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Reverse voltage protection with PMOS

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Kian

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Hi all,

I am trying to implement a reverse current/voltage battery protection circuit. My circuit uses a 3V coin cell battery to power up a MCU and I want to put a protection circuit in case someone puts the battery in the wrong direction.

I found this application note by texas instruments (http://www.ti.com/lit/an/slva139/slva139.pdf) and I followed their circuit using a PMOS FET (figure 3), using a Si2323.

When I connected up this circuit, it works fine when the battery is inserted in the right direction and I am getting 3V out from the Source of the Si2323. However, when I connect the battery in the reverse direction, I am getting a -0.6V from the source. Is this normal? Will it cause any damage to the rest of my circuits?

Thanks in advance!
 

crutschow

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However, when I connect the battery in the reverse direction, I am getting a -0.6V from the source. Is this normal? Will it cause any damage to the rest of my circuits?
Is anything connected to the MOSFET when you make the measurement other than a voltmeter?
If not, you are likely just measuring leakage current.
0.6V / 10meg equals 60nA leakage (assuming the common 10megΩ input impedance of a typical digital multimeter. The value in the data sheet is here:
upload_2018-3-26_22-6-37.png
so 60nA is well withing the maximum.

In any case -0.6V should not harm your circuits.
 

Kian

Member
Is anything connected to the MOSFET when you make the measurement other than a voltmeter?
If not, you are likely just measuring leakage current.
0.6V / 10meg equals 60nA leakage (assuming the common 10megΩ input impedance of a typical digital multimeter. The value in the data sheet is here:
View attachment 111883
so 60nA is well withing the maximum.

In any case -0.6V should not harm your circuits.
Thanks for the reply. There was nothing connected to the MOSFET when I made the measurement other than a voltmeter.
I just tried connecting to a load (the MCU and another sensor) and with the battery connected in reverse, I am getting a -0.2V. But then something weird happens after that. My CR2032 battery had a voltage of 3V initially. However, after I connected it in the reverse direction, I took out the battery and measured its voltage and it had drop to 2.2V, and after a minute or so, the voltage on the battery increased to 2.8V.

Is this normal?
 

Kian

Member
I think I know what is the problem. Its the construct of the battery holder. When the battery is inserted in the wrong direction, the battery holder actually sort circuit the positive and negative terminals! Looks like I should look for another battery holder.
 

crutschow

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When the battery is inserted in the wrong direction, the battery holder actually sort circuit the positive and negative terminals!
Perhaps that was intentional to prevent a reverse battery installation from frying the circuit.
Better a dead battery than a dead circuit.
 

dknguyen

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Perhaps that was intentional to prevent a reverse battery installation from frying the circuit.
Better a dead battery than a dead circuit.
Unless it's a lithium battery...smok'n'flames.

I guess just putting a diode anti-parallel to the battery is the cheapest way of having 0V drop reverse polarity protection. Certainly cheaper than a PMOS and potentially matched transistors and resistors or the crazy circuits out there dedicate a charge pump to the reverse polarity transistor. I'm not sure what low voltage circuit would be so sensitive to input voltage drop that would warrant a charge pump just for reverse polarity protection.
 

crutschow

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I guess just putting a diode anti-parallel to the battery is the cheapest way of having 0V drop reverse polarity protection.
That would short the battery, the same as the battery holder he has.
Certainly cheaper than a PMOS and potentially matched transistors and resistors or the crazy circuits out there dedicate a charge pump to the reverse polarity transistor.
I see no reason for a charge-pump circuit.
All the circuit requires is a small MOSFET with a low minimum Vgs(th), which the Si2323 has.
No other parts required.
 

dknguyen

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That would short the battery, the same as the battery holder he has.
I see no reason for a charge-pump circuit.
All the circuit requires is a small MOSFET with a low minimum Vgs(th), which the Si2323 has.
No other parts required.
I saw it in an app note so I assume it must have some use somewhere.

https://www.maximintegrated.com/en/app-notes/index.mvp/id/636

Maybe it was written at a time (1996?) where low logic level FETs, or maybe even logic level FETs did not exist yet.
 

crutschow

Well-Known Member
Most Helpful Member
The charge pump is only needed if you want to use an N-MOSFET as a high side diode in the plus line.
A P-MOSFET works in the high side without the need for the charge pump.

That Maxim article you posted does mention logic-level MOSFETs.
 
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