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Thread: 7900GT over voltage protection?

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  1. #1
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    Quote Originally Posted by Ebola
    Congratulation men, you did it.

    Did anyone check where via lead on the other side of the PCB, because soldering directly on ISL legs sounds quite hard.

    Is 1.5V and higher not too high for the regulator embedded on the GT ? (that are different from GTX )
    I'm afraid it won't be much easier...



    BUT: If Pin 32 and pin 22 is *already* '0', why not just connect
    Only pin 32 to pin 30 and 31 to get 1,350V or
    Only pin 22 to pin 21 to get 1,4V or
    pin 32 to pin 30 and 31 AND pin 22 to pin 21 to get 1,55V

    Would that work?
    Last edited by Belarnion; 03-23-2006 at 01:30 PM.

  2. #2
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    If you guys encounter problems soldering those tiny legs of the intersil chip just isolate the closests legs near the one to solder with seal string for example...


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  3. #3
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    Quote Originally Posted by Belarnion
    I




    BUT: If Pin 32 and pin 22 is *already* '0', why not just connect
    Only pin 32 to pin 30 and 31 to get 1,350V or
    Only pin 22 to pin 21 to get 1,4V or
    pin 32 to pin 30 and 31 AND pin 22 to pin 21 to get 1,55V

    Would that work?
    Depends of what is '0' in this case. If pin #32 (and pin #22) of the ISL6568 are connected directly to ground (hardwire), then your idea works.

    But if pin #32 (and #22) of the ISL6568 are connected to another chip and that chip sets the logic "0" on those pins, then there can be a big current going into pin #32 if you connect pins #30 and #31 to pin #32. It all depens of how much is VCC (pin #4) and what pull up resistors those pins have.

    I'm assuming that the chip that is connected to the ISL6568 has open collector outputs. If we can figure out where are the pull up resistors and how much is VCC (pin #4), I can tell if it is safe or not to connect between pins.

    It's easy to know if the pins #32 and #22 of the ISL6568 are hardwired to ground. With the card in your hands, measure the resistance between pin #32 and ground.

    My 7900GT will arrive next week, so I can not do the test myself. I can only research looking at the pictures and using what people say here.
    Last edited by TimeOut; 03-23-2006 at 02:52 PM.

  4. #4
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    Quote Originally Posted by TimeOut
    Depends of what is '0' in this case. If pin #32 (and pin #22) of the ISL6568 are connected directly to ground (hardwire), then your idea works.

    But if pin #32 (and #22) of the ISL6568 are connected to another chip and that chip sets the logic "0" on those pins, then there can be a big current going into pin #32 if you connect pins #30 and #31 to pin #32. It all depens of how much is VCC (pin #4) and what pull up resistors those pins have.

    I'm assuming that the chip that is connected to the ISL6568 has open collector outputs. If we can figure out where are the pull up resistors and how much is VCC (pin #4), I can tell if it is safe or not to connect between pins.

    It's easy to know if the pins #32 and #22 of the ISL6568 are hardwired to ground. With the card in your hands, measure the resistance between pin #32 and ground.

    My 7900GT will arrive next week, so I can not do the test myself. I can only research looking at the pictures and using what people say here.
    thanks for clearing up what I guess I was trying to say about tying the pins together and pull up resistors...
    thanks

    Offhand, do you know if one of the solder pads for nearby capacitor or some other solder pad, via etc, are good easy "ground"?

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  5. #5
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    Quote Originally Posted by Belarnion
    I'm afraid it won't be much easier...

    BUT: If Pin 32 and pin 22 is *already* '0', why not just connect
    Only pin 32 to pin 30 and 31 to get 1,350V or
    Only pin 22 to pin 21 to get 1,4V or
    pin 32 to pin 30 and 31 AND pin 22 to pin 21 to get 1,55V

    Would that work?
    Yes, this will work. Without changing the Resistor, you will have 1.55Volt.
    Do not worry about burning pull up resistors as somebody suggested, these pull up resistors are very large in value.
    To be more specific, the internal pull up resistors are in fact internal current sources, supplying 40UA ( 40exp-6 Amp) of current wen the input is connected to ground.
    Last edited by t024484; 03-24-2006 at 12:04 AM.

  6. #6
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    Quote Originally Posted by t024484
    Yes, this will work. Without changing the Resistor, you will have 1.55Volt.
    Do not worry about burning pull up resistors as somebody suggested, these pull up resistors are very large in value.
    so there's no need to remove any resistors at all?
    all that's needed is to bridge those contacts for up to 1.55v?

    i'm stilll waiting for my xfx card, but if the mod is really this easy i might actually try my first hardmod
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  7. #7
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    Quote Originally Posted by Pyr0
    so there's no need to remove any resistors at all?
    all that's needed is to bridge those contacts for up to 1.55v?

    i'm stilll waiting for my xfx card, but if the mod is really this easy i might actually try my first hardmod
    Yes, that's all there is.
    On the other hand, if you are not a skilled solderer, you can, as an alternative, solder the two larger points (more on on the edge of the board) to ground, that LardArse found to be in direct connection with point A (+0.2Volt) and B (+0.1 Volt) . That will give you up to 1.5 Volt

  8. #8
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    LardArse,
    Why do you think that the two resistors indicated in some picture from you, remove the current limit. Where are these resistors connected to?
    According to the chip supplier the current limit is set with a resistor between point 10 and point 11.
    This is just one resistor instead of two.
    On the other hand, there is no mentioning until now if the current limit comes into action at all with the higher Vgpu.
    I would have expected that to happen with 1.5Volt and 750 Mhz.

  9. #9
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    Quote Originally Posted by t024484
    Yes, that's all there is.
    On the other hand, if you are not a skilled solderer, you can, as an alternative, solder the two larger points (more on on the edge of the board) to ground, that LardArse found to be in direct connection with point A (+0.2Volt) and B (+0.1 Volt) . That will give you up to 1.5 Volt
    On his diagram he has 1.6V and 1.7V. I assume thats because he's using the VRM9 codes (aka replace resitor with 50K). Also, since I dont have a 7900GT (yet), could we get a better zoom shot of that area... where exactly is that in relation to the IC pic of the front?

    Quote Originally Posted by Pyr0
    thanks t024484, and of course everybody who all helped defeat the OVP
    I dont think Cronic did the over-current protection, and he's well past 650Mhz. I dont think its needed up to 700Mhz.. not sure when? Not defeated over-volt protection really, since cant disable it.. just different way of over-volt mod... now in 2006 its digital :p

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