# L.E.D's in parallel

Discussion in 'The Projects Forum' started by Neil Groves, Aug 3, 2012.

1. ### Neil Groves Thread Starter Member

Sep 14, 2011
125
3
i have a thorny problem in as much as i am trying to run 2 leds in parallel but the two leds in question have different ratings, one is a bog standard green and the other a high brightness Blue both having different voltage/current ratings, what is the best way to wire these guys up so they both come together when power is applied?

Neil.

2. ### #12 Expert

Nov 30, 2010
18,082
9,646
Use a different resistor in series with each LED.

3. ### wmodavis Well-Known Member

Oct 23, 2010
739
151
+1 to #12's comment. In other words you'll have to use a combination series/parallel circuit not actually having the LEDs directly in parallel with each other.

What voltage are you using to supply them?

4. ### Neil Groves Thread Starter Member

Sep 14, 2011
125
3
one takes 3v the other 1.7 so yes thats a good idea, i can do a different value resistor for each led,

thanks.

Neil.

Mar 24, 2008
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6. ### #12 Expert

Nov 30, 2010
18,082
9,646
You can either tell us what the supply voltage is or work it out from Bills tutorial.

7. ### Neil Groves Thread Starter Member

Sep 14, 2011
125
3
i am having trouble calculating the resistor values, i am using a supply of 9v, one led needs 1.7v at 20mA, the other needs 3.0V at 30mA, i guess i could use 20mA for both though since the blue led is VERY bright, so if i use 20mA for both, how do i calculate the voltage drop across the resistor in each case?

Neil.

8. ### Neil Groves Thread Starter Member

Sep 14, 2011
125
3
I just took a gander at that link Bill, very interesting read, i will find time next week to experiment i hope.

thankyou.

Can someone give me the basics to get me started please?

Neil.

9. ### KJ6EAD Senior Member

Apr 30, 2011
1,570
443
You have plenty of voltage to run both LEDs in series with one resistor. A 270Ω resistor should put the current around 16-18mA.

Last edited: Aug 4, 2012
10. ### #12 Expert

Nov 30, 2010
18,082
9,646
(9-1.7)/.02 = any resistor with more than 365 ohms (390)
(9-3)/.03 = any resistor with more than 200 ohms (220)

or,

(9-1.7-2.7)/.02 = any resistance more than 230 ohms (240)
to give both of them .0192A
give or take the tolerance of the voltage ratings on the LEDs and the accuracy of the 9 volt supply.

Edited to comply with KJ6 in post#11

Last edited: Aug 4, 2012
11. ### KJ6EAD Senior Member

Apr 30, 2011
1,570
443
The Vf of the blue LED will be less than 3 at 20mA, probably by 2 or 3 tenths which makes 220Ω potentially dangerous to the health of the green LED. 240Ω is the lowest I'd risk without a test.

#12 likes this.
12. ### Neil Groves Thread Starter Member

Sep 14, 2011
125
3
(9-1.7-2.7)/.02 = any resistance more than 230 ohms (240)
to give both of them .0192A

doesn't that equate to 37.5mA each which is way too much for the green led?

9v/240R=37.5.

Neil.

13. ### #12 Expert

Nov 30, 2010
18,082
9,646
Not if you put the LEDs in series as described by KJ6 in post # 9

"i guess i could use 20mA for both though since the blue led is VERY bright"

Putting both LEDs in series and allowing 20 ma for both of them allows the use of only 1 resistor.

Last edited: Aug 4, 2012
14. ### Neil Groves Thread Starter Member

Sep 14, 2011
125
3
oh yes you are right, i misread the equation. :x

well anyway i got the resistance thing right and the leds are working ok.

thanks.

Neil.

Last edited: Aug 5, 2012