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Voltage Stabilization?

Discussion in 'General Electronics Discussion' started by icor1031, Apr 27, 2010.

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  1. icor1031

    icor1031

    65
    0
    Apr 27, 2010
    At 3.1v and 150ohm I have .1441..?
     
  2. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

    25,411
    2,779
    Jan 21, 2010
    Aaah, no.

    Try using the values I gave you.

    25mA (0.025A) and the various resistors.
     
  3. icor1031

    icor1031

    65
    0
    Apr 27, 2010
    31.875
    29.375
    16.875
    10.000

    25ma and 510o, 470o, 270o, 160o...?

    I probably don't understand what ^ means in math. :D
     
  4. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

    25,411
    2,779
    Jan 21, 2010
    I^2 means I squared (I x I)

    That should get you more reasonable answers :)
     
  5. icor1031

    icor1031

    65
    0
    Apr 27, 2010
    That's what I thought I did..

    Is the first one 19921875?
     
  6. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

    25,411
    2,779
    Jan 21, 2010
    let me work through one with you.

    P = I^2 R
    = 0.025 * 0.025 * 150 (for 25mA through a 150 ohm resistor)
    = 0.000625 * 150
    = 0.09375

    So that resistor at 25mA is dissipating 0.09 Watts. Unless other constraints existed I would use a 1/4 W resistor.
     
  7. icor1031

    icor1031

    65
    0
    Apr 27, 2010
    ~~Try connecting 1 in series with a 510 ohm resistor,
    ...
    So, my first answer was right.. just had the wrong decimal placement.

    I don't see how to do it with 2 or 3 LEDs, though. :S
     
    Last edited: Apr 28, 2010
  8. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

    25,411
    2,779
    Jan 21, 2010
    I didn't see a correct answer from you at all. Neither did I see the results of your calculation for the other resistor values.

    Did you ever read this?

    I know you didn't bother to answer me.
     
  9. icor1031

    icor1031

    65
    0
    Apr 27, 2010
    ~~Did you wire them in series or parallel?

    *I wired 3 LEDs in series... pos power to pos led, neg led to pos led, neg led to pos led, neg led to neg wire.

    I did answer.


    ~~~Try connecting 1 in series with a 510 ohm resistor,

    **31.875

    ~I didn't see a correct answer from you at all. Neither did I see the results of your calculation for the other resistor values.

    I calculated that the same way you calculated this (which I came up with the same answer)

    ~P = I^2 R
    ~= 0.025 * 0.025 * 150 (for 25mA through a 150 ohm resistor)
    ~= 0.000625 * 150
    ~= 0.09375
     
  10. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

    25,411
    2,779
    Jan 21, 2010
    OK, so the answer is YES you wired them in series, but NO, not like the link showed (i.e. with a resistor)

    So the wattage for each of the 4 resistors would be?
     
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