I am looking to protect my circuit from a reverse polarity event. I found suggestions for using a MOSFET, particularly a p-channel type. Does anyone have a specific recommendation for a device to use for the LiFePO4 battery (about 3.2V nominal) technology? I am not totally sure about this, but would it be possible to select a MOSFET that has a Vgs that would also protect the battery from being over-drained?

Edit: Max current 500mA. Also looking to save electrons, so the device needs to be efficient both in on and off state.

Edit 2: Looking for a specific P-MOSFET or spec requirements to check in the data sheet when ordering parts to try out. Thanks.

  • \$\begingroup\$ What max current? \$\endgroup\$ – Russell McMahon Sep 28 '16 at 6:27
  • \$\begingroup\$ Sorry, 500mA. Added to the OP. \$\endgroup\$ – bbglazer Sep 28 '16 at 19:36

The two requirements (reverse protection & reverse polarity) are not easily met in the same single device.
By itself a MOSFET's Vgs is too poorly defined between parts and not sharp enough yto make a good low battery cutout.
By adding the world's lowest cost IC and two or three resistors you can meet both needs.

A P Channel MOSFET in the +ve lead, drain (not source) to B+, source (not drain) to Load+ and gate to B-, will provide FET which is turned on with correct battery polarity.

Now add a TLV431 shunt regulator from FET gate to ground and ise two resistors to program it to turn om when Vbat is above a preset limit. Add a 3rd resistor gate to B+ to turn the FET off when the TLV431 is off.
I nominated a 1.25V TLV431 rather than a 2.5V TL431 as the device cathode pulls down to Vref - 0.6V at best (based on my experiences and a logical result of the internal structure) and nominally to about Vref.
With 1.25V as Vref in the TLV431 and say 3 volts Vbat that leaves (3-1.25v) = 1.75V Vgs for the MOSFET, meaning it needs Vgsth of more like 1V to work well with 1.75V on the gate. MOSFETS that meet this spec are not common but do exist.
If desired an NChannel FET can be used but the level switching circuit will probably be slightly more complex.

A single MOSFET will protect against reverse polarity but, as Bruce Abbot noted, does not fully protect against deep discharge due to the body diode conducting. To protect against both over-discharge and reverse polarity two back to back MOSFETs are needed

The first diagram below shows the 1 FET version with reverse polarity protection and limited low voltage protection.
For reverse polarity only omit R1, R2, R3, D1, and connect the the FET gate to groumnd. This is the standard reverse polarity protection circuit which turned up decades ago - I don't know who first thought of it but somebody attempted to parent the concept long long after it was commonplace.
Direct gate grounding is OK for Vbattery << Vgsmax.
For combined reverse polarity and low voltage cutout: When polarity is correct and battery voltage falls below set voltage the FET turns off and the body diode will be placed in series with the load - so available voltage drops from say 3V1 to maybe 2V4 (0.7V drop due to inferior diode structure). If the load was eg a white LED with Vf ~= 3.0V in normal operation then current drain would drop to a very small @ of normal.

In the second diagram the 2nd FET (both P Channel), with opposite d-s polarity to the first stops conduction via the FETS. A sbefore the R2 R3 discharge path remains and R2 should be as large as other design parameters allow.

Fig 1 - Reverse polarity protection, low voltage adds a 0.7V+ drop diode in discharge path.


simulate this circuit – Schematic created using CircuitLab

Fig 2 - Reverse polarity protection, low voltage disconnects load.

Rx is notionally required to turn on back-to-back FET pair but in practice tends to be unneeded. Use Rx = 10 megohm if desired.


simulate this circuit

Which on reflection, can be slightly simplified as shown below. The dual floating gate is eliminated. The top MOSFET now deals solely with polarity reversal and the bottom is the undervoltage cutout. The TLV431 draws essentially no current when off (typically 0.05 uA max) so the value of R3 can be set to ensure that the TLV431 draws minimum regulating current when on - this varies with brand but is typically 80 uA MAX.
R2 R3 still draw current at all times but as Iref is 0.3 uA max , 0.15 uA typical a 100k resistor gives an error of V=IR = 0.3 uA x 100k = 0.03V.
With R2 = 100k, Vref = 1.25V and Vbat = 3V I_R2 = (3-1.25)/100k =~ 18 uA.
A 1% charged cell with 2000 mAh capacity would last (2000 mAh x 1%)/18 uA > 1000 hours.



simulate this circuit

Minumum discharge voltage for a LiFePO4 cell umder loads of under C are usually in the 3.0 - 3.1V range. Under heavy loads. 1C 10C - ... lower voltages are usually due to voltage dropp from internal impedance. 3.2V is extremely conservative.

The circuit above drains the battery with R2 + R3 when the FET is off. So to minimise off mode drain R2 & R3 should be as large as possible while minmising Vset due to bias currents in the resistors.

A possibly suitable part - depends on desired max current NTJS3151P - example only.

| improve this answer | |
  • \$\begingroup\$ ...and the body diode? To protect against both over-discharge and reverse polarity I think you will need two PMOSFETs, wired back-to-back. \$\endgroup\$ – Bruce Abbott Sep 28 '16 at 7:06
  • \$\begingroup\$ @BruceAbbott Good point. Why doesn't somebody make a dual device with opto driven gate in a single package? Maybe they do :-). || (1) FET will provide reverse polarity protection as shown. (2) When polarity is correct, when voltage falls below set voltage the body diode will be placed in series with the load - so available voltage drops from say 3V1 to maybe 2V4 (0.7V drop due to inferior diode structure). If the load was eg a white LED with Vf ~= 3.0V in normal operation then current drain would drop to a very small @ of normal. Adding a second FET in 'normal' polarity of d s would fix this. \$\endgroup\$ – Russell McMahon Sep 28 '16 at 10:33
  • \$\begingroup\$ @BruceAbbott Howzat? \$\endgroup\$ – Russell McMahon Sep 28 '16 at 10:52
  • \$\begingroup\$ Thanks for the extensive answer! How much power does the TLV431A consume in the off state? Would a SI2333DDS-T1-GE3 work in this application? It has a lower Rds on at 28mOhms. \$\endgroup\$ – bbglazer Sep 28 '16 at 19:57
  • \$\begingroup\$ I am kind of confused. But for reverse polarity protection only, all you need is the FET, with no other circuitry. Source to load, gate to GND, and drain to battery. Just make sure it is guaranteed to be on hard at 2.7V (a requirement that is easily met). \$\endgroup\$ – mkeith Sep 29 '16 at 5:04

Here is an easy reverse polarity circuit. It has basically zero quiescent current and acts fast enough to protect the load completely. Reverse current will basically be zero. You may want to put ESD/Zener diode protection from gate to source to make sure the FET does not fail due to ESD.

In the normal case, when the battery is connected correctly, the FET will be on unless the battery (V1) voltage drops very low. This circuit is really not useful for low voltage shutdown due to imprecise voltage threshold, and body diode conduction.


simulate this circuit – Schematic created using CircuitLab

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  • \$\begingroup\$ Thanks for re-affirming the quiescent current, etc. I am looking for a particular device recommendation, well suited for the application. Or I guess specs in particular that need to be met in this scenario, to be more confident in parts I order to try out, thanks. \$\endgroup\$ – bbglazer Sep 29 '16 at 8:20
  • \$\begingroup\$ @bbglazer, I updated my answer. \$\endgroup\$ – mkeith Sep 29 '16 at 16:34

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