Wiring several leds

[HailToTheBlues]

Hey guys,
Just wanted to ask if it is possible to wire several leds to the same 3pdt, so that they lighten all at the same time whenever i turn the pedal on. If it is, how would i need wire them?

Best regards

[RobA]

It depends on how many you want and what their forward voltages are. If you want a small enough number that the total voltage drop is less than your supply voltage, you can wire them in series with one resistor. You'll need to change the resistor value to set the current to what you want to maintain the brightness. If you have too many to do this way, you can wire them in parallel with a resistor per LED. You could also do a combination of the two.

[HailToTheBlues]

well, i'm thinking of wiring 9x 3mm leds, so it may be too much to wire them in series

[GermanCdn]

I wired up my wife's Deathstar Jack-o-lantern with 30 LEDs, and I think I tapped out at series loops of 5 before I ran into problems.

If you're looking to do 9, I'd probably suggest doing three parallel loops of three.  But it's easy and cheap enough to do a dry run test on it.  Just don't forget your CLR.

[HailToTheBlues]

So, how would i wire them? If i get this straight, i would need to wire 3 clr in the dc jack, each connected to one led, that is connected to other 2 leds, and these 3 runs would go to the 3pdt, is that right?

[RobA]

So, how would i wire them? If i get this straight, i would need to wire 3 clr in the dc jack, each connected to one led, that is connected to other 2 leds, and these 3 runs would go to the 3pdt, is that right?

Yes that would work. Just to be clear, in each of the three runs you would have V+ (R +led- +led- +led-) switch_in switch_out ground, with what's inside () in parallel three times. 

[HailToTheBlues]

So, how would i wire them? If i get this straight, i would need to wire 3 clr in the dc jack, each connected to one led, that is connected to other 2 leds, and these 3 runs would go to the 3pdt, is that right?

Yes that would work. Just to be clear, in each of the three runs you would have V+ (R +led- +led- +led-) switch_in switch_out ground, with what's inside () in parallel three times.

Ok, thanks! And what about the value of the clr? I would need to use a different value for the clr, would'n I? What would be the right value for each 3 leds?

[GermanCdn]

So, how would i wire them? If i get this straight, i would need to wire 3 clr in the dc jack, each connected to one led, that is connected to other 2 leds, and these 3 runs would go to the 3pdt, is that right?

It would be easier if you have a piece of veroboard, mount your CLRs to the vero (all common 9V source, then stagger your resistors so they each terminate on different strips) then run your +ve to the parallel loops off of those strips, and have your loops terminate on a common ground strip on the vero, then run a wire from the common ground strip to your 3PDT.

[RobA]

Yes, you'll need to change the value. It's going to depend on the LED's that you use. You could compute it based on the current you have when you are using just one LED. V = I R ==> I = (V+ - Vf)/CLR. If all of the LED's are the same, then the new setup would have I' = (V+ - 3 Vf)/CLR' and you want the currents to be the same. So, I = I' ==> CLR' = CLR * (V+ - 3 Vf)/(V+ - Vf). Which for a 9V supply and a Vf of about 2.2V turns out to be about 1/3 of the original CLR. If you were using a 12V supply, it'd be closer to 1/2.

[slimtriggers]

I use THIS all the time for doing multiple LEDs.  Never done more than 5, though, so YMMV 

[thesameage]

I use THIS all the time for doing multiple LEDs.  Never done more than 5, though, so YMMV 

very cool!

[HailToTheBlues]

Yes, you'll need to change the value. It's going to depend on the LED's that you use. You could compute it based on the current you have when you are using just one LED. V = I R ==> I = (V+ - Vf)/CLR. If all of the LED's are the same, then the new setup would have I' = (V+ - 3 Vf)/CLR' and you want the currents to be the same. So, I = I' ==> CLR' = CLR * (V+ - 3 Vf)/(V+ - Vf). Which for a 9V supply and a Vf of about 2.2V turns out to be about 1/3 of the original CLR. If you were using a 12V supply, it'd be closer to 1/2.

Could you make an example of that with a blue led - "Blue LED, 3mm. Current 3,2V-3,4V. 3,3V at 20mA. Clear case, about 6000mcd.". If you could, don't just tell me the result, please explain the way you do it, because it would be very usefull to know that.

[RobA]

Yes, you'll need to change the value. It's going to depend on the LED's that you use. You could compute it based on the current you have when you are using just one LED. V = I R ==> I = (V+ - Vf)/CLR. If all of the LED's are the same, then the new setup would have I' = (V+ - 3 Vf)/CLR' and you want the currents to be the same. So, I = I' ==> CLR' = CLR * (V+ - 3 Vf)/(V+ - Vf). Which for a 9V supply and a Vf of about 2.2V turns out to be about 1/3 of the original CLR. If you were using a 12V supply, it'd be closer to 1/2.

Could you make an example of that with a blue led - "Blue LED, 3mm. Current 3,2V-3,4V. 3,3V at 20mA. Clear case, about 6000mcd.". If you could, don't just tell me the result, please explain the way you do it, because it would be very usefull to know that.

Sure, what CLR would you normally use for this LED? I have to go do some work for a couple of hours, but I'll try to remember to respond after that.

The link that Slimtriggers put up comes up with exactly what I would do for a final result in the fully symmetric layout (3 x 3).

[RobA]

I'm going to assume that you normally use a 4.7k CLR just to go through the calcs with. You can adjust it for what you would normally use.

The two parameters that determine what you are going to do are the Vf of the diode and the current you use to get the brightness you want. The brightness is essentially linearly proportional to the current that flows through the diode. The current is set by the CLR. if we take a supply voltage of 9V, then the voltage drop across the resistor and diode is 9V. 3.3V is dropped across the diode leaving 5.7V to be dropped across the resistor. So, the current you normally have flowing through the diode is 5.7V/4.7kΩ = 1.21mA. We want to shoot for about the same current in each of the branches for the array of LED's.

To figure out the number of branches, we just have to figure out how many Vf's we can fit in the supply voltage. We need to leave a bit of voltage so that we don't get into trouble if the supply goes a bit low. I'd guess that having about a volt leftover after all the diode drops would be OK. In this case, it's not going to matter because 3 diodes is too many and 2 diodes is OK. So, we are going to end up with 4 branches of 2 diodes each and one with 1 diode.

All we need to do now is figure out the CLR's. For the two diode branches, the voltage dropped across the resistor is going to be Vr = Vsupply - 2 * Vf = 9.0V - 2 * 3.3V = 2.4V. Then from Ohm's law we have R = V/I = 2.4V/1.2mA = 2kΩ. For the branch that has one LED, Vr = 9.0V - 3.3V = 5.7V. So, R = 5.7V/1.2mA = 4.7kΩ. Which is the same as the CLR for the case when you only have one diode which it should be because this branch is completely independent of the other branches and does behave exactly the same as the case when we only have one diode total.

2kΩ is a bit of a weird value for a resistor. A 2.2kΩ would be close enough and is easier to find. So, in the assumption of a 4.7kΩ CLR in the normal one diode setup, a 9 diode setup would have four branches of 2 LED's in series with a 2.2kΩ CLR and a fifth branch with a single diode and a 4.7kΩ CLR

[HailToTheBlues]

Thank you very much! I think i'm going to use less leds,
maybe 7 or 6, because i think that if i use 9, the pedal will be eating to much mA of the power supply.
But anyway, thanks to all the help guys, and thanks for the explanation RobA.

Best regards