35 replies to this topic

[lowlancer]

I want to run a red, blue and green led off the same 9v power source. Given that they are all different voltage drop and mA ratings, I am going to run each led and suitable resistor as seperate assemblies, then connect the assemblies in parallel.

My question is regarding where the + of each assembly connects to... Does it run back to the power source, or simply from the + leg of the LED assembly preceding it?

LED ratings are

1 x red @ 2.1v drop and 50mA
1 x Blue @ 3.2v and 30mA
1 x green @ 3.5v and 30mA

[lowlancer]

May I add I am in a massive hurry so help is appreciated

[k-trips]

I have 2 LED's running in parallel both from the battery but each with a seperate switch (ones a toggle and the other is just the ignition switch) - when I have one lit up with the ignition switch and I turn the toggle switch on for the other - the first LED goes out.. (this actually works perfectly for my setup - albeit by fluke, not planning)

I'm not an electrical wizard by any means but I assume that if you do the same thing then you will have the same result.

[Evan]

if its in parallel there wil be nothing preceeding in. effectively they will all come from the + side of the battery (+ side of LEDs contacts + side of battery, negative side of LED contacts negative side of battery) . run one wire off the + side and peel each of the 3 led/resistors off that, do the same to the negative side

running one led after another is in series not parallel.

does that help?

EDITED ABOVE

[k-trips]

Have a re-read, Eliot said he is going to connect the +ve legs together not +ve to -ve..

[lowlancer]

Nah KK I think he's right, so they all pick up direct from the positive and negative seperately? The reason I'm doing them in parallel is because in series they'd have too much voltage drop for the power source.


So wiring them all up in parallel will keep them in their own "curcuit, unaffected by the voltage and current demands of the others, right?

[Evan]

kk = ????
yea i should draw you a pic but i dont have time haha

yea the voltage across all 3 seperate parts will be te same but the current will divide among the 3 circuits (I think from memory thats how it works)

[Ned]

Do you really want/need the red one to run brighter? or just all the same, at 30mA?

[k-trips]

Am I losing the plot??

top is series - bottom is parallel




The way I read Eliot is you want to connect the positive legs of thr LED's together as per the bottom diagram..

[slacker.cam]

This is a horrible 2 sec paint drawing so hopefully it makes sense.

Each LED and resistor combo is connected to the battery independently. You can view the circuit as 3 seperate circuits that just happened to be connected to the same battery.

[Evan]

yea both of those was what i was trying to say haha.

[slacker.cam]

KK, yeah you're pretty much on to it now. The bottom one is the correct technique. Except you need to add the current limiting resistors inline with each LED before it joins back up to the main wire.

You could do it like your top drawing but due to the differing specification of each LED you'll end up with different brightnesses. Ie. different colour LEDs need different amounts of current to achieve the same brightness.

[Ned]

But red LEDs are gay as they appear brighter regardless as the yare easier to see by humans

Here, pick one


If you are using a battery, don't use option 1 (or any of them really)

[Ned]

or read this; http://www.pcbheaven... ... g_methods/ if you want to do it properly...

[k-trips]

KK, yeah you're pretty much on to it now. The bottom one is the correct technique. Except you need to add the current limiting resistors inline with each LED before it joins back up to the main wire.

Yeah the bottom one was what I was trying to explain in my first post - I suck at explaining myself well.. sorry guys..

[lowlancer]

Knew Ned would suss it.

I'm using a 9v battery but it is only for a couple minutes max. I'm using number 3 in any case. And would number 1 not work given the voltage drop would be too much?

[Ned]

If your voltages change, so if you use a battery or are planning on possibly using a couple different supplies, then you have to use a constant current driver. The easiest way to do that is with an LM317 (you can buy these from jaycar/rs and some dick smith stores) and a resistor. Then you can use a wide range of supplies, batteries.

Number 1 would work, but if your battery was at 9.6V instead of 9V, you would have 100mA through the red led and 60 through the green ones.

Number 3 will be fine with the current resistor values. If your battery is at 10V, you will be running 32.5mA through the green or blue led instead of 30mA... aka nobody cares.

[lowlancer]

Sooooo, will this not work with a battery?

[Ned]

Will be fine, just use setup number 3 but maybe use a slightly larger resistor value for each as your battery will be above 9V.

Or just use roughly the correct value (can go up or down) as you are only running it for 3 minutes and they will happily run at slightly higher current for a few minutes

[lowlancer]

Sweet, yeah am using 220r but still 150r. Thanks for the help dudes