how to power array of leds

Thread Starter

rahulb

Joined Jun 5, 2019
20
Hi,

I am trying to make a decorating led light.

I am using 675 leds. I will divide them in three parts of 225 leds each.

I am trying to add them in parallel so that one bad led doesn't effect the others.

All the three array will be operated through a led chaser circuit which I will build in next phase.

Assuming that the voltage drop of led is 2.0 v and current consumption(5mm) is 20 mA, the total current consumption is around 14 Amp.

My question is , what will be best way to connect them. Should I connect them in parallel individually like in the image attached below. Or Connecting 3-4 leds in series through a resistor in each array to bring down the current consumption.

I am trying to use 12v smps power supply for it.

led array.jpg

thanks
 

oz93666

Joined Sep 7, 2010
673
Well... things are determined by your power supply , which is 12V

Normally for 12V ... three leds and a resister , then repeat these units in parallel ....

The very best way would be to have more than 3 ... see how many you need so that no power wasting resistor is needed , but drive the leds at 10mA to give a big safety margin ....

Try 5 in series , measure the current , if it's about 5 to 15 mA that's good , just put many of these 5 's in parallel , watch out you don't exceed the max current of the supply.
 

panic mode

Joined Oct 10, 2011
1,783
agreed, with 12V supply you may want to connect 3 or 4 LEDs in series using just one series resistor such as 180-220Ohm for approx 10mA per LED group.
this way you not only save on resistors but power consumption will be much lower. 2A power supply would do. even if you want 20mA per LED, that is still only 3.3A instead of 14A.

about chasing lights - how many channels? with or without microcontroller? microcontroller can add some fancy effects (dimming etc.). for simple on/off chaser, just use 555 and 4017. of course need transistor on each 3017 output...

here are some examples:
https://www.nutsvolts.com/magazine/article/led-chaser-sequencer-circuits
 

Thread Starter

rahulb

Joined Jun 5, 2019
20
thanks a lot for the advice. 3 or 4 leds is series will be the best way.

But, I have a confusion yet. Do the 5mm led consumes 10mA or 20mA. I was calculating resistor value according to 20mA current.

Please advise.
 

oz93666

Joined Sep 7, 2010
673
thanks a lot for the advice. 3 or 4 leds is series will be the best way.

But, I have a confusion yet. Do the 5mm led consumes 10mA or 20mA. I was calculating resistor value according to 20mA current.

Please advise.
All leds start emitting light at the slightest current ... the figure you have for your 5mm led of 20mA is the Maximum recommend current . Life expectancy is dependent on current ... the higher the current the shorter the life so it's advisable not to run at maximum current , 10mA will still look very bright .

For larger leds the maximum current rating is only permissible with a very good heat sink , otherwise it will blow quickly ...

Your 5mm ones are encapsulated in plastic and don't require a heat sink.
 

djsfantasi

Joined Apr 11, 2010
5,704
First, 3-4 LEDS can’t be split evenly into 225 or 675 LEDs.

For a quick check, let’s use 5 LEDs in series and 135 of these strings in parallel.

The 5 LEDs at 2V per will drop 10V from your 12V supply. That leaves 2V at 10 mA, requiring a 200Ω current limiting resistor.

Your supply must supply 1.35A at 12V.

Make sense?
 

Thread Starter

rahulb

Joined Jun 5, 2019
20
First, 3-4 LEDS can’t be split evenly into 225 or 675 LEDs.

For a quick check, let’s use 5 LEDs in series and 135 of these strings in parallel.

The 5 LEDs at 2V per will drop 10V from your 12V supply. That leaves 2V at 10 mA, requiring a 200Ω current limiting resistor.

Your supply must supply 1.35A at 12V.

Make sense?

thanks.

But , Each array have different color of leds.

One is red. second is green. third is blue.

red and green can be added 5 leds in series but since blue led voltage drop is around 3.3 v, I can add mostly 3 leds in series. Am I correct?
 

djsfantasi

Joined Apr 11, 2010
5,704
thanks.

But , Each array have different color of leds.

One is red. second is green. third is blue.

red and green can be added 5 leds in series but since blue led voltage drop is around 3.3 v, I can add mostly 3 leds in series. Am I correct?
Not a problem if you do the math... start with 3 blue LEDs in series. Their forward voltage is (3 x 3.3) = 9.9V. That leaves 2.1V across a current limiting resistor. At 10mA, that requires a 210Ω resistor. Note a couple of things. The blue LED needs to have its datasheet checked to confirm that 10mA is the appropriate current. Secondly, note that 4 x 3.3 is 13.2v, hence won’t work with a 12V supply.

The red and green LEDs may also differ in their specs, so you should perform calculations for each color. And calculate the total current draw for each color. Then add them together for the total required current.

The base formula for n LEDs in series, requiring a current of i amps with a supply voltage of b to calculate the value of a current limiting resistor is as follows.

First,
i LEDs current
Vf LEDs forward voltage
Vs Supply voltage
n # of LEDs
RC current limiting resistor in Ω

Rc = (Vs -(n*Vf)) / i​

With this formula, you should be able to design series/parallel LED strings of any color, forward voltage and current...
 

WBahn

Joined Mar 31, 2012
24,854
In your first post you mentioned that you are going to turn this into an LED chaser circuit next. So keep in mind that LEDs in series are either all one or all off. As long as you can define groupings in your chaser configuration that allow for this while accomplishing the visual effect you want, you will be okay.
 

oz93666

Joined Sep 7, 2010
673
It's not quiet that easy .... back voltage is not fixed but depends on current ....

This is very handy ... you can do it all without resistors and make sure all the power goes to the leds and not heating up resistors .... You have to experiment , try 5 in series first ... a current anywhere between 5 and 15 mA will be fine ... at lower currents the leds are up to 35% more efficient (lumens out per W in)
Remember, as you put more load on the power supply the voltage will drop , also as the leds get warm this will make them draw slightly more current . that's why it's best not to push to close to the max. 20mA.

The only reason it's common practice to use resistors is to allow for variability in power supply ( a strip sold as 12V needs to accommodate 10.5-13.5) also the manufacturer wants to drive leds close to max to save costs on components , he doesn't care about efficiency or long life.
 
Last edited:

panic mode

Joined Oct 10, 2011
1,783
First, 3-4 LEDS can’t be split evenly into 225 or 675 LEDs.
I think 675 and 225 both divide by three... And if that was not the case, resistor for remaining LEDs would need to be different.

As we everything, it is a (calculated) tradeoff... the less voltage drop on resistor, the less losses so better efficiency. But this means lower margin and less linearized system so harder to adjust when using different LEDs. There was plenty of good info so far. If you have LEDs, follow advice and try some tests - see what you get.
 

djsfantasi

Joined Apr 11, 2010
5,704
I think 675 and 225 both divide by three... And if that was not the case, resistor for remaining LEDs would need to be different.

As we everything, it is a (calculated) tradeoff... the less voltage drop on resistor, the less losses so better efficiency. But this means lower margin and less linearized system so harder to adjust when using different LEDs. There was plenty of good info so far. If you have LEDs, follow advice and try some tests - see what you get.
Why in fact they are divisible by 3. My math is declining with age.
 

Bernard

Joined Aug 7, 2008
5,291
I would measure a hand full of each color for Vf @ 10 ma and use the average for each color.
We want a resistor equivalent for every 4 to 5 LEDs, so take supply V divide by LED V, 12 V /
1.8 v measured red LED @ 10 ma = 6.6. Take whole no. 6 = 5 LEDs & 1 R. 5 x 1.8 V = 9 v. 12 v - 9 V = 3 / .01 = 300 ohms. This makes more sense with longer series strings.
 
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