The darlington chip is a ULN2803APG and looks like this:
I've connected it to my Arduino uno like this:
Am I safe if I leave the COM unconnected?
Yes.
Is there anything in a 12 V LED strip that kicks back and needs the COM connected?
No.
However your big problem is the power ratings of that chip. While it might seem that each driver can handle 0.5A each ( and so they can ) thermal considerations show that the total current that can be switched at any one time is only about 600mA for the whole chip.
See:-http://www.thebox.myzen.co.uk/Tutorial/Power_Examples.html
You are correct about the "kickback" matter - it is not a problem with LEDs.
However the ULN2803 (or for only 9 outputs used, ULN2003 would have been sufficient as you have to use two of them anyway) is essentially obsolete; you are losing a Volt or two as well.
The concern is - how much current are your LED strips going to require? A TPIC6A595 can switch 350 mA each output with a continuous total up to more than one Amp (more again if heat-sinked) with less than 350 mV loss and two of these would require only 3 Arduino pins to control.
But your proposed circuit is concerning. Supplying 12 V to the "barrel jack" and drawing the LED current from "Vin" has two obvious problems; one being the rating of the diode between these which is perhaps one Amp and another 700 mV loss.
The other problem as always, is that the on-board regulator is capable only of supplying sufficient current for the ATmega chip itself; nothing else really. If your design has nothing else connected to it (just a "pretty pattern" generator), then the ULN2803 does not draw much current and you are probably OK, but in general, it is a bad idea to use "Vin" or the barrel jack.
And a Nano is a more practical form factor.
Ok, I might leave the VIN alone and connect the leds directly to the 12 V.
In my circuit, as I can see, the regulator only runs the ATmega and the nine digital pins going to the ULN2803. I think the pin current was some 5 mA per pin. Each of the three led strips contains six 5050SMD RGB LEDs. So there is a lot of current, when everything is on full.
I have one or two spare ULN2803, if the total current per chip gets too big.
However the ULN2803 (or for only 9 outputs used, ULN2003 would have been sufficient as you have to use two of them anyway)
Each of the three led strips contains six 5050SMD RGB LEDs. So there is a lot of current, when everything is on full.
So 6 LEDs @ 20mA per LED colour gives you 0.12 A per strip to switch. The most worked ULN2003 is the one switching 5 strip colours so the total current for that chip is 0.6A which is just on the limit. It will get warm.
Careful with the A:s and mA:s there...
Johan_Ha:
Careful with the A:s and mA:s there...
Thanks for spotting that. +1
Now corrected.
Grumpy_Mike:
So 6 LEDs @ 20mA per LED colour gives you 0.12 mA per strip to switch. The most worked ULN2003 is the one switching 5 strip colours so the total current for that chip is 0.6A which is just on the limit. It will get warm.
Grumpy_Mike:
Thanks for spotting that. +1
Now corrected.
Err, no!
Paul__B:
Err, no!
Well yes, I did correct one but not the other.
Grumpy_Mike:
So 6 LEDs @ 20mA per LED colour gives you 0.12 A per strip to switch. The most worked ULN2003 is the one switching 5 strip colours so the total current for that chip is 0.6A which is just on the limit. It will get warm.
A datasheet of ULN2003 mentions 500 mA as a max continuous collector current. A datasheet of ULN2803 however mentions max 500 mA per channel. Of course, if all eight channels would handle a continuous 500 mA, it would get hot. But a single channel would never exceed 120 mA.
Of course, if all eight channels would handle a continuous 500 mA, it would get hot.
No it would be destroyed in a very short time.
But a single channel would never exceed 120 mA.
And like I said the worse case is 120mA * 5 = 600mA, which is on the limit and would get hot.
Did you not read the link in reply #1? Did you understand it?
If not read the build up to the example at:-
http://www.thebox.myzen.co.uk/Tutorial/Power.html
It is nothing to do with the individual current driving capacity of each of the drivers in the chip, but the thermal consequences of the collective current from all the drivers.
I redad the link, thanks for that.
Did I get it right? The max current per darlington is 120 mA, that's six LEDs, each draws 20 mA.
According to the datasheet, the collector-emitter voltage drop would be 2.4 V for a collector current of 200 mA. Now the collector current would be 120 mA for one darlington. What would the drop be? Maybe 2 V. That makes 0.24 W. Five darlingtons working full would be 1.2 W. The datasheet says the total power dissipation for all 8 darlingtons is 2.25 W.
The thermal resistance is 55 degrees per watt. So I'd get some 86 degrees. Hot but quite ok.
[edit]
Actually the voltage drop is only 1.2 V at 120 mA according to this diagram. That would give 1.2 V * 0.12 A * 5 channels = 0.72 W:
Though I don't quite follow how the rising temperature will affect these values. Will the power dissipation escalate when everything gets hotter?
