LED + Resistor question!!

[Docaroo]

Hi guys,

I'm using the LED calculator on Metku mods site to figure out what resistor to use for some LEDs and I need a quick check...

I have 12 LEDs wired in PARALLEL as they are going to be running from 3 AA or AAA batteries (1.5v batt times 3 = 4.5v total power supply).

The LEDs have a forward voltage of 3.2 - 3.6v so I guess suppling 3.5 v to them would do me.

Now here's the question, do I use a resistor to take the current down to 3.5v as if there were only 1 LED? This would require a 26ohm resistor.

OR do I believe what the LED calculator says and use a 2.2 ohm resister - the figure changes depending on how many LEDs are in the parallel connection.

This doesn't make sense to me as I thought the voltage would be shared across all the LEDs meaning you would need the 26 ohm?

Also, if the calculator is correct, can I get away with having no resistor since it's such a tiny value (2.2 ohm)?

thanks in advance guys!

 

[bitslice]

Vsource= 4.5
Vfwd = 3.5
led forward current = 20mA

load resistor = 56ohm 1/4 watt (req 12 off)


(this is way too early for me...)
current doesn't get shared over all leds, one would suck it all,
each led needs a resistor

 

[Docaroo]

but in parallel wiring this is different!

I'm putting 50mA through the LEDs so when I use the metku calculator I get this:

Vsource = 4.5v
Vfwd = 3.5v
current = 50mA
No. of LEDs = 12

resistor = 1.8 Ohm

However, when I use a different number of LED's:

Vsource = 4.5v
Vfwd = 3.5v
current = 50mA
No. of LEDs = 1


resistor = 22 Ohm

another calculator said use 26 Ohm! This is the part confusing me...

Do I need one 22 ohm at the start of the circuit, one 1.8 ohm at the start of the circuit or one 1.8 ohm before ever LED in PARALLEL wiring?? :S :S :S

eek!

 

[bitslice]

Vsource= 4.5
Vfwd = 3.5
led forward current = 50mA

load resistor = 22ohm 1/4 watt (req 12 off)


it should look like this:

+Ve ---///----|>|---- -Ve
+Ve ---///----|>|---- -Ve
+Ve ---///----|>|---- -Ve
+Ve ---///----|>|---- -Ve
....
.... etc
....
+Ve ---///----|>|---- -Ve



post link to your calc, it looks wrong



If you had one LED, it would still use the same value resistor

V=I x R
R= V/I

4.5-3.5/0.05 = ~ 22ohm



.

 

[Docaroo]

here's the calculator:

http://metku.net/?sect=view&n=1&path=mods/ledcalc/index_eng

The LEDs in parallel one is at the bottom ...

so I would need a resistor before every LED in the parallel loop? damn!

 

[Docaroo]

thanks for the explanation though, thinking about it - it makes sense to need a resistor before ever LED, as you need to take the voltage down to 3.5 every time you have an LED there...

Is there a way to take the overall voltage down so you only need 1 "master" resistor??

thanks again

 

[bitslice]

ah, I see, (stick + wrong end)

sorry, the one resistor thing is not normally a circuit that's used.
It will work but the LED's may have uneven brightness.


If the LEDs require slightly different voltages only the lowest voltage LED will light and it may be destroyed by the larger current flowing through it


btw,
the forward current of 50mA seems a tad high, normally they are about 20-30mA.?
...not a problem, just something to double check

hth
.

.

 

[BenST]

Yep, it's not good practice to share one resistor with multiple LED's as you can never guarentee that all LED's will use the same amount of current, resulting in brighter/duller LED's.

In theory if you wanted one 'master' resistor you would simply add up the current of all LED's, take the voltage that needs to be dropped across the resistor and divide, i.e 12x30mA = 0.36A , 4.5 - 3.5 = 1v , 1/0.36A = 2.7ohms. Then you would need to work out the power dissipated across the resistor i.e V2 / R , 1/2.7 = 0.37W. You would need a 2.7ohm 1/2w or greater resistor.

 

[Docaroo]

so the calculator is essentially correct?

all the LEDs are the same make/model/type/fwrd voltage (as only LEDs of the same voltage can be used in paralllel as you said)...


So if I used a "Master" resistor of around 3 Ohms I'd be ok?

 

[bitslice]

yep.

about 1 watt 1R8

(you can put 4 x 1/4W resistors in parallel if you quadruple their resistance)

4 x 6R8