BATTERY REVERSE PROTECTION USING P CHANNEL MOSFET

Thread Starter

ir_tronik

Joined Jul 12, 2017
15
Im having a problem to understand the battery reverse polarity protection below:
rvr.JPG
It says that when the battery charger voltage is above battery voltage, the MP1 will disables the MP2. So, how this circuit is gonna charging the battery?
Could someone help me and explain how this reverse protection circuit work? and how does it will let the charger to charge the battery? Thank you
 

Irving

Joined Jan 30, 2016
2,308
I think it means if the battery is reversed it shuts off...

In normal use, charger voltage is o/c or below battery volts (assumes a diode in output of charger) so current initially flows from battery through reverse diode of MP2 and load making the drain of MP2 more +ve than its gate and turning it on, driving the load, MP1 being held off as its gate is essentially at the same voltage as its drain. As charger volts increases above battery volts the charger starts to drive current through the load and the current through MP2 reverses, which is still held on by its gate being even more -ve than its drain, charging the battery. This illustrates an oft-misunderstood property of a MOSFET compared to a BJT. Once its switched on by an appropriate gate-source voltage the drain-source channel is effectively a bidirectional resistor, there is no right or wrong way round (as long as it is sufficently turned on so that Vsd < Vf of its internal diode).

1635256342946.png

However, if battery is connected reversed and charger is 0v a tiny current flows through R2 and load (only uA so non-damaging) and if charger switched on, it drives current through load. MP1 gate is held -ve wrt its drain by the reversed battery, turning it on effectively shorting MP2 gate to its drain, holding it off and preventing damage to the battery.

1635256386533.png
 
Last edited:

MrSalts

Joined Apr 2, 2020
712
All mosfets behave as reversed biased diodes. That effect is used when the battery is properly connected because M1 and M2 are "upside down" compared to how one would expect in a "normal" circuit. Along with being upside down, when the battery is out in upside down, the gates of the mosfet are not allowing current to flow. So it is a good blocking system.

My question, why didn't they just use a diode? Is a mosfet cheaper than a diode of comparable current carrying capacity? Who knows.
 

Irving

Joined Jan 30, 2016
2,308
My question, why didn't they just use a diode? Is a mosfet cheaper than a diode of comparable current carrying capacity? Who knows.
Reduced losses. A good turned-on PMOS has an Rds of say, <20mOhm, so at 1A its dissipating <20mW, at 10A, <2W. A good schottky diode at Vf = 0.4v would dissipate at least 0.4W and >5W at 10A as its Vf will be typically >= 0.5v. Pus the loss of that 0.5v represents reduced run time off batteries, esp single cell solutions.
 

crutschow

Joined Mar 14, 2008
28,170
why didn't they just use a diode?
Simple.
A diode will not protect against reverse battery connection in a charger since the output current direction is the same for both battery polarities.

Below is an alternate reverse-battery protection circuit for battery chargers which uses a MOSFET and a BJT, shown for both transistor polarity options:

1635268221635.png
 
Last edited:

Thread Starter

ir_tronik

Joined Jul 12, 2017
15
I think it means if the battery is reversed it shuts off...

In normal use, charger voltage is o/c or below battery volts (assumes a diode in output of charger) so current initially flows from battery through reverse diode of MP2 and load making the drain of MP2 more +ve than its gate and turning it on, driving the load, MP1 being held off as its gate is essentially at the same voltage as its drain. As charger volts increases above battery volts the charger starts to drive current through the load and the current through MP2 reverses, which is still held on by its gate being even more -ve than its drain, charging the battery. This illustrates an oft-misunderstood property of a MOSFET compared to a BJT. Once its switched on by an appropriate gate-source voltage the drain-source channel is effectively a bidirectional resistor, there is no right or wrong way round (as long as it is sufficently turned on so that Vsd < Vf of its internal diode).

View attachment 251176

However, if battery is connected reversed and charger is 0v a tiny current flows through R2 and load (only uA so non-damaging) and if charger switched on, it drives current through load. MP1 gate is held -ve wrt its drain by the reversed battery, turning it on effectively shorting MP2 gate to its drain, holding it off and preventing damage to the battery.

View attachment 251177
Thank you Irving, now I understand how it works. Very nice of you showing me the LTSpice simulation
 
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