FriedFred
Senior Member
Last Activity: 04-12-2025, 01:15 PM
Joined: 05-22-2024
Location: Munich
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I forgot: With a multimeter, you can measure activity at the clock lines. This is usually enough to know. The measurements can be something like 0,2V or so. At least, higher than 0V and lower than 5V, it must be. Then there is activity at those lines. -
Okay, let's check the power supply first. Here is the Pin-Out:
P8-Connector:
Orange: Power Good (PG)
Red: +5V
Yellow: +12V
Blue: -12V
Black: Ground
Black: Ground
P9-Connector:
Black: Ground
Black: Ground
White: -5V
Red: +5V
Red: +5V
Red: +5V
The Power Good signal (PG) is very important, without getting this PG signal from the Power Supply, the mainboard will not start at all. So check this signal, it shall be 5V and stable.
Then I suggest to check all quartz / clock generator...Leave a comment:
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It's possible that some of the pins or contact surfaces have become somewhat "blind" due to dust, oxidation, etc. Installing and removing the RAM a few times should make the surfaces electrically conductive again. If that doesn't solve the problem, you'll probably need to perform pin-by-pin comparison measurements between the three slots without the RAM installed to see if all three slots behave identically. So, measure pin by pin in diode mode. If two slots are identical, but a different reading is displayed for the slot in question, there may be a defect.Leave a comment:
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The dark-grey components are inductors. The brown components are capacitors. The inductors need to show lower-ohm. The capacitors need to show higher-ohm to GND and highest ohm to any VDDs. Usually something between several ohms and up. Also, there shall be no connection between the input rails and the output rails.
I think it's necessary to check the charging circuit without the battery first. Usually, the tablet should work without the battery. But nowadays some designers considered to assume a working battery. Let's see.Leave a comment:
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Probably it gets too hot, maybe the thermal paste is dryed out.
Do all coils show some voltage?Leave a comment:
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The EC needs the bootloader to fetch the firmware from a FLASH memory. If the EC's internal memory is transparent to the SPI/QSPI interface pins, then any other third-party programmer can flash this memory and you can buy a bare/blank EC chip. You can get access to the SPI/QSPI interface through the keyboard connector, or TPM module connector, or LPC connector. I understand you need to wait for 4 weeks to get something from China, so I tried to give you some sophisticated informations about the EC chip. The final solution depends on the EC chip vendor and mainboard design.Leave a comment:
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This error is a kind of BSOD-error, the "blue screen of death". Is there hardware diagnosis available, with pressing "F2"? Any reaction from the BIOS?Leave a comment:
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I think 1.7V for a mobile SoC is far too high. I would expect something between 0.65V and 1.2V at maximum. It's quite difficult to identify a dead-short on a CPU. But if the current consumption at 0.6V is still low, the SoC (and connected compoents in this area) might be okay.
I think, to judge the 1.7V whether this voltage is acceptable or not, you need to identify the used SoC and check it out for the used or recommended voltages.Leave a comment:
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Often times a pre-programmed component is marked with a colored dot on the surface. But the missing colored dot on this particular EC here does not necessarely mean that the EC chip can re-program itself. To be able for a re-program, a bootloader is necessary.
The point is now, if you buy this EC, does it contain the bootloader on default? And does the boot loading procedure fit to the content of the FLASH/EEPROM where the BIOS firmware resides? Maybe it needs a test.
Is there any FLASH nearby this EC? Then it's likely, that the EC can re-program itself. It's also possible that the...Leave a comment:
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The first steps of booting are provided by the EC chip. The Embedded Controller chip handles the startup-sequences of the mainboard. Usually, the current consumption is about 60mA or so. But 140mA is not a dead-short. So something needs some higher current, but it doesn't lead to a booting of the entire system.
I suggest to check-out the EC chip and the signals surrounding it. Especially the I²C-bus and SPI-bus. If there is not any traffic on those data lines, then maybe the EC chip has an issue. The EC chip itself can be damaged, or the memory content of the connected SPI-EEPROM can be...Leave a comment:
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I think 9 Ampere is absolutely not normal as a charging current. Usually, it starts with a testing sequence, and a little current. After the raising voltage is detected, the current will be raised up to a nominal charging current. I would expect a pre-charge current of let's say 100mA or 200mA or so. Then, after some seconds, the charging current can go high to 2A or 3A or maybe 4A. But 9A sounds to me like a dead-short anywhere.
But it depends on the current demand of the product. What says the spec label on the bottom of the case about the required current and voltage?Leave a comment:
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The easiest solution might be, if the battery is just full charged. Another reason can be, that an over-temperature issue occures and forces the protection. So I think there can be an issue inside of the battery.
What is the behaviour of the tablet after just switch-on? With and without charger connected?Leave a comment:
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The U1200 gets very hot because it has an internal short circuit. Or this IC is being short-circuited by a downstream component. I don't know what the U1200 IC's function is, but you can find out, especially if you have a thermal imaging camera, where exactly the short circuit is coming from. Apply a small voltage to various points on the U1200. If every time the short-circuit current set on the power supply flows, only the U1200 gets hot, the U1200 itself is damaged. However, if the short-circuit current flows and the U1200 doesn't get hot, the short circuit must be in a downstream component...Leave a comment:
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I suggest to add those removed, damaged capacitors and not only just remove them. Usually, the boost circuit needs the capacitor to boost an input voltage to a desired output voltage.Leave a comment:
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Well, if the original value of the resistor is unknown, I might advise to use a higher value, like 10k or 1k. Because usually it's just a pull-up or pull-down resistor. If this card works well together with an USB interface card, it might be that there is such a resistor existing on the USB interface card. It's worth to make some measurements on those solder pads dismounted and mounted in the USB adapter card.
If you mount a 10R and the behaviour of the board is not as expected, the value might be too low.
You can also measure where those pads are going to. It looks like they are connected...Leave a comment:
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Unfortunately, I think it needs a programming. Most of those special EC chips need some software. But you can usually program them over I²C or SPI or LPC. This task needs several steps overall. a) Buy the correct replacement part. b) Find the correct Firmware for the EC chip and Mainboard. c) Get a programming tool. d) Program this thing. I have never done this, but I know several sucessful trials.Leave a comment:
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I think it's on the low-side of the DR-MOS. But it can be proved. Boardview, Schematics, or just a simple measurement with a meter.Leave a comment:
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What is the GATE voltage at those MOSFETs? Needs to be about 25V or so.Leave a comment:
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I remember, there is an issue with shared function of those SATA slots. I read it in the User's Manual for a board. Can be the same trouble here.Leave a comment:
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Is there a connection from Pin 21 to the VCC power rail?Leave a comment:
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