0
\$\begingroup\$

I have the following circuit meant for reverse polarity battery protection. The battery is a 2s li-ion (8.4 V to 7.0 V), but it could also be a 12 V max PSU. I'm not sure what's wrong on the PMOS gate, why is it only getting 3.1 V from the Zener?

Falstad circuit

My expectation was to have a gate voltage increasing together with the battery voltage, until it reaches the Zener voltage. Then the Zener would regulate it and keep it at about 7.4 V.

The Zener is the MM5Z7V5T1G. The MOSFET is the YJQ1216A. Link for the circuit simulation: Falstad simulation.

\$\endgroup\$
1
  • \$\begingroup\$ Isn't the diode enough for reverse polarity battery protection? \$\endgroup\$
    – sai
    Commented May 16, 2023 at 14:19

1 Answer 1

1
\$\begingroup\$

Your zener is in the wrong place. When reverse-biased, the zener will initially block like a normal diode, but eventually break down at it's zener voltage and begin conducting. For this zener (in your simulation), that's around 5V. If 5V were to drop across the zener that would leave 3V across the resistor,
3V / 10k Ohm = 50uA
This small current isn't actually enough to open the zener, which needs around 1mA to operate.
So what you're seeing is the zener isn't operating properly, and is in the region before it breaks down, and hence passes 307.2uA, and drops 4.9V

You need to swap the resistor and zener around so that the regulation from the zener is between the gate and ground, rather than between the gate and Vin like current. You should also decrease the resistor to allow enough current for the zener to operate.

enter image description here
I've used values for the diode in your simulation, so I could adjust the simulation. The diode you linked has 7.5V breakdown voltage, but the math is much the same

You could also do away with the zener and just drop a resistor divider in. The Gate threshold voltage is -1V, so even a pair of 10k's to make a divider would be enough. You could also increase the value in order to waste less power.

\$\endgroup\$
1
  • \$\begingroup\$ Thanks! That was actually very helpful. I modified the circuit and used 2x 2k resistors in parallel to avoid heat problems in the worst case scenario: -12V applied. \$\endgroup\$
    – lupolucio
    Commented May 17, 2023 at 14:29

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.