USB Power Isolation HELP PLEASE!!!! - Page 1
Forum section: electronics / beginners

author: XaeroDemon
author_rank: Contributor
timestamp: August 15, 2016, 10:45:54 am
content: So I have to power my MCU through 3.3V regulator with 5V and when usb connected through Bat54C schottky diode.My regulator is capable of 500ma but the maximum forward current of Bat54C is 300ma.if I want to use full 500ma of the reg ..can I just connect a single schottky diode between VBUS and IN(please see the schematics attached) or just directly connect the 5V to VIN of bat54c?? this way VBUS is still Isolated from the 5V rail right? Ultimately the Isolation is what we want right??But when I connect the USB the VBUS is connect to 5V rail ...Is this acceptable?

author: biggerB
author_rank: Newbie
timestamp: August 15, 2016, 04:32:30 pm
content: From what I understand, you're saying you basically wanting to bypass the top schottky to allow for greater current to flow.Can't you just connect another BAT54C in parallel? I guess this would work if you got both of them from the same lot and the differences between them would be small.

author: MosherIV
author_rank: Super Contributor
timestamp: August 15, 2016, 04:43:24 pm
content: QuoteQuoteYes but it would not work as expected, one of them will start to conduct before the other and probably blow.QuoteYes but you would loose the benefit of the BAT54C schottky diodes.It is not 100% clear why they are there because you have not posted the complete schematic but it looks to be like they are simply to protect against wrong polarity of the power.You could replace with a different Schottky with a higher current rating - maybe a BAT720

author: Kilrah
author_rank: Unknown
timestamp: August 15, 2016, 06:14:47 pm
content: The half on the VBUS side is pretty obviously to prevent the 5V supply from leaking into the USB host which is a big no-no, the other would do the opposite but nothing the OP mentioned allows to find out whether that's a requirement or could be ommitted.So OP... you're asking us if removing the diode is OK, but really the answer to that depends on the circuit that supplies the 5V, of which we know nothing.

author: XaeroDemon
author_rank: Contributor
timestamp: August 16, 2016, 07:43:47 am
content: @MosherIVQuotethis is what I should probably do but the BAT720 isn't a dual barrier.5v should not be connected to the VBUS as @Kilrah said but connectinng VBUS to 5V through a diode to 5V(one way) might actually be a benefit in this case cus that would power the 5V rail when the USB is connected and battery is not.so If I want to power my 5V peripherals from USB ..I can just connect a diode from VBUS to 5V right so that way The Vbus is isolated(5v can't backfeed the VBUS) but Vbus can power 5v when battery is not connected...but when both usb and battery is connected is there any complications.? I don't see any ..whats ur thoughts ??

author: Kilrah
author_rank: Unknown
timestamp: August 16, 2016, 07:55:18 am
content: If your 5V supply circuit can safely have its output powered when its input is not it's OK, but we don't know if it can. The limit is then that officially you may only draw up to 100mA from USB without negotiation.

author: XaeroDemon
author_rank: Contributor
timestamp: August 16, 2016, 07:59:13 am
content: if the current limit is 100ma then I have to can this idea but its okay I got my answer ..thanks guys!