# LED Parallel with with independent switches

Discussion in 'The Projects Forum' started by CeltMagic, Mar 2, 2014.

1. ### CeltMagic Thread Starter New Member

Mar 2, 2014
5
0
Hello All

I haven't had any experience in circuits since school so I've come looking for help, sorry if my questions have been answered, I have searched but maybe my I have missed the answer in my lack of knowledge.

I would like to know if the diagram supplied would work to operate each LED independently in various arrays of on/off states.
Since I may only want a single LED on at a time, my initial thought is that each resistor should have the same resistance value as it would in a single LED series circuit.
Please can you let me know if any of this is correct or if there is a better solution i.e; is there such a thing as a variable resistor that will let me to use just one resistor allowing only the minimum current required but also allowing more when the circuit requires it?

If it makes a difference I imagine making a single enclosure, housing the power, resistors and switches supplying the LED's at various distances (some long). The project is to supply lighting for my son's collection of Lego buildings etc. for which I intend to make a large play table for the Lego and fix the circuits.

Many thanks for reading, the big kid inside me is taking over

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2. ### bertz Active Member

Nov 11, 2013
327
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That will work quite nicely. Use a 330 ohm, 1/4 watt resistor in each leg.

(9 - 2)/ 330 = 21 mA

I think you have an error in your schematic, I doubt that the LED is drawing 0.2 amps (200 mA).

3. ### CeltMagic Thread Starter New Member

Mar 2, 2014
5
0
Thanks for the quick response.

So just to confirm I need the full required resistance per leg of the circuit as if it were a single LED series?

If I put 2 LED's into a leg, adjust the resistance to that leg whilst keeping the single LED legs with a 330 ohm, 1/4 watt resistor?

Thanks again

4. ### bertz Active Member

Nov 11, 2013
327
60
If you put 2 LEDs into a leg, change the resistor to 270 ohms.

(9 - 4) / 270 = 19 mA

You want to keep the current in each leg around 20 mA.

5. ### MrChips Moderator

Oct 2, 2009
18,207
5,722
A typical 9V battery will not last very long for lighting LEDs. Use a wall adapter instead it that is an option.

6. ### CeltMagic Thread Starter New Member

Mar 2, 2014
5
0
Thanks bertz for helping me get my head around this resistance thing, I'm fairly confident with it now (the 0.2 Amps was me being a bit green, I did mean that to be 20 mA)

MrChips, I haven't thought outside the battery yet but it's more than an option, the 4 LED array was more an example of what I was going to do and to check my thoughts on resistance were correct.
Could you let me know or even link to what kit I would need use?
I also have a high capacity solar battery that has several connections for laptops/phones and usb devices etc that I use when I can't get to a mains supply so if I had the option to use that as well it would be a bonus.

Regards

7. ### #12 Expert

Nov 30, 2010
18,076
9,678
There is no law that says you have to give the LEDs the highest current they are rated for. 10 ma will work nicely in many situations.

You might put (3) LEDs per string at 10 ma to make the battery last longer.
(9V - 6V worth of LEDS)/.01 amp = 300 ohms
(same as above)/.02 amps is 150 ohms.
Use something between 150 and 300 ohms.
As the battery gets tired, the LEDs will get dimmer.

Now you know the math.

8. ### bertz Active Member

Nov 11, 2013
327
60
Here is a nice little power supply that works of household voltage. Of course you would have to adjust the resistors since the supply voltage is 12V. You can also put two LEDs and a resistor in each branch. Just remember that the current in each branch is a function of the voltage drop across the LEDs plus the current limiting resistor. The total current draw will be the sum of the currents in each branch.

Have fun!

http://www.ebay.com/itm/12V-1A-Powe...837?pt=LH_DefaultDomain_0&hash=item2ecb387695

9. ### CeltMagic Thread Starter New Member

Mar 2, 2014
5
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Cheers bertz, you're a star.

Its true, the more I learn the less I realize I know. I'm beginning to feel like a right pain with these questions, sorry.

So essentially I can power up to 12v per branch of the circuit allowing for the voltage drop (if the LED was 2v then in theory I would not be able to add 6 due to the voltage drop but 5 could be possible).
Following that theory I would be able to add as many 5 LED (12v) branches providing the total current of all the LED's does not exceed the 1A rating of the Power supply?

Obviously I'll change the resistor ratings but should I also change the watt rating or is 1/4 enough.

Also found these to save cutting the power cable

http://www.ebay.co.uk/itm/10-x-Holl...D-Strip-Male-Female-/200944676429?refid=store

10. ### bertz Active Member

Nov 11, 2013
327
60
Remember, in a series circuit the current is the same at all points in that series branch. Let's analyze the following circuit. It has three parallel branches. Branch 1 has 3 LEDs, Branch 2 has 2 LEDs and branch 3 has 1 LED. Furthermore, let us assume a forward voltage drop of 2 volts for each LED and 15 mA as a target current in each branch.

For Branch 1:
R = (12 - 6) / .015 = 400 ohms, but this is not a standard value so use 390 ohms.
I = (12 - 6) /390 = 15 mA
W = (12 - 6) x .015 = .09 watts, so 1/4 watt is good

See if you can complete the analysis for the other two branches. Note that the total current for the circuit would be 47 mA. By now you should have figured out that you could install 6 LEDs in a branch without any current limiting resistor.

Finally I have attached a typical LED datasheet. Don't assume that all LEDs are the same. You have to be aware of the maximum current the LED can tolerate as well as the forward voltage drop. The data sheet will also give you an indication of typical operating current. Data Sheets - don't leave home without them.

Good luck!

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11. ### CeltMagic Thread Starter New Member

Mar 2, 2014
5
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If this is wrong I'm going to use candles to light the Lego buildings

R = (12-4) / .015 = 533.3 (510)
I = (12-4) / 510 = 15.6 = 16 mA
W = (12-4) x .015 = 0.12

R = (12-2) / .015 = 666.6 (620)
I = (12-2) / 620 = 16 mA
W = (12-2) x .016 = 0.16

So I gather the total current for the whole circuit is the sum of I 15+16+16

Thanks for the time, it's much appreciated.

12. ### bertz Active Member

Nov 11, 2013
327
60
Well Done! Wasn't so bad was it?

You may or may not find the values of 510 and 620 ohms a bit hard to come by. If you have any difficulty finding them you can use 470 ohm and 680 ohm respectively.

13. ### #12 Expert

Nov 30, 2010
18,076
9,678
I think I found a mistake. This sentence will cause smoke if you don't qualify it with an explanation.

14. ### poolman966554 Member

Jan 22, 2014
38
0
6 leds are possible, if they were red or yellow, as they require 1.8v - 2.2v respectively. If OP wanted to use white, then 3.0-3.5v are typical. Then 6 would not be feasible.

Also, something that hasn't been mentioned is the voltage to amperage on the led.
Note it gives a variance of voltage in the datasheet. Thats because the led will draw less voltage as current is lower and more as current raises.

Ex: running white/blue/green leds at 10ma may run at 3.1v, vs the same led running at 20ma may run 3.4v. Might be something to factor, maybe not.

Edit: You should ALWAYS have a current limiting device! such as resistor, or constant current driver preferably

Last edited: Mar 3, 2014
15. ### #12 Expert

Nov 30, 2010
18,076
9,678
Right. It's the, "no resistor" part that worries me.
If you let an LED into the voltage cabinet without a resistor for a chaperon, they get drunk and start smoking.

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16. ### burger2227 Member

Feb 3, 2014
194
25
Well you need to know the LED voltage drop and current to figure the resistor so if the voltage fits or is less then a resistor may not be needed, just like any other load.

17. ### bertz Active Member

Nov 11, 2013
327
60
Don't leave home without them!

18. ### #12 Expert

Nov 30, 2010
18,076
9,678
True, but that's not the kind of thing you can just demonstrate with the usual math. You have to have each LED and measure their start voltage and how they are going to act under power. LEDs are so inconsistent about their breakover voltage that you often can't get the answer close enough without measuring the LEDs you have on hand.

19. ### burger2227 Member

Feb 3, 2014
194
25
Most of the time bright LED's are very forgiving. The red and yellow ones are the hardest to work with because they use less voltage normally and current ramps up pretty fast.

I use a Joule thief circuit to test LED's in and out of circuit. It does not harm any kind of LED's no matter what the voltage rating. Lower volt ones will light when in parallel with brighter ones but I can run as many as six at a time when comparing brightness for lighting.

I seldom use red or green ones anymore. But I just got some bright bipolar red/green 2 leg ones I'm going to use for tool charging indicators on the output of an Op Amp. They seem to get brighter and safer to mess with every year!

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