Identifying component?

[wilsoe]

This came out of an entry unit. A wire harness got closed in the door and blew this part. I was wondering if this is a repair I could get done locally or should I send it off to the manufacture? Also is this a common part?
Thanks

 

[(*steve*)]

Looks like you've damaged three tantalum capacitors and 2 or maybe 3 resistor networks.

Nothing that looks immediately difficult to source or replace as long as there has been no other damage.

edit: in fact the obvious damage may even be more limited than that. Looks like a single resistor in the network got *very* hot and the other discolouration may just be due to that smoking a lot. At first I thought it was caused by external heat from a nearby damaged cable (I concentrated on the second image) however from the first it's clear that this was not the case. I'd also check for damages to tracks.

 

[wilsoe]

That side of the board it covered and has nothing on top of it.

This is a picture from more straight down.

The board did smoke when the wire harness was crimped. So how would someone figure out what resistor is needed?

Flickr Set

Thanks

 

[davenn]

chances are that the burnt resistor network is the same as the other ones around it

from the angle of your photo I cant quite read the value as an example it will be something like 103X9 meaning the resistors in the package are 10kOhms and there are 9 of them

on that cooked resistor pack you can just see some numbers b4 the burnt part... see if they are the same as on the other resistor packs.

The cooked tantalum capacitor will also most likely be the same value as the ones around it

cheers
Dave

 

[Resqueline]

The carbonized parts is the easy part, but it's highly likely that an IC has taken a hit too without it showing. You'll have to trace the tracks from the pins where there's burns.

 

[(*steve*)]

Do you know what voltages were present on the cable that was damaged?

The MC1489's are rated for an input voltage up to 30V, so as long as the short was from a rail <= 30V to the inputs, the line receivers may have escaped damage.

It seems likely that the short was from a data line to some higher voltage as the damage seems to have started on a single pull-up (or pull-down) resistor.

If you are *really* lucky, you may be really lucky.

 

[davenn]

The carbonized parts is the easy part, but it's highly likely that an IC has taken a hit too without it showing. You'll have to trace the tracks from the pins where there's burns.

Yes possibly, but at least its an easy IC to source must have at least a dozen lying around here hahhaha
MC1489 = Quad line Receiver

I used to use 1488 and 1489 for RS232 interfacing of 300/300 baud phone modems back in the good ol' days of computer club bulletin boards

 

[wilsoe]

Thanks for all the help guys!
I took it to a tv repair shop today to get repaired and he said that the part was a proprietary part. I didn't argue with him but he said he could put the part on if I had it.

Where would be a good place to get a new piece soldered on? A tv repair place? Also where could I purchase this part? I believe I need this part "10x-2-151 B99351"? What do you guys think?
Thanks

Flickr Set

 

[Resqueline]

A resistor network is hardly proprietary, but it's not something that a TV repair shop usually stocks (they may occasionally be found in TV's but they never fail).
The B is the Bourns trademark. 10 is the # of pins, 2 is a code for the variety with individual resistors, & 151 means 150 Ohms. So # 2 of the 5 resistors in it burned.
Here is the datasheet. Any large well-stocked company should have it. Try Mouser, Farnell, etc.
A TV repair shop is as qualified as any to exchange that part. The capacitor may have survived but needs to be ohmed to be sure (at least it needs a good clean).
If 24V was what found its way there then 3.8W could have been dissipated in that resistor and would account for its look, and the IC could have survived (but not for sure).

 

[(*steve*)]

If you have a multimeter, measure the resistance from the left pin (the one with the square solder pad) to each of the other pads.

The first one certainly looks OK, the others may just be "interesting".

Do the same for the one market 10X-1-472. I suspect you'll find the resistance of those to be 4700 ohms (+/- tolerance%).

These resistors are not too hard to find, nor particularly expensive ($0.39).

Here is a datasheet that may actually be for the component in question!

See if you can wipe the soot off the capacitors. They *may* be undamaged. However if you can give us a clear photo of the markings we may be able to suggest a replacement part.

When you remove the bussed resistor, clean up the board. I'd use isopropyl alcohol.

edit: Yeah, that's a 2. Measure the resistance between pairs of pins (start with the end 2 furthest from the burn) -- that will indicate the correct value.

edit2: The isolated resistors are no more expensive

 

[wilsoe]

I believe that I tested the resistor bank. I put my multimeter on the last two pins to the far right (far left being the square) and got 149ohms?

I also think I figured out what the capacitor is, after cleaning everything off. It looks like it says: 105 35 (B) +020
But I am not really sure?
Thanks for all the help!
Thanks

Flickr Set

 

[(*steve*)]

My guess is that the capacitor is a 1uF 25V 20% tantalum, something like this.

The damage to them may be completely due to the nearby smoke and flames. If the capacitor doesn't appear burnt, it may not need replacing.

I'd say the resistor network is one of these.

 

[wilsoe]

Official parts from the manufacture or at least that is what they sent me:

I called the manufacture twice and they said that I had to send the board in. Then I emailed them and they let me purchase the parts but they wouldn't tell me the parts to order. I forgot the prices but they were around $1.25 and $2.50 for each capacitor.

So is there a way to check the capacitors on the board or do you have to remove them? I have a Ideal 61-340 multimeter.
Thanks for everything!

 

[Resqueline]

So you tested the first resistor in the bank is ok. Not that it matters but what about the other four resistors in that bank? The second resistor is obviously "gone".
The tantalums on the board are 1uF 35V but the electrolytics you got are 10uF 25V. I wouldn't want to put those electrolytics in unless I had no other choice.
Your multimeter is fully equipped to check out the cap's. First you Ohm them for shorts, and if they're not shorted then you try to do a capacitance check on them.
If the values are "off" then you simply do a comparison with the other "known good" parts.

 

[(*steve*)]

The tantalums on the board are 1uF 35V

Hmmm, partially my fault. I said 25V. And that was a typo as I *knew* it was 35V :-(

Sorry wilsoe

 

[wilsoe]

(*steve*) you linked to the correct product. The picture I posted are the parts the manufacture sent me to fix the problem. They sent different parts then what were originally there. I was emailing the plant manager I assume he knows what he is talking about? But the product that the board goes in is obsolete and quite old.

Resqueline after seeing the differences in the capacitors I am just going to change out the resistor bank first and hope everything works. Then I was going to test the capacitors. But not really sure how to do that. Is there a guide or something somewhere to help? I have read a little online but I just don't really understand. I also looked at the instruction manual and did that but again I don't really know.

 

[(*steve*)]

Start by removing the resistor network and cleaning up the board to remove the muck.

then measure the resistance across all the capacitors. If the resistance across the blackened ones appears significantly different, that may be a bad sign.

If you have a capacitance meter, try removing the most damaged looking capacitor and testing it. If the leakage is low and the capacitance is OK (at least 80% of what it's supposed to be, then put it back).

If you had replacement capacitors laying around, I'd just replace them -- probably less effort.

Check that the resistance of the removed network matches the new one, and solder in the new one.

Then do a smoke test.

Then see if it works.

Then try to figure out what ICs may have been damaged.

 

[davenn]

Official parts from the manufacture or at least that is what they sent me:

I called the manufacture twice and they said that I had to send the board in. Then I emailed them and they let me purchase the parts but they wouldn't tell me the parts to order. I forgot the prices but they were around $1.25 and $2.50 for each capacitor.

So is there a way to check the capacitors on the board or do you have to remove them? I have a Ideal 61-340 multimeter.
Thanks for everything!

ok one problem I see for a start....
for a moment, assuming the cooked resistor pack value is the same as the others around it... then the new one you have is a different value its a 150 Ohm instead of a 4.7k Ohm one 150 Ohm may be too low and seriously load down a power or signal rail

and why did you order or they supply a electrolytic instead of a tantalum cap. ?
The electrolytic will work ok but wont have the life expectancy of the tantalum specially if that area of the cct gets very warm and dries out the electro. cap.

my 2cents + tax thoughts

Dave

 

[(*steve*)]

the new one you have is a different value its a 150 Ohm instead of a 4.7k Ohm one

The one that has been consumed by fire had a remaining resistor measure 151 ohms. It's quite likely that this is the correct part.

 

[wilsoe]

I got the board back today with the new resistor bank and it still doesn't work. So I should probably just send it in now? What do you think I should try next?
Thanks for all the help!