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### Topic: [Solved][Follow-up question on page 2] Resistor placement - 30 LEDs (Read 3829 times)previous topic - next topic

#### retrolefty

#15
##### Jul 10, 2011, 01:54 amLast Edit: Jul 10, 2011, 01:59 am by retrolefty Reason: 1
Quote
Is there any simple way of powering this circuit, except for using an external powersupply and transistors for turning it on and off?

No, you will require an external +5vdc power source rated at or above 600ma. You will have to use some form of external switching for the led array as it's current draw is too much for direct control from a output pin, however a simple switching transistor is easy to use. Your arrangement of leds in series/parallel is fine, however that lone 1.5 ohm resistor serves no useful purpose, lose it.

Keep in mind that different color LEDs have different forward voltage drop values and your circuit would not work with say blue leds.

Quote
Parallel LEDs is dodgy, becouse if one fails, then all are bound to fail relativly soon.

That is a false conclusion. There is nothing about one parallel led string failing (somehow?) that would effect the other strings at all.

PS: By my estimation (if using red leds) each parallel leg will draw around 10ma for a total of 100ma, how did you get your 600ma estimation?

Lefty

#### fkeel

#16
##### Jul 10, 2011, 02:03 amLast Edit: Jul 10, 2011, 02:07 am by fkeel Reason: 1

That is a false conclusion. There is nothing about one parallel led string failing (somehow?) that would effect the other strings at all.

There are enough differences in LEDs that hooking them up in parallel, one will draw more current ending up burning it out faster than the others and once that happens, because your circuit's current limiting resistor was designed to supply current to 3 LEDs the amount of current draw on the remaining two LEDs will increase. This increase will quickly burn out your remaining LEDs in a snowball effect.

The above is only true, if I have one resistor limiting the current for each parrallel LED.
If I have a limiting resistor for each parallel circuit, then I dont run into this problem, correct?

*

Thanks for teaching random people on the internet electronics, Lefty :-)

EDIT:
I am using IR LEDS. Each LED requires 20mA
Hm. That means that I need 200mA and not 600, correct?
(I was thinking 20x30 = 600 but it doesnt work like that, right?)
http://embodimentlabs.tumblr.com/
http://paulstrohmeier.info/

#### retrolefty

#17
##### Jul 10, 2011, 02:12 am

Quote
Parallel LEDs is dodgy, becouse if one fails, then all are bound to fail relativly soon.

That is a false conclusion. There is nothing about one parallel led string failing (somehow?) that would effect the other strings at all.

@Lefty, the original reference was to LEDs hard-wired in parallel with a single current limiting resistor. Since LEDs are not exactly perfect they will tend to draw slightly different currents. Under worst-case an LED could hog the current and if that causes it to burn-out open, then too much current will be presented to the remaining LEDs, causing them to fail prematurely, also.  NOT a good circuit design practice.

I was just commenting on the latest drawing he posted where there was no sharing of resistor. Except for that strange 1.5 ohm one.

#### fkeel

#18
##### Jul 10, 2011, 02:33 amLast Edit: Jul 12, 2011, 07:58 am by fkeel Reason: 1
Thanks, KEG7GKP & Lefty

My project is going forward, thanks to you guys.

EDIT:

Anyone following this with similar questions: I just discovered this here

http://ledcalculator.net/

http://embodimentlabs.tumblr.com/
http://paulstrohmeier.info/

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