What sets the inrush current

At the instant a supply is switched on, its bulk capacitor is discharged and looks like a short circuit. EPCOS's application note says exactly that: the high currents "are caused by the extremely low impedance of smoothing capacitors or coils which almost produce short circuits at the moment of switching on". The first peak is then the peak voltage over whatever resistance happens to be in the loop — source, wiring, bridge, ESR — and with half an ohm on a 120 V line that is over 300 A. It lasts a few milliseconds and it is enough to blow a fuse that would carry the running current for ever, to exceed a bridge rectifier's surge rating, and to weld the contacts of the switch that closed.

An inrush current limiter puts resistance in the loop for the first few cycles and takes it out again. The NTC thermistor does that by itself: a high resistance cold, and as the current heats it, a resistance that drops "by a factor of 10 to 50" (TDK) to a few percent of its cold value. Ametherm's example: a 10 A supply that would draw 100 A at turn-on is held to 35 A by a 10 Ω part that then settles at 0.05 Ω. A fixed resistor of the same value would burn 500 W at 10 A, which is why the alternative to an NTC is not a resistor but a resistor with a relay across it, or a MOSFET that does the same job without contacts.

Three numbers pick the NTC, and both vendors list the same three. The cold resistance sets the peak: Ametherm's rule is the peak voltage over the highest current the fuse or bridge allows. The energy per turn-on, ½CV² at the peak voltage, has to be below the part's rating — TDK expresses it as a maximum capacitance the part can be discharged through from 375 V, and the tool converts. And the steady-state current has to be below the part's Imax, which the datasheets rate up to 65 °C; Ametherm's transformer example applies 90 % at 75 °C. The cold resistance says how much limiting, the other two say how big a disc.

Inrush energy chart: what the limiter absorbs

The energy that charges the bulk capacitor, ½CV² at the peak of the supply, computed by the calculator above. It is the number an NTC is rated by, and it is the number that surprises: it goes as the square of the voltage, so the same capacitor on a 230 V line stores nearly four times what it stores on 120 V.

Bulk C24 V DC48 V DC120 V AC230 V AC
470 µF135 mJ541 mJ6.77 J24.9 J
1000 µF288 mJ1.15 J14.4 J52.9 J
2200 µF634 mJ2.53 J31.7 J116 J
4700 µF1.35 J5.41 J67.7 J249 J
10000 µF2.88 J11.5 J144 J529 J

Minimum cold resistance chart

Ametherm's first criterion, the cold resistance that holds the first peak to what the fuse or bridge allows, Vpeak / Iallowed, with nothing else in the loop. Resistance already there, in the source, the wiring and the capacitor, comes off this figure.

Allowed peak24 V DC48 V DC120 V AC230 V AC
5 A4.80 Ω9.60 Ω33.9 Ω65.1 Ω
10 A2.40 Ω4.80 Ω17.0 Ω32.5 Ω
20 A1.20 Ω2.40 Ω8.49 Ω16.3 Ω
50 A480 mΩ960 mΩ3.39 Ω6.51 Ω

Worked example: 120 V line, 4700 µF, 20 A fuse

Ametherm's own selection example, which the calculator's defaults reproduce: a 120 V RMS input, 4700 µF of reservoir capacitance, a 20 A limit set by the fuse, 3 A running.

V_peak   = 120 × √2                        = 169.7 V
R25      ≥ 169.7 / 20                      = 8.5 Ω     → next stock value up, 10 Ω
E        = ½ × 4700 µF × 169.7²            = 67.7 J    → a part rated 70 J or more
I_ss     = 3 A at 25 °C                    → I_max ≥ 3 A (no derating below 65 °C)

with 0.5 Ω already in the loop:
peak without the NTC       169.7 / 0.5      = 339 A
peak with 8.5 Ω cold       169.7 / 9.0      = 18.9 A
τ, part cold               9.0 × 4700 µF    = 42 ms;  95 % at 127 ms

Ametherm rounds up on every axis — "8.4 Ω, 3 A, 6.65 J" becomes 10 Ω, 3 A, 7 J in the text, though the 4700 µF example computes to 67.7 J (the paper prints 67.6) as the paper's own arithmetic shows two lines earlier — and lands on a 10 Ω, 3 A disc. The charging time constant is the worst case, with the part held cold; in practice the disc heats within the first cycles and the capacitor charges faster than the curve shows. What does not get faster is the cool-down: TDK gives "1 to 2 minutes" for the resistance to return, EPCOS "30 seconds to two minutes" depending on the disc. Switch the supply off and on within that window and the limiter is warm, low, and limits less.

On the MOSFET side, the calculator's defaults are a 12 V rail with 1000 µF behind the FET ramped over 10 ms: the load charges at a constant 1.2 A, and the FET dissipates ½CV² = 72 mJ during the ramp, 7.2 W average. The ramp time trades the current against how long the FET spends in its linear region with the full rail across it, which is what its safe-operating-area chart is for.

Where the inrush limiter model stops being valid

Common inrush limiter mistakes

Further reading