# Rectifying reversible polarity battery with minimal power loss

I'd like to build a battery-powered DC power supply that will accept a single-cell battery in either polarity (i.e. it doesn't matter which way you put the battery in). I'm thinking of using the circuit below to rectify the voltage.

Are there any gotchas? Is there a better design for my constraints (listed below)?

You'll notice that Q1 and Q2 are N-channel MOSFETs. Q3 and Q4 are P-channel. Let's assume they all support positive and negative gate voltages of more than the battery voltage.

The resistor labeled LOAD stands in the place of the rest of the power supply and everything powered by it.

My design constraints for the rectifier include:

• Very low power loss (less than 1uA quiescent current; less than 1Ohm resistance)
• Support small loads (average of 130uA, peak 20mA, minimum 5uA)
• Support single cell Li-ion (3.7 - 4.2V)
• Support battery in both polarities

Perhaps importantly, I won't be supporting the case where the battery is inserted one way and then quickly inserted the other way. In other words, I don't need to rectify square waves. For that matter I don't need to rectify AC of any kind.

• You will introduce leakage currents on par with your minimum current usuage I think with any sort of configuration like this... – MadHatter Jul 4 at 16:32
• @MattThomas How do you define quiescent currents? – Huisman Jul 4 at 19:31
• Now also draw in the diode that every MOSFET has between Drain and Source. Does this circuit still work? – Bimpelrekkie Jul 4 at 19:45
• @Bimpelrekkie yes, it still works. In fact this circuit would just be a full wave bridge rectifier if the FETs' gates were disconnected and only the body diodes were considered. But driving the FETs reduces voltage drop, giving current a low impedance path other than through the body diodes – Matt Thomas Jul 4 at 23:35
• @Huisman My intention was not to suggest that the body diodes would prevent the circuit from working. I just mentioned them so that they're taken into account (not forgotten). The fact that the diodes would form a bridge rectifier is a good point, – Bimpelrekkie Jul 5 at 8:42

As mentioned above, mosfets have leakage currents that might be in the same range as your minimum current requirements. The blocking mosfets have a Drain-to-Source Leakage Current and the conducting mosfets will have a Gate-to-Source Leakage Current.

If these issues can be tackled, then your suggested circuit will work nicely. Therefore:
search internet for (ultra) low leakage mosfets.

I found that way the nice CSD23381F4 12V P-Channel FemtoFET TM MOSFET.

It has:

• really less than 100 nA IDSS (because VDS is max -4.2 V instead of -9.6 V)
• really less than 50 nA IGSS (because VGS is max -4.2 V instead of -8 V)
• less than 300 mΩ ON resistance (because VGS < -2.5V and IDS is max -0.020 A)
• Support single cell Li-ion (3.7 - 4.2V) because of its low VGS and a VDS of -12 V.
• costs only 0.05 / 1ku

DISCLAIMER
I don't know if you want to make a PCB or even a breadboard (please don't do the latter), and were planning hand soldering: then this package might be a problem.
This answer is more to point out what keywords to search on and what parameters to consider. It's also the reason only a PMOS is suggested.

• Will I also need to consider Gate-to-Source leakage currents for the blocking MOSFETs? Those will also have a voltage applied to their gates (negative for the N-channel, positive for the P-channel) – Matt Thomas Jul 5 at 13:22
• @MattThomas I'm not sure how the gate-to-source leakage current runs, I think you should read scientific papers about that. The name suggest from gate to source: then in shown circuit above, this leakage current in Q4 will run through and therefore contribute to the load. Moreover, I_GSS is defined with V_DS =0 which is not the case. – Huisman Jul 6 at 6:28
• @MattThomas I don't see the benefit of using more mosfets in parallel. What voltage drop will 300 mohm give with 130uA?! Moreover, the battery contacts and other components and maybe wiring will give you resistance too in the same order or even higher. So halving the R_DSON contributes maybe to a quarter less power losses. – Huisman Jul 6 at 6:37
• I found the prices on the website of TI – Huisman Jul 6 at 6:39
• So halving the R_DSON contributes maybe to a quarter less power losses. -> So halving the R_DSON contributes maybe to a (way) less relative power losses. – Huisman Jul 6 at 6:44