# Reverse polarity protection using MOSFET - what is the voltage drop after it?

I am using this schematic for reverse polarity protection of an IC (bq78412). However the same IC has an ADC that senses the source voltage and appears not to be tolerant to reverse voltage either. Since the IC is measuring the voltage with +-0.5% at 12V I assume that it would not harm if I measure the voltage after the MOSFET since according to the data sheet its Rds(on) is about 0,082 Ohm at Id = - 3A. The total current consumption of the IC is 3,2 mA. So I assume a voltage drop of 0,082*0,0032= 0,2624 mV which is way below 0,5%*12V = 60 mV.

Is this assumption correct?

Vinput = 12v +-3v Q1 - SI2319DS

Thanks!

The MOSFET is connected in an unusual but entirely valid manner in this circuit.
Some while ago somebody patented this principle - while attempting to ignore the substantial and well documented 'prior art' - people have done this for many decades.
The drain and source are reversed in polarity to "normal" so that the body diode conducts when correct power supply polarity is applied. While the drain - source polarity is reversed to normal, gate to source is forward biased in the normal manner when correct polarity is applied. As a MOSFET is a 2 quadrant device it works as usual with Vds reversed. This is not usually a useful polarity as the body diode passes current when reverse Vds is applies, bypassing the FET channel when the FET is turned off. In this case this is exactly what is required.

The MOSFET is a P Channel device, which is correct for the application

**SI2319 data sheet*

RDson TYPICAL can be seen in the graph at top left of page 3. At high gate drive voltage (near 12V) the slope of the lines is about 0.6V at 10A so Rdson TYPICAL = V/I = .6/10 = 60- milliOhm.

The curve makes no sudden foray near 0 current so the Rdson probably more or less applies at 3.2 mA s= V_FET_On ~~+ IR = 0.0032 x .060 = ~~ 0.2 mV.

So, yes, your assumption is in the order of right. Typical values differ from worst case and temperature has an effect and ... but well under 1.0 mV and probably under 0.5 mV can be expected in most cases.

Interest only: This is an extremely useful and effective circuit and works well in practice. I have used this in 100,000++ devices when battery reversal was a potential issue and I could not tolerate the voltage drop of a forward biased diode in the battery lead.

If Vin can be certain to always be much less than Vgsmax(which is -20V for the device yu have specified) then the gate can be connected directly to ground, eliminating the zener diode and both resistors - and giving reverse battery protection with a MOSFET alone.

Digikey wants \$US0.15 for these even in 150,000 quantity , but in volume Asian production a suitable MOSFET would cost a cent or two.

Note: Despite having used this circuit extensively I initially incorrectly commented on the mode of operation. Must be bed time :-). While I was fixing that I see I had a typo in the mV drop figures - not a good day :-)

• The circuit is perfectly fine - the body diode is meant to conduct until sufficient voltage is applied at the input for the FET to turn on. reversing the input voltage reverses the diode and turns off the MOSFET. Commented Mar 26, 2015 at 11:33
• Swapping the drain and source would entirely defeat the object of this circuit. Doing that would allow a reverse polarity supply voltage to cause current flow through the body diode. Commented Mar 26, 2015 at 13:16
• @Russell McMahon. as Andy aka says I believe that the orientation is correct. The Vin is between 10,8 and 14,4v (car battery). Can I in this case omit the resistors and the zener? Commented Mar 26, 2015 at 14:01
• Yes. I agree. Brain fade on my part for some reason - I knew (of course) that the body diode worke in the "correct" direction and was augmented by the FET. The 100,000++ that I have had built were wired correctly :-). Commented Mar 26, 2015 at 15:09

There's nothing wrong with the circuit (see this TI article for verification) and with a device that has such a low Rds(on) with such a small drain current load, the volt drop error is going to be much less than a milli volt.

See also this Stack exchange question for same circuit reverse protection