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I'm trying to understand how the following over-current protection circuit works:

enter image description here

enter image description here the way I understand this, as long as +VBUS_USB3_P0 = 5V and it's equal to +5V_ALW, there is no current current going through F2. but In over current situation, +VBUS_USB3_P0 can go below 5V (and higher current), where a current must go through F2, and that will drag Cathode voltage of D88 low, and that will drag USB_OCP_ALL# down and send an indication to CPU/Controller.

Is my analysis right?

Firas Abd El Gani
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  • Do you mean "detection" instead of "protection"? Unless there's something else important about `+5V_ALW`, it seems the fuse provides OCP. Note that the output might not be fully shorted out, so an analog input, or a threshold close to 5V, should be used to detect partial drop. – Tim Williams Apr 11 '23 at 08:13
  • Hi @TimWilliams, Yes I meant detection. But how does the circuit work? how is USB_OCP_ALL# is eventually pulled low? – Firas Abd El Gani Apr 11 '23 at 08:22
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    Your reasoning is correct, it will be pulled low as the output voltage drops (presumably because the fuse has opened). – Tim Williams Apr 11 '23 at 09:04

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