Apparently a recent thunder-storm has burned out my flatbed scanner (no thanks to my fancy-schmancy mid-size APC surge protector).
I checked to see if it was the power-adapter, but that's fine. I opened the scanner and looked at the circuit board and it is indeed burned. It seems that a large resistor next to the power connector has fried and the coating crumbled onto the board. The board in that area is a little discolored as well. I've attached a couple of snaps to show the damage.
Anyway, I am hoping that I can salvage the scanner by replacing the resistor, but I cannot determine what value it is since the coating is gone. Can anyone give me some advice on this?
You can't just 'attach' pictures on this type of newsgroup. You need to put them up on some webspace somewhere, and post a link. For what it's worth, I think it unlikely that the burnt resistor, which may or may not be your problem, was caused by storm damage. If the pcb is discoloured under it, then this resistor is just a hot-running device as a normal situation, which given the location that you suggest for it, is quite likely. It is not at all uncommon for the cement coating of such thermally stressed resistors, to crumble away as you describe, leaving a very distressed looking component, which often is not electrically faulty at all. Are you quite sure that the power adapter is "fine" ? Have you measured its output voltage with it connected to the scanner ?
Arfa Daily wrote (in news:nOnrk.6889$ snipped-for-privacy@newsfe19.ams):
I don?t know if attachments are blocked in this newsgroup, but I forgot to attach them, so they weren?t there anyway. Here?s the pics:
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Not really, when I did the ?quick test??aka licked it to see if there?s juice?it didn?t have the same kick as the third party adapter. By fine I meant the DMM said it had the right voltage.
the scanner ?
No, I measured it on its own. It is a 12V, 1A brick. It read ~13.3V, and slowly rising (I got tried ~14V and stopped). I also tried an 800mA variable adapter set to 12V (which read ~14V). Neither one causes the LED power indicator to light up or the scanner to be recognized by the computer.
I just tried to read it while connected, but connecting it caused that resistor to get REALLY hot; it burned my fingers. I checked the resistance across the power connector?s terminals to see if it?s shorted. There?s three terminals (which I don?t understand, there?s only two conductors in the wire). I get 65 Ohms between front and back, 0.5 Ohms between front and side, and 65 Ohms between back and side (in the picture, the ?back? is facing us).
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Alec S.
news/alec->synetech/cjb/net
> "Alec S." wrote in message news:g8ki4h$es8$1@aioe.org...
> > Apparently a recent thunder-storm has burned out my flatbed scanner (no
> > thanks to my fancy-schmancy mid-size APC surge protector).
> >
> > I checked to see if it was the power-adapter, but that\'s fine. I opened the
> > scanner and looked at the circuit board and it is indeed burned. It seems
> > that a large resistor next to the power connector has fried and the coating
> > crumbled onto the board. The board in that area is a little discolored as
well.
> > I?ve attached a couple of snaps to show the damage.
> >
> > Anyway, I am hoping that I can salvage the scanner by replacing the
resistor,
> > but I cannot determine what value it is since the coating is gone. Can
> > anyone give me some advice on this?
Methinks there's more blown than just the resistor. Judging by the devastation, it got very very very hot over a fairly short period of time. To me, that means a short or high current draw somewhere else on the board. My guess(tm) is that the burned resistor is in series with the input of the nearby 3 terminal regulator. It's intended to reduce the heat dissipation of the 3 terminal regulator (because there's no heat sink). One lead goes to the power supply connector hot +12V pin. No clue where the other end goes.
Even if the resistor were replaced with the proper value, my guess(tm) is that it will smoke again. The most obvious shorted culprit is a fried 3 terminal regulator, but I'm guessing.
Anyway, judging by the internal (spiral) construction, my guess would be a 3-5 watt resistor, with a value of less than 1 ohms. All the overheating has done is crack off the ceramic coating. The resistor is just fine and the correct value will be whatever it measures. It's fairly difficult to blow up such a big resistor.
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# Jeff Liebermann 150 Felker St #D Santa Cruz CA 95060
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I pretty much agree with all Jeff L says in his reply. You can ignore the fact that the socket has 3 terminals - they generally all do. The barrel connection is a 'sprung' one which not only serves to 'anchor' the plug when it's inserted, but also forms a switch to the remaining terminal, when there is no plug inserted, which allows it to spring back and make contact with that third terminal. It allows for the batteries to be disconnected on portable gear, when an external power supply is used. This is important to prevent dry batteries being 'charged' by an external supply. You can also use a diode to isolate batteries, but that loses you around 0.6v of the battery's terminal voltage, so the switched socket is generally preferred. In the case of your scanner, it makes no odds, as you will see, if you look at the print carefully, the side and back terminals are just shorted together, which is why your meter reads 0.5R. It's the best it can manage, to display what is effectively the resistance of the meter leads and probes.
A reading of 65R both ways across the 'active' connector pins, does seem a little low for a scanner in an idle state. Assuming a 12v supply, that equates to getting on for a 200mA draw. I agree with Jeff that although you can't actually see where the burnt resistor goes, it probably is straight to that 3-term regulator at the back. That being the case, I would not expect to get an ohms reading that low between its input pin and ground. What is that device's number ? 7808 or 7805 perhaps ? If so, you need to measure the voltage between its tab and its left hand pin, to confirm that you have the same voltage as on the side of the resistor remote from the input socket, and then measure between the tab and the right hand pin, to see if you have whatever the numbers on the reg say (78- 05 = 5v, 78- 08 = 8v). If you do, then the problem is downstream of the regulator. If you don't, then the problem may well be the regulator itself. These regulators do have overload foldback protection on them, but you have to be up at over an amp of draw, before that kicks in. Another possibility for the problem, could be C44 being leaky. It appears to be connected straight across the incoming supply.
Beyond checks finding a fault at this level, it is unlikely that you will get to the bottom of the problem if it's elsewhere. Once you are past the regulator, you're into the realms of faulty LSIs ...
On Fri, 22 Aug 2008 09:13:35 +0100, "Arfa Daily" put finger to keyboard and composed:
If you look at the print around L5, there are at least two sections of the PCB that see the full unregulated input. IMHO that's potentially a big concern, although it could be just supplying the motor(s) and lamp.
- Franc Zabkar
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Thanks for the explanation. I actually figured out the anchor reason shortly after posting, but the other one makes sense too.
Yes it does. I?ve posted an overhead shot of it here:
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The resistor goes from the + terminal to the input pin on the 7805, apparently passing through an SMT resistor or something (B32?is there a site with a list of standard PCB text identifiers?)
It seems that the input voltage pin is shorted to the ground pin and tab at 1.8 Ohms?the ground pin is connected to the tab at 0.3 Ohms. The power adapter that came with the scanner does indicate a current of 1000mA and it is an always-connected/on device (the power LED is always on, but the light is only on when in use).
So it?s probably the voltage regulator that died? I don?t have another 7805 on hand (all I?ve got at the moment is a 7905, oh so close). I?ll keep my eye open while scavenging and try it out.
It?s a UMAX 2100U (with a PowerVision PV6098F IC).
It goes to the 7805?s input voltage pin
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through SMT component B32?which seems a little crispy. When I connect the power adapter, the resistor does get extremely hot. Also, a 470µF, 16V cap towards the bottom of the board also gets really hot. When I pull the power, the resistor cools down quickly, but the cap stays hot for quite a while. The voltage regulator does not get discernably hot.
It very well could be. It is only 1.8 Ohms across the input voltage to the ground (and tab).
I?ll try to find another 7805 to replace this one and see if that works. All I have at the moment is a 7905. Doh!
Is the regulator itself (the 7805) getting hot? If not, the problem is before the regulator. Your statement that the 470uF capacitor gets hot is your big clue here. Capacitors in a linear power supply shouldn't get hot. I'd wager that the capacitor is shorted (or very leaky). If you trace the PCB wiring, you might find that the capacitor is the input filter from the wall wart. If it's connected to the input pin of the regulator, it's definitely the input filter. Replace the capacitor!!!!
As a precaution, you could also replace the regulator, but I'd wager that it's probably OK.
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Dave M
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address)
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DaveM wrote (in news:A4SdncLmlOBJ1CnVnZ2dnUVZ snipped-for-privacy@comcast.com):
The voltage regulator itself does not get hotter. I have tried using both the
12V, 1000mA adapter that came with it and a 500mA variable adapter set to 12V. The stock adapter causes the resistor to get really hot, really quickly, and the capactor warms up a bit. Using the 500mA variable adapter causes the resistor to quickly heat up a lot and the capacitor gets much hotter than with the 1000mA adapter. Yes, I let it cool down between. :P
I?ve taken a couple more snaps to show the the whole left (right?) side of the PCB from top and bottom. The capacitor in question is C1 at the top of the first pic.
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It seems to be a multi-layer PCB, so I can?t identify all the paths. I don?t have DMM with cap-testing ability, but I did the simple resistance test and it seems to check out.
filter.
Do you suppose it could be capacitor plague? It is only slightly bluging at the top; it doesn?t seem to have vented at all. Here?s a closeup of the cap:
Franc Zabkar wrote (in news: snipped-for-privacy@4ax.com):
Thanks. I tried some Google queries like ?d1 u1 l1 q1 r1 c1 resistor capacitor inductor diode transistor? and got a fairly good list of PCB labels. I still wish I had a proper list of standard labels. I?ll have to check a couple of electronics engineering texts later, maybe they give them in the schematrics sections.
On Tue, 26 Aug 2008 17:45:08 -0400, "Alec S." put finger to keyboard and composed:
I don't understand why you are having such a hard time of it. You have measured a dead short at the input to the regulator. Desolder the regulator's input pin and test for a short again. If the short remains after desoldering the regulator, then desolder other connected components until the short goes away. The last component will be your culprit.
- Franc Zabkar
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Franc Zabkar wrote (in news: snipped-for-privacy@4ax.com):
keyboard and composed:
Sounds good, thanks. I don?t have an electronics bench, so I can?t _just_ do that. I have to get out my iron and wait for it to heat up and so on. It?s a whole production, which is why I am saving up steps/tasks to do together (including getting replacement components). I?ll try your suggestion the next chance I get.
Alec, Forget about trying to identify a component by its reference designation. Those designations are for schematic and parts list identification only; they have absolutely no bearing on the component's value. The only "standard" thing about them is the use of Cxx for capacitors, Rxx for resistors, Uxx for ICs, Qxx for transistors, etc. And even that's not standard... Some manufacturers use Qxx for transistors, others might use Txx or Trxx; I've even seen some use Vxx. Some use Uxx for ICs, others might use ICxx. The Bxx designation is usually used to designate a blower, fan, or some other kind of cooling device. Don't know why they used Bxx on your board. As I said, it's up to the manufacturer.
A capacitor doesn't have to burst open and ooze its guts all over the place to be defective... If it's getting hot, replace it. If it's bulging, replace it. If it's bulging and getting hot, REPLACE IT!!!
As Franc suggested, start unsoldering components and check resistance/voltage until the problem goes away. Replace the component that made the problem go away. If I were you, I'd start with C1.
Cheers!!!
--
Dave M
MasonDG44 at comcast dot net (Just substitute the appropriate characters in the
address)
Life is like a roll of toilet paper; the closer it gets to the end, the faster
it goes.
DaveM wrote (in news:v5Wdnc7UH9OOESnVnZ2dnUVZ snipped-for-privacy@comcast.com):
That?s what I meant.
Yes, unfortunately that?s what I?ve read. I was trying to find a list of common lables because a lot of surface mount components look identical, so it can be difficult figuring out what a component is. Even bigger components look alike. For example on this one, L5 looks an awful lot like a resistor to me. ;)
Really? C1 is all the way down there on the other end of the board. It doesn?t look like it?s directly connected to the power area. The negative terminal is connected to pin 11 of the IC U2, and I can?t tell where the positive terminal is connected to (it connects on one of the middle layers).
Anyway, I?ll get break out my iron and check those three main ones (the cap at C1, the regulator a Q2, and the resistor at R58) tonight or tomorrow. Hopefully it is indeed the capacitor because I?ve got a ton of spare caps, so it shouldn?t be too hard to find a replacement.
Looks like B32 is directly across the input of the regulator and ground. Seems logical that B32 is causing the problem. Remove it from the board and recheck. A cleanup is definitely in order. The charred area should be carefully excavated and cleaned. Rubbing alcohol isn't a good cleaning agent here, nor is nail polish remover. Both contain oils (skin conditioners) that will be left behind after evaporation. A good cleaner if you can't easily find 99.99% isopropyl alcohol is lacquer thinner. You can get this at your local home improvement store, or even Wal-Mart. A small can will cost much less than 99% alcohol, and it will leave no residue. For sure, replacing C1 is a good idea. It isn't supposed to be getting hot, and it isn't supposed to bulge as it is. It needs to be replaced.
--
Dave M
MasonDG44 at comcast dot net (Just substitute the appropriate characters in the
address)
Life is like a roll of toilet paper; the closer it gets to the end, the faster
it goes.
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