Unknown component

Hello,

I have the circuit in the photo attatched and I beieve the component marked is burnt, but don’t know what it is. In this case the circuit is used to ilght up the LED’s on the back when feeded with a power source between 7.4 nd 12.8 Volts.
Does anyone have a clue on what it could be?

Thankk you

8 LEDs on a PCB.

Perhaps a simple lighting board, a light source for a detector, a source of light for a reader, a progress indicator.

What’s on the other side?


Current driver perhaps?

The yellow rectangle is around is a zero ohm resistor.

Thank you. I really didn't know this existe. Now I am shure it burned out. So, if I understood it right, the zero-ohm link is used as a fuse? Or is it just a jumper?

“ if I understood it right, the zero-ohm link is used as a fuse? Or is it just a jumper?”

Or both.

By the looks of the soldering, it's been replaced several times.

Paul

A zero ohm resistor is simply a jumper, but a zero ohm resistor is so much easier for pick and place machines to handle.

It is going to be difficult to burn a zero Ohm resistor. Why do you think it is burnt? If it is burnt, then the other components are likely to be damaged as well.

Do tell us for starters, the purpose of the module in question, where it is used, from where did you get it and so on?

Zero ohm resistors have various uses, typically as jumpers or to allow the same PCB to be populated in several ways for circuit variations, perhaps in a filter network. Sometimes they are used to allow prototypes to have current shunts that allow each chip's current consumption to be monitored (not needed in production). They also function as zero inductors. A zero capacitor is an unpopulated capacitor of course!

The chip is LSC1117, +ve linear voltage regulator.
It looks like the 0ohm link is just connecting pin 1 to gnd.

If it’s not working it’s more likely that the regulator is bad not that 0R. Did you know the 0R come with a tolerance. Mine says 1% lol

IamFof:
The chip is LSC1117, +ve linear voltage regulator.

Which would be consistent with it being used as a constant-current driver.

A better photograph would be rather nice. :roll_eyes:

Paul__B:


Current driver perhaps?

Really?
That's a zero ohm resistor (i.e. a jumper). And to the OP: it's not burned.

wolframore:
If it’s not working it’s more likely that the regulator is bad not that 0R. Did you know the 0R come with a tolerance. Mine says 1% lol

That IS funny! LOL!

Plus or minus 1%? Awesome! It could be a negative resistor!

(hmmm wait a minute... 1 percent of zero... oh boy my head hurts again...)

Just measured one, it is .017Ω :frowning:

Had to use current.

krupski:
That's a zero ohm resistor (i.e. a jumper). And to the OP: it's not burned.

I never suggested otherwise.

What it is doing there is certainly unclear as it is obviously a double sided board and so a jumper as such should not be needed.

There are however two other resistors whose value I cannot quite discern, which would be entirely consistent with it implementing a constant-current driver.

larryd:
Just measured one, it is .017Ω :frowning:

Had to use current.

In a four-terminal configuration of course. :cold_sweat:

Could you let me know what tolerance that is for 0 Ohm? I skipped math.

larryd:
Just measured one, it is .017Ω :frowning:

17 mOhms looks like quite high value for 0R. Are you sure you did not measure contact resistance instead? This is comparable to good MOSFETs. But it is true I have never bothered measure it.

I was using the voltage drop across the zero ohm resistor for (I think) 1.00mA current flow.

R = V/I