MS-16J91 help in identifying a component

[Fronczek]

Hey there,
I am trying to troubleshoot a no screen issue on my MSI laptop (as a training thing), and identified that there is no +3V_LCD line on the eDP connector. This line led me back to this component where it leads to +3VRUN that goes in via pin 5 (lower right pin), and goes out to +3V_LCD and ground via pin 1 (directly opposite of it). I have two questions:

Does the +3V_LCD exist only if the power switch is flipped on or is it always on?

What is this component called and how can I test it with a multimeter?

Thank you in advance!

[mon2]

U37 is a load switch (solid state relay) that will supply power to the target load which is the display in your case. Only if the ENable pin is logic high (~3v3) to ground, then this load switch will take the input voltage from pin #5 and pass it through onto pin # 1 to enable the LCD display. This enable should only be active when the lid is OPEN so that the battery does not drain when not in use / OFF mode.

Meter in DC volts mode -> red meter probe onto pin #3 of this component (be careful to not short to ground = pin # 2) -> black meter probe to a ground somewhere else on the board (metal shield?) - do not use the ground from pin #2 here as it is too close for a short circuit.

When pin #3 is a logic high ('1' = ~3v3); then pin #1 should follow pin #5. The LCD must be connected as well. If pin #1 does not show ~3v3 and ENABLE is high, then this switch could be defective. Remove all power and check the resistance to ground of pin # 5 on this part. Also check pin #5 in DIODE mode. Meter in DIODE mode. Red meter probe to ground; black meter probe onto pin #1. Repeat for pin #5 and also for pin #1 so we can check the INPUT and OUTPUT of this load switch.

Post each measurement.

[Fronczek]

Thank you for the quick response!
Here are the measurements you asked for:

DC - 0V (power supplied without the LCD connected)

Resistance - 82.5 kiloohms (power cut with LCD connected)

Diode - both pins show 853 (power cut with LCD connected)

Also, sorry if this is out of topic but mind telling me what the suffixes of the lines mean? (e.g. +5VSUS or +3VALW)

Cheers!

[mon2]

+5VSUS = +5 volts suspend mode.
+3VALW = +3V always mode - this voltage should always be present.

[mon2]

Please confirm:

1) what is the voltage to ground between pin #5 (red meter probe) & ground of your board? This is the INPUT voltage to this load switch and should be present to power the LCD.
2) The actual voltage to power the LCD display should appear on pin #1 of this part AFTER the voltage on ENable pin # 3. So pin #3 is acting like the switch. Either way, you should not have 0 volts on pin #5.

[Fronczek]

Always learning something new. Thanks.

[Fronczek]

Please confirm:

1) what is the voltage to ground between pin #5 (red meter probe) & ground of your board? This is the INPUT voltage to this load switch and should be present to power the LCD.
2) The actual voltage to power the LCD display should appear on pin #1 of this part AFTER the voltage on ENable pin # 3. So pin #3 is acting like the switch. Either way, you should not have 0 volts on pin #5.

Pin #1, #3 and #5 all have 0V on them with the LCD connected. Could this indicate an issue with +3VRUN?

[mon2]

You do not have power to this load switch so start with the basics. Connect your power adapter and measure the voltage to ground of the mosfet @ PQ6. As per the schematic, pins 1-2-3 are shorted together by design. So select any of these pins for the SOURCE, gate pin #4, drain pins 5-6-7-8.

For each, red probe onto pin #1 (or 2 or 3); black to ground. What is the voltage?
Repeat with red probe on #4 and ground.
Finally with red probe on #5 (or 6 or 7 or 8).

Post each measurement. Checking if the power adapter is not dead and if this mosfet is enabling to pass the power onto the logic board.

[Fronczek]

19.92V on pin #1, same with pin #8 and 8.32V on pin #4. Is almost 20VDC if the power adapter is rated for 19.5VDC an acceptable margin?

[mon2]

The adapter voltage is fine to power this board. The mosfet (acts like a relay / switch) @ PQ6 is part # AO4447AL. From the datasheet for this part, we know that it is a P-channel mosfet. This means that the GATE voltage (acts like the ON/OFF switch) must be LOWER than the voltage being passed between source and drain pins to ENABLE this mosfet. That is, the gate voltage is OK since it is lower than the ~19v from the adapter. This confirms that the logic board is trying to enable this mosfet. The drain pins however do not have the same as the adapter voltage.

Remove all power. Meter in resistance mode. Check to see if this PQ6 mosfet is leaky / defective by measuring the resistance.

One meter probe on source (1-2-3 - any of them will do) & other meter probe on drain (5-6-7-8) ; what is the resistance measurement? (no power on any resistance check)
Then check source & gate (4)
Then check gate (4) & drain (5-6-7-8)

Post each measurement.

Repeat for the next DCin mosfet @ PQ8.

Post each measurement.

[Fronczek]

PQ6:
883k on source & drain
238k on source & gate
640k on gate & drain
PQ8:
510k on source & drain
238k on source & gate
750k on gate & drain

From the datasheet for this part, we know that it is a P-channel mosfet. This means that the GATE voltage (acts like the ON/OFF switch) must be LOWER than the voltage being passed between source and drain pins.

Does that mean that an N-channel mosfet has to have the gate voltage higher?

Also sorry that I'm asking so many questions, but how do I know what pin is the gate when I don't have a schematic? Is it marked somehow?

[mon2]

Does that mean that an N-channel mosfet has to have the gate voltage higher?

>> Correct. For N-channel mosfet, the gate voltage has to be a boosted value and is supplied by the charger IC. The charger IC (often are the 'BQ' series from Texas Instruments) and uses the ACDRV pin.

To enable the N-channel mosfet the ACDRV pin voltage = adapter_voltage + REGN voltage = ~19v + 6 volts = ~25 volts which is higher than the adapter voltage so the N-channel mosfet will enable.

Each mosfet and IC and most active parts (parts that are powered) are with a polarity. So are polarized capacitor if they have a '-' (negative) or '+' symbol.

This mosfet has a dimple or slanted cut on the top side of the part to denote that the TOP LEFT of this region = pin # 1. Then you count the pins in a 'U' pattern. So next pin # 2 is the 2nd pin below the TOP LEFT pin. ICs with pins on 2 side are the same.

Best to always download the datasheet of the part to study such details - if the datasheets are available. When you replace the part, be cautious to replace exactly in the same way as the original.

Remove all power. Meter in resistance mode.

Measure the resistance to ground of the following pins:

PQ6 (pins 5-6-7-8)
PQ8 (pins 1-2-3)

Post each measurement. These measurements are different than before. Need to check if someone is tanking this voltage rail to the wrong value with a short.

One meter probe on the requested pin; other pin to ground. (need to grab some food / groceries - bbl)

[Fronczek]

Very good advice and useful knowledge.
PQ6 - 430k
PQ8 - 87k (something I noticed is that on my manual range multimeter when set to max 200k it slowly goes up and stops at around 87k unlike PQ6 which sets it's value almost instantly)

Be safe out there!

[Fronczek]

A little update, when poking around I accidentaly shorted 2 pins on a component marked as PU4 and I think I fried it, since my ALW lines on it don't show voltage anymore.

I always have to make it harder on myself...

Oh well all part of the learning process, now I know that these things require precise instruments and a steady hand At least I'll learn how to solder components onto the board.

Anyway, I identified the component as TPS51225CRUKR_QFN20-HF, and according to the Texas Instruments website it's a "5.5-V to 24-V, dual-synchronous buck controller with VREG5 always-on LDO".

I'm guessing I will have to solder a new one in it's place, but when I searched it only TPS51225CRUKR came up. Is the rest essential and will I have to search for this specific one or will this be compatible? Also do you know if this component is common and if I can grab it from another board?

Sorry if you're a bit disappointed, it's on me. I still learned a lot so thanks for that!

[mon2]

Remove all power. Meter in diode mode. Red meter probe, yes red to ground. Black meter probe to this always on ldo rail. What is the diode mode reading?

do you still have a voltage from the adapter present? Yes it is possible this part is defective and my have to be replaced. The part is available through mainstream distributors like Digikey and Mouser.

TPS51225CRUKR is the full part number. The schematics show extra information for the package details. Before replacing, let us check the mosfets so. You can order in a single batch. AliExpress is another option.

[Fronczek]

Good morning!

PU4:
Beeping on pins #3, #10, #13-15 and #18
Visual damage on pin #12 but meter reads 404/405
Pin #11 shows infinite

Power from the adapter is still present:

Mosfet PQ6:
19.9V on source
19.9V on drain
8.4V on gate

Mosfet PQ8:
19.4V on source
19.9V on drain
19V on gate

[mon2]

Post a clear pic of this regulator. With beeping in diode mode, this usually means the pin is shorted. Be sure that red probe is to ground and it is the black probe being used to test the actual pin.

Can you double check these measurements? This does not look correct:

Mosfet PQ8:
19.4V on source
19.9V on drain
19V on gate

If the gate is at the same voltage as the source / drain pins then the mosfet should be OFF. Perhaps this mosfet is defective and needs to be replaced. PQ8 = P-channel mosfet so for this mosfet to enable, the GATE voltage must be LOWER than the source / drain voltages yet there is voltage on each side of the mosfet.

Also measure the voltage to ground of the ACOK signal. If this signal is high then the adapter is deemed to be OK.

[Fronczek]

Measurements are the same. ACOK at PQ10 shows 0V, double checked at PQ9 and it is in fact 0V. Does that mean that the MOSFET is defective or adapter?

[mon2]

See attached.

Interested to know the voltage to ground @ ACDET pin #6 on the charger IC. Measure carefully.

On ACDET are 2 resistors @ PR82 and PR86. These 2 parts work to create a voltage divider to accept the high ~19v from the power adapter to lower it to a safe value for the ACDET pin. If the voltage on this ACDET pin is above a threshold (exact value is posted in their data sheet), then the ACOK pin will be high impedance (hi-z = floating). This is called a power good signal and are often open-drain. If the power (measured on ACDET pin) is GOOD, then this pin will be floating and therefore we MUST have an external pull-up resistor to park this floating line to a logic '1' = ~3v3. In tracing the circuit, there is a resistor @ PR83 for this function but the +3VALW is missing. So, it makes sense that this line is @ 0 volts because the pull-up voltage is absent. Post the measurement @ ACDET.

[Fronczek]

This is much more interesting than I anticipated when starting to learn electronics!

2.73V at ACDET pin. (thank god my vision isn't bad )