# Powering Arduino + ~50 LED's

Hi,

I was wondering how to power the Arduino and my LED's. The LED's are connected to multiple TLC5940's.

My calculation:

LED's average power drain: 2.7A / 13.5 Watt Arduino maximum power drain: 0.5A / at 9V power supply: 4.5 Watt

Can I just plug a 3.5A power adaptor into the arduino power-jack? Or will components burn out with current on its way to the last LED?

and should this be the case which I higly suspect, how should I power this? Maybe a 5V power adaptor parallel connected with arduinos USB and the TLC5940's?

p.s. should I use a capacitor to filter power-grid-noise or rapid changing of currentflow that happens when LED's are turned on and off?

You could seperately power the LEDs from an external power source, and then control them via Transistors from the Arduino?

You'll NEED to connect a separate supply, the Arduino's voltage regulator is only rated at 1 amp. It starts getting very hot much before that, but it also depends on the voltage used.

But as far as hooking up all the LED's go.. how many TLC5940's are you using? and are they rated high enough to handle as many LED's as you're trying to run off em?

@TchnclFl

You mean one 9V power supply and then one transistor for each TLC parallel? As pointed out by CaptainObvious this should dissipate the heat.

@CaptainObvious

I'm using 5 TLC5940's which are according to TI able to sink >3.6V up to 120mA of current.

my setup:

TLC1: 11 red LED's      ---> 40mA
TLC2: 11 blue LED's     ---> 40mA
TLC3: 11 green LED's    ---> 40mA

TLC4: 11 white LED's    ---> 100mA

TLC5: 10 random LED's   ---> 20mA


what I want to build is a livingroom RGB-LED lamp...

it would be nice if I could operate this using only a single power adaptor.

cheers!

120ma sounds about right, but if that's the case.. aren't you going to be overdoing them with your LED setup?

11 LED's running at about 20ma, would equal.. well, 220 milliamps. That would require 2 5940's. (you MAY be able to piggy back them, I'm not 100% sure, but worth checking into.)

All though, you could use transistors on the outputs of the 5940s instead, so you won't have to worry about sourcing so much current from them.

Well you can run the Arduino off the same power supply, just need to make sure it's regulated into a good range. Also, making your own Arduino with some better voltage regulators could save you some worry on the heat issue. (btw, the Arduino itself is rated at 200mA max, the board is actually rated at 1 amp max, but only 500ma through the USB cable.)

Ah ok, I missread that part, I thought the Arduino would make full use of the 500mA from USB.

I understand the TLC5940 can do a sink of 120mA for each channel.

In the datasheet it says: Constant output current: OUT0 to OUT15, Vcc>3.6V: 120mA

Sorry I'm not familiar with data-sheet-language ;-)

btw. here’s the link to the datasheet, translation considering maximum current drain per channel greatly appreciated

http://focus.ti.com/general/docs/lit/getliterature.tsp?genericPartNumber=tlc5940&fileType=pdf

cheers!

edit:
ok, just found out on the ti-homepage, it is indeed 120mA per channel!

Further questions:

Until now I figured out I will use a 9V Power supply to connect parallel to the arduino and each TLC. Between the 9V and a TLC I will use a voltage Regulator, most likely a 7805 or 7905.

So, does it matter if I use a positive voltage regulator:

Input:  +9V
Ground: -9V
Output: +5V

or negative:

Input:  -9V
Ground: +9V
Output: -5V


Since the TLC sinks a current of 120mA it just seems natural to use the negative one, but then I have a common kathode(+) :-?

On the other hand it shouldn't make a difference if I use the positive one since I have a closed circuit with common grounds.

cheers

So, does it matter if I use a positive voltage regulator

Yes a positive regulator is what you want, it will not work with a negative one.

Since the TLC sinks a current of 120mA it just seems natural to use the negative one

You are mixing things up, this is nothing to do with the polarity of the voltage regulator.
With a TLC you NEED a common anode LED otherwise it will not work.

it is indeed 120mA per channel!

Yes but don’t go thinking you can take that on all 16 channels at the same time. look at the power dissipation calculations.

Ah, thanks a lot for explaining(s)!

problems solved :-)

Dang! Not solved!

Ok, I looked through datasheet and found aformula for calculating power dissipation:

For LaTeXs ;)
P_{D}=(V_{cc}\cdot I_{cc})+(V_{OUT}\cdot I_{MAX}\cdot d_{PWM}\cdot N)


Pd=(Vcc*Icc)+(VOUT*IMAX*dPWM*N)

P = Power dissipation in Watts
Vcc = supplied voltage to chip
Icc = supplied current to chip
Imax = max currrent supplied to LED
Vout = LED Voltage
dpwm =duty cycle
N = numbers of LEDs

so:

P=(5V4A)+(3.5V0.1A113)=24.55 Watts :-?

Ok (3.5V0.1A113) equals 4.55 watts, which is high but ok, but this part (5V4A) would mean 20W which I don’t understand because I just supply 4A but the chip doesn’t need 4A so actually there would only be a flow equal to LEDs power consumption which is 4.55 Watts…
Any physicists aboard?

Cheers

Any physicists aboard?

Yep ;)

(Vcc*Icc) where:- Vcc = supplied voltage to chip - in your case 5V Icc = supplied current to chip - this is the current the chip takes independent of any LED current. You will find this in the data sheet and is in the mA region. The chip is not taking 4A that is just your maximum current from the supply.

^^

In my case that would be:

data-transfer + all outputs on:

0.025A with Iref as 1.3 kOhm
0.060A with Iref as 640 Ohm
mine:
~0.120A with Iref as 390 Ohm ← to let ~100mA pass through

would mean: total of 5.15W
maximum power dissipation without powerpad soldered(HTSSOP): 2W

For RGB-Channels ~1.8W at least that is working…

So it as well you found out by calculating rather than burning out the device. :)

One solution is to use more than one TLC chip. If you use two and chain them and don't connect every channel then the problem is solved. Alternatively you can try clamping an external heat sink on top of the chip to improve (reduce) the thermal resistance of case to ambient.