LED wiring questions, resistors

Discussion in 'General Modeling' started by OPPI, Nov 29, 2005.

  1. OPPI

    OPPI Well-Known Member

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    Ok, looks like i need electrical guru help again. I want to power the following 2 leds with one 9v battery. Both on at the same time. What resistors do i need to keep them real bright and not explode? Thanks in advance

    white - 20mA (max) Typical Voltage: 3.6V Typical MCD: 1100

    Blue - Typical MCD: 2600 20mA (max.) Typical Wavelength: 468mm Size: T-1-3/4 (5mm) Viewing Angle: 30° Sold in package of 1 Typical voltage: 3.7, with a maximum voltage of 4.5V

    thanks
    dan
     
  2. Fetthunter

    Fetthunter Sr Member

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    You'll need 1 82 Ohm resistor, not usually available at Radio Shack. Try digikey.com Type "82QBK-ND" into the search box. I think you have to order multiples of 5. :)

    Here's a wiring diagram for you:

    [​IMG]
     
  3. OPPI

    OPPI Well-Known Member

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    so the resitor goes in the negative line back to the battery???
    I could always go with a slightly higher resistor too right, just cause i have a few lying around here and i might be able to get close to 82ohms

    thanks
    dan
     
  4. Fetthunter

    Fetthunter Sr Member

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    Higher the resistance, the dimmer the light. Lower the resistance, the brighter the light (until it burns up). ;)

    Yes, you could use, say, a 100 Ohm resistor.
     
  5. OPPI

    OPPI Well-Known Member

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    this would also work the same if i were to use one of thse wall warts that puts out 9v right?

    thanks again for all the help. I just soldered it up to th ac/dc 9v wall wart. So once i hear from you i will plug it in. Got the red to the + of led 1, neg of led 1 to the pos of led 2, the neg of led 2 to the resistor and the resistor back to the neg of the power

    hopefully all correct

    thanks
    dan
     
  6. Fetthunter

    Fetthunter Sr Member

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    Your wiring is correct, however I can't recommend using the wall wart. I don't know how much current it's putting out, and could fry the circuit. A 9V would be easier and safer (to the circuit). You can buy a wall transformer with a variable current setting, however they're pricey compared to a 9V battery. Depending on your application, you may just be better off going with the battery. ;)
     
  7. OPPI

    OPPI Well-Known Member

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    Hey thanks again. I think my wall wart is busted anyway. I tried to plug it in real quick and nothing happened. Hooked it to a 9v batttery and its bright as can be.

    Whats the reasoning for putting the resistor in the negative lead? I would think you would want to "resist" the current in the pos lead before it gets to the leds...no?

    thanks for everything

    dan
     
  8. Fetthunter

    Fetthunter Sr Member

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    Current travels from negative to positive, believe it or not. Shocking, I know. (pun not intended) :lol
     
  9. Nexus6

    Nexus6 Sr Member RPF PREMIUM MEMBER

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    I have a project in the works that will require a similar setup. Not knowing much about electronics, won't the 1st LED be brighter than the second? Also, plz correct me if I'm wrong, but is this setup wired in series? And if so, why is that preferable to parallel?

    Thanks in advance.
     
  10. exoray

    exoray Master Member

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    No both LEDs will be the same brightness they will both pass the same current in the above diagram, so it's a good idea to use balance LEDs (the same kind and values), and yes they are wired in series...

    Well in this case wiring them in series will only pull 20mA... If you wired them parallel you would pull 40mA and the battery would go dead twice as fast... It's just more efficient to do it this way in this case... The limitaion that a series wiring like this has is the supply voltage... Since each LED in this case want's 3.6 volts you can only run 2 in series off a 9volt supply, since 3 would want 10.8volts... So if you want to run 3 or more in this example a series/parallel arangement would be best... In a series/parallel setup you would make several (depending on the number of LEDs) of the above circuit and wire them circuits in series... The example below shows this being done with 7 LEDs...
    [​IMG]
     
  11. exoray

    exoray Master Member

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    As long as you use a regulated power supply the currentof the supply isn't important, the circuit will draw what it needs... The voltage is more important in this instance and the resisitors will need to be changed depending on the wall warts voltage rating, but a 250mA or 1000mA wall wart won't make a difference...

    In series wiring it makes no difference, it's mostly a matter of preference...
     
  12. OPPI

    OPPI Well-Known Member

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    So as far as wall warts go just get one that says 9volts right? As volts is the important thing. I am looking at the ones at radio shack, for about 10-15 bucks each. These should be ok

    thanks everyone for all the help. One of these days i will learn to figure it out on my own, and actually trust what i come up with

    dan
     
  13. WookieeGunner

    WookieeGunner Well-Known Member

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    Not exactly. You will want to get a 9V DC power supply. Power supplies can be either DC (the current flows one way), or AC (the current flows back and forth (this is what actually comes out of the wall). If you get a 9V AC it will work, but it will flicker at 60 times per second (assuming US, I believe Europe uses 50Hz power instead of 60)

    LEDs have the property where they will only allow current to pass one way, so if you use AC power, it will turn on when the power is running the correct way and turn off when it is running the wrong way.
     
  14. exoray

    exoray Master Member

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    True, but for all practical purposes the human eye won't notice the flicker also when you are dealing with cheapy DC wall warts they are generally nothing more then rectified AC versions with a few diodes to rectify the current to DC and maybe a capacitor to smooth things out... So you really are not getting away from the flicker...

    Yep, they will work fine as stated above it would be better to get a DC version just because...
     

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