Couple quick questions.Slightly confused.Need refresher.

Discussion in 'General Electronics Chat' started by Grail, Apr 14, 2011.

  1. Grail

    Thread Starter New Member

    Apr 14, 2011
    2
    0
    Hi. I'm new so please forgive any noob opps stuff on this thread. I think I'm posting in the right place?
    Here's the scoop:
    I've been doing some commissioned work that I need to add LED's to.They didn't come with many stats,so I'm just going on what I know.
    I'm using 16 flat top blue 5mm LED's at 3.4V and I believe at 20ma?
    1 White round 5mm LED at 3.4V again, I believe 20ma?
    and 1 regular round red at 3.0V and 20ma.
    I need to decide on a power supply. and I have some questions.

    1) 12V or 9V power supply. Provided I use the proper resistors,will there be any difference in brightness? I'm guessing no?
    2) can I connect the array like this?
    [​IMG]
    *I made this diagram,it doesn't come out this way.
    My last question
    3)Are the LED's connected resistor(I'm aware that this can be on the cathode or anode side)Then anode to cathode to anode to cathode like the diagram? Seems like they should be anode to anode and cathode to cathode? Am I on glue here? :)

    I am slightly electrical saavy,but I still get confused and I definitely have not done any for awhile,so I appreciate any input and I thank you for your time.
     
  2. someonesdad

    Senior Member

    Jul 7, 2009
    1,585
    141
    1) Doesn't matter with the correct resistors. The LED's brightness is controlled by how much current flows through it.

    2) The way to check each leg is to assume 20 mA of current in the leg, then add up the voltage drops. If the sum isn't equal to the power supply voltage, the resistor is the wrong value (multiply the resistance in ohms by 0.02 to get the voltage across the resistor).

    3) It's irrelevant where you put the resistor with respect to the LEDs as long as it is in series with them.

    Make sure your power supply can a) supply the needed current (20 mA times the number of legs) and b) is well-regulated enough to handle the load without dropping too much in voltage (otherwise you won't get the brightness you want).
     
  3. Wendy

    Moderator

    Mar 24, 2008
    20,765
    2,535
    LEDs, 555s, Flashers, and Light Chasers

    A tutoral on LEDs, focus on chapters 1 and first half of 2.

    You can measure the real voltage drop (Vf) with a 470Ω ¼W 5% resistor and a 9V battery. Just light it up and measure the voltage, that is the real Vf.
     
  4. Adjuster

    Well-Known Member

    Dec 26, 2010
    2,147
    300
    LEDs in a series string are indeed connected anode, cathode, anode, cathode... and so on, so that the current flows through all of them in the forward direction. To do otherwise would leave some LEDs reverse biased, which would not work.
     
  5. fadelo

    New Member

    Apr 12, 2011
    22
    0
    if you want to use a 12 v battery .. u rather should connect the LED with 1 k ohms

    resistor in series
     
  6. Wendy

    Moderator

    Mar 24, 2008
    20,765
    2,535
    Uhhh, no. Tis better to measure the Vf and calculate the correct resistor for full (or nearly full) LED brightness.

    If you have 3 white LEDs that drop 3.3 volts in series, and a 12V battery, you would need:

    (12VDC -(3 X 3.3Vf)) / 0.02A = 105Ω ≈ 100Ω or 110Ω, which ever you can get your hands on. Even a 120Ω would work well.
     
  7. fadelo

    New Member

    Apr 12, 2011
    22
    0
    so accurate .. thumbs up

    thnx
     
  8. Grail

    Thread Starter New Member

    Apr 14, 2011
    2
    0
    Bill M: I totally see where you are coming from,thanks.

    Adjuster:thanks too. That was totally getting me confused for some reason.
    Thanks someonesdad too .. hilarious by the way.

    12V batt is pretty much the hugest overkill ever for my purposes in this case. I see the validity though. I just meant a 12Vac or dc wall plug supply from the store that you can buy?

    I haven't had time to read the tutorials yet,(I work a lot), I guess I'm just really stuck on getting my head around having the last resistor or 2 in the array being a different measure of Ω? (I'm trying to wrap my head around the math for someonesdad's answer for #2,so I'm just trying to figure out if that is in fact the answer I need:) Sorry,I'm a little dim at the moment. :)

    Thanks guys. That helps. I really appreciate it. I have a little homework here.
     
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