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Topic: Multiple LEDs dimming using one PWM pin (Read 118 times) previous topic - next topic

haxan

Hi. I am working on a project in which i need to control the brightness of multiple LEDS (about 8) using only one pin of PWM. I am using Arduino DUE. Can anyone please suggest a simple way to do this.

By the way - I have a buffer IC 74LS244 too. Was wondering if i could use it somehow?

The challenge i feel is DUE's low amps. If i connect 8 LEDs on a single pin - i am pretty sure the pin would fry (as its suited for only 3ma only)

Please help me.

haxan

Sorry i forgot to mention, each LED can be in a different state (ON/OFF) from a digital I/O pin but the dimming can be the same for all.

RayLivingston

If you buffer it adequately, you can drive 10,000W of LEDs with a single PWM output.

Regards,
Ray L.

MorganS

What is the output current of your 74LS244? The first datasheet I found online shows it's not any better than the Due's pins.  Other than that, it does look like an appropriate device to buffer the Due outputs and PWM them with the enable pin.

haxan

thanks guys.

sorry i didnt check the output current of the buffer. I assumed it would be able to drive 8 leds if the baising current was high like 800ma (from DUE's) 5V pin.

connecting PWM pin with EN pin is a great idea. I thought it would latch the data initially.


RayLivingston

An LS244 is a buffer, not a latch.  Depending on WHICH LS244 (LS, AS, C, HCT, etc.), output current will be from 10 to perhaps 60mA.  The Due's VCC pin is not good for anywhere NEAR 800mA.  More like 80, if that.

Regards,
Ray L.

haxan

sorry now i am confused. From http://arduino.cc/en/Main/ArduinoBoardDue it says 3.3V and 5V each have 800mA. I have based my entire circuit with that assumptions.

According to this datasheet http://www.uni-kl.de/elektronik-lager/417791

ICC is 46 - 54 mA or am i looking at something else ?

haxan

In another datasheet it says:

Typical IOL (sink current) 24 mA
Typical IOH (source current) −15 mA

So if IOL for each pin is 15mA that's enough to drive an LED on right (if the current limiting resistance is about 380 ohms or round about) ?

RayLivingston

Just because the regulator chip itself is rated for a max of 800mA does not mean it will deliver that on a particular board.  Given the thermal design of the regulator on the Due, I'd be amazed if it came anywhere near that.  Nor is there any guarantee the PCB traces can handle that much current.  Highly unlikely they can.  Even if the PCB and regulator were up to the task, there would only be that much current available when powered by an external DC supply connected to the DC IN jack.  With USB, you're limited to 100mA total current for the entire board and anything connected to it.  If you're going to be powering high-powered peripherals, they should have their own power supply, with suitable buffering of all control signals.

Regards,
Ray L.

haxan

Thank Ray for the insight. I totally agree with you with what you said. I am going to be powering up the board with 9V external supply. Regarding the traces on the board - I believe the 5V and 3.3V (on the power socket should have thicker traces). I have checked and the traces on my DUE for all the 5V pins and 3.3V are pretty thick.

dlloyd

You'll get the most current (500mA) using USB power i.e. from USB Hub with DC Adapter.

The next best solution would be a 7.5V DC adapter connected to Vin. Haven't tested this but I'm quite sure you'll get at least 300mA.

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