Iphone 6 no display diode d1501 burning

[h-run08]

Hello I received iphone 6 which boot perfectly but that displays nothing I noticed that the diode d1501 was burning I desolder the diode and test the resistance of the line pp_lcm_bl_anode and my multimeter displays 200 ohm while on 1 iphone that works I have no value Do you think the capacitor c1505 c1531 c1530 could be in question? Can 1 capacitor be half short and cause this low 200 ohm value? Or is it the chip u1502? What do you think ? thank you

[zarbourne]

Originally Posted by h-run08

Hello I received iphone 6 which boot perfectly but that displays nothing I noticed that the diode d1501 was burning I desolder the diode and test the resistance of the line pp_lcm_bl_anode and my multimeter displays 200 ohm while on 1 iphone that works I have no value Do you think the capacitor c1505 c1531 c1530 could be in question? Can 1 capacitor be half short and cause this low 200 ohm value? Or is it the chip u1502? What do you think ? thank you

D1501 heating up indicates your backlight system is down..

1.Check voltage make sure:

- VCC_MAIN = 4v
- PP1V8_SDRAM= 1.8V
- PP1V8 = 1.8V
- TEST

2.Check if any of these are shorted or damaged:

- C1513
- C1552
- C1597
- C1530
- C1531
- C1505
- C2017
- C2018
- C2019
- TEST

3.Change following:

- U1502
- L1503
- D1501

Let me know how it goes!

[h-run08]

I wanted to know if 190 ohm can be considered short circuit?
Can a capacitor not be completely short and create a small resistance of 190 ohm?

VCC_MAIN = 3.75 - 3.90 Its rise and fall
- PP1V8_SDRAM= 1.8V
- PP1V8 = 1.8V

C1513 = 190 ohm
- C1552 = 190 ohm
- C1597 = 1400 ohm
- C1530 = 1400 ohm
- C1531 = 190 ohm
- C1505 = 190 ohm
- C2017 = 190 ohm
- C2018 = no short
- C2019 = no short
-

[short_toground]

Originally Posted by h-run08

I wanted to know if 190 ohm can be considered short circuit?
Can a capacitor not be completely short and create a small resistance of 190 ohm?

VCC_MAIN = 3.75 - 3.90 Its rise and fall
- PP1V8_SDRAM= 1.8V
- PP1V8 = 1.8V

C1513 = 190 ohm
- C1552 = 190 ohm
- C1597 = 1400 ohm
- C1530 = 1400 ohm
- C1531 = 190 ohm
- C1505 = 190 ohm
- C2017 = 190 ohm
- C2018 = no short
- C2019 = no short
-

Do a diode mode check with the diode removed. Positive pin on ground, negative pin on each side where diode goes.

If you get a low reading <.3v drop then there's a short in the direction of the pin that measured low.

If it's after the diode check the caps. One could be leaky causing a partial short. If it's before filtering caps check the ic or supply to ic

[AYSER]

Originally Posted by h-run08

I wanted to know if 190 ohm can be considered short circuit?
Can a capacitor not be completely short and create a small resistance of 190 ohm?

VCC_MAIN = 3.75 - 3.90 Its rise and fall
- PP1V8_SDRAM= 1.8V
- PP1V8 = 1.8V

C1513 = 190 ohm
- C1552 = 190 ohm
- C1597 = 1400 ohm
- C1530 = 1400 ohm
- C1531 = 190 ohm
- C1505 = 190 ohm
- C2017 = 190 ohm
- C2018 = no short
- C2019 = no short
-

Hello Dear.
Having a 190 Ohms on the capacitors ( C1530, C1531, C1505) means that at least one of these capacitors is damaged, could be that is not totally shorted but it's damaged and will create an over consuming of the current which will burn the diode (D1501).
Please remove the capacitors one by one and test.

Thanks for Solution source
( www.tsa-mtech.co.uk ) SFS software.


Sent from my iPhone using Tapatalk

[h-run08]

I plugged a power supply with 2 volt + on most of the line pp_lcm_bl_anode and the - to ground and I put the freez bomb and the capacitor c1505 melted very quickly.
Now I will replace it and re-solder the diode and I would tell you if it works

[HDccam]

I had the same problem.. and fixed it.

Your problem is almost for sure C1530 short to ground.
And yes 600 Ohm shortcut means, your diode will burn too!*

So just remove c1530 and your diode will not burn anymore

**