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EEVblog #1334 – Mystery Dumpster Teardown

Mystery dumpster teardown time! With the most amazing mechanical mains power switch you’ll ever see! ...


  1. Well done, Dave. I think you showed very well the importance of studying and understanding information in datasheets, even for simple parts like fuses and diodes.

    When I was first learning electronics, I heard about a basic “rule” to avoid incompetent designs: “Never depend on a fuse to protect a transistor.” This is not bad general advice, but as the Fluke design shows, sometimes you need protection and you can use a fuse as part of that.

  2. What would happen if you measure an power supply with a constant current of 1.5A?

    • The multimeter shows out of range for 1 second until the fuse blows.

      Then if the current source is a real life one its output will drop to 0A because its not capable of providing enugh voltage.

      If you have an ideal current source its voltage will shoot up to a few kV at what point something inside the multimeter will arc over and set the multimeter on fire. Luckuly these current sources arent real.

  3. I just wanted to point out that the manufacturer is “littelfuse” not “little”. I came across this couple years ago when finding those smd fuses.

  4. Thanks for pointing that out, Dave. I would never have thought to look at the dynamic aspect of fuses or of diodes.

    Not sure what the diode
    bridge has to do with R20, since they aren’t connected.

  5. I believe that one of the L1 functions is to reduse the current rise rate DI/DT in order to reduce the voltage spike during the diode turn on time.
    That spike can be problematic with high voltage regular silicon diodes like the 1N4007.

  6. Dave ,
    As Usual a Excellent Video To Show Us The Multimeter Protection .Forever Yours Fan

    Amarbir From India

  7. If anyone wants to know – the 10A range has a 5 milliohm sense resistor. 10A is 50mV through that. even 100A is only 500mV – barely above the threshold of ONE diode!

    On the milliamp range, 1A would induce a 5V across the sense resistor, well above the 3.5V needed by the diode protection.

    That’s why there’s no protection for the 10A range.

  8. It seems that you linked the wrong video, #374 instead of #373 😉

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