Hi
You could diode isolate the vin to a buck boost to isolate it's input and you can diode isolate its output to create a pseduo isolated regulator that would prevent reverse bias from being a big problem.
Typically, the output stage of a buck/buck-boost regulator is an inductor- capacitor. The voltage on the cap is the regulated rippled/filtered output. Then there usually is a resistor feedback network right at the capacitor to feedback the cap voltage to the regulator IC and close the loop; so you can actually regulate the output.
You can usually put a diode after the inductor and before the capacitor to pseudo isolate the output. The feedback network should still be located at the capacitor though. This way, the regulator will achieve your desired output voltage (that you set through the resistor divider network) regardless of the diode drops.
Steve Spano
From: Joseph Smith joe@settoplinux.org Date: 2008/11/15 Sat PM 04:33:24 EST To: steve@fl-eng.com CC: Peter Stuge peter@stuge.se, coreboot coreboot@coreboot.org Subject: Re: [coreboot] LPCflasher Project
On Sat, 15 Nov 2008 16:09:58 -0500, steve@fl-eng.com wrote:
Host USB ports are supposed to use a MosFET with internal current sense
for
power protection. The FET signals the O/S when its current threshold is being exceeded and then the O/S suspends the port and removes power.
If you apply +5v INTO the host port, you can create current spikes across the FET which would damage the FET, trip the O/S into thinking a power
fault
has occured and shut down the port. Even in this state, the reverse bias
diode
in the FET will continue to allow the external voltage to be fed into the
the
motherboard.
The FET/Motherboard external voltage will keep fighting each other as
the
ripple in the motherboard +5v and your external +5v criss-cross each other.
This would probably be bad....
That's what I was thinking, thanks.
You can diode OR all of your mother board/external supplies together
and
then run them into a buck/boost regulator. THat way the buck/boost will maintain a local +5v supply at your board "regadless" of external voltages.
Thoughts?
Ah, so I can use a Buck-Boost regulator to compensate for the diode
voltage
drop. That makes sense. In that case I could just cut out the simple LED circut, use the Buck-Boost regulator to boost the 5V to 8V and put a LED inline as my diode and the output should be a solid one way 5V power line right?
-- Thanks, Joseph Smith Set-Top-Linux www.settoplinux.org