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I'm designing a battery-powered circuit in which a power MOSFET, controlled by an ATtiny85 MCU, switches a resistive load (very, very low frequency). The battery is Li-Ion, fully charged at 4.2V. Maximum load current is about 16A. Board real estate is really limited, so I'm trying to keep the component count to a minimum. I need overcurrent, overdischarge and reverse polarity protection. I handle the first two directly in the MCU (with calibrated internal reference) to reduce board size: overcurrent is sensed over differential ADC from a shunt resistor, and overdischarge is measured by reading the internal reference and back-calculating the supply voltage. This is the relevant part of the schematic (GATE/SCK is the MCU output, the resistive load is between +V and LOAD_GND):

Circuit

The problem arises with reverse polarity protection. I need to deliver as much voltage as possible to the load, dropping at most 50mV (I chose a PSMN0R9-25YLC MOSFET, which has 1mOhm Rds, i.e. 16mV @ 16A). Of course, even a series Schottky diode drops too much, plus it has to stand the load current (= bigger, and board size is important). I'd usually do this with another MOSFET to break the power line based on polarity. But in this board adding another MOSFET is a problem, due to its size.

On the other hand I can stand the drop of a Schottky diode on the MCU supply, since it'll run down to 1.8V at the clock I set it to and the overdischarge protection will keep the battery from going under 2.7V or so. I thought about protecting the MCU with a diode and leaving the MOSFET unprotected, since it can stand -20V. The problem with this is that, once the polarity is reversed, no damage will occur but the load will always be powered through the MOSFET body diode.

So here comes the question: does somebody know of a creative solution to protect this circuit that drops at most 50mV @ 16A and uses very little board space? To clarify on how much space is available, I can probably get another PSMN0R9 in there, but it would be really crammed and I'd prefer a smaller solution. Cost is not too much of an issue.

UPDATE: a few things that may not be clear and to which I replied in the comments:

  • There is no charger involved. The battery is a user-removable 18500 or 26650. The user will charge it with an external separate charger.
  • The circuit is already 3mOhm, but I can drop the shunt (R1) to 1mOhm by sacrificing some accuracy in the overcurrent protection.
  • The load is purely resistive. Reverse polarity protection is needed in case the user wrongly inserts the battery (since the battery form factor makes this possible).
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  • \$\begingroup\$ You want 3mohm at 16A? That's really pushing it... \$\endgroup\$
    – Ignacio Vazquez-Abrams
    Commented Oct 11, 2015 at 23:15
  • \$\begingroup\$ Is the load purely resistive, or is there some other reason for thinking it would ever reverse-supply the body diode? \$\endgroup\$
    – rdtsc
    Commented Oct 11, 2015 at 23:50
  • \$\begingroup\$ @rdtsc Maybe I wasn't very clear in the question. This is to protect from the user accidentaly putting in the battery the wrong way. It'a 18500 o 26650 battery. \$\endgroup\$
    – Andrea Biondo
    Commented Oct 12, 2015 at 4:36
  • \$\begingroup\$ @IgnacioVazquez-Abrams I know. I can trade some accuracy in the overcurrent sensing and reduce the shunt to 1mOhm if needed. Of course low resistance is taken into consideration when laying out the PCB. \$\endgroup\$
    – Andrea Biondo
    Commented Oct 12, 2015 at 5:59

1 Answer 1

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I would go ahead and add a second one of the PSMN0R9's in series with the first. When connecting it up put the two MOSFETs drain to drain and the second ones source to the "LOAD_GND". The gates of the two FETs tie together. This is the simplest way to work around reverse polarity protection with MOSFET body diodes.

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  • \$\begingroup\$ Thank you, I am starting to think adding another MOSFET is the only solution. Wouldn't it be better to just break the ground return path with the second MOSFET? Source to circuit ground, drain to battery ground, gate to V+ (battery positive). On correct polarity, the body diode conducts and pulls the source down, switching the MOSFET on. On wrong polarity, the ground path is broken. This way I'd save the extra protection diode on the MCU. \$\endgroup\$
    – Andrea Biondo
    Commented Oct 12, 2015 at 6:41
  • \$\begingroup\$ It's just that I have this feeling that there must be a more elegant solution than just throwing in another MOSFET. I don't know, probably I'm wrong. \$\endgroup\$
    – Andrea Biondo
    Commented Oct 12, 2015 at 6:44

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