Toshiba Satellite L555-10W Issue with "+1.05VSP" power supply line

Motherboard: Compal LA-4982P, Rev.: 1.0

Hello!
Listed laptop normally turns ON, but the screen stays black (no backlight, no picture).
I found out that the problem is on the "+1.05VSP" power supply line. Voltage on this line is only 0,82V instead 1,05V. The strangest thing is that the voltage on "+1.05VSP" line becomes ok, if I heat up the switching controller chip "PU7" (TPS51117), and everything works ok.
The chip must be heated on about 40°C or more to function properly. It is necessary to heat up only the chip (not surrounding components). I tried to precisely heat it up with the tip of the soldering iron.

Pictures of the top side of the board, bottom side of the board and part of the schematic:



Datasheet of the chip "PU7" (TPS51117) is in the attachment.

Voltages, when the chip "PU7" is not working properly:
Output voltage (on capacitor "PC63"): 0.82V
Enable (pin 1 of "PU7"): 3.23V
5V logic supply (pin 4 of "PU7"): 5.06V
Feedback (pin 5 of "PU7"): 0.55V
Power supply (on booth ends of coil "PL21"): 19.15V

What I've tried up to now:
-Chip "PU7" replaced three times with different chips. One pulled from scraped board, the other two was brand new ones. -> The same issue
-Replaced surrounding components: resistors PR92, PR93, PR96, PR97, PR98, PR99, capacitors PC61, PC62, PC64, and PC67 -> The same issue
-Removed jumpers "PJ6" and "PJ9" -> In this case the output voltage 1.05V is ok regardless of whether the chip "PU7" is warm or not.

I'm out out of ideas now...
Any advice and suggestions would be greatly appreciated.

ICH shorted? Check the current flow through PJ6 with PJ9 split. If you heat it and it outputs 1.05V does the screen remain black?

Check PR98 = 8.25K, PR99 = 20.5K

rosak napisał(a):a) ICH shorted?
b)Check the current flow through PJ6 with PJ9 split.
c) If you heat it and it outputs 1.05V does the screen remain black?

a) What do you mean with ICH?

b) It is very tricky to measure the current through jumpers, because the voltage is only 1.05V, and voltage drop on internal resistance of ammeter will reduce the output voltage.
An example: If internal resistance of ammeter is 20mΩ, and current through it is 5A the voltage drop will be 100mV (0.02Ω*5A=0.1V). So in my case the output voltage will be only 0.95V (1.05V-0.1V=0.95V), which is too little for normally operation of laptop.

c) No, it doesn't. If I heat up the switching controller chip "PU7" I get picture on the screen, and everything is working ok.

Machine napisał(a):Check PR98 = 8.25K, PR99 = 20.5K

I've already replaced it.
I didn't have the original values, so I replaced it with the followed values, which was pulled from scraped motherboard with the same controller chip (TPS51117).
PR98 (8.25kΩ) -> 10.5kΩ
PR99 (20.5kΩ) -> 25.5kΩ

All is exactly the same like with the original resistors. When the chip is "cold" output voltage is only about 0.82V, and if I heat it up the output voltage increases at 1.05V.

I was pulled replacement resistors from motherboard LA-4101P. Part of the schematic is on the following picture:



Thank you both for reply!
Any more ideas?

matic napisał(a):a) What do you mean with ICH?

ICH = I/O Controller Hub

b) It is very tricky to measure the current through jumpers, because the voltage is only 1.05V, and voltage drop on internal resistance of ammeter will reduce the output voltage.
An example: If internal resistance of ammeter is 20mΩ, and current through it is 5A the voltage drop will be 100mV (0.02Ω*5A=0.1V). So in my case the output voltage will be only 0.95V (1.05V-0.1V=0.95V), which is too little for normally operation of laptop.

Internal resistance of ammeter is usually 10mΩ BUT if you already have 5A on power supply during boot and without display/backlight then you definitely have a short somewhere. Check all capacitors and resistors at PCH side on +1.05VS rail.

which is too little for normally operation of laptop.

You are only supposed to measure the current, not to solder the ammeter in permanently for normal operation.

rosak napisał(a):ICH = I/O Controller Hub

Ok, so you mean the southbridge BGA chip.
No, it is not shorted. If it would be, it won't work, when the power supply controller chip is warm.
Resistance between +1.05VSP power line and ground is 28Ω, no matter if the controller chip is warm or not. It is internal resistance of northbridge BGA chip, which is ok.

rosak napisał(a):Internal resistance of ammeter is usually 10mΩ BUT if you already have 5A on power supply during boot and without display/backlight then you definitely have a short somewhere. Check all capacitors and resistors at PCH side on +1.05VS rail.

Yes, 10mΩ of ammeter + 20mΩ or probably more of test leads.
I measure the current now, and it is about 2A, when the controller chip is cold. I'm pretty sure that the real current is probably 2* higher (if we ignore the internal resistance of ammeter).

Thanks for reply!

Hello!

I finally succeeded to repair this laptop.
I soldered 33pF ceramic capacitor in the place of PC66, which was not placed in the factory.
The voltage "+1.05VSP" is ok now, regardless if the chip PU7 is warm or not.