Simple LED question (definately leading to more)

[Mike - K8LH]

Poor Angie1199, subjected Ohm's Law and Kirchoff's Law within 24 hours (grin)...

Regards, Mike

 

[ljcox]

Ok, so graph 1 and 2 are from the datasheet.

Graph one shows Fv at 15mA. So resistor required is (12 – 4 * 1.92) / 0.015 for red which is 288 ohm. Resistor for green is (12 – 4 * 2.1) / 0.015 which is 240 ohm. Excellent!

So in the circuit there would be a common resistor of 240 ohm which would be used by all LEDs and each red would require an additional 12 ohm resistor run in parallel to the red LED so that particular red LED is then using 15mA when lit. As Audio wrote, this is wrong. You must have miscalculated somewhere.

This would then give the Luminous intensity of each LED colour as shown in Graph 2 which, at 15mA, are virtually the same. As Audio wrote, this is not necessarily correct. You will have to experiment with the LEDs to see how bright they are for various currents.


Then, as was said earlier, another resistor is placed in parallel to the red LEDs as the voltage is different therefore it has to be allowed for. This is the circuit diagram I did in visio with the extra resistors in parallel. I think this is where you meant them to go. Yes, we just have to find the right resistance value.

This is a bit tricky, so I'll post a spread sheet later

 

[angie1199]

I don't think that Luminous Intensity is the same as Visual Intensity because your vision is more sensitive to green. Therefore even though the green operates with less current it might appear brighter than the red.

Oh. That's now why I have to test.... Maplins this morning then. Thanks AG

Graph one shows Fv at 15mA. So resistor required is (12 – 4 * 1.92) / 0.015 for red which is 288 <a href="#">ohm</a>. Resistor for green is (12 – 4 * 2.1) / 0.015 which is 240 <a href="#">ohm</a>. Excellent! See, nearly that elusive Eureka moment

So in the circuit there would be a common resistor of 240 <a href="#">ohm</a> which would be used by all LEDs and each red would require an additional 12 <a href="#">ohm</a> resistor run in parallel to the red LED so that particular red LED is then using 15mA when lit. As Audio wrote, this is wrong. You must have miscalculated somewhere. Oh boy. here I go again. Doesn't electricity like logic? if red needs 288 Ohm for 4, and green needs 240 for 4 to both have 15mA then that's 48 difference which, divided by the 4 red LEDs is 12 per LED.

This would then give the Luminous intensity of each LED colour as shown in Graph 2 which, at 15mA, are virtually the same. As Audio wrote, this is not necessarily correct. You will have to experiment with the LEDs to see how bright they are for various currents. Ok, so Maplins tomorrow.

Then, as was said earlier, another resistor is placed in parallel to the red LEDs as the voltage is different therefore it has to be allowed for. This is the circuit diagram I did in visio with the extra resistors in parallel. I think this is where you meant them to go. Yes, we just have to find the right resistance value. Ya-hey... Eur...

This is a bit tricky, so I'll post a spread sheet later. You're not kidding! :?

 

[audioguru]

You don't need to calculate all this stuff. Just connect four LEDs in series with a 220 ohm or 240 ohm resistor and look at them. They'll light.

Maplin's cheap Chinese LEDs are probably all different anyway.
I bought a bag of 100 Fairchild LEDs. I've used and tested about 75 of them and they are closely the same. Of the cheap Chinese LEDs that work (many didn't) that I bought locally, they were all different. :cry:

 

[ljcox]

Here it is. The resistor across the red LED (assuming you need one) (ie. one per red LED) has to bypass the excess current.

You can vary the parameters in the spread sheet to see the effects.

 

[angie1199]

Being a newbie to small electrics I hadn't a clue about any of this. I appreciate it would be easy to plug the things in and look but I do like to know what's going on and why. I hate things that 'just do'.

Between all the replies I've got on here from AG and LJCox I think I can safely say I know more than I did three/four days ago and for that I'm greatful. If all goes bang I'll be back soon, but if not thanks for all your help.

Angie