1 Gig Celeron on BX-20??

[corrupt]

Hi all,

I'm not if this is possible... but I'm trying to put a 1 Gig Celeron (socket 370 in a slot 1 card) (1000/128/100/1.75V) processor into a Abit BX-20 motherboard. The motherboard does work (tested with a p2 400). The motherboard does not seem to have a 10 x multiplier by jumper settings (max 9.5 as far as I can tell) but the frequency can be set to 100 (and unofficially 133 but 7.5 x 133 didn't work). Any setting I try doesn't seem to post on startup though. Anyone have any suggestions? Is this a bios issue? mobo issue? processor issue?

The bios for 128 Gig HD does work great though. Thanks

[ajzchips]

Forget the multiplier, it's factory-set and will work at its rated setting, so no problem there.

a) You might have an ordinary PPGA-Slot1 converter... you'll need an FCPGA-Slot1 converter.

b) Your onboard voltage regulator needs to be able to generate what your processor's VID pins request: 1.75V. If it can't, it just won't boot. You'll have to try 1.70V or 1.80V, 1.85V, 1.90V.... (try not to exceed 1.90V). Very likely one of these settings will run your processor.

Note: (b) is only applicable if you have a SoftBIOS or your slotket has specific jumpers for VID adjustment.

[corrupt]

a) You might have an ordinary PPGA-Slot1 converter... you'll need an FCPGA-Slot1 converter.

The converter came with the processor and I'm pretty sure the PC shop said that they had tested the processor with this board... I'm not too sure about the specs on the converter board or where to start looking. It has no manufacurer's name printed on it that I can find. Only a couple stickers saying Made in China and F9943008501 370SP. It does have 1 jumper on it but is only marked JP1 so I'm not sure what it does (having not played around with these converter boards before now).

I'll see if I can do anything about the voltage...

Forget the multiplier, it's factory-set and will work at its rated setting, so no problem there.

So the motherboard should be able to handle the multiplier even though it wasn't designed to go that high?


Thanks for the reply

[corrupt]

The card looks almost like this: (pic url: http://www.kikumaru.com/pc/s370tos1/ ) It has the same layout of components and etching (gold board though) but some components are slightly different (not present) although the spaces for them do exist (the transistor on the right for example).

[ajzchips]

Stuff you might want to read:
http://www.hardocp.com/article.html?art=MjIx
http://membres.lycos.fr/ngc/voltman.html
http://www.plasma-online.de/english/upg ... ltage.html

Your VID pins for 1.75V are set as follows:
VID3=0
VID2=1
VID1=1
VID0=0

The only possible setting change I can see is 1.70V, but there's no guarantee your voltage regulators will detect this.

Note: to transform a 0 into a 1, all you need is to cover the contact with a piece of tape. [0 = closed circuit = connected; 1 = open circuit = no contact]
Note2: it's impossible to transform a 1 into a 0, because the connection is internally open. Nothing to do there.

[corrupt]

Note: to transform a 0 into a 1, all you need is to cover the contact with a piece of tape. [0 = closed circuit = connected; 1 = open circuit = no contact]
Note2: it's impossible to transform a 1 into a 0, because the connection is internally open. Nothing to do there.

Wouldn't this be the opposite? To change a 0 to a 1 wouldn't be possible since the pins are internally connected in the processor and to change a 1 to a 0 connect the pin to the vss pin... or am I misunderstanding? This was my understanding after reading: http://www.hardocp.com/article.html?art=MjIx

If I'm understanding correctly that would leave me with: 1.85v, 1.95v, and 2.05v

Do you think the celeron would handle 1.85v or 1.95v?

It does have a pretty good fan installed (please excuse the bad pic):

[ajzchips]

The hardocp.com site refers to the processor connected directly to the motherboard's socket370, and refers to soldering wires to permanently change the VID of that particular processor.

And no... the zero state means that there's a physical connection closing the circuit. Logical zero would represent the voltage difference between both sides of that point. If it's zero, it's a closed circuit, therefore it's "undoable".
If it's a one, then it's open inside the processor itself, so there's no way of ever closing the circuit. There's nothing you can do to that pin that'll close the circuit.

You might prefer experimenting with a more sophisticated slotket (full of jumpers for lots of settings).

[corrupt]

I tried to find a better board today but I wasn't able to find anything at all locally unfortunately. I could probably order something (and I might) but I don't mind altering the board or processor. I got them really cheap and it's not a critical system to me . I have a bit of experience with electronics and altering PC equipment but not specifically with processors (yet). I guess the part that I'm a little unlear with is how/where the connections are made to determine the voltage used by the processor.

Does a combination of shorted pins at the processor determine the voltage that is sent to the vcc pin by the mobo? If not, how does twisting wire around pins at the processor (when they are not internally connected in the processor) raise the voltage in the example here: http://www.hardocp.com/article.html?art=MjIx ?

Sorry to be a pain... I'm trying to understand the theory behind how this works

[Denniss]

Try to get a converter card with jumpers to change VCore - many BX-based mainboard have 1.8V as minimum voltage - even my very old Asus P2B Rev 1.02. is able to deliver 1.8V as Minimum

[ajzchips]

Exactly as Dennis says... although most early Slot1 BX boards where designed to operate at a minimum of 2.0V, there was usually a +-0.2V tolerance, so their lowest voltage generation was at 1.8V.

[Rainbow]

My Asus P2B rev. 1.10 has exactly the same limitation - so I'm frying (modified) Tualatin Celeron with 1,8V The temperature is a bit high but it works fine