is it possible with this Variable resitor to change VGPU to be not only 1,5 or to adjust another voltage
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is it possible with this Variable resitor to change VGPU to be not only 1,5 or to adjust another voltage
Use a 62.5K VR? Assuming lowering resistance raises the voltage.Quote:
Originally Posted by milen_bg
:stick:
Please explain, this mod AND vcore mod, or just this mod?
both i guess.Quote:
Originally Posted by DesertShooter
i guess a 100kohm for ovp will be safe conseding that lowering resistance gives more voltage
With this mod you get only 1.5V
With this + vmod you can get 1.57V. After that--> black screen. (for me)
DO NOT use anything else than 50Kohm! This mod change VID table what ISL6568 read, nothing more.
So there isn't any way of going up a little more?Quote:
Originally Posted by SF3D
At the moment no..maybe later.
What exactly did you do, I do not see anythingQuote:
Originally Posted by SF3D
Post 49. Thank you also! I thought this mod week ago, but your opinion today convinced me.
^
Sorry my mistake..you don't see the pic?
Very smart solution SF3D.
The pin 1 is the VID12.5 and the ISL6568 datasheet says:
"The VID12.5 pin also servers as the internal DAC compliance selector. The way this pin is connected selects which of the three internal DAC codes will be used.
- For VRM10 codes, this pin must be less than 3V.
- To encode the DAC with Intel VRM9.0 codes, connect the VID12.5 pin to a +5V source through a 50K Ohm resitor.
- To encode the DAC with AMD Hammer VID codes, connect this pin to a +5V source through a 5k Ohm resitor."
Now, looking at the close picture in this thread page 1, we can see that the resitor connected to pin 1 is 68b. This is a 5K resistor, meaning that the ISL6568 in the 7900GT cards is using AMD Hammer VID codes.
The 1.2V in in Hammer VID codes is "01110". Changing resitor to 50K Ohm will cause this chip to use VRM9.0 codes and the "01110" is 1.5V.
original DAC version was AMD HAMMER (5kOhm) with 01110 = 1.2VQuote:
Originally Posted by DesertShooter
50kOhm switch to VRM9 with same VID logic 01110, but voltage increased to 1.5V (see table http://www.pctuning.cz/ilustrace2/Pirs/7800GT/VIDs.gif )
How are your temps at that vgpu?
The attachment was not there, now it is.
I can see that you changed the table from AMD Hammer to VRM9.0.
This means that the code on the 5 pins is "01110" which means in AMD mode 1.2Volt and in VRM9.0 Mode 1.5 Volt.
If you want to have more output voltage with this VRM9.0 setting (with the 50Kohm resistor) you can do the following, starting from 1.5Volt.
-------1.500-1.525-1.550-1.575-1.600-1.625-1.650-1.675-1.700Volt
Pin22----0-----0----0------0------0-----0------0----0-----0
pin21----1-----1----1------1------1-----1------1----0-----0
pin30----1-----1----1------0------0-----0------0----1-----1
pin31----1-----0----0------1------1-----0------0----1-----1
pin32----0-----1----0------1------0-----1------0----1-----0
To turn a 1 into a 0, connect this pin to ground.
To turn a 0 into a 1, let this pin float without a connection
When you connect for instance pin 30 to ground, you have 1.6Volt
When you connect pin 21 to ground you have 1.7 Volt
If you do not want to replace the 5Kohm resistor into a 50Kohm resistor,you have a fewer amount of possibilities.
I will outline this in a next mail
Temps are 37/47. I have normal watercooling. 705 @ 1800 is totally stable.
Did you measure the voltage on the 50k resistor?
Very nice.Quote:
Originally Posted by SF3D
And just to clarify as I am no where near EEE inclined, all you ahd to do was replace that resistor for the 1.5v? :cool:
i knew it was only time before someone had teh brains/balls to figure this one out, good work mate :)
If you want to have a higher Vgpu but without changing the standard 5Kohm resistor into a 50 Kohm resistor, you have the following options starting at 1.200 Volt.
-------1.200-1.250-1.300-1.350-1.400-1.450-1.500-1.550
Pin22----0-----0----0------0------0-----0------0----0
pin21----1-----1----1------1------0-----0------0----0
pin30----1-----1----0------0------1-----1------0----0
pin31----1-----0----1------0------1-----0------1----0
pin32----0-----0----0------0------0-----0------0----0
To turn a 1 into a 0, connect this pin to ground (easy)
To turn a 0 into a 1, let this pin float without a connection (not so easy)
When you connect for instance pin 21 to ground, you have 1.4Volt
When you connect pin 21 plus pin 30 to ground to ground you have 1.5 Volt
1.4 is realy easy to achieve, and 1.5 Volt is also quite easy.
Ok so pin 21 to a ground is 1.4V, my question is (sorry if this was answered) what pins are we talking about on the actual card. :p) And what would eb the ideal ground, or does it not matter?
http://www.xtremesystems.org/forums/...8&d=1142891065
So...after having the 50kohm resistance connected instead of the stock 5kohms if we connect the ISL6568 21pin to ground we will read on the vgpu measuring point 1,7v am i right ?
So without removing the 5kohm we can have 1,5v max...after replacing it by 50kohms we can raise till 1,7v ?
I can see in the 7900GT picture on this thread page1 that pins 22, 21, 30 31 and 32 go somewhere else, meaning that the current 1.2V on the 7900GT is not hardwired.
Does anybody have a clue of where those pins are going? GPU?
Thanks for that pic, helps me.
And yes that is how I understand it as well.
We are talking about the pins of the voltage regulator, a square chip with 32 pins, type nr ISL6568.Quote:
Originally Posted by Chr0n1c
The counting is counterclockwise, starting from the small hole in the chip housing The first side goes from 1-8 the next side from 9-16 etc.
Any ground will do, it is not critical but do not make the wire too long.
Thank you very much, was a lot of help. :DQuote:
Originally Posted by t024484