Hi, i get some old laptop. There is short cut on board but i cannot find where. I have 19V on ground. There is no heating components, or any indicator. I found some burned dual zener diode.
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HP Compaq nx7400
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Re: HP Compaq nx7400
You should apply a 5V hight power amperage voltage from an external power supply that has no protection for short-circuit (like from a basic coils core transformer) to the motherboard 19V line and see which component is heating. That will be the faulty one, or if is a mosfet connected directly from the 19v line and not to the ground, then it is the component next to it, mostly like a capacitor or a bus ic.
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Re: HP Compaq nx7400
Don't put 5V. Start with 1V. If a high-side MOSFET for the CPU/GMCH/ICH is shorted, you are sure to kill it by sending 5V down the line if it's not dead already.
EDIT: also thread is 5 years old…OpenBoardView — https://github.com/OpenBoardView/OpenBoardView
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Re: HP Compaq nx7400
Voltage will drop depending on the output impedance of the power supply, said differently if it cannot provide enough current the voltage will drop. The voltage will also drop through the resistance of the wire.
But that's not what I was talking about. If the short is due to a shorted high-side MOSFET on GMCH VCore for example, you'll measure resistance lower than 80ohms on this rail usually, so you'll measure less than 80 ohms on B+ as well. The voltage you inject on B+ will be sent directly to GMCH VCore, which means you're sending 5V instead of the 1.05V it expects.
Heat is related to power, not voltage. Power is voltage times current, so you have to monitor the current consumed and the actual voltage (if it's dropping) to estimate if it's going to heat up.
For example when you inject 0.5V, if the current consumed is 6A, it means dissipating 3W which is usually enough to clearly make something heat up on the board. If it's consuming only 0.5A, it means 0.25W of power which will be harder to detect.
So you should start at 1V, check actual voltage and current consumption, if the resulting power is too low and no sensitive component is heating up (CPU/GPU/GMCH/PCH/ICH/etc… depending on the board) then you can increase the voltage.
Note that if the power supply has reached its maximum output current, the voltage will drop. For example if you have a 3A supply and you try to inject 5V into a 1ohm short, it won't keep up, so the voltage will drop to 3V. This is Ohm's law, note that it doesn't really apply like that to semiconductors since they're not linear devices but doesn't really matter for the explanation.Last edited by piernov; 12-18-2018, 02:40 AM.OpenBoardView — https://github.com/OpenBoardView/OpenBoardView
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Re: HP Compaq nx7400
Originally posted by piernov View PostDon't put 5V. Start with 1V. If a high-side MOSFET for the CPU/GMCH/ICH is shorted, you are sure to kill it by sending 5V down the line if it's not dead already.
EDIT: also thread is 5 years old…
Many trick and good suggestion also we could found on old threads.
Originally posted by piernov View PostSo you should start at 1V, check actual voltage and current consumption, if the resulting power is too low and no sensitive component is heating up (CPU/GPU/GMCH/PCH/ICH/etc… depending on the board) then you can increase the voltage.Last edited by on/off; 12-18-2018, 06:16 PM.
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Re: HP Compaq nx7400
If you have a high power Ah transformer then yes you can use 1V. I do with 5V for current of 1 Amp and never had a problem to burn something else as the voltage it will drop significantly. Anyway that the Idea, having a high Amps current related to low voltage send to the short circuit.
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