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Thread: General question about powering an arduino

  1. #81
    Resident 100HP water-cannon operator SXRguyinMA's Avatar
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    Default Re: General question about powering an arduino

    ok so I was trying to figure out how to make the circuit with 12v power and using a 5v regulator, but the simulator doesn't have a LM7805 regulator in it! hmmm....

  2. #82
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    Default Re: General question about powering an arduino

    alright I came up with this. Except I'll be replacing the 5 ohm resistor with a 5v regulator. what do you think?

    Code:
    $ 1 0.0050 1.0751013186076355 40 10.0 62
    g 208 96 160 96 0
    R 208 32 160 32 0 0 40.0 12.0 0.0 0.0 0.5
    w 368 176 368 208 0
    w 368 208 464 208 2
    r 448 288 448 336 0 15.0
    s 208 32 288 32 0 0 false
    z 416 336 416 288 1 0.805904783 5.6
    w 416 288 448 288 0
    w 448 288 512 288 0
    c 416 336 416 384 0 10.0 0.20502644242779824
    c 416 384 416 432 0 10.0 0.20502644242779625
    w 416 432 464 432 0
    w 416 336 448 336 0
    w 464 432 512 432 0
    178 352 48 400 48 0 1 0.2 0.6 0.05 1000000.0 0.02 20.0
    w 352 80 288 80 0
    w 288 80 288 32 0
    w 352 96 208 96 0
    w 512 288 512 208 0
    w 512 208 464 208 0
    w 400 128 368 128 0
    r 368 128 368 176 0 5.0
    w 416 432 224 432 0
    w 224 432 208 96 0
    162 512 288 512 432 1 5.0 1.0 0.0 0.0
    w 352 48 352 80 0
    z 400 64 400 128 1 0.805904783 5.6
    o 9 64 0 35 0.625 0.4 0 -1
    o 10 64 0 35 0.625 0.4 1 -1
    o 24 64 0 35 10.0 6.4 2 -1

  3. #83
    Fox Furry crenn's Avatar
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    Default Re: General question about powering an arduino

    My understanding is that it's using the reverse breakdown voltage to disrupt a voltage divider that could occur.

    EDIT: That will work, just you don't need a zener diode before the simulated 'voltage regulator' if I'm correct.
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  4. #84
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    Default Re: General question about powering an arduino

    I have no idea what you're saying there lol. did some more messing though and came up with this:

    Code:
    $ 1 0.0050 0.625470095193633 43 10.0 62
    g 208 96 160 96 0
    R 208 32 160 32 0 0 40.0 5.0 0.0 0.0 0.5
    w 368 176 368 208 0
    w 368 208 464 208 2
    r 448 288 448 336 0 15.0
    s 208 32 288 32 0 0 false
    z 416 336 416 288 1 0.805904783 5.6
    w 416 288 448 288 0
    w 448 288 512 288 0
    c 416 336 416 384 0 10.0 9.9810148918879E-4
    c 416 384 416 432 0 10.0 9.98101489188794E-4
    w 416 432 464 432 0
    w 416 336 448 336 0
    w 464 432 512 432 0
    178 352 48 400 48 0 1 0.2 -3.653919108052395E-20 0.05 1000000.0 0.02 20.0
    w 352 80 288 80 0
    w 288 80 288 32 0
    w 352 96 208 96 0
    w 512 288 512 208 0
    w 512 208 464 208 0
    w 400 128 368 128 0
    w 416 432 224 432 0
    w 224 432 208 96 0
    w 352 48 352 80 0
    w 368 128 368 176 0
    d 400 64 400 128 1 0.205904783
    181 512 288 512 432 0 300.0037876461304 0.025 5.0 0.4 0.4
    w 352 48 352 16 0
    w 352 16 448 16 0
    r 448 16 448 64 0 10000.0
    g 448 64 448 96 0
    x 468 86 538 89 0 10 5v to sense pin
    x 573 362 610 365 0 10 Arduino
    x 45 149 190 155 0 24 IMPORTANT!
    x 91 176 183 179 0 10 Combined grounds!
    o 9 64 0 35 2.5 0.1 0 -1
    o 10 64 0 35 2.5 0.1 1 -1
    o 26 64 0 35 5.0 0.025 2 -1
    Now if you take out the diode right after the relay and replace it with wire, you'll see how the relay stays open until voltage drops enough for it to close. Not that it only does this when you connect the relay ground and capacitor grounds together (as is the setup in a normal psu - common grounds) If you make each have it's own ground it won't do it

    :EDIT: the grounds don't have to be combined - it still does it without the diode

  5. #85
    Fox Furry crenn's Avatar
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    Default Re: General question about powering an arduino

    Look up above again. But I just thought of something, give the 12v rail a diode after it passes through the relay. That should disrupt the circuit.
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  6. #86
    Resident 100HP water-cannon operator SXRguyinMA's Avatar
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    Default Re: General question about powering an arduino

    like this?

    :EDIT: just saw the part about the zener. but if I take it out the relay won't shut. will that possibly change with the vreg in place?

    Code:
    $ 1 0.0050 0.30802168489180315 61 10.0 62
    g 208 96 160 96 0
    R 208 32 160 32 0 0 40.0 12.0 0.0 0.0 0.5
    w 368 176 368 208 0
    w 368 208 464 208 2
    r 416 288 416 336 0 15.0
    s 208 32 288 32 0 1 false
    w 448 288 512 288 0
    c 416 336 416 384 0 10.0 9.999999728595806E-4
    c 416 384 416 432 0 10.0 9.999999728595758E-4
    w 416 432 464 432 0
    w 464 432 512 432 0
    178 352 48 400 48 0 1 0.2 1.973851188894976E-9 0.05 1000000.0 0.02 20.0
    w 352 80 288 80 0
    w 288 80 288 32 0
    w 352 96 208 96 0
    w 512 288 512 208 0
    w 512 208 464 208 0
    w 400 128 368 128 0
    r 368 128 368 176 0 5.0
    w 416 432 224 432 0
    w 224 432 208 96 0
    162 512 288 512 432 1 5.0 1.0 0.0 0.0
    w 352 48 352 80 0
    z 400 64 400 128 1 0.805904783 5.6
    w 448 288 416 288 0
    o 7 64 0 35 2.5 0.4 0 -1
    o 8 64 0 35 2.5 0.4 1 -1

  7. #87
    Fox Furry crenn's Avatar
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    Default Re: General question about powering an arduino

    Yes, because I don't think 5V or lower can pass back through the voltage regulator.
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  8. #88
    Resident 100HP water-cannon operator SXRguyinMA's Avatar
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    Default Re: General question about powering an arduino

    ahh I see. I'll breadboard it up tomorrow after work and check it out!!

  9. #89
    Resident 100HP water-cannon operator SXRguyinMA's Avatar
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    Default Re: General question about powering an arduino

    alright, since I ruined my 12v relay de-soldering it, all I'm left with is the 5v (and don't feel like another trip to RS to pick up another 12v one lol). This is what I'm going to breadboard up this afternoon after work:

    Code:
    $ 1 0.0050 1.6308177459886661 40 10.0 62
    g 128 96 80 96 0
    R 128 32 80 32 0 0 40.0 12.0 0.0 0.0 0.5
    w 288 176 288 208 0
    w 288 208 384 208 2
    r 368 288 368 336 0 15.0
    s 112 64 192 64 0 1 false
    z 336 336 336 288 1 0.805904783 5.6
    w 336 288 368 288 0
    w 368 288 432 288 0
    c 336 336 336 384 0 10.0 0.0010010099896179414
    c 336 384 336 432 0 10.0 0.0010010099896179475
    w 336 432 384 432 0
    w 336 336 368 336 0
    w 384 432 432 432 0
    178 272 48 320 48 0 1 0.2 0.0 0.05 1000000.0 0.02 20.0
    w 272 80 208 80 0
    w 272 96 128 96 0
    w 432 288 432 208 0
    w 432 208 384 208 0
    w 320 128 288 128 0
    r 288 128 288 176 0 5.0
    w 336 432 144 432 0
    w 144 432 128 96 0
    162 432 288 432 432 1 5.0 1.0 0.0 0.0
    z 320 64 320 128 1 0.805904783 5.6
    R 112 64 80 64 0 0 40.0 5.0 0.0 0.0 0.5
    w 208 80 208 64 0
    w 208 64 192 64 0
    w 128 32 272 32 0
    w 272 32 272 48 0
    x 467 359 550 365 0 24 Arduino
    x 320 152 572 155 0 10 diode and resistor will be replaced with a 5v regulator
    x 259 66 296 69 0 10 5v relay
    o 9 64 0 35 5.0 0.05 0 -1
    o 10 64 0 35 5.0 0.05 1 -1
    o 23 64 0 35 10.0 1.6 2 -1
    o 4 64 0 35 1.25 0.05 3 -1
    o 6 64 0 35 1.25 0.0015625 4 -1

  10. #90
    Resident 100HP water-cannon operator SXRguyinMA's Avatar
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    Default Re: General question about powering an arduino

    well I'm happy to report that it works perfectly!! thanks again crenn for all your help!!

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