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Possible short circuit after applying the wire trick?

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celmail

New Member
Joined
Mar 23, 2004
Hi!

I'm planning to apply the wire trick to be able to use my unlocked 2600+ in a chaintech 7vjda mobo (where the 5th multiplier bit is not accessible).
I still have one question. On the unmodified mobo the 5th bit is constant low. Doesn't this mean that AJ27 is connected to the ground?
If I connect AJ27 to VCC with the wire won't I make a short circuit on the board (as AJ27 is already connected)? Or AJ27 is always connected through a resistor to the ground by the board designers?

cheers,
C.
 
Welcome to the forums


The pin AJ27 is one of the L1 pin, it does not DIRECTLY connect to GROUND (VSS). It connects through a resistor (1 K Ohm I think) to GROUND, as shown below.


GROUND ----- 1K ----- L1 ---x--- L3 ----- AJ27


There is also a pullup path to VCC (Vcore) at the node x. The node x also connects to a bridge voltage detecting circuit for multiplier control.

The 1K Ohm to GROUND and the pullup path to VCC form a voltage divider between VCC and GROUND. It sets the default voltage for the node x.

Vx = R_pulldown / (R_pullup + R_pulldown)

By designing proper vaules for resistors R_pullup and R_pulldown (1 K Ohm), under default condition, the voltage at x is LOGIC LOW (less than certain voltage threshold) for the mutilplier control circuit.


Two ways to alter default bridge voltage (how it works)

1. Blowing bridge: To alter default bridge voltage, if a lower bridge (the bridge that is closer to GROUND) is cut, the voltage at node x will go to LOGIC HIGH (pulled up by the pullup path to VCC). Hence the bridge detecting circuit will detect a different logic value, LOGIC HIGH instead of LOW, altering the default setting.

Similarly, an open bridge can be closed to alter the default logic value at node x.

2. Wiring trick: A high voltge (typically VCC) can be forced to a bridge pin (such as AJ27) at the CPU socket by wire bridging VCC to the pin, hence forcing VCC to node x to alter its default logic value (if is normally LOW). Hence the bridge detecting circuit will detect a different logic value, altering the default setting. Even if the brdige of a node (x) had already been open, forcing a high voltage would do no harm (since it was at HIGH).

Similarly, the same can be done by forcing VSS (GROUND) to a socket pin using the wire bridge, altering default LOGIC HIGH at a bridge node to LOGIC LOW.

There is a minimal disadvantage of wiring trick, it wastes a little additional power, about a few mW (milli-watt). Comparing to a typical CPU power dissipation of ~100 W, it is considered acceptable.

This alters the bridge voltage, whether it can change the CPU configuration (for multiplier, default Vcore, ...) depends on the whether the chip is LOCKED or not (internally).


This link suggests more details about how the bridges look electrically and logically,
http://fab51.fc2web.com/pc/barton/athlon-e20.html#index


VCC is the voltage pin for Vcore.
VSS is the voltage pin for ground.
If they are shorted, the voltage regulator of the motherboard can be damaged (instantly).

There are lots of VCC and VSS pins in the CPU socket, if any pair of them are shorted, motherboard can be damaged. So check the work very carefully.
 
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