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Topic: High power LEDs and Arduino Nano on the same 12V power supply (Read 4176 times) previous topic - next topic

jacobs

I have several high power LEDs controlled by Arduino Nano.

These LEDs are powered by a dedicated 12V/3A regulated power supply. The Arduino itself is powered separately via the mini-USB from regulated 5V power supply and it controls the LEDs via TIP122 transistors. So far so good.

Now 2 power supplies seem a bit inefficient to me so I wondered if I could power everything from the same 12V regulated power supply (in case of Arduino directly to the VIN pin).

This works fine if I connect only the Arduino to the 12V power supply (no LEDs connected). In this case I see the Arduino drawing about 20mA.

However once I connect the LEDs in parallel with the Arduino, the Arduino starts to overheat quite quickly and I see it suddenly drawing about 120mA (the LEDs draw about 140mA on their own).

My electronics background is very limited - could someone please explain why this is happening? And ideally how to prevent it from happening so that I can power both from the same power supply?

Any hints would be much appreciated!

retrolefty

Could you provide links to the specific DC power supply and 'LEDs' being used and how they are wired together?

Lefty

jacobs

The power supply looks very much like this: http://www.alibaba.com/product-gs/651910096/LPV_35_12_12V_3A_LED.html?s=p

The LEDs are probably these: http://www.electron.com/non-waterproof-led-strip-5050-warm-white-strf5-5050-w3-12v.html

The LED strip consists of blocks of always 3 LEDs connected in series (resistors included). Individual blocks are then connected in parallel and can be cut and used individually (which is what I did).

I have 5 such blocks connected to 5 Arduino's pins via 5 TIP122 transistors - the base of the transistor is always connected to one PWM pin on the Arduino, emitters go to the common ground, collectors go through the LED block to +12V.

When I try to connect the Arduino to the 12V power supply, I just connect GND to the common ground and VIN to +12V.

retrolefty

Well as that is sold as a "LED driver" power supply I would assume it can't be assumed to be just a simple regulated 12 vdc power supply. I wouldn't power an arduino with it until I understood how it might differ in operation from a standard regulated 12 VDC power supply. There doesn't seem to be enough technical information in the link for me to make such a judgement, so if it was my arduino I would say hands off until I understood it better.

Lefty

jacobs

Well, what confuses me is that when I connect _only_ the Arduino to it, it works well.

Only when I also connect the LEDs, the current on the Arduino increases dramatically. Would this also happen with a simple regulated 12V power supply?

As I said, my electronics background is very poor, so I'm thinking I'm missing something essential here. Based on 1st kirchhoff's law, I'd think that adding the LEDs would increase the power draw of the whole power supply (Arduino's draw+LEDs draw), but not that of the Arduino itself. But I'm probably quite off with this logic :-)


seanz2003

You should have current limiting resistors between your pins and the base of the transistors to protect your pins, but without a clear picture or diagram I am only guessing at how you have connected it.

jacobs

That was exactly it! I forgot the resistors between my pins and the transistor base. I though those resistors are there for the LEDs and I can scrap them as my LED strips already had them included. That was obviously not my sharpest moment :-)

Thanks a lot Seanz!

So just for my understanding, what happened is that I've basically short-cirucited my pins to the ground (going from the pin to the base and then via emitter to the ground) as the transistor itself has little or no resistance at all, right?

seanz2003

right. always limit a pin's output to less than 40milliamps. calculate it using ohms law: r=v/I =5/.04=125 Ohms for a minimum value. I would recommend 1k resistor since a lot  transistors do not usually  benefit much from the extra current.

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