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Idea shootout: ideal diodes.

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technix:
I have detected a certain powered USB hub feeding power back to my motherboard, causing me problems in booting. Now it seem to me that I will need some USB 3.0 power diodes to protect my motherboard against that. Using Schottky diodes leads to a 0.3V drop, which can cause problems in some loads...

So here is the shootout: an ideal diode circuitry. Specs: 5.5V, 2A.

My entry is the following:


Ignore the diode and the cap, as this is excerpt from one of my projects. The power comes in from top left and leaves top right, and the bottom rail is ground. Basically the same circuitry taken from the Raspberry Pi 2 schematics, but with some component selection changes.

exe:
"ideal diode controller" , such as http://www.ti.com/power-management/power-switches/ideal-diodes-oring-controllers/overview.html .

richard.cs:
I've been staring at that circuit trying to understand it and I have to say I am struggling to get my head around how it works, though throwing it into simulation it does seem to behave quite nicely. It doesn't switch on until there's half a volt or so forward across the MOSFET but then goes to about 50 mV drop and turns off before the current reverses though I am not really sure what causes that - is it triggered by the fall in the forward drop as the current falls?

It's quite neat anyway, I like it.

magic:
Any problem with simply letting the hub power itself?

not1xor1:

--- Quote from: richard.cs on April 18, 2020, 12:54:01 pm ---I've been staring at that circuit trying to understand it and I have to say I am struggling to get my head around how it works, though throwing it into simulation it does seem to behave quite nicely. It doesn't switch on until there's half a volt or so forward across the MOSFET but then goes to about 50 mV drop and turns off before the current reverses though I am not really sure what causes that - is it triggered by the fall in the forward drop as the current falls?

It's quite neat anyway, I like it.

--- End quote ---

when there is an input voltage the BJT on the right is OFF and so once the capacitor on the right get charged via the MOSFET body diode above the threshold voltage the mosfet goes ON. Once the input voltage is removed, the base of the right BJT is at 0V and the BJT goes ON shorting the MOSFET gate and source.

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