# Feasibility of reverse polarity feature

If I were to design a circuit for a small appliance that works in DC and I would want to allow it to work with the batteries in reverse polarity could I feasibly use a diode bridge?

Would it make sense? Would the energy efficiency drop significantly?

Have a look at the LT3240 Ideal Diode Bridge Controller.

The LTĀ®4320/LT4320-1 are ideal diode bridge controllers that drive four N-channel MOSFETs, supporting voltage rectification from DC to 600Hz typical. By maximizing available voltage and reducing power dissipation (see thermograph comparison below), the ideal diode bridge simplifies power supply design and reduces power supply cost, especially in low voltage applications. Source.

Using MOSFETs rather than diodes reduces voltage drop.

• Those things looked interesting when they came out - have you ever used one? Like to know what they're like in practice, anything to watch out for that came to light if you had. – TonyM Jun 18 '17 at 21:31
• Just at a gut level they strike me as having a lot of smoke generating potential. – Ian Bland Jun 18 '17 at 21:35
• Nope, I have not used them. I read about them here first. – Transistor Jun 18 '17 at 21:48

It would work.

The loss in efficiency would be given by the diode drop divided by the operating voltage. Obviously, efficiency would be maximised for a low-drop diode, and a high operating voltage. To give a couple of examples:

Using a standard 1n4148 diode, with a forward voltage drop of 0.65V, and a 1.5V battery, efficiency is only 55%.

Using a 0.4V diode with a 48V lead acid would give 99% efficiency.

These numbers are to provide protection only - plugging it in backwards wont work but won't damage anything either.

For it to work both ways, you need two diode drops, and efficiency is 10% and 98% respectively.

• Thank you! May I stretch things a bit and ask about any speculation regarding the loss of energy for a 5V, 1A operation? – Andrei Rînea Jun 18 '17 at 20:12
• You would have to look at available diodes, but roughly speaking... somewhere about 85-95% would be achievable, using one diode to give basic protection, or 70-90% using two diodes to make it work either way round. – Jack B Jun 18 '17 at 20:18
• Don't forget, for a diode bridge you need to double the diode voltage drop! – DoxyLover Jun 18 '17 at 20:36
• Hopefully that's clearer now. – Jack B Jun 18 '17 at 20:39
• Yes, it's clear! Thanks! @JackB : I am greedy and look beyond a simple protection. I intend to make it work with the battery reversed.. – Andrei Rînea Jun 18 '17 at 20:44

Old School

simulate this circuit – Schematic created using CircuitLab

Figure 1. Old school relay solution.

• If V1 is positive RLY1 picks. If V2 is positive RLY2 picks.
• Voltage drop is zero but the penalty is a little current for the relays.

You can think of this as a very slow full wave mechanical rectifier.

You could use a diode bridge. A silicon bridge would lose you nominally 1.4v, a schottky bridge 0.7v.

A better solution would be to use a MOSFET bridge, which being resistive when on would result in near zero drop.

For a range of input voltages > Vgs(th) but < Vgs(max), the following should work ...

simulate this circuit – Schematic created using CircuitLab

Four complementary FETs are shown, with their intrinsic body diodes. Neglecting any FET operation, the body diodes form a normal bridge rectifier.

Now consider Va high, Vb low. Va high turns on M1. Vb low turns on M4. M1 shorts D1, which is conducting anyway to connect Vb to the output. M4 shorts D4, which is conducting anyway to connect Va to the output. Result, happiness, as long as Vin exceeds Vgs(th) for both types. Obviously the complementary thing happens for Vb high.

Another way to look at it is to notice we have two complementary FET inverters.

• I am in favour of the MOSFET bridge. For battery operation I arranged two N channel and two P channel devices so that the body diodes are in a bridge rectifier configuration and then connected the gates of the MOSFETs to the inputs so that the MOSFETs are turned on if their body diode were to be conducting. This works if the battery voltage is high enough to turn the device on but low enough not to exceed the Vgs max of the FETs. I'll try to draw it up tomorrow and add it as an answer if someone doesn't get there first! – Matthew Jun 18 '17 at 20:25
• I'd like to see that, Matthew. – Ian Bland Jun 18 '17 at 21:13