Parallel LED Circuit

Discussion in 'LEDs and Optoelectronics' started by mjc, Jan 30, 2014.

1. mjc

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Jan 30, 2014
Hi Everybody,
I'm a newbe to this forum, Electrician by trade but dabble with Electronics as a hobby. I am making LED post cap lights for the posts on my deck. I intend to put 3 LEDs next to each other on the side of a post cap. The LEDs are 5mm white thru-hole type with a Forward voltage of 3.2vdc and a Forward Current of 20ma each. My supply will be 10.28vdc and I will be wiring each post in parallel with each other to maintain a constant voltage. Lets use just two posts (19 total) one post will utilize 6 LEDs and post two will utilize 9 LEDs. 15 LEDs exactly can be Used. According to ohms law using 3 LEDs with a 39 ohm resistor in series with them will provide an array of 15 LEDs. My question is: can I put a Parallel array of 6 LEDs in PARALLEL with another parallel array of 9 LEDs and so on for 19 post caps? How will the current from one array effect the other? and what would the current total for the circuit be? Thanks in Advance for the answer.

2. (*steve*)¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥdModerator

25,191
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Jan 21, 2010
You should not place LEDs in parallel unless they each have their own resistor, or they are carefully matched and maintained at the same junction temperature. The only real way to do the latter is by having multiple LEDs on a die, or the dies in close thermal proximity (in a single package or COB). Separate 5mm LEDs do not fall into either category.

3. mjc

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Jan 30, 2014
Thanks Steve – for the clarification of the 1/29/14 post.
Same Thread with the LEDs, but let’s try this: Putting together nine 5mm white thru hole LEDs.
There specs are Vf=3.2vdc, If=20ma My supply voltage is 16.4vdc
Three LEDs or an array will be in series with each other and a current limiting resistor of 340 ohms.
There will be three arrays in parallel with each other to form a Total Resistance of 113.3ohms.
Each series Array is limited to 20ma the total current used by all three arrays is 60ma.
Using Ohms Law Et/Rt=It 16.4/113.3=.1447a or 144ma.
Question: If my circuit only Draws 60ma from the power supply and the CAlCULATED Circuit current=144ma, (144ma-60ma = 84ma) Where does the additional 84ma go to?

4. BobK

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Jan 5, 2010
Your calculation is wrong. The voltage across the resistors is not the full voltage, it is the fulll voltage minus the forward drops of the 3 LEDs. So the corrected calcuation is:

I = (16.4 - 3*3.2) / 113.3 = 60mA

Bob

5. mjc

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Jan 30, 2014
Thanks, I see what I was doing. It makes so much more sense now. Just have to remember that.  