Dell XPS 13 9370 no power USB amp meter reads 5V , 0.11A

[mon2]

From my understanding, there should be a voltage of +5v here. If any of the PD controllers are mated to a USB Type C power adapter, then this common output should offer +5v.

Remove all power. Meter in resistance mode. Measure the resistance to ground of +5VALW = same as the last point of interest. We are checking if there is a shorted part onboard that may cause this rail to turn off / short to ground.

[waakye]

A11, C11, B11, D11 and H10 all reads 51.7ohms

[waakye]

This is the PD controller I ordered: TPS65982DMCZBHR. Hopefully it is compatible.
Thanks

[mon2]

I think this replacement is fine. The PD firmware is stored externally so it should function like the original.

The +5VALW rail resistance to ground is low but not a complete short which makes it difficult to locate the root cause.

+5VALW is mated to +5VALWP through PJP501 pcb jumper trace. +5VALWP is created by PU501.

Confirm the voltage to ground of the following points of interest:

B+ @ PL502 (either side of the inductor is ok)
ENLDO_3V5V = EN2 on PU501
Pin 15 LDO pin on PU501
PG (ALW_PWRGD_3V_5V) on PU501

[waakye]

Good morning, mon2. Happy Sunday. I really appreciate your support on this.
I took the measurement. The board view I have for the XPS 13 9370 does not show PL502 but I took the measurement from PR323 which reads 0V
I took all the measurements from PU501 pins and all pins reads 0V except pin 9 that keeps cycling from .642V to 0V.
Hope that will help you further assist me, I am willing to take more measurements.
Thanks again.

[mon2]

There are 2 of these PD controllers onboard - U19 & U21. Be sure to test each and also flip them 180 degrees to confirm the voltage and current draw for each permutation.

Be sure that you are using the official power adapter although a quality 20V 5A (100W) unit with a qualified USB Type C 100W rated cable should be ample to power this unit.

Do you see 0A current draw for each test of the type C ports?

[waakye]

Hi, I flipped the power connector and the voltage reads 5V 0A. One of the USB C connector when flipped does not measure anything.
Thanks again

[waakye]

Btw I am using the original power adapter. Which is 20V 2.25A

[mon2]

Confirm the voltage to ground of:

PU300, pin 17 DCin
PU300, pin 17 ADP
PU300, pin 24 ACOK

PBAT_CHARGER_SMBDAT
PBAT_CHARGER_SMBCLK

[waakye]

Good morning mon2,
PU300, pin 17 DCin = 0V
PU300, pin 16 ADP = 0V
PU300, pin 24 ACOK = 0V

PBAT_CHARGER_SMBDAT =0V
PBAT_CHARGER_SMBCLK = .345V and cycling to 0V

Thanks

[mon2]

See attached. Check if you have a voltage of ~3v on pin 5 of PU101.

If not, proceed to check the voltage to ground of the yellow highlighted input rails on the left side of this partial schematic.

[waakye]

Pin 5 of PU101 reads 3.27V.

[mon2]

Each USB Type C connector is mated with 2 x DCin mosfets in its power path. Select the USB Type C connector you are using -> measure the voltage to ground of the 1st Dcin mosfet - measure the source (1-2-3); gate (4) and then drain (5-6-7-8) to ground voltage.

Repeat for the 2nd DCin mosfet.

Be sure to use only the mosfets that are in the path of your USB C connector used by the adapter. At this time, one or both of these DCin mosfets are not enabling to supply the same +5 volts onto the B+ pin.

Example power path is attached.

[waakye]

Thank you so much mon2!
Below are the measurements:
JUSBC1
PQ100 pin 1, 2, 3 = 5V
Pin 4 = 0V
Pin 5, 6, 7, 8 = -63mV

PQ101 pin 1, 2, 3 = 0V
Pin 4 = 0V
Pin 5, 6, 7, 8 = -63mV

JUSBC2
PQ113 pin 1, 2, 3 = 5V
Pin 4 = 0V
Pin 5, 6, 7, 8 = -63mV

PQ1114 pin 1, 2, 3 = 0V
Pin 4 = 0V
Pin 5, 6, 7, 8 = -63mV

JUSBC3
PQ153 pin 1, 2, 3 = 5V
Pin 4 = 0V
Pin 5, 6, 7, 8 = -0.233V

PQ155 pin 1, 2, 3 = 00V
Pin 4 = 0V
Pin 5, 6, 7, 8 = -0.250V

[mon2]

Each of these DCin mosfets are P-channel type. If the gate voltage is lower than the voltage being passed through the source / drain pins, the mosfet will enable. Yet, the drain pins (5-6-7-8) are each at 0v on each of the first DCin mosfets.

Remove all power. Meter in resistance mode. Measure the resistance to ground of CHG_VIN_20V. What is the resistance in ohms? Checking if there is a shorted part on this main power rail.

If the resistance is normal and not too low, then proceed to test the same mosfets above but using resistance mode checks.

Remove all power. Meter in resistance mode.

Measure the resistance across:

source (1-2-3) & drain (5-6-7-8)
source (1-2-3) & gate (4)
gate (4) & drain (5-6-7-8)

Post each measurement in ohms.

[waakye]

Hi,
CHG_VIN_20V measures 122.3K ohm

Is this normal for main power rail. It starts off in the mega ohms and slowly settle in 122.3K ohm


PQ100
Source to drain= .625M
Source to gate = .494M
Gate to drain = 2.23M

PQ101
Source to drain= .342M
Source to gate = .495M
Gate to drain = 2.02M




PQ113
Source to drain= .649M
Source to gate = .493M
Gate to drain = 2.12M

PQ114
Source to drain= .323M
Source to gate = .495M
Gate to drain = 1.90M





PQ153
Source to drain= .649M
Source to gate = .495M
Gate to drain = 2.3M

PQ155
Source to drain= .339M
Source to gate = .494M
Gate to drain = 2.02M


Thanks

[mon2]

These measurements look ok. They are not low values so not shorted.

Suspecting the gate voltage of the above mosfets from post # 34. Can you confirm them again.

Given that the drain pin(s) are at 0V implies the mosfets are OFF so this means the gate voltage is most likely the same voltage as the source pins = ~5v.

Not clear as to why the gate voltage is at 0v. What should happen is that when these mosfets are expected to be ON, then PQ106 will enable (as an example).

When PQ106 is enabled, the resistor at PR106 will be connected to ground. When this occurs, resistors PR100 & PR106 form a voltage divider to lower the Vin from the power adapter (~5 volts at this time) to a lower voltage and this lower voltage is presented onto the gate pin. For a P-channel, this is required and then the mosfet will enable due to the lower gate voltage.

Confirm again if the gate voltage is truly at 0v but suspecting it will be ~5v to ground.

[waakye]

Below are the gate voltages of the midgets

JUSBC1 PQ100
Pin 4 = 5V

PQ101
Pin 4 = 0V


JUSBC2
PQ113
Pin 4 = 5V

Q1114
Pin 4 = 0V

JUSBC3
PQ153
Pin 4 = 5V


PQ155
Pin 4 = 0V

Thanks

[mon2]

Take a piece of wire and connect one side onto the gate pin # 4 of the mosfet. The other side of the wire to ground. This will force the gate voltage to be shunted to ground to force enable the mosfet. Apply onto any of the first mosfets of the above group. That is, the 5v on the gate pin is due to the pull-up resistor on each local mosfet and now you are forcing the P-channel mosfet to enable by shunting the gate to ground.

[waakye]

I short the gate of the mosfet PQ100 pin 4 to ground.
JUSBC1
PQ100 pin 1, 2, 3 = 5V
Pin 4 = .474V
Pin 5, 6, 7, 8 = 5V

PQ101 pin 1, 2, 3 = 5V
Pin 4 = 5V
Pin 5, 6, 7, 8 = 5v