hp omen 15-dh1015nl does not turn on
				
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 I thing I find something strange.
 Here in the pics I report the V at those resistors (I think they are resistors). If I measure the resistance of the resistor inside the circuit I get 150KOhm ish. If I measure against GND the second gives me 150KOhm again but the top one gave me 0.4V!!! Is it possibile that after the smd I have a short?
 I tried to inject 1V and it takes all the A i gave (I stayed low around 0.5A) but I didnt see nothing becoming hot so i stop the injection.Last edited by giulio93; 04-04-2025, 11:56 AM.Comment
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 As per post#20, i've given you hint to get charging ic pinout by referring to another schematic. You need to do your homework and stop playing with voltage injection and instead focus on troubleshooting.
 
 Resistor needs to be measured out of circuit. Just de solder them and put it on one pd of the resistor.Now you can measure true resistance.Comment
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 I am not very good in reading schematics. Anyway I found the schematic of the laptop you suggest me and to understand better what is the ACOK i look at a datasheet of a BQ735. Anyway now I measure again the pin2 and i get 2.8V, I don't know if the first reading was wrong or if something has change (most likely the firs hipotesy). 2.8V is high or still low?As per post#20, i've given you hint to get charging ic pinout by referring to another schematic. You need to do your homework and stop playing with voltage injection and instead focus on troubleshooting.
 
 Resistor needs to be measured out of circuit. Just de solder them and put it on one pd of the resistor.Now you can measure true resistance.Comment
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 It is a bit low but may be ok. This same voltage is used to enable / disable the mosfet @ PQ4410 using the gate pin linked to a resistor based voltage divider.
 
 Carefully, measure the voltage to ground of pin # 4 (gate pin) of PQ4410 with the power adapter connected.
 
 If this gate voltage is = ~19v (power adapter voltage) then ACOK is ok since
 
 * this voltage has turned on the mosfet @ PQ4405. When PQ4405 is ON (enabled), the resistor @ PR4422 mates to ground, building a resistor based voltage divider using PR4422 and PR4441.
 
 Together, these 2 resistors will offer a lower voltage than the power adapter voltage of 19v. Being a P-channel mosfet @ PQ4410, the gate voltage needs to be lower (than the 19v) to fully turn on.
 
 Without a load, the expected voltage for ACOK is ~3v9.
 
 
 
 
 
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 How can I find those mosfets and resistors on the board?It is a bit low but may be ok. This same voltage is used to enable / disable the mosfet @ PQ4410 using the gate pin linked to a resistor based voltage divider.
 
 Carefully, measure the voltage to ground of pin # 4 (gate pin) of PQ4410 with the power adapter connected.
 
 If this gate voltage is = ~19v (power adapter voltage) then ACOK is ok since
 
 * this voltage has turned on the mosfet @ PQ4405. When PQ4405 is ON (enabled), the resistor @ PR4422 mates to ground, building a resistor based voltage divider using PR4422 and PR4441.
 
 Together, these 2 resistors will offer a lower voltage than the power adapter voltage of 19v. Being a P-channel mosfet @ PQ4410, the gate voltage needs to be lower (than the 19v) to fully turn on.
 
 Without a load, the expected voltage for ACOK is ~3v9.
 
 
 
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 The DCIN mosfets are those I already measure? 19.9 on D e S and 25V on the gate?Comment
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 Ok I'll do It. But reguarding the previus questione? You ask me for a top marking chip. I answer to your question? Can we continue in the ACOK checking?Comment
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 This looks almost completely "cold"
 
 However, is the Main power rail present or not? The approach has almost no structure for some reason.1 PhotoComment
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 It is normal that I have 2.8V ad the right node of the resistor (the black chip) and 0 on the left side? Measuring in resistance mode I have 0 Ohm to ground on the left and 100KOhm on the right. It is normal that one side of the resistor is connected to ground? And I don't understand if the ACOK is a INPUT or OUTPUT for the ISL chip.
 ThanksComment
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