Re: [coreboot] LPCflasher Project

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....

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?

Steve Spano
FLE
> From: Joseph Smith 
> Date: 2008/11/15 Sat PM 03:53:43 EST
> To: Peter Stuge 
> CC: coreboot 
> Subject: Re: [coreboot] LPCflasher Project
>
> On Sat, 15 Nov 2008 21:26:05 +0100, Peter Stuge

wrote:
> > Joseph Smith wrote:
> > > Anyone know if USB ports have power feedback protection.
> >
> > All the details are in the USB spec. IIRC electrical is a separate
> > chapter.
> > > What I mean by that is if I use the LPCflasher as a inline flasher
> > > with a PLCC32 socket plug,
> >
> > I don't understand this. Do you mean a socket or a plug?
>
> Both, a socket on top of a plug that is also connected to the LPCflasher.
> If I power on the motherboard it is going to supply voltage back to the
> LPCflasher, right?
> > > flash the chip
> >
> > Flash which chip?
>
> Any LPC or FWH.
> > > and then power up the motherboard, I don't want the VCC's from the
> > > motherboard to short out the flasher and or the USB port from the
> > > host PC.
> >
> > Make sure GND is connected across and you should be OK.
>
> Yup.
> > > The 3.3V regulator I am using has a internal diode so I am not so
> > > worried about that. I am just worried about the power source 5V
> > > USB. I could always use a Schottky diode that will only drop the
> > > voltage by .6V, but I don't know if 4.4V will suffice???
> >
> > Suffice for what? Look at the documentation for the regulator you're
> > using to see what input voltage it needs.
>
> I wasn't really ready to release this yet, because the power part is still
> a work in progress but it will be the easiest way to explain this. See the
> diode I have above the regulator? I put that there so power is not flowing
> back to the USB port of the PC doing the flashing when you power on the
> motherboard that just got flashed. My question is if I take out that diode
> and you power on the motherboard that just got flashed is power going to
> flow back into the USB port and damage it? And if I leave the diode it is
> going to drop the 5V by approx. half a volt. Will I still be able to flash
> 5V chips at 4.5V? Does that make sense?

LPCflasher Project

[Joseph Smith]

On Sat, 15 Nov 2008 16:09:58 -0500,  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?