Does this power selector circuit work?

[Thane of Cawdor]

Hi,

Just wanted to verify whether this circuit could work as a power selector (charge the battery and run the output from the input power line or run off battery power when the input power line is low) using an ideal diode.

Would the enable pin need to be inverted on one of the sources to prevent reverse leakage current into the source/battery or will this setup work well enough?

Thanks

[Thane of Cawdor]

Update: the resistors in the diagram would be around 10-100K. The input power line should always be higher than the battery voltage - facilitating reverse bias condition on the inline battery ideal diode when powered up.

[Peabody]

I've never used those.  It's not clear whether the load would automatically be powered by the highest voltage, which is what you need.  Also, you can kinda do the same thing with a standard load sharing circuit - a P-channel mosfet, a Schottky diode, and a resistor.

[Thane of Cawdor]

Thanks for that. I think the only concern with the load sharing circuit would be the much higher forward voltage drop across the diode at 500-1000mA. For the 'ideal diode' ICs these are quoted to be sub 100mV.

Would inverting the enable pin input (with respect to the other) on one of the ideal diodes make it more like the load sharing circuit without the forward voltage drop perhaps?

[kjr18]

Not really, Shottky diodes have smaller voltage drop than other, and it's not really important in this design, as voltage is boosted anyway to 5V. SB220 in this example has about 400mV @1A. Not so much i think.

[Peabody]

In the load sharing circuit it's assumed the charging source is ultimately mains powered, so dissipating a little heat in the diode doesn't really matter.  But the mosfet switches the battery with very little drop.

If I read the datasheet correctly, you can have both ENables turned on all the time, and the highest voltage source will always power the output.  I don't think you have to invert one of the enables because it says the diode has to be forward biased to conduct.  So if V1 is higher than Vbat, then it would always power the output if it's plugged in, even if both diodes are enabled.

It also says you need capacitors on input and output which you wouldn't need in the load sharing circuit.  And there would be some current needed to power the diode logic when running on battery - a few microamps which you wouldn't have with the plain load sharing circuit.

[Thane of Cawdor]

Thanks all for the responses