100nF

Reverse polarity protection calculator

A Schottky diode, a self-driven P-channel MOSFET or an N-channel FET with an ideal-diode controller: each stops a reversed supply, and each costs something in the forward direction. TI's SLVAE57B measured all three at 10 A on 12 V — 465 mV and 4.65 W for the diode, 35 mV and 0.35 W for the MOSFET, 20 mV regulated for the controller — and the calculator scales those to your current and supply: the drop, the power, the share of the supply lost and the temperature rise for each, the current above which the ideal diode is no worse than a plain FET, the P-channel's gate-drive and gate-rating checks, and what a hot Schottky dissipates when the supply is actually reversed.

0.0 W1.9 W3.8 W5.8 W7.7 W0.0 A3.8 A7.5 A11 A15 Aload current — power dissipated in the protection elementload 10.0 Ared: Schottky, 465 mV × Iblack: P-channel FET on, I² × 3.50 mΩblue: ideal diode, 20.0 mV regulated then I² × 3.50 mΩ
Fig 1 — What each element burns against load current. The Schottky's loss is linear in current at its forward drop; the MOSFET's is I²R, tiny at low current; the ideal-diode controller holds 20.0 mV across its FET to keep reverse-current detection, so below 5.71 A it costs more than a plain FET and above it the two are the same. SLVAE57B's figure 6-4 generalised from its 10 A measurements.
Schottky: drop · power · loss · rise
465 mV · 4.65 W · 3.88 % · +186 °C
P-channel 3.50 mΩ on: drop · power · loss · rise
35.0 mV · 350 mW · 0.29 % · +18 °C
Ideal diode, 20.0 mV regulated: drop · power · loss · rise
35.0 mV · 350 mW · 0.29 % · +18 °C
Ideal diode crossover V_reg / R_DS(on) · Schottky leakage loss when reversed
5.71 A · 6.00 W
Voltage left for the load from 12.0 V: Schottky · P-FET · ideal
11.5 V · 12.0 V · 12.0 V

The Schottky burns 4.65 W at 10.0 A against 350 mW for the MOSFET — SLVAE57B's "forward conduction results in significant efficiency loss at higher load currents", with a heat sink "to manage power dissipation, increasing cost and space". Its 465 mV also comes off the supply: at a cold-crank "3 V or 4 V" that is headroom the downstream converter does not get.

Reversed, the Schottky's 100 mA of leakage at 60.0 V is 6.00 W — SLVAE57B's example, "which amounts to 6 W of power dissipation at −60 V". Leakage "increase[s] drastically with temperature", so a hot diode in a reversed supply runs away; the FET solutions dissipate nothing when off.

How this is calculated

Standard: TI SLVAE57B

PSchottky=VF I,PPFET=I2RDS(on)P_{Schottky} = V_F\,I, \qquad P_{PFET} = I^2 R_{DS(on)}
SLVAE57B's 10 A measurements: 465 mV → 4.65 W; 35 mV → 0.35 W, i.e. 3.5 mΩ.
Videal=max⁡(Vreg, IRDS(on)),Icross=VregRDS(on)V_{ideal} = \max\left(V_{reg},\ I R_{DS(on)}\right), \qquad I_{cross} = \frac{V_{reg}}{R_{DS(on)}}
The controller regulates V_reg (20 mV) across the FET until I·R exceeds it: 5.7 A for 3.5 mΩ.
ΔT=P θJA,loss=VdropVsupply\Delta T = P\,\theta_{JA}, \qquad \text{loss} = \frac{V_{drop}}{V_{supply}}
Temperature rise in the package as mounted, and the share of the supply the element takes.
Preverse=IR VRP_{reverse} = I_R\,V_R
The Schottky's leakage dissipation with the supply reversed: SLVAE57B's 100 mA at 150 °C and −60 V is 6 W.

Assumptions

What sets the cost of reverse polarity protection

Reverse polarity protection is a one-way valve in the supply, and the price of a valve is its drop in the forward direction. SLVAE57B lists the three ways to build one and measures them at 10 A on 12 V. The Schottky diode is "one part, one voltage drop": "when the battery is installed correctly, load current flows through the forward biased schottky diode. When the battery polarity is reversed, the schottky diode is reverse biased and blocks reverse current." Its cost is that "forward conduction results in significant efficiency loss at higher load currents", "heat sink is needed to manage power dissipation", the drop "reduces subsequent power converter head-room" at a cold-crank "3 V or 4 V", and in reverse its leakage "increase[s] dramatically with junction temperature" — the note's 60 V STPS20M60S "has a 100 mA reverse leakage current at 150 °C, which amounts to 6 W of power dissipation at −60 V".

The P-channel MOSFET replaces the drop with I·RDS(on) and drives itself from the supply: "the body diode from MOSFET is forward biased and conducts for a very short time until the MOSFET is turned ON when gate voltage is pulled below source. When the battery polarity is reversed, gate-source voltage swings positive and the MOSFET is turned off." Its conditions are the gate's: the supply must exceed the VGS that gives the datasheet RDS(on), and must not exceed the VGS rating without a Zener. And it "does not block reverse current from flowing back into the input" — an output held up by a capacitor or a second supply flows backwards through the on channel.

The ideal-diode controller drives an N-channel FET and regulates the drop across it — SLVAE57B's 20 mV — so that a reversal of current is a reversal of that voltage, which it detects and turns the FET off on. Below Vreg / RDS(on) it therefore costs more than a plain FET; above, the FET is fully on and the two are the same. The calculator computes the drop, the power, the efficiency loss and the temperature rise for all three at the entered current, the crossover, the Schottky's reverse-state dissipation, and the P-channel's gate checks.

Worked example: SLVAE57B's 10 A on 12 V

The defaults are the note's measured parts: a Schottky at 465 mV, a P-channel at 3.5 mΩ (its 35 mV at 10 A), an ideal diode regulating 20 mV across a 3.5 mΩ FET, 40 and 50 °C/W packages, and the 60 V diode's 100 mA of hot leakage.

Schottky        0.465 V × 10 A     = 4.65 W    3.9 % of the supply   +186 °C at 40 °C/W: a heat sink
P-channel on    10 A × 3.5 mΩ      = 35 mV     0.35 W  0.29 %        +18 °C at 50 °C/W
ideal diode     max(20 mV, 35 mV)  = 35 mV     0.35 W                  same FET, same loss at 10 A
crossover       20 mV / 3.5 mΩ     = 5.7 A     below it the ideal diode holds 20 mV: 60 mW at 3 A, the FET 32 mW
reversed        100 mA × 60 V      = 6 W       in the Schottky; nothing in the FETs
headroom        11.54 V  ·  11.97 V  ·  11.97 V

The 4.65 W is the whole argument: a 12 V, 10 A system spends 4 % of its input in the diode and needs a heat sink to survive it, where the FET spends 0.3 % and a footprint. SLVAE57B's board-area comparison for the FET solutions above 6 A — 140 mm² for the P-channel against 37.1 mm² for an N-channel with its controller, "three times smaller" — is the other half of why serious designs use the N-channel and a controller rather than the self-driven P-channel.

Where the comparison stops being valid

Common reverse polarity mistakes

Further reading