Mates bread maker stopped working so I had a look inside and saw this burned resistor.

I’m guessing the heat changed the colors a bit so wondering if anyone has experience in reading cooked resistor values.

I removed it from the PCB and measured it at 403 Ohms.

Thanks for any help.

  • @[email protected]
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    181 year ago

    It looks like 1GΩ (black-brown-white gold). But that doesn’t sound likely unless you have a very high voltage bread maker.

    If we treat the black band as discolored-brown, and read it the other way, we get (yellow - white - brown - brown) which is 490Ω and closer to your measured value. I wouldn’t rule out (yellow - white red - brown) either at 4.9 kΩ, although that doesn’t match closely to your measurement.

    A good question is ‘why did the resistor burn?’. If I didn’t know why, then I would assume that replacing it will just result in it burning again, although maybe not immediately.

    • @WaltzingKea
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      61 year ago

      Thanks! I’ll try replacing it with a 490 ohm resistor and see if it works again.

      The element in the bread maker looks like it came loose a bit and made slight contact with the internal metal housing. I wonder if that caused the resistive element to sink more current than the PCB was designed for, burning out the resistor.

      • @[email protected]
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        41 year ago

        That sounds like a possible fail state. Also shitty design. It should use a resettable thermal fuse or something to detect faults without parts burning.

        Consider maybe adding a fuse to the design?

      • @[email protected]
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        11 year ago

        You’ll need a fairly high wattage resistor. That one looks like maybe 1W but it might be more.

    • @[email protected]
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      21 year ago

      Isn’t it safer/better to start by replacing and testing with a higher resistor? Or is my thinking too simple?

      • @[email protected]
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        41 year ago

        Well, arguably keeping the resistor the same value would result in a somewhat known state, and changing it would put it in an unknown state. The unknown state could be better or worse. I can’t see enough to know what the circuit does to say.

        What you could do instead, is set the resistor to the same value, but rated for higher thermal dissipation. Then measure how hot it gets to identify if the real problem is somewhere else. Another part might burn/explode instead though, so I’d consider carefully how to proceed, and probably wear goggles + have a fire extinguisher in the room.

        My main concern is by ‘fixing’ it with a resistor with higher thermal dissipation, I’ve created a fire hazard because that dissipated heat now has to ‘go somewhere’, which may be the plastic case. A thermal camera is handy to see if some part of the board gets unacceptably hot during normal operation.

        • @WaltzingKea
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          31 year ago

          The case is basically fully metal, just a bit of plastic inside for mounting the PCB to and a few other bits of plastic outside. Plus there is a temperature fuse in the case also.

          From the resistor size (11.5 x 4.5mm) I think it would have been a 2W resistor when comparing to sizes on Digikey. I made a 500 Ohm 2W resistor from 8 1/4W 1K resistors then put a larger resistor in parallel to that to bring it down, measured it to 489 Ohms.

          I’m going to run it a few times then open it up again to see if there is any new damage to the board before returning it.

            • @WaltzingKea
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              111 months ago

              Four loafs of bread later with no issues, opened it up and everything looks fine :)

              • @[email protected]
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                111 months ago

                Excellent news!

                The science gets done, and we bake a fresh bun, for the people who are still alive!

        • @[email protected]
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          11 year ago

          Thank you for the detailed insight! I miss some basics in electronics but am eager to learn how to test and fix circuits.

          Years ago I tried to repair an old keyboard/synthesizer by cleaning it and replacing leaked/bloated capacitors. Unfortunately the onboard sound memory could not be loaded anymore or was wiped entirely as far as I understood. But due to lack of knowledge (me and community that time) it was too complex to got the keyboard up and running again. It’s sometimes sad to loose good hardware…

          Back to the resistor/thread: I can’t imagine a resistor to be the source of the problem. Isn’t it more possible that a capacitor wears out or a transistor cooling fails?

          • @[email protected]
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            31 year ago

            Those things are indeed more common!

            However, if the circuit was in an abnormal state (e.g. the contact with the case), then a resistor could very well blow. It would not be surprising if it took some other components down with it, and that this damage is not obvious yet. “The transistor blows to protect the fuse” is a common fail-state, facetiously stated.

            Another possibility is just… bad design. You could call me adequate at circuit design (I mostly design prototypes, not finished systems that have to last thousands of hours), but regularly see commercial products designed poorly with some stupid point of failure. For example, using a 1 watt resistor that is dissipating close to 1 watt, instead of designing a more efficient system that doesn’t require dissipating heat at all.

            I spend a lot of time answering questions for people just getting started. Probably 75% of them boil down to a few things. Here is that list in case amusing / useful:

            1. Relays are not a great solution in general, and there are many better alternatives (MOSFET, SSR, etc).
            2. Output impedance matters: you can’t power a huge motor off a microcontroller pin.
            3. Back-EMF from inductive loads can burn out your control system unless you add a protection diode.
            4. Lead acid batteries aren’t a magic solution to power everything. Especially automotive ones. Understand and use lithium ion.
            5. Connecting LEDs in parallel then adding a single resistor will lead to failure pretty quickly.
            6. Generally, don’t pass significant power through a switch. Use the switch to control the state of a power MOSFET or similar.
            7. Button debouncing.

            Most of the rest is refusing to do other people’s homework, help people build weapons, or do unwise things with mains power / high voltage / centrifuges. Occasionally people ask me really interesting questions though, so I don’t mind that the interactions are a bit scripted the rest of the time! I’ve noticed on Lemmy I’ve gotten much more interesting questions so far!

  • @[email protected]
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    41 year ago

    You measured the resistor at 403 Ohms? That would qualify it as “not failed” then. Resistors pretty much exclusively fail open, or on rare occasions, out-of-tolerance on the high side. After 5 years of doing electronics diagnostics for USAF aircraft, I never say any other type of resistor failures.

  • @[email protected]
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    21 year ago

    I’m old school. I’d reverse engineer that part of the board, work out what the resistor was doing and then choose a value, much as the original designer did.

    From that small section of the board - I’d guess that it is a resistor in the CR voltage dropper, used to power the electronics.

    • @[email protected]
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      121 year ago

      That looks correct to me, by the color codes…

      …but how in the world do you burn a 1 GΩ resistor? That looks sort of like it could be a 1 watt resistor too. So back-of-the-napkin this would have to be from over a 30kV supply. So that sounds a bit off.

      Unless it isn’t. One hell of a bread maker then. I want one.

      • @[email protected]
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        21 year ago

        Only thing I can think of, maybe it’s a bleeder resistor for that cap, and it failed by some kind of internal short which reduced its resistance (and increased its heat dissipation hence the blackened board)? But fails-short is an unusual failure mode for a resistor and 1 GΩ is pretty high even for a bleeder, so maybe we’re misreading something.