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My system is powered from a wall adapter and backup SLA battery.

schematic

simulate this circuit – Schematic created using CircuitLab

Since the voltage of the wall adapter is higher than the SLA, I have ORed the two so that the battery will be used only when wall is not available. In the event that wall is not available and the battery is discharged, I would like to disconnect the battery from the load to prevent further discharge. I could simply place a high side switch before the ORing diode, however I thought using an ORing mosfet controller would be smarter since it can do the ORing with better efficiency and act as a switch at the same time (win win!!).

I found the TPS2419 ORing contoller with enable and LM5050. The enable pin on these controllers allow me to turn the high side N-mosfet off, but to my surprise this does not disconnect the load as the mosfet is connected such that its body diode will conduct (see figure below). Every other ORing controller I have seen also uses the same configuration. The data sheet of the LM5050 even mentions that current will still conduct when the external mosfet is off. LM5050 external Mosfet configuration

I have also looked at power multiplexer ICs but they are rated for lower voltages.

My question is, is it even possible to use the ORing mosfet as a load switch? If not, why is the mosfet connected that way and what is the purpose of the enable pin?

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It's similar to the ideal diode controller from LT (LTC4359). If the load voltage is above the Vin voltage, then the MOSFET's intrinsic diode won't conduct. If you want a switch + ideal diode, then you have to put two transitors. enter image description here

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    \$\begingroup\$ Seems like the two back to back mosfets is the solution I'm looking for but for anyone looking to implement this, the TI chips cannot be used in this configuration according to this thread it also explains why the LTC4359 is 3 times as expensive \$\endgroup\$ – Fiebbo May 8 '17 at 7:56
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    \$\begingroup\$ Luckily there is a LTC4359 board, china made, on ebay and others . Just in case you want to experiment, unfortunately it is just up to 15A. I have also found some other company that sells ideal diode (no second transistor) with much larger MOSFET, 40V 600A. It seems that the LTC4359 can drive also MOSFET with larger gate charge (or two smaller?). IMO you should use avalanche rated MOSFETs, just in case of large kick-back voltage pulse. \$\endgroup\$ – Marko Buršič May 8 '17 at 8:13
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The point here is that the MOSFET is not being used as a switch, but as an "ideal" diode (almost zero forward voltage). This is achieved by operating the MOSFET in reverse mode.

If you simply used a regular diode instead of a reverse-operated MOSFET, you would get an undesired voltage drop and a (potentially high) power dissipation in that diode.

You can avoid that by using a reverse-operated MOSFET as an "ideal" diode, trading-off complexity for performance. But the operation fundamentals of the circuit will be the same: diode ORing of the power supplies.

The key to understand what's happening here is the fact that the conduction channel doesn't have any intrinsic polarity, it acts pretty much like a short circuit (low resistance path) that we can control.

Now recall that the conduction channel is shunted with the body diode of the MOSFET. So we can take advantage of this to build a controller (and TI, Linear, and so many others did) like the LM5050, that monitors the drain and source voltages of the MOSFET and choses when to short circuit the body diode or when to leave it do its "diode role". This way an "ideal" diode can be synthesised because we can get the effective forward voltage as close to zero as needed (or as much as is practical to do it) .

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    \$\begingroup\$ Good explanation of the concept, I'm now convinced that the mosfet cannot be used in the other polarity since its body diode would be forward biased and allow current back into the battery (power source). Should have noticed this earlier. \$\endgroup\$ – Fiebbo May 8 '17 at 9:24

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