9700 vgpu equation questions

[Ragnarok]

http://www.xbitlabs.com/images/video/titanium-overclocking/scheme-1.gif

I am an aspiring electrical engineer n00b... (well gimme time, still @ uni)

Looking the xbitlab's vgpu mod equation absolutely made no sense to me, they seem to be completely wrong... and from other guides i know 8Kohm should give u 1.6VGPU..

but if u plug into the equation clearly u get jibberish values

well anyone wanna come up with correct equation?

[CATCH22ATPLAY]

Bah those guys are complete idiots. They don't even do thier own mods. They steal em from other people. To include from me. Try this formula for that chip. It works btw

Vout = [1.25 X (R12 + R1)] / R1

If i do the math it looks like this

[1.25 X (150 + 124)] / 124
(1.25 X 274) / 124
342.5 / 124 = 2.762 Vout

As you can see Semtech's math is a little bit off. But if you read thier data sheets they claim an estimated voltage drop of 0.10 volts. So i guess Semtechs 150 and 124 ohm resistors are close enough. I just wish xbits would stop posting complete bull.

Now since you're at UNI i'm gonna give you a test. Gimme the equation for figuring out current aka I symbol. Lemme start it for you.

I out = ?????????????????????

Hey don't take me too serious. I'm just having fun here. But if you want that answer i can give. Hey not bad since i've never actually taken any sort of electronic courses. However i do design electronic stuff. Must be why i work for an engineering company. Hmm nah that can't be it. Maybe i just read too much. LOL

I'm amazed nobody has answered this either. Doh!!!

[CATCH22ATPLAY]

Arg i did the Semtech equation. But never did complete it for adding that resistor for voltage modding it. So hmm parallel resistance now on R1 which i'll call the one being added R1' as well. Try this

R1* = (R1 X R1') / (R1 + R1')
Vout = [1.25 X (R12 + R1*)]/ R1*

I'm sure you can take it from there :) Correct termonology is nice. But i'd prefer R1' to be given a real number like R50 and R* to be named RT (Resistance total or total resistance). But that's just me. I like to make it simple stupid. So my math would have looked like this. Now this moron aka me can figure it out. :)

RT = (R1 X R50) / (R1 + R50)
Vout = [1.25 X (R12 + RT)] / RT

Damn i hope i didn't mess this math up. It's late and i'm falling asleep in my seat. Night! ZzZzZzzzZzz

[Ragnarok]

cheers man, that really helped! I guess a lot of ppl dont really care about all the schemeatics behind volt modding and just blindly solder and plug in random values :(

what a pity...

[CATCH22ATPLAY]

I agree and it's a shame. I think everyone should at least learn something constructive from theese mods. Me i love trying to figure this stuff out. I'm not always right and i fry stuff doing it. But when it works it's sweetness. Like when i just tried to do this equation for you. It's like a challenge that i gotta solve. I think i enjoy making up theese mods more then i do trying to stay within the top 10 in 3dmark. A sense of accomplishment etc. I'm glad you're interested in expanding your knowledge. :) I'm afraid that although the equations look simplistic to me and you it may be way over most peoples heads LOL

[Nohto]

I=E/R

[Per Hansson]

If you remember a while ago I wrote a TI4200 volt-mod article (http://www.techspot.com/tweaks/geforce4-vmod/voltmod-2.shtml) (thanks to the great help of all the great members at this board)

Right now I'm in the works of making a similar article, however after reading my own article and the resistance formula part I still can't for the life of me figure it out... Neither did the light shine through when I read the above resistance formula...

I.e. what I need explained is how you go from finding the base resistance (220ohm if I measured correctly on the 9800) to figuring out what ohm level the pot needs to be so you don't end up with a too weak pot and kill the board due to a voltage zap...

I would highly appreciate any help anyone can give me on this.

[Per Hansson]

Ok, I'm starting to believe I actually figured this out now:
---
Total Resistance=(124x10000)/(124+10000)
1240000/1012400
=1,2248
---
Vout=(1,22x(R12+R1*))/R1*
(1,22x(150+220))/220
(1,22x370)/220
451,4/220=2,0599v

But if I'm correct about that a 5kohm pot would give 2,0350v; which means this formula was for calculating the max voltage?

What I'd like to know is what the minimum voltage will be after you add a pot of X k ohm... So if anyone knows the formula for that I'd gladly have a look at it ;-)

[Per Hansson]

Alright, Macci helped explain this too me now over ICQ

Vgpumodded = Vgpu * [1+(base resistance/VR resistance)]

vgpu = your standard voltage (1,70v)
base resistance = measured resistance when card removed from PC
VR resistance = pot resistance you want to use

Therefore:

1.7 x [1+(220/10000)]
1.7 x [1+0,022]
1.7 x 1,022 = 1,7374v

So with a 10k ohm pot our startup voltage will be 1,74v.

And our maximum voltage (formula in post above) with a 10k ohm pot will be 2,06v

[CATCH22ATPLAY]

Although the origanal question was "well anyone wanna come up with correct equations?" based on xbitlabs effort which i provided. I like your equations very much. In fact that's the equations i use for Vmem and GPU etc etc mods. Although it doesn't give exact values, it will actually give you a bit less than what you think you'll get. So in a way you can say in answer to your earlier question that yes with your equation that will be max volts. Because it won't exceed what you think you'll be getting based on your equations. I like your equation because it's just so close and easy to use that it's an excellent equation and it's fairly safe. Thanks for adding that equation. That may be the best way for people to do theese mods :)

And now i must put egg on my face. It seems there are more ways then one to get the real equations. I just did xbits math and it works too. Damn and i thought they were wrong for sure. I also did it a 3rd way and came up with correct results too. Geez i'm blowing it. I give up for now. I think i'll just stick to the old way of doing it. The 4th way and easiest aka Per Hansson's way and to hell with the real math :)

[Per Hansson]

Originally posted by Per Hansson
Alright, Macci helped explain this too me now over ICQ
Originally posted by CATCH22ATPLAY
I think i'll just stick to the old way of doing it. The 4th way and easiest aka Per Hansson's way and to hell with the real math
Please don't give me cred for this equation, it was Macci who explained it to me over ICQ...

I only posted it here for everyone's convenience...

[\/icious]

Originally posted by Ragnarok
cheers man, that really helped! I guess a lot of ppl dont really care about all the schemeatics behind volt modding and just blindly solder and plug in random values :(

what a pity...

hey now im one of those fools who blindly solder :lol:
ill learn what all that bs means sooner or later just like ive learned everything else that i know about computers i just need a starting point.