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?