My thought on the reverse polarity protection

Question

I am now designing a power reverse polarity protection circuit for my system. I found that most methods used for the protection would cause voltage drop in the power path. Even the MOSFET, which has maybe 0.025ohm Rdon, would cause about 5mV drop(0.2A). Some may not cause voltage drop but would be a little complex or not resettable. http://www.wa0itp.com/revpro.html

Then I have a new thought(maybe new or maybe not). My design is showed below. My circuit has three power sources(+-12V and 0V). I add two diodes between +12 and 0V and between 0V and -12V with big enough series resistors.

Normally, the two diodes would turn off.No current would go through R1 and R2. So no voltage drop when normal operation. If the polarity is reverse,D1 and D2 would turn on. Current would go through the diodes.

Is there any problem in this circuit? Can it work?

^{simulate this circuit – Schematic created using CircuitLab}

Answer 1

This DPST NO (2 form "A") relay solution will work:

Answer 2

The diodes will have no effect to speak of on reverse polarity, and will provide no protection. If the device is a chip, it will act like a forward biased diode, and will take almost all of the current compared to your parallel diode with 1K in series.

Edit: You can play with simulations below. With the left one I've simulated the 12V source (reversed) with 10m\$\Omega\$ source resistance (2mV drop at 200mA), and no load. The right circuit is the same except D6 represents a chip that you're trying to protect.

Voltage on the left circuit is -12.0V, with or without D1+R1

Voltage on the right circuit is -11.8V, with or without D1+R1, but now D6 (the chip that's supposed to be protected) is conducting about 19 amperes. It is not long for this world. In both cases D1 conducts little current (about 11mA).

^{simulate this circuit – Schematic created using CircuitLab}

There is no reason generally to worry about a few tens of mV drop on a power supply rail- usually the tolerances are much more. The MOSFET method is the best way for very low drop unless you know the source has limited current (in which case, consider a shunt Schottky), a series diode can work in some cases.

Answer 3

Thinking only the positive side of your circuit: assuming you use a normal diode with a voltage drop of 0.7V, GND and +12 reversed, the voltage across thq 1k resistor is U=RI => all remaining voltage will be over the resistor. Thus the protection doesnt work.

One way of doing this (better or worse) would be remove he 1k resistor and using a diode with as small voltage drop as possible, eg schottky diode. Voltagedrop across the "reverse protection line" has to be lower than voltage drop across reverse connected IC. Also you need a fuse wich prevents diode from burning up.

Answer 4

Could I interest you in building a bridge rectifier with Schottky diodes instead? Then polarity would be immaterial.