LED parallel circuits and how they function?

AI Thread Summary
When adding LEDs to an iPhone 4s with a 2.8V supply, it's essential to connect them in parallel, as the voltage is insufficient for a series arrangement. Each LED, such as red, green, and blue, should have its own resistor to ensure consistent brightness and prevent variations in intensity. The power dissipation in the resistors must be calculated to avoid exceeding their specifications, ideally keeping it below one-third of the maximum rating. For example, a 100Ω resistor can be used to achieve a desired current of 10mA, resulting in a power dissipation of 0.01W. Proper resistor selection and configuration are crucial for optimal LED performance.
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First off, I'm looking at adding in a few LED's into my iPhone 4s (I'm very limited on space, but there is a pretty open area.) I have found a connection point that gives off 2.8v and is commonly used for lighting up LED's with the screen.

I plan to add in a red, green and blue 0603 (or even 0402) SMD LED in a parallel set up, but I don't know how the voltage would work out. Should I expect for all 3 of the lights to use different resistors and that they won't be affected by the circuit? (Using this to figure out resistance: http://www.muzique.com/schem/led.htm )

I've done lots of small scale soldering before so SMD is no big deal to me, but I would assume that I should also use SMD resistors. What exactly should I pay attention to in the resistors?(I'm new here :D)
 
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You have only 2.8V, you don't have enough voltage to put all three is series, putting two in series and one stand alone, then using resistor might end it have one LED giving different light intensity unless you adjust the resistor to give even intensity. The easiest way is to have three individual LED/resistor circuits. With this, you want to do a quick calculation on the power dissipation in the resistor. Make sure the power does not exceed the specification of the 0603 or the 0402 SMD resistor.
 
yungman said:
You have only 2.8V, you don't have enough voltage to put all three is series, putting two in series and one stand alone, then using resistor might end it have one LED giving different light intensity unless you adjust the resistor to give even intensity. The easiest way is to have three individual LED/resistor circuits. With this, you want to do a quick calculation on the power dissipation in the resistor. Make sure the power does not exceed the specification of the 0603 or the 0402 SMD resistor.

I planned on putting them parallel, I was able to realize I did not have enough voltage for a series arrangement. I didn't think about putting the blue one on one circuit and then the red and green on another, but the red and green together would be about 3.6V needed to light both.

What exactly is the power dissipation? I haven't gone too far into electrical engineering to fully understand the terms.
 
Say your LED drop 1.8V as you indicated, you have a 2.8V supply, so the resistor in series is going to drop 1V. Say you want to draw 10mA through the LED to give you the brightness you want, so you put a 100Ω resistor in series with the LED. The power dissipated in the resistor is \;W=IV\; where I=10mA and V=1V. So the power dissipation is 0.01W. Use this as the guide line. In your case, you should keep the power below the specified maximum power dissipation allow for the resistor used. Preferably 1/3 or less the specified power of the resistor.
 
yungman said:
Say your LED drop 1.8V as you indicated, you have a 2.8V supply, so the resistor in series is going to drop 1V. Say you want to draw 10mA through the LED to give you the brightness you want, so you put a 100Ω resistor in series with the LED. The power dissipated in the resistor is \;W=IV\; where I=10mA and V=1V. So the power dissipation is 0.01W. Use this as the guide line. In your case, you should keep the power below the specified maximum power dissipation allow for the resistor used. Preferably 1/3 or less the specified power of the resistor.


Alright, thank you very much for your help :).
 
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