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

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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Sorry, I don't know what went wrong with my email there.

What you need is a resistor ideally around 5 ohms rated at 5 watts or more, this will test your booster at the full output (I'm assuming your booster can provide around 1 amp). You connect it between the positive and zero volts connections at the output of the booster. Then get your multimeter and measure the voltage by placing the multimeter leads on either side of the resistor.

You might have made a mistake when you first set up your meter, as setting it to the 20V DC range should show be fine. Make sure you have turned it to the DC measurement range as you will get strange readings if set to the AC.

You could use a higher value resistor, but as you increase the value of the resistor the amount of current drawn from the power supply will go down. This can be important, as the booster might work fine at low currents and show around 5V, but as more current is needed by whatever it is connected to the voltage will drop considerably if the booster is faulty. Which is why you should try to test the booster near to the maximum current it might need to provide.

Colin

Sorry, I don't know what went wrong with my email there.

What you need is a resistor ideally around 5 ohms rated at 5 watts or more, this will test your booster at the full output (I'm assuming your booster can provide around 1 amp). You connect it between the positive and zero volts connections at the output of the booster. Then get your multimeter and measure the voltage by placing the multimeter leads on either side of the resistor.

You might have made a mistake when you first set up your meter, as setting it to the 20V DC range should show be fine. Make sure you have turned it to the DC measurement range as you will get strange readings if set to the AC.

You could use a higher value resistor, but as you increase the value of the resistor the amount of current drawn from the power supply will go down. This can be important, as the booster might work fine at low currents and show around 5V, but as more current is needed by whatever it is connected to the voltage will drop considerably if the booster is faulty. Which is why you should try to test the booster near to the maximum current it might need to provide.

Colin

To substantiate Collins point about testing under load... A friend dropped by yesterday evening and asked me why his system didn't work. He has a DCDC converter (actually two) set for 13.6V, but they dropped to 10V under the intended load. (one always did so immediately, the other would sometimes hold out at 13.6V for a short while before dropping).

@Colin: resistors can be used outside their stated wattage without problems. In an experiment last week I needed a load for a 20V system. So grabbed the "cheap resistors" bag and the first one I pulled out was 100 ohms. 20V/100Ohms  = 200mA. That'll be fine for testing at first. But 200mA * 20V = 4W. So I grabbed 4 of the resistors, twisted them into 2S/2P configuration for quadruple the wattage = 4*0.25=1W. I'll stay below 1W by PWMing the thing at most 25%... Well you know how things go. When PWMing at 25% works you work your way up to 90 or 100%. Then you get the smell of burnt-electronics. Darn! Blew something up anway? Nope, just the resistors getting a tan.