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Please help Electronic (problem in power bank) Answered

one of a resistance it's blow up in the power bank and i don't have the number of it !! and I could not find any scheme for this Motherboard of power bank

and chip that use in this power bank is mt4606

and name of this Motherboard is yd178_b

Discussions

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IgnacioN8

5 weeks ago

Hi maatoubi, you could start drawing a simple schematic trying to see what the nearby components are. Then, it could be maybe easier to know or calculate a resistor value.

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maatoubiIgnacioN8

Reply 5 weeks ago

thank you I who destroyed this resistance !!! I forgot the positive wire without welding and he Touch it so it's carbonization :(
It is connected directly to the negative And give energy to the chip mt4606

hfgh.jpg
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IgnacioN8maatoubi

Reply 5 weeks ago

Hi again @maatoubi.
As you show us on the picture, it seems pin 1 and 2 of this "mt4606" are been used.

This pins are from a N-Channel Mosfet (IMAGE 1),
- Pin 1 = S1 (Source pin).
- Pin 2 = G1 (Gate pin).

It seems the connection between those capacitors and resistors from the board is GND (Ground), so you have the S1 pin connected to GND and G1 seems to have a pull down resistor (IMAGE 2).

Maybe this can help you continuing with your investigation. Maybe I am wrong with my advises, so keep going. I will be happy to continue with this discussion. =)

AACmaatoubi.JPGAACmaatoubi2.JPG
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maatoubiIgnacioN8

Reply 5 weeks ago

oh thank you I found a solution i will using a variable resistor and i Start reducing the resistance value until i get the correct values maybe it's will work


Potentiometer.jpg
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IgnacioN8maatoubi

Reply 5 weeks ago

If I was you I Would continuing trying to resolve what that resistor function is, and where is it connected. Maybe its risky to probe the variable resistor without knowing that.

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IgnacioN8IgnacioN8

Reply 5 weeks ago

Where is the other therminal of the resistor connected?

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maatoubiIgnacioN8

Reply 4 weeks ago

The results are not guaranteed I know but i must try it i see some video in youtube do that and it's work

the therminal of the other resistor you ask R1 is 103

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IgnacioN8maatoubi

Reply 4 weeks ago

Ok, so when you start the testing, it would be fine to know the results you obtain =)

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maatoubiIgnacioN8

Reply 26 days ago

hey The method of variable resistor it's work I got the required values to the resistor 300 ohm ... I've used variable resistor 1k ohm And i start to reduce values until I got the required values so now it's work

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maatoubiIgnacioN8

Reply 4 weeks ago

ok i will tell you the results lol when i testing it lol

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Jack A LopezIgnacioN8

Reply 4 weeks ago

Looking at this thing a second time, I think the other terminal of the burned "resistor" is connected to only one thing, namely the wire to B-.

If that is true, then I think the "resistor" was what Downunder35m suggested: it was a fuse, meaning its resistance should be zero.

Or in other words: replacing this "resistor" with a piece of wire will fix this thing, if it does not have any other problems.

Previously, I was thinking this resistor was a current sensing resistor, Rsense, with one terminal connected to the ground node, the other connected to B-. However that hypothesis requires some other component (like big resistor e.g. 10K, 100K, 1M, etc, trailing away to an IC, or just tiny wire trace trailing away to an IC) connected to B-, to sense the voltage difference between B- (one side of Rsense) and ground (the other side of Rsense)

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Jack A Lopezmaatoubi

Reply 4 weeks ago

You're welcome. Did I guess correctly, and putting a piece of wire in place of the burned thing, makes it work? If so, I had help, since that was DU35m's hypothesis, that little thing was a fuse, or a jumper.

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Downunder35m

5 weeks ago

I highly doubt it should be a resistor!
A fuse would make more sense.
Having said that, a resistor in this location would most likely act as a current limiter.
From a 3.7V cell to get about 2 Amps out that resistor would need to be below 2 Ohm already.
More like a shunt really.
If the power bank has a higher battery voltage then the resistor would be a bit higher in value.

For a 3.7V battery I and assuming the output is only good for 1A on the USB ports I would start with a 10 Ohm wired resistor of at least 1W.
Check the output and if way below specs reduce the resistor to about 5 Ohm )2 10 Ohm parallel for example).
From there same procedure until you end up wih the stated output current.

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maatoubi

Reply 5 weeks ago

it's Apple but i think it's not lol ( China)

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Jack A Lopez

5 weeks ago

Sometimes if you can read the numbers on the ICs, and then find the data sheets for those ICs, the data sheets will contain circuit diagrams for "suggested" or "typical" applications, and that can be a real help to figuring out what on the board does what.

Of course that advice has to be combined with with IgnacioN8's advice, to follow what wire goes where, and draw a schematic diagram of your own, to help visualize what wire goes where.

By the way, is this thing cheap? Is it just using one Li-ion cell? I mean, if it is cheap to buy another one just like it, you could do that, and that circuit board would have a good, undestroyed, resistor on it, that you could examine.

Although there is risk of the exact same thing happening, if you simply buy another one just like the previous one. Seems like there is some English idiom that captures the concept. Like maybe, "doubling down", on a bad bet, or "throwing good money after bad"

Regarding the actual function of that destroyed resistor, my best guess was it was there as a current sensing resistor, since it was tied on one side to B-, the negative side of the battery (or single cell, whichever it is). In other words, it was a very small resistance, intentionally placed in series with the battery, so that all the battery's current had to flow through that tiny resistor, and this current could be measured, by measuring the voltage across that tiny resistor, since I=V/Rsense, where Rsense is the tiny resistor.

That hypothesis could be corroborated, if we could see what other things are attached to that resistor; i.e. something that looks like it could be sensing, watching, the voltage across it.

I have no idea, exactly what size Rsense is supposed to be for this circuit, but it kind of has to be small, typically less than 1 ohm. Guessing something in the range 0.01 ohm, or 0.1 ohm. Its resistance could not be too large, for a current sense resistor, because a resistor necessarily dissipates power, P=I^2*R. Make R too large, or I too large, and the darned thing is in danger of burning itself up due to the power it dissipates.
;-)