[coreboot] DL145G3

samuel samuel.verstraete at gmail.com
Fri Mar 27 09:05:47 CET 2009


On Fri, Mar 27, 2009 at 8:59 AM, samuel <samuel.verstraete at gmail.com> wrote:
> All this talking about a single pin that is causing this is making me
> think about something...
> This mainboard has 2 jumpers. Each next to each other...
> One is "BIOS RESET" and the other is "CRISIS PARK"... Could that
> CRISIS PARK maybe connect to pin 8 to disable the flashrom writes??? I
> guess if i measure the voltage and/or ohm that should tell me?
>

reference:
http://bizsupport1.austin.hp.com/bc/docs/support/SupportManual/c00856243/c00856243.pdf
page 63 and page 64 (it's component 18)
on page 65:

PARK: "Normal boot" or "Boot block enabled" doesn't sount like bios
write protection... so this prolly is nothing...




>
>
> On Fri, Mar 27, 2009 at 8:55 AM, Peter Stuge <peter at stuge.se> wrote:
>> samuel wrote:
>>> >>Do you have access to a continuity tester? Please check this
>>> >>connection.
>>>
>>> I do not have access to a continuity tester.
>>
>> ..
>>
>>
>>> >> * You can insulate the #TBL pin of the flash chip from the
>>> >>   corresponding socket contact, and connect #TBL to Vdd. According
>>> >>   to the datasheet of SST49LF080A, providing Vih(=Vdd) to #TBL
>>> >>   disables write protection on the boot block.
>>>
>>> I guess i can try this... I suppose it won't hurt?
>>
>> Yep, it's a good idea. Bend pin 8 up and make sure it doesn't have
>> contact with the socket, then connect it to pin 32 or 25. (Both VDD)
>>
>>
>>> >>Correct. It is most likely the cause. But - it would also be
>>> >>interesting if you could measure the voltage Samuel, on pin 8
>>> >>after having run flashrom in any mode.
>>>
>>> I will get a volt meter... What pin should i use as ground? I tried
>>> having a look at page 7 of the LPC data sheet but i didn't see a
>>> ground pin...
>>
>> 16, VSS is ground.
>>
>> Maybe that volt meter also has an ohm meter function. That works
>> well as a continuity tester too - if two points are connected the
>> resistance will be 0.00 ohm.
>>
>>
>> //Peter
>>
>> --
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>>
>




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