[coreboot] LPCflasher Project
steve at fl-eng.com
steve at fl-eng.com
Sat Nov 15 23:18:24 CET 2008
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
> From: Joseph Smith <joe at settoplinux.org>
> Date: 2008/11/15 Sat PM 04:33:24 EST
> To: steve at fl-eng.com
> CC: Peter Stuge <peter at stuge.se>, coreboot <coreboot at coreboot.org>
> Subject: Re: [coreboot] LPCflasher Project
> On Sat, 15 Nov 2008 16:09:58 -0500, <steve at fl-eng.com> wrote:
> > Host USB ports are supposed to use a MosFET with internal current sense
> > 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
> > has
> > occured and shut down the port. Even in this state, the reverse bias
> > in
> > the FET will continue to allow the external voltage to be fed into the
> > motherboard.
> > The FET/Motherboard external voltage will keep fighting each other as
> > 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
> > 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
> 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
> Joseph Smith
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