Is my raspberry unrepairable?

May I interject and offer an opinion here?

Consider the time taken to monitor this thread, probe the board and buy all the parts necessary to follow advice offered here.

How long before your costs (including an hourly rate) outweigh the value of what amounts to a throwaway computer?

One of the aims of the Raspberry Pi project was to put computers in the hands of children and were designed to be cheap to replace when careless hands did something that resulted in a dead board.

Greg

Greg, the time vs real dollars tradeoff can be very different for other people.

On the other hand, there is a learning experience which will be useful at a  later date but the value currently is unknown.

Mauricio: The LM1117-1.8 is a reaonsable LDO to make 1.8V from the 5V. Similarly the LM1117-3.3 is a good one for the 3.3V rail. In fact rpi-1 used one for the 3.3V rail. [update: And for the 1.8V rail as well!). Take care that you put the recommended capacitors near the LM1117. Check the datasheet for the VENDOR that you're using. TI might have other recommendations than ST for example.

Finding a place to inject 3.3V is easy: The GPIO will do fine.

Finding a place to inject 1.8V is more difficult. Hmm... Can you find C108, R57 and R58? Either side of R57 or C108 will do, but R58 being "not fitted" may be easier to solder to. (the other side of R58 is GND. Can you find C8 near there? One side is GND the other is 1.8V. Might be a lot bigger than the other ones.

I think for Mauricio, this is a good learning experience. Even if the board turns out in the end to be fried, at least he will remain with some valuable learned lessons.

The bad thing about RPI2 is that the schematic is not available.

I made some rough sketches over a high res photo of the RPI2 power supply area.

U3 and the famous U16 (the flash camera shy chip) are switching regulators.

U3 takes 5V (from the USB receptacle) and provides 3.3V and 1.8V power rails. The camera shy chip (U16) provides the 1.2V power rail.

And now the debugging part. Before going any further, I want you to know you will need to desolder some parts to be 100% sure the bad parts are removed. Rather than throw away a defective board, it can still provide you some electrical debugging lessons.

Since both U3 and U16 are buck regulator chips, you might be able to avoid desoldering them.

Take a look at the attached picture and make some measurements while powering the board from an USB adapter.

1. Check the voltage on the point where you should have 3.3V

2. do the same for the point where you should have 1.8V and 1.2V

Please come back and tell what voltages did you measured. Based on your findings, the next steps will probably involve desoldering a few parts from the board.

Wow - great detective work!

Just to add to that, based on your photo, it looks like the U3 device is PAM2306A, because on a hunch I checked the Pi Compute Module schematic (which is published) and the pins on U3 appear to match that schematic. So it is probably a near-identical schematic for that portion (extract of the Pi CM schematic under fair use):

Interesting how the part in your photo is clearly not marked as that device though (the datasheet specifically indicates the package marking). However, it is highly likely to be the same part, or one in the same family.

I think it is important for use to know this area of the circuit, because so many questions come up related to this topic in some way.

Just goes to show how important schematics are for those who want to explore more.

Wow - great detective work!

Just to add to that, based on your photo, it looks like the U3 device is PAM2306A, because on a hunch I checked the Pi Compute Module schematic (which is published) and the pins on U3 appear to match that schematic. So it is probably a near-identical schematic for that portion (extract of the Pi CM schematic under fair use):

Interesting how the part in your photo is clearly not marked as that device though (the datasheet specifically indicates the package marking). However, it is highly likely to be the same part, or one in the same family.

I think it is important for use to know this area of the circuit, because so many questions come up related to this topic in some way.

Just goes to show how important schematics are for those who want to explore more.

Thank you! I think you did a great job too. I wasn't able to find the mysterious U3, dual regulator chip.

BTW, I used the RPI2 hi res photo available on wikipedia. I don't think I would be able to take a better picture

Now, having your input, I think we should have at least some good knowledge about RPI2's power supply.

Since I was assuming the U3 is a dual buck regulator and if in Mauricio's case, the FETs are broken in this chip, he might get along with only desoldering the L1 and L2 inductors (assuming the U16 - camera shy chip is not broken). If U16 is broken as well, then desoldering L3 as well would at least isolate the CPU's power rails from the regulators.

If the 3 power rails (3.3, 1.8 and 1.2V) are not drawing excessive amounts of current, then there is still hope.

Hi!

Sorry for late responding, I was doing a final exam. x.x Anyways mine is different when it comes to the voltages for the 5v.

And by the way, when I plugged the raspberry pi to my computer it showed me a message saying the usb is taking a lot of power; I don't know if it's an important information.

This are my measurements:

Thank you for taking your time to teach me

Hi Maurico,

Did you measure the orange location? That one is very

Important to measure. There is a good chance it may be higher than the expected 3.3V : ( which would be bad news. Anyway, it needs checking.

Thank you for the reminder! It's at 0.40v.

Hi Mauricio,

This is an interesting finding.

It seems both the 1.2 and 1.8V rails are quite ok in your case. However, the 3.3V rail being so low is not good. There are many chips on the board using the 3.3V power rail. If the CPU is getting really hot when powering the board, then it is probably gone. What about the ethernet and USB hub chip? This the one with the SMSC logo onto it. Is this chip getting hot too?

Other ideas would include to desolder L1, which is the inductor used by the 3.3V regulator. By doing so, you will disconnect the 3.3V rail. If afterwards, the 5V rail is quite normal (the voltmeter readings should show a value close to 5V), then this means the fault is only in the 3.3V rail and it can be in one or more chips using it.

To definitely know which chips are faulty, you can power the board from USB (having the L1 inductor disconnected, thus the board's 3.3V switching regulator won't interfere) and then connect an external 3.3V power supply (you can connect it at the point where you should have 3.3V) capable of delivering a few amps (at least). Then, all the faulty chips should get very hot.

With the 3.3V at 0.4V the SMSC chip (the lan chip) will not be powered. It is powered only from the 3.3V rail. So with the power rail at 0.4V, it doesn't have the power to become hot.

The 3.3V is used by:

* the lan chip. (main power).

* the SD card (main power)

* the CPU (as IO voltage).

If you have a lab power supply, setting the voltage to 3.3V and the current to say 40mA, remove the SD card  and power the 3.3V rails through the GPIO pins. If it goes into current limit that might be the LAN chip which might not even be broken. Increase the current to about 100mA. It should stay below that. All this of course afteryou've disconnected the regulator from the 3.3V by removing L1. as bbolo1 suggests.

Ok so, I tried to power it through my home plug and the 5v (red ones) are now at 4.85v, and all the others stayed the same voltages. All chips gets hot( they feel pretty much the same temperature). I didn't tried to desolder anything since the new voltage for the 5v changed. Should I still desolder anything or plans changed?

@Roger Wolff I don't have a lab power supply, for now I can use a battery right? I can perhaps buy a low mA battery. Are lab power supply really usefull?

Yes, lab power supplies are really useful!

(note that you mentioned the word "really" and I answered with the word "really" in my response. Those are different meanings of a word spelled and written the same. Think about it. :-) )

A lab powersupply has a configurable current limit. This means that you can set that limit at say 30% above what you expect your circuit to use, and then not blow stuff up if there is a mistake in a circuit.

Example: I'm working on a motor-control-PCB. During startup, the CPU apparently tells the powerstage: connect V+ to motor pin A, while at the same time: connect GND to motor pin A. Of course that should never happen, but apparently it happens during startup. So far the "current limit" led of the powersupply blinks an instant and nothing blows up. If I connect a good LIPO battery as the powersupply I'm pretty sure that the FET switches would blow up.

If you say that your chips get hot, that means that something is seriously wrong. A hint that nothing more can be done.

If you still measure 0.4V on the 3.3V line, then it is very difficult for the LAN chip to become hot: It doesn't have any power. I question the observation....

If the 3.3V line is "shorted" to GND (about 0.4V) by the regulator, then I can imagine the CPU getting hot. It MIGHT have a diode from its 1.2V supply to the 3.3V IO voltage. Normally not conductive, but now suddenly a current can flow.

Many people confuse mAh ratings and mA ratings of batteries. I have a 3Ah LIPO battery rated at 120A. It can deliver 3A for an hour, or say 0.3A for 10 hours. But it can also deliver 120A for about 1/40th of an hour (90 seconds). Take a 10x smaller version of that battery and it says "300mAh", and you'd think: find that won't harm my 'pi. But it can deliver 12A, quite a lot end easily enough to damage a 'pi.

Roger Wolff wrote:

 

A lab powersupply has a configurable current limit. This means that you can set that limit at say 30% above what you expect your circuit to use, and then not blow stuff up if there is a mistake in a circuit.

Yes, that is an extremely useful feature.  Another fun thing you can do with a lab supply is to clear a thin conductive bridge that's shorting together two signals because of a PC board defect.  You set your lab supply voltage to 0.5V or so and set the current way up to 10A or so.  Then you connect the leads to the two signals.  If the bridge is thin enough, the 10A will blow the bridge like a fuse and the 0.5V makes sure you don't apply too much voltage when the short clears.  This only works for some shorts, but when it works it rescues a board.

Absolutely. Did exactly that a week or two ago. Although my lab powersupply does not go any further than 3A, but it worked. :-)