BQ24735 Problem (HP 255 G8)

[ianbak]

Hi,

I have an HP 255 G8 that runs for as long as its battery holds charge but won't accept AC power.

I have partially diagnosed the problem and the issue seems to be that there is no ACDRV output from the BQ24735 to turn on the two input MOSFETs. The 24735 has a VCC of 19.57 V, the ACDET input is at 2.93 V and the REGN 6V LDO is outputting 5.98 V, so everything looks good - but ACDRV is stuck at 28 mV.

I've already replaced the 24735 and both input MOSFETs and nothing has changed.

I'd appreciate some suiggestions as to what to try next.

[mon2]

Link the schematics.

Share the measurements of the charger IC. Is the charger IC the original that was on the board from the vendor or an equivalent?

A number of conditions must be met for the ACDRV to enable. Also investigate a possible reverse voltage crowbar circuit that often triggers from the barrel / coax power adapter connector.

[ianbak]

I have attached a schematic for the board. The replacement charger IC is a genuine BQ24735, as originally used by HP. There is a reverse voltage protection circuit shown on the circuit diagram, but on my board the components are not present.

LINK to schematic -> https://www.badcaps.net/forum/troubl...ardview-please

[mon2]

Carefully flux and remove PRB9. Do not lose this ~4k resistor. Power up again and check the voltage to ground on the PRB9 PCB pad (charger side) that drives the ACDRV pin.

What is the measurement now ?

[ianbak]

Here is a full set of voltages for the BQ chip:

Pin 1 - ACN - 347 mV
Pin 2 - ACP - 913 mV
Pin 3 - CMSRC - 34 mv
Pin 4 - ACDRV - 24.3 mV
Pin 5 - ACOK - 0 V
Pin 6 - ACDET - 2.93 V
Pin 7 - IOUT - 0.7 V
Pin 8 - SDA - 2.1 mV
Pin 9 - SCL - 1.9 mV
Pin 10 - ILIM - 0 V
Pin 11 - BATDRV - 2.07 V
Pin 12 - SRN - 1.07 V
Pin 13 - SRP - 1.07 V
Pin 14 - GND - 0 V
Pin 15 - LODRV - 0 V
Pin 16 - REGN - 5.98 V
Pin 17 - BTST - 3.86 V
Pin 18 - HIDRV - 1.08 V
Pin 19 - PHASE - 1.07 V
Pin 20 - VCC - 19.57 V

[mon2]

Which mosfets were used for the DCin replacements?

With no power, meter in resistance mode, measure the resistance across the mosfets as follows:

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

Share the measurements.

[ianbak]

The MOSFETs were replaced like for like from a donor board.

PQB11 Source/Drain - 3.2 MOhm, Source/Gate - 3.7 MOhm, Gate/Drain - 6.3 MOhm

PQB12 Source/Drain - 4.3 MOhm, Source/Gate - 3.7 MOhm, Gate/Drain - 6.0 MOhm

[mon2]

Were the measurements in post # 5 taken directly at the charger IC pins?

Flux and reflow the IC. The conditions look correct for the ACDRV to enable. The DCin mosfets are not shorted.

Update : Replace the cap that is on REGN pin with another. Must be 2.2uf or higher.

reference:

BQ24780S: ACDRV question - Power management forum - Power management - TI E2E support forums

[ianbak]

With PRB9 removed (it took me a while to identify it!) ACDRV is still at 26 mV, he same as before

[ianbak]

Yes, directly on the charger IC pads.
I tried reflowing the 24735, and it's all still as before with ~27 mV on ACDRV. I tried swapping the capacitor too - the one on the donor board measured 1.2 uF before I put it in, which is consistent with the 1.0 uF specified in the schematic.

[mon2]

Do you have a battery connected to the board?

Pin 11 - BATDRV - 2.07 V
Pin 12 - SRN - 1.07 V
Pin 13 - SRP - 1.07 V

Review the battery charge mosfets to see if they are leaking.

With no power, meter in resistance mode, measure the resistance across the mosfets as follows:

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

Share the measurements of the high side and low side battery charge mosfets.

As a quick test, flux and remove the high side battery charge and also the battery pass-thru mosfets to disconnect each from the main power rail that is being monitored by the ACP / ACN current loop circuit. Want to remove all influences on this rail.

Then test again.

[ianbak]

Sorry for the delay in replying - busy day yesterday.
There was no battery attached when those measurements were made. With the battery attached and power connected I get the following:

Pin 1 - ACN - 10.6 V
Pin 2 - ACP - 11.3 V
Pin 3 - CMSRC - 24 mv
Pin 4 - ACDRV - 18.3 mV
Pin 5 - ACOK - 3.3 V
Pin 6 - ACDET - 2.9 V
Pin 7 - IOUT - 0.8 V
Pin 8 - SDA - 0.1 V
Pin 9 - SCL - 3.2 V
Pin 10 - ILIM - 0.6 V
Pin 11 - BATDRV - 17.6 V
Pin 12 - SRN - 11.3 V
Pin 13 - SRP - 11.3 V
Pin 14 - GND - 0 V
Pin 15 - LODRV - 0 V
Pin 16 - REGN - 6.0 V
Pin 17 - BTST - 11.0 V
Pin 18 - HIDRV - 11.3 V
Pin 19 - PHASE - 11.3 V
Pin 20 - VCC - 19.5 V

Again, these look pretty OK to me.

PQB13 isn't shorted - I was getting at least 50 MOhm every way when I measured it.
I tried removing PQB1, which is a double package that I think combines the two battery charge MOSFETs you are referring to, and it made no difference.

[mon2]

Pin 1 - ACN - 10.6 V
Pin 2 - ACP - 11.3 V

Can you confirm these measurements? These do not look correct for a normal board.

[ianbak]

I've just re-measured it, with and without the battery present, and the numbers are all still as before.

[mon2]

Remove all power. Meter in (lowest) resistance mode. Measure the resistance between pins 1 (ACP) & 2 (ACN) of the charger IC @ PUB1. Short your meter probes to review the lowest measurement capable by your meter.

Next, measure the resistance to ground of pin # 2 (ACN) of PUB1.

The voltage drop across this current sense resistor @ PRB1 is too large. Post each measurement.

[ianbak]

We might finally be onto something here! ACN to ACP is open circuit, which explains how there can be a 0.7V drop between them when they should be bridged by the current sense resistor! The current sense resistor itself has continuity (measured across terminals 1 and 4) and both ACN and ACP are MOhms to ground.
I'll continue to investigate this tomorrow.

[ianbak]

That was the problem - a broken trace between ACN and the current sense resistor. I located the break, bridged it, and now have a working laptop.
Thanks for your help - I think I'd have given up before finding the fault without it!

[mon2]

Np. Glad to have helped. It was a good case to learn from for future readers.