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Thread: cap dump schematic questions

  1. #51
    Hi Cmor,

    Quote Originally Posted by Cmor View Post
    Hi guys, ...........A quick question. The schematic shows a diode between the top of the power windings/battery positive and the 555 chip. Do all the power windings go through one diode or is there a separate diode for each winding?
    All the power windings connect to the primary battery positive. There is only one diode between the 555 chip and the primary battery positive.
    Gary Hammond,

  2. #52
    Hi Gary,
    Thanks.

    I’m assuming the battery is never the source of current to the 555 chip, but that instead it's powered by current from the power windings (produced by the approach of each magnet). Is that right?

  3. #53
    Hi Cmor,

    Quote Originally Posted by Cmor View Post
    Hi Gary,
    Thanks.

    I’m assuming the battery is never the source of current to the 555 chip, but that instead it's powered by current from the power windings (produced by the approach of each magnet). Is that right?
    NO! The 555 is powered by the primary battery.
    Gary Hammond,

  4. #54
    Thanks Gary.

    I'll confess, originally I thought it was the battery, but then I checked the max. supply current on the 555 datasheet and it was only 200 uA, not 100's of amps like the battery can deliver. But... I also read "HIGH INPUT IMPEDANCE : 10 Ω", and thought maybe that drops the amps down enough.

    But...being a beginner (posting on the intermediate forum because cap dumps aren't a beginner topic) without a good grasp on current direction, I thought, if battery current flows from - to + perhaps it's not the source (though I also couldn't see a return path from the power windings).

    Lots of room for error at my level.

    So how is it that the battery doesn't cook the 555?

  5. #55
    Hi Cmor,

    I checked the max. supply current on the 555 datasheet and it was only 200 uA, not 100's of amps like the battery can deliver.
    Check again. It's 200 ma, not 200ua. And it also says the following.
    The output circuit is capable of sinking or sourcing current up to 200 mA. Operation is specified for supplies of
    5 V to 15 V.
    So how is it that the battery doesn't cook the 555?
    The battery, at only 12.0 to 13.0 volts, is less than the 15 volt max supply voltage for the device. The output current of the device is controlled by the load (resistance) applied to it.

    I=E/R P=EI
    Gary Hammond,

  6. #56
    Thanks, that helps.

    Is the current powering the 555 flowing from the battery’s positive to negative post, or the reverse?

    What is the function of the capacitor after the diode, before the chip?

  7. #57
    Hi Cmor,

    Quote Originally Posted by Cmor View Post
    Thanks, that helps.

    Is the current powering the 555 flowing from the batteryís positive to negative post, or the reverse?

    What is the function of the capacitor after the diode, before the chip?
    The "conventional" or Heavyside flow is from positive to negative. The "electron" current flow is from negative to positive.

    I'm not sure the diode and capacitor are even really needed. I think they are there to filter out any voltage fluctuations or spikes from affecting the 555 input voltage.
    Gary Hammond,

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