[radwerx]

Hi all. I have this later generation quicksilver G4 733 that I'm trying to do a voltage mod on. However, it has a different voltage regulator from the earlier versions and the resistors are arranged differently. I have attached some pics below for reference. I have googled all over and all articles cater to the older chip. Hence, here i am requesting for assistance from any electrical whiz....Just tell me how to get it to deliver 2.0Volts to teh CPU core. Thanks in advance for all help rendered.

Here are the pictures.

NEW VOLTAGE REGULATOR





OLD VOLTAGE REGULATOR

[Spliffdaddy]

Link to page for LTC1709EC with a link to the datasheet >

http://www.linear.com/search/searchR...eria=LTC1709EC


Direct link to .pdf datasheet (may not work) > http://www.linear.com/pc/downloadDoc...83,P1819,D2985

See page 17 and it shows which of the 5 'VID' sens pins must be pulled to ground in order to yield 2.0v at the output.

Looking at the picture, the 5 'VID' pins are the last 2 pins on the bottom right (vid 0 & vid 1 left to right), and the last 3 pins on the top right (vid 2, vid 3, & vid 4, from right to left). Currently, it appears that vid 2, vid 3, and vid 4 are pulled to ground - which would yield 2.3v at the output. But don't quote me on that - I ain't no expert.

Note that pulling all VID pins to ground also yields 2.0v output (see notes on .pdf page 17).

Looks simple enough.

If you need more assistance, just ask. I'll check back later.

[cfm]

aw man i need someone like u at the moment.....lol
ive killed my dvplayer by taking the back off (ripped a ribbon Connector off.....)
ive got a soldering iron and a dremmel and in my usual state am about to murder it better

sorry for being a newb by the way.

[Spliffdaddy]

Good luck on the ribbon cable. Repairing electronics at the circuit board level is difficult. Too damned small.

Anyhow, for the original poster:

See the row of 3 resistors (their value is 40K, probably) that ground the VID pins 2,3,&4 ??

See how there is room for 2 more resistors just to the right of those 3?

I'm guessing that those empty spots are connected to the VID pins 0 & 1 located at the bottom right on the LTC1709 regulator in your picture. Check those connections and see if I'm right. Makes sense because there are 5 VID pins that are used set the output voltage...and there just happen to be 5 places in a row for resistors.

What I would do is remove the 3 resistors. Then take one of them and solder it in the empty spot that corresponds to the VID-0 input. According to the datasheet, grounding *only* the VID-0 pin should yield a 2.0v output.

Otherwise, if you have 2 extra resistors laying around - solder them onto the 2 empty spots. Again, according to the datasheet, grounding all 5 VID pins will also yield 2.0v.

Please remember I am not an electronics expert. I'm barely a hobbyist. While I've done this sort of modding on my own CPUs, my efforts have not been met with 100% success.
If you can't risk killing your CPU, then don't do anything to it. Consult somebody who has some expertise in this sort of thing - because I don't.

If it were *my* CPU, I would probably just slap a kick-ass heatsink on the sucker and run it at 2.3V - and forget about modding resistors altogether. Over-volting a CPU by 10-15% usually won't kill it. It can help stabilize a flaky overclock, assuming you address the increase in heat associated with the over-volt. Wait, nevermind, I've killed too much hardware in my overclocking days to be giving you advice.

[radwerx]

Thanks a lot, SpliffDaddy, that was extremely insightful. May I ask how u get the thing to operate at 2.3V without touching any resistors at all? BTW, I really don't think its running at 2.3V now like you theorised since its a stock chip currently Stock voltage is 1.80 V or 1.85V i think.....I may be wrong too though.... : )

[radwerx]

Ok. You're rite bout vids 0 and 1. They correspond to the 2 empty jumpers above and are arranged in the same left-right order as well.

[Spliffdaddy]

oops

OK, this is why I'm not an expert.

Looking at the datasheet (pg 17) there are "0" and "1" which show where the resistors should go. I was guessing that "1" meant you need a resistor on that pin.

The 1.9v setting shows a "0" at VID locations 2,3,&4 - so if we assume, instead, that "0" means a resistor goes there, then that would correspond to the existing setting.

IF your current voltage is 1.9v (find a way to make sure) then your 2.0v CPU will probably work just fine. And *if* your current voltage is 1.9v then that means the datasheet is showing a "0" in the locations of the grounded VID pins. Sooo, in order to get 2.0v, you would simply remove all the resistors which would yield a 1,1,1,1,1 - which defaults the voltage regulator to 2.0v.

To put it simply, you need to determine whether you currently have a voltage of 2.3v or 1.9v.

If your voltage is 2.3, then my original post is correct. If your voltage is 1.9, then this post is correct - and you can simply remove the 3 existing resistors to get 2.0v output.

[radwerx]

Hmm... does anyone here know how to determine the voltage currently running across the CPU core?

[radwerx]

I tried filling up all 5 jumpers and emptying all 5. The computer just chimes and gives me a blank screen. ( at a clock speed of 1Ghz that is....)

[Spliffdaddy]

update:

"0" does represent resistor locations on the VID pins.

That means your current voltage is 1.9v

So all you need to do to get a 2.0v output is to remove the 3 existing resistors.

This turned out to be pretty simple.

[Spliffdaddy]

Originally Posted by radwerx

I tried filling up all 5 jumpers and emptying all 5. The computer just chimes and gives me a blank screen. ( at a clock speed of 1Ghz that is....)

regardless of whether you have 5 resistors connected (2.05v) or 5 resistors removed (2.0v) your CPU is getting the voltage it needs.

The clockspeed issue is handled in a similar manner - that is, the multiplier is set via resistors. The formula is FSB X multiplier. If your FSB was 133MHz and your multipler was 7.5X then it would result in a core speed of 997.5MHz.

FSB of 100MHz X multiplier of 10X would be 1,000MHz

If the CPU speed is wrong, it's because the multiplier is still set for the old CPU.

I've never played with Mac hardware, so I can't tell you how to adjust the multiplier. But I could probably google the information.

Also, it's likely that the multiplier setting is on the CPU circuit board itself - not the motherboard. So it would be impossible for the multiplier to be set wrong.

The FSB setting is generally unchangeable and is handled by the motherboard chipset.

Does the old CPU and the new CPU share the same 133MHz FSB speed?

worst case: the CPU pin assignments are different between the old CPU and the new CPU. Which means the motherboard can't properly support the new CPU. This can probably be corrected. Depends on how far you're willing to go.

This reminds me of a P3 Xeon project I worked on years ago. One problem after another.

[radwerx]

Hmm...doesnt work either ways ( I mean it doesnt want to boot). Perhaps it just can't run at 1Ghz then. Don't bother googling about the multiplier and bus stuff. I'm not new to overclocking....haha, just can't figure out this voltage regulator. I'm sure I got the multiplier right. Cos it worked at 933. Just cant geddit to boot at 1Ghz.

[Spliffdaddy]

Oh, cool. The fact that a 733 booted at 933 is pretty impressive. Getting it all the way to 1GHz would be quite an achievement. Hell, up the voltage to 2.3, bolt on a waterblock and liquid cooling system and go for broke.

If you want to try for 2GHz maybe I'll let you borrow my 'Deliverance' machine. -70F has a way of kicking CPUs asses into submission.

[radwerx]

ERM.....can you explain a bit what this machine is about? Cos it looks too large for a compressor from a fridge and too complicated for a water cooling system or peltier cooler.... It does look...er...powerful and purposeful though... HAHA

[Spliffdaddy]

basically, it's a pair of R134a compressors with liquid heat exchangers. The antifreeze mixture is circulated through the chiller's heat exchangers where it's cooled to about 29F (-1C) - and pumped to a waterblock that cools the hot side of a cascaded (stacked) TEC (aka, peltier) array. The cold side of the TECs has a quarter inch thick copper 'cold plate' that is placed against the face of the CPU. The TEC array consumes about 1500watts @ 10vdc. The power supply for the TECs is the big rectangular aluminum block with the 8gauge wires running in front of it. It was a fairly expensive component, made by Lamda. The output voltage is adjustable via knobs on the green boxes to the right. Just for kicks, I used 1960's era Nixie numerical display tubes to show output voltage. Also, there is a air compressor onboard that operates the pneumatic motherboard tray - which flips outward for easy access.

The machine weighs about 160 pounds and requires about 35 amps of AC power. I have to plug it into 2 separate outlets else it trips the circuit breaker.

After about 15 minutes of operation, most of the motherboard is covered with a layer of frost. Almost 2 years ago I was able to get 4GHz from a P4 2.8 - which was the limit of the motherboard's FSB settings (1066MHz).

Under load, the cold plate will maintain -70F temperatures. The temp drop from room temperature (72F) to -70F takes about a half second - and you can actually hear a sizzle like bacon frying, as the humidity in the air freezes on the copper cold plate.

I left it running for about half an hour while I went to Wendy's to get a burger. When I got back home I found this:

[radwerx]

I don't know what to say. Its a real engineering masterpiece. But to have to use up so much power to make a chip go that little bit faster seems like a case of diminishing returns... Well at least when your fridge runs out of ice and u need to make some real quick you know who to turn to...heh heh

Thanks for all your help Spiffdaddy, really enjoyed your pictures too. Seems like a prometia is in my Christmas wishlist.