Eu sei Ricky tb penso da mesma forma ( quem melhor do k tu k já o fizeste para o dizer) só referi a kestao dos pot's po vgpu por causa desta afirmação de alguem k já o tinha feito:
"Here's more information on the 9800XT Vgpu vmod. As mentioned above, the 5240 controller has 5 voltage select pins, VID4-VID0. The 9800XT has them hardwired: VID0,VID1 and VID4 are grounded and VID2 and VID3 are high. This normally gives an output voltage of 1.40V. However, unlike the example circuit in the data sheet, the 9800XT has a resistor between pin 17 and pin18 (VcoreD and Vcore+), as well as the series resistor between pin 18 and the Vgpu voltage. The resistor between pins 17 and 18 results in a voltage drop from Vgpu to pin 18. On the 9800XT the resistor between pins 17 and 18 is R1591 and the series resistor is R1592. The Vgpu voltage is thus:
Vgpu=1.40x(1+R1592/R1591)
The original resistor values are:
R1592= 432 ohms (62A code)
R1591=1620 ohms (21B code)
This give a default Vgpu=1.4x(1+432/1620)=1.77V. The measured value is 1.78V, which is excellent agreement. To increase Vgpu the easiest approach is to reduce the resistance between pins 17 and 18 by carefully connecting a pot across R1591. I'd suggest using a 20K pot. When the pot is at 20K, Vgpu would be 1.80V. The pot would be at 7.9K for 1.85V and 4.8K for 1.90V. If you want to go above 1.90V, a 10K pot might be a better choice."
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Vortan