Is my raspberry unrepairable?

You have most lightly blown something on the PI

The 4.3V battery should NOT have been connected between the 3.3V pin and the 5V pin, these are both seperate power + rails for parts of the PI board and you just put effectivly 8V into the 5V pin or caused excessive current to flow through pins not designed to have a difference of the battery or booster voltage.

What you should have done is connect the batter or booster output to 0V for the black wire and 5V for the red one (Using the booster to give you the 5V

you best option is actually to feed the 5V into the PI via the USB connection though, so battery to booster, booster to PI USB power connector

the schamatics have not been published but based on your description of how you connected a battery capapable of providing many amps between the 3V3 and 5V pins I would expect something to have failed

Sorry for the bad news, may the PI RIP.

Peter

So I cannot replace anything (nothing has physical damage except the pins, so I thought I may be able to replace them or even replace the led xd)? And one more question how can I put the black cable to 0v (because both cable where fused with the + and - of the boost output) ?

Thank for the fast responding! :O

Reaistically no, there is no real standard for soldered connections format, so even if they are removed, they may not fit other boards for different products.

And they don't often go wrong they should normally outlive the life of the remainder parts on the 'pi, so won't be useful for other pi's either.

The 40-pin connector is useful, but is very hard to remove with basic tools.

You could perhaps use the board to practice desoldering maybe, but apart from that not much.

Holy guacamole - It takes quite a bit off current to sublimate metal.  That's what Li++ batteries are made to do, however.  When you feed too much current on pins it will most likely burn out that subcircuit, as well as connected and adjacent ones.  Looking at the die for the BCM2835 (the CPU for a RPi) there are more than 40 peripherals hooked directly to power.

The likely reason you're measuring 4.12VDC at points where you expect 5.2VDC is that when the overcurrent blew a subcircuit it created a short to ground somewhere.  That's draining off the "missing" 1.08V to ground.  The fact that the micro USB port gets so warm is that a short (maybe the "missing" power subcircuit, maybe not) is pulling a lot of current out of the power supply.  The RPi also doesn't have a voltage regulator. 

The fact that the power LED isn't coming on is that the voltage was pushed beyond the breakdown voltage.  According to the Shockley equation, applied to diodes (you can Google it, if you're interested), a small rise in voltage can result in a large increase in current.  In plain terms: the LED is dead.  Just for information: The LED only reflects a voltage rise in the control circuit it's attached to - Board power isn't actually routed through the LED.

You're unlikely to find what exactly was damaged; unless you have access to a microtome, to cut layers off the CPU and PCB, and a microscope to view features smaller than 0.1mm.  You'd likely just see where the metallic parts and adjacent PCB melted.  It would be a good learning experience, however, to use a magnifying glass or something to visually locate where the copper traces were damaged, and what accessories on the board were killed.

The bottom line is that the RPi is dead.  You might be able to get some lights to come on (the network LED's, for instance) and feel hot spots where power is going through.  Even if all the peripherals were intact (highly unlikely) the "brain" of the BCM2835 is dead.  It's common with integrated boards (microcontroller, SoC, ASIC, etc.) when they get too much power to pins, to have a few things that will still work (at least for a little while).  It will never be repairable back to normal function; not would I use anything off the board in something else.

Hi Brian,

As you say, likely the board is fried. But just as an aside (not that it matters to the end result) the LED may be intact, because it has a resistance in series. There is a DC-DC converter (well, more than one) and by connecting to +5V and +3v3 the MOSFET which is part of the buck topology inside that integrated circuit saw the supply across it with nothing to protect it : ( anyway I'm speculating any further. I could be totally wrong.

So, I cannot use my boost converter since it also might be damaged? And Thank you, it's very informative!

That is probably fine, but might be worth testing it before you connect it to anything. You need a multimeter to see if it gives 5V.

Before I joined the cables to the gpio pins, I adjusted the booster to 5.2v but after, the next day, I retest it and it showed 4.12v output.. this means it got affected?

If its behaviour has changed then yes. Not all have protection. Anyway you have to use your own judgement and experimentation here, since I'm speculating. If you're measuring 4.2v when it was originally 5 or 5.2V then all other things being equal it looks like a fault.

alright, Thanks!

Understood Shabaz - Thanks, much!

I was trying to give him general reasons how/why things fail.  I know something of the circuitry in the RPi, but certainly not that much.  Thanks for the insight!

shabaz wrote:

 

That is probably fine, but might be worth testing it before you connect it to anything. You need a multimeter to see if it gives 5V.

To be really sure you should test the regulator under load (a resistor of a few watts will do) to test that the regulator does indeed deliver the current at the voltage specified even under a realistic load

shabaz wrote:

 

That is probably fine, but might be worth testing it before you connect it to anything. You need a multimeter to see if it gives 5V.

To be really sure you should test the regulator under load (a resistor of a few watts will do) to test that the regulator does indeed deliver the current at the voltage specified even under a realistic load

Huhhhh.... I don't get it. I should buy a resistor and connect it to the booster (including the battery) and then I should test the volts ? I'm confused how I'm supposed to know if it works cuz first time I bought it, I connected my voltmeter(at 20v) to the the output and it showed 1. So, I lower it until it says 5.2v. But what am I supposed to see now if it works, it has to say 1 again? Or put it more higher the voltmeter and see a specifc number ? And also do any resistor work or is there an especific one ?

thanks!

Hi Mauricio,

John is saying that your boost converter needs testing, and that to test it you can use a resistor (see ohms law on wikipedia) to confirm it can support the 1A or so the pi may need. If you've bought some low-cost boost converter from eBay then it may need testing basically. Another option is a ready-made battery pack intended for charging mobile phones. It has a built in battery and boost converter and is a fixed voltage - no need for any adjustments. 

Maricio: I have measured one of my converters (I don't remember if it was a net-power adapter, a boost or a buck converter) to increase voltage if the output current goes up. This compensates a bit for the losses is bad cables, but if you're using good ones, the 5.2V may increase too much. So... that needs checking....

That said: The 'pi should work from the battery voltage directly. It's just that peripherals like USB mouse and HDMI->VGA converters are more picky about the voltage....

The "under load" that was proposed should be about "1A", so with ohms law we calculate 5V/1A = 5 Ohms. At 1A and 5V the power is 5W. Normally I would recommend that you use a resistor with a higher rating than strictly necessary. In  this case 10W. But if you restrict your testing to say a few minutes, a 5W or even 2.5W model will do. Just don't leave it "testing" overnight.

If I understood I have to plug the convector with a resistor of 5w-10w and test the amperes and it has to be 1A?

If you can find a 5 ohm resistor, Ohms law will hold. If you measure the voltage, you can calculate the current. But measuring the current would be fine too.

Actually, the voltage is way more important. You want the adapter to deliver 5V EVEN when the current is 1A. So: trust the resistor to draw "about 1A" when you connect it, and then check the voltage.

If you can order at farnell:

CP00105R000JE14 - VISHAY DALE - Through Hole Resistor, CP Series, 5 ohm, 10 W, ± 5%, Axial Leaded | Farnell element14

is a matching resistor. If you want to save about half a euro, MCKNP03UJ050JB00 - MULTICOMP - Through Hole Resistor, MCKNP Series, 5 ohm, 3 W, ± 5%, 500 V, Axial Leaded | Farnell elem… will last long enough for the experiment.

Mauricio;

What John and Shabaz are saying is that you have to put your boost converter (or anything that supplies power: an actual power supply, buck/boost converter, transformer, etc.) under load (connect them into something that actually uses the power).  This can be a resistor (this is great as you know what the resistance is) or any kind of "black box" that uses a known combination of any two: voltage, current, or resistance.  It depends upon what the unknown is.  If you don't know current, you want to use a known voltage and resistance, for unknown voltage you use known current and resistance, for unknown resistance you use known current and voltage.  These are just the algebraic permutations of Ohm's Law: voltage=current x resistance (E=IR or V=IR, depending upon how you were taught).

I've got several nominal 12V/2A switching power supplies that are 15-16V without a load.  They drop to 11.8-12.5V under a minimum 220ohm load.  You won't get reliable measurement of what your power source is doing without putting it in the same conditions as expected when it's operating.  I once had a nominal 20VDC power converter that I didn't test - Just took the manufacturers specs.  It was at 28VDC with no load.  When I plugged the hot power supply into my circuit there was enough juice to let the magic smoke out of my 25V filter cap before the voltage dropped to the rated level.

You're welcome!

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