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2 Different LEDs on One Circuit? Answered

If I have a Red Led that takes 2v and a green led that takes 3.2 volts...then can i use 5.2 volts to power them?? Will that work without giving too much juice to the Red LED?? THANKS!

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Joe426 (author)2007-10-10

SOMEONE HELP PLEASE!?!?! OR SHOW ME A DIAGRAM?? THANKS!!!

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Patrick Pending (author)Joe4262007-10-10

Why, do you not like the help you have been given? Pat. Pending

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Joe426 (author)Patrick Pending2007-10-10

I just didn't understand it fully. if i wire it like this - will it work? THANKS!

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Joe426 (author)Joe4262007-10-10

or would i need another resistor between the power and 3v led?

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Patrick Pending (author)Joe4262007-10-11

Yes another resistor between power and the 3V LED would be better. You need to calculate the appropriate values for each of these resistors. By including a series resistor it helps protect the LED from excess current flowing due to variations in the power source.

Resitance = V(source)-V(LED) / I(LED)

Do you know the forward current for each of the LEDs?

Pat. Pending

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Joe426 (author)Patrick Pending2007-10-11

one is 3.5 volts, the other is 2 volts. I tried wiring them in a series with resistors - they haven't burned out and seem to be as bright as ever.

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Patrick Pending (author)Joe4262007-10-11

I'm not sure if you misread my question, as you have given me the forward voltage drop of the LEDs instead of the forward current. How have you calculated the value of resistor required? Pat. Pending

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Joe426 (author)Patrick Pending2007-10-11

using the metku mod calculator - i just added 3.5 volts +2 volts = 5 volts. then i said the power source was 9 volts and then calculated a resistor value. so - I said i had one 5.5v led in a circuit powered by 9 volts...and it spit out a resistance.

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Patrick Pending (author)Joe4262007-10-11

It looks like you used 20mA suggested by the calculator for the forward current. So it looks like we have come full circle back to my original post, where I said it would work fine as long as they have the same current requirement. Cheers, Pat. Pending

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Joe426 (author)Patrick Pending2007-10-11

thx for the paitence and help!

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Patrick Pending (author)Joe4262007-10-11

Anytime, just ask. Pat. Pending

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Joe426 (author)Joe4262007-10-11

the power source is 9v battery.

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Patrick Pending (author)Joe4262007-10-10

Assuming the power-box is a stable 3 Volts and the resistor is calculated to drop 1Volt at the specified forward current of the 2Volt LED, then yes it will work. However, it does have some design shortcomings. I will have a bash at explaining them later - I have to sleep now |-I Pat. Pending

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whatsisface (author)2007-10-10

You could always just pull the old "Potential Divider" Trick. Just need to get the right value resistors.

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Goodhart (author)2007-10-10

In this case, the LED's in parallel would be receiving the same wattage and so calculating for the resistance would be fairly straightforward, but in series, it could be "touch and go", if your pwr in varies at all.

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Patrick Pending (author)2007-10-10

If the two LEDs have the same forward current at those particular voltages it would be OK. If they have significantly different current requirements then you may be overdriving one whilst underdriving the other (i.e., one would be bright while the other dim). If they have different current requirements then a parallel connection with individual current limiting resistors would be best solution. If you connect them in series then a current limiting resistor will help protect them. However , if you are sure that there will be no significant voltage rise in the source, then you can drive them directly without a resistor. Cheers, Pat. Pending

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schorhr (author)2007-10-10

Led-circuit calculator, for leds, leds in series, leds in parallel, and leds everywhere

http://www.hebeiltd.com.cn/?p=zz.led.resistor.calculator

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CameronSS (author)2007-10-09

But in parallel you can hook up separate resistors for each LED... How would that cause problems? I'm confused...

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Goodhart (author)CameronSS2007-10-09

Sorry, I am sleepy and should not be making statements when I am tired. I realized as soon as I posted that I was somewhat in error, as series situation may cause worse problems in this case. In parallel, as Cameron posts, one would just figure the resistance needed for that LED, and add the resistor to that line. One for each leg. 'Tis around 0 dark 23 here....good night.

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Joe426 (author)Goodhart2007-10-09

can i see a diagram of this please??!??!!??! THANKS!

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