# LED Resistors

Discussion in 'General Electronics Chat' started by englewood, Jul 24, 2015.

1. ### englewood Thread Starter New Member

Jul 24, 2015
25
0
Oaky I have 11 LEDS.
I have 9V supply
I have 1V Drop Across a Transistor

I have 11 LEDS
All LEDS have a fV of 3.2V
I want all LEDS run around 20mA
To make a letter

I want one current resistor for the 10 LEDS
I want one current resistor for the 1 LED

I have 5 STRINGS OF 2 LEDS in para to make the 10.

8-3.2/0.02/240/10 = 24 OHMS (27)

Will one 27OHM resistor be enough for 10 LEDS?

8-3.2/0.02 = 240 OHMS SINGLE LED

SEE ATTCHED IMAGE

File size:
57.3 KB
Views:
18

Jul 24, 2015
25
0
27 OHM

3. ### mcgyvr AAC Fanatic!

Oct 15, 2009
5,082
1,104
Nope... LEDs in parallel without their own current limiting is a NO-NO as they do not share current nicely and will eventually die..
Since your LEDs have a Vf of 3.2 you can do 5 strings of 2 on each string with a resistor for each string.
Then the single LED with its own resistor too.

150 ohm 1/4W resistor for each series string.
330 ohm 1/4W for the single LED

cmartinez likes this.
4. ### englewood Thread Starter New Member

Jul 24, 2015
25
0
So then

8v - 3.2 / 0.02 = 240 / 2 = 120 (Closet highest resistor = 150) One for each string of 2 LEDS.

Why 330 ohm for the single? :-S

8v - 3.2 / 0.02 = 240 (Closet Highest Res = 270)

Apr 30, 2011
1,426
364
For the calculation of the limiting resistor, you keep showing 8V - 3.2V but series strings of 2 would be 8V - ( 2 * 3.2V ) = 8V - 6.4V = 1.6V.

1.6V / 0.02A = 80Ω. The preferred value would be 82Ω.

For the single LED, ( 8V - 3.2V ) / 0.02A = 240Ω.

Your total current is going to be 120mA which is not going to be sustainable for long with a standard 9V battery. The brightness of the LEDs will drop continually over a period of approximately 4 hours.

Last edited: Jul 24, 2015
6. ### mcgyvr AAC Fanatic!

Oct 15, 2009
5,082
1,104
Oh sorry.. I used 9V.. forgot about the 1v drop part..

7. ### englewood Thread Starter New Member

Jul 24, 2015
25
0
So whos right ha. Im confused now.

That does make sense:

8 - 6.4 (2x3.2 VD) / 0.02 = 82

8 - 3.2 / 0.02 = 240

Its running from a 9V supply .

Apr 30, 2011
1,426
364
I'm right, of course.

The BU806 data sheets I've seen don't show Vce for Ic less than 500mA so your 8V figure may be a little bit off, but if you supply the circuit with a single battery, the voltage drop at the source will be the most noticeable factor. If your supply is sustainable without drop (not a battery), no problem.

9. ### englewood Thread Starter New Member

Jul 24, 2015
25
0
Haha, I think you are just done the maths myself.

Oh sorry, ive changed that its a BC517 Darlington

10. ### englewood Thread Starter New Member

Jul 24, 2015
25
0
How can I make a 4017, skip Q2 -Q8?

Apr 30, 2011
1,426
364
So your transistor voltage drop will be 0.8V now. It probably won't make much difference but you should run the numbers to be sure.

Apr 30, 2011
1,426
364
Now you're hijacking your own thread with this off-topic post. You can't "skip" any of the states on the counter but you can decode the output or shorten the sequence by resetting before it completes all ten counts.

13. ### englewood Thread Starter New Member

Jul 24, 2015
25
0
Hold on,

Why does the calc say this for 5 x 2 LEDS?

Oh sorry, I want to do this should I make a new thread?

8 - 6.4 / 0.02 = 80 / 2 cause there 2 LEDS = 40 OHMS Makes Sense.

File size:
69.8 KB
Views:
6
14. ### djsfantasi AAC Fanatic!

Apr 11, 2010
3,496
1,253
Where do you get the 6.4 voltage drop? Parallel circuits have the same voltage across each leg. If the LED Vf is 3.2 V, the voltage drop across the two legs is also 3.2V.

15. ### englewood Thread Starter New Member

Jul 24, 2015
25
0
Each LED has a 3.2 VD, 2 x 3.2 = 6.4.

8-6.4VD / 0.04mA = 40 ( Next res 47 OHM)

2 LEDS WITH 3.2 VD
2 LEDS WITH 0.02mA

This is want the end result is.

14.63mA through 270 OHM Res
14.55mA through 47 OHM Res cant get any closer than the .

The calculations are correct, I have checked the current flow in Proteus through each resistor and its almost the same

Try this

http://www.quickar.com/bestledcalc.php?session=

Last edited: Jul 24, 2015
16. ### djsfantasi AAC Fanatic!

Apr 11, 2010
3,496
1,253
You add the voltage drops in a SERIES circuit. Your LEDs are in a PARALLEL circuit where the voltage is the same in each leg. In your case 3.2V.

The calculations are correct; you are using the wrong equation.

To add to the confusion, LEDs in a parallel circuit need a resistor in each leg...

cmartinez likes this.
17. ### englewood Thread Starter New Member

Jul 24, 2015
25
0
Oh idiot,

They are in Series in 2 LEDS in my Schematic. Doh!!!!!!

But why does the software show the correct mA flowing through each resistor?

• ###### euro.pdf
File size:
143.3 KB
Views:
4
Last edited: Jul 24, 2015
18. ### englewood Thread Starter New Member

Jul 24, 2015
25
0
I think ive messed up on the first Letter DOH

19. ### englewood Thread Starter New Member

Jul 24, 2015
25
0
All sorted , I think.

Last edited: Jul 25, 2015