SONY VAIO SVE141D11L dead

[JaroTech]

Re: SONY VAIO SVE141D11L dead

Measure PQ269 and PU112.

[Romulo]

Re: SONY VAIO SVE141D11L dead

PQ269 both 3pins have 0v PU112 PIN1=0V PIN2=2.9V PIN3=0V PIN4=0V PIN5=2.9V

[Romulo]

Re: SONY VAIO SVE141D11L dead

anyone can help on this please

[Romulo]

Re: SONY VAIO SVE141D11L dead

anyone please help on this

[Romulo]

Re: SONY VAIO SVE141D11L dead

anyone please help on this

[BlueMidnight]

Re: SONY VAIO SVE141D11L dead

There are some inconsistencies in your measurements. However, as best as I can tell:

- PQ222 is off because the gate is 19V.
- That's because PQ266B is switched on due to 0V on its gate (-VAOFF).
- That 0V might be due to PQ270. That's part of a protection circuit that won't switch on PQ222 and let the 19V through if it detects that VIN never rises up high enough (i.e. there may be a short on that rail).

I recommend measuring all pins on PQ270 (to confirm the hypothesis).

Also, measure VIN to ground in diode mode (e.g. at PR265). I saw you measured the 3V/5V outputs, but didn't see any mention of measuring for a short on VIN.

It sounds like you have a bench power supply, so you could also inject voltage on VIN (which is a more definitive test than the meter for higher resistance shorts or more unstable shorts). Set the current limit to something small like 10mA-100mA, then slowly raise the voltage to 19V. See if the rail voltage will rise to 19V before it hits the current limit.

[Romulo]

Re: SONY VAIO SVE141D11L dead

thanks bluemidnight for the response I think i found the problem i double checked visually around PQ266 and found this chip burnt i think this is PQ270 because pin3 is connected to PC538. should i replaced this and what is the replacement value for this?i attached the image.i forgot to mentioned PR465 both sides have 19v and resistance have 0.656v

[BlueMidnight]

Re: SONY VAIO SVE141D11L dead

Ahhh yes that is PQ270. So PQ270 itself has a problem. You can replace it and the part number on the schematic is 2N7002DW. I would definitely try to get more than one. It's still not clear why it failed so spectacularly in the first place..

[Romulo]

Re: SONY VAIO SVE141D11L dead

ok thanks again for the response i will try to find the same value on my scrapboard here and replaced and see what happens.seems to be difficult to find this transistor

[Romulo]

Re: SONY VAIO SVE141D11L dead

already done replace the burnt chip with the same 2N7002DW, but i have problem pin1 the pad is broken where should i make jumper to connect this pin1?I have attached the image

[diif]

Re: SONY VAIO SVE141D11L dead

If it is pin one, then looking at the schematic it goes to ground, same as pin 4.

[Romulo]

Re: SONY VAIO SVE141D11L dead

yes it is pin1 should i connect a jumper to the ground for this?and also i checked after replacing without this pin connected still there is no voltages on this pq270 and pq269.anyone please help on this

[Romulo]

Re: SONY VAIO SVE141D11L dead

when i replace again with another chip and jumper the pin1 to the ground since the pads is broken, now i measured pin2 and pin3 of PQ270 it shows only 9.5v and same voltage on PR445 but PR447 have 19v on one side other side have 0.anyone please help on this

[BlueMidnight]

Re: SONY VAIO SVE141D11L dead

9.5V isn't bad. It's not the voltage we want there right now, but it shows that part of the circuit is working. We do want it to be 0V eventually though.

That must not have been PR447 you measured, however. PR447 is right above PQ270 in your last photo and directly connected to pins 2/3 of PQ270 on one side, and ground on the other. It must read 0V on one side and the same voltage as PQ270 pins 2/3 on the other side.

The FET on the right half of PQ270 on the schematic is responsible for causing the gate of the left FET to be pulled to 0V, turning off the left FET. It only does this when VIN is high enough. Otherwise pins 2 and 3 of PQ270 will be roughly half of VA+ because they are in the middle of a resistor divider made of two 1M ohm resistors (ignoring the insignificant 10k ohm one). 9.5V is perfectly half of 19V, so no problems there. We just need to know why the right FET is turned off at the moment.

It would be helpful to know the measurements on all pins of PQ270 and not just a few. And because there is always a delay when speaking on a forum, it would save time to just measure that whole part of the circuit: PQ270, PR452, PR450, PD16, and VIN.

[Romulo]

Re: SONY VAIO SVE141D11L dead

thanks for the response bluemidnight you are right sorry my mistake i measure not pr447,by the way pr447 above PQ270 on the other side that is connected to pin3 measured 9v and on the other side connected to pin2 measured 0v. and here is the below measurements

PQ270
PIN1=0V which is the pad is broken ive made a jumper wire to the ground
PIN2=9V
PIN3=9V
PIN4=0V
PIN5=0V
PIN6=0V

PR452 one side 19v and other side 0v
PR450 both sides 0v
PD16 both sides 0v
VIN=0V

[BlueMidnight]

Re: SONY VAIO SVE141D11L dead

PR450 and PR452 are connected at one end and have one of the voltages in common. And 19V isn't possible there. So I don't think that was PR452 that was measured.

I highly recommend writing your measurements on the PDF (either in your PDF program or by printing the page and writing on the paper). That way you will immediately spot these kinds of inconsistencies and know to double-check your measurements.

And now I am again suspicious of VIN because it is 0V. I see you wrote PR465 in a previous post after I mentioned PR265 and I misread that and didn't realize it before. So I'm not sure if that was a typo or you measured the wrong resistor. But we need to be very very sure about VIN so I'm going to ask you to confirm this. At PR265 (or anywhere else where you can find VIN with absolute certainty) please measure the voltage with the board powered. Also, with the board unplugged, measure from there to ground in diode mode.

PR265 is the white component indicated in the attached photo (the color may vary). I like doing the measurement from the sense resistor because they look so different from other resistors (making them easy to find), and you don't have to worry about which pin is which.

[Romulo]

Re: SONY VAIO SVE141D11L dead

Thanks for the response i checked again PR450 and PR452 have 0v.i already checkef PR465 both sides have 19v but PR265 has no 19v both sides when i try injecting voltage in PR265 bench lab not taking ampere but when i plugged in charger and measured again PR265 it becomes 3v dropping to 0v.

[BlueMidnight]

Re: SONY VAIO SVE141D11L dead

I had originally written PR265, so PR465 never needed to be checked. This new information is very helpful. It looks like there really is a short on VIN (or downstream from it).

I also suspect that PQ270 was damaged when you saw the spark trying to measure pin 2. If you had accidentally bridged pins 1 and 2, there would have been a spark (which should never ever happen while measuring voltages) and the 19V on pin 1 would have been shorted to ground through PQ270, destroying it. It happens. We've all done that at least once in our lives. But now that is out of the way, so we can continue finding that next problem.

The next step is to inject power at PR265, but this time you can slowly let the current go up to ~1A (with the max voltage set to 19V). Then feel to see if any components are getting hot. The most likely components to be bad are capacitors, ICs, and FETs. Hopefully 1A is enough to notice something, though you can go a little higher if needed.

[Romulo]

Re: SONY VAIO SVE141D11L dead

thanks for the response you are right maybe that pq270 damaged due to probes short pin1 and 2 during my measurements.Ill try injecting 19v on Pr265 not taking ampere and by the way my lab bench is switching bench power supply the cheap one. usually i set voltage to 19v when i inject it automatically taking amperage when there is a short but this time when i try on PR265 not taking any amperage

[BlueMidnight]

Re: SONY VAIO SVE141D11L dead

Oh your power supply isn't the kind with a current limit control? If you can output 12V instead, then just use a 12W automotive light bulb in series with the output and it will limit the current to 1 amp. A 24W bulb (or two 12W bulbs in parallel) will draw 2 amps. And so on..

If you can't find the bad component that way, then you'll have to do it the slow way. But it can still be done.