Hey guys,
i have a Meanwell SP-320-48 power supply set to 40 V output. I use it to power my DM556 stepper driver. An Arduino Due controls the DM556 Stepper driver. Right now, this Arduino is powered with a separate 7.5 V wall wart power supply.
Is there a way to get rid of the wall wart power supply and power the Arduino Due from the Meanwell power supply, with something like a step down coverter?
If so, what step down converter would you recommend?
Thanks,
Julia
Hi Julia,
unless you have some specific need I would go for any step down buck dc to dc converter rated for your input and output requirements having the converter output something between 6 and 9 V so you can use the Arduino voltage regulator to provide a less noisy 5V power line (buck converters produce a quite noisy output).
You seem to have no specific current requirement for your Due as you said it's only driving the step motor driver so anything rated for even 0.5A would do... the one I've posted seems to be rated for 5A, but I never used one of those and I do not have a Due.
Hi,
thanks for the answer.
a HC-05 Bluetooth module (i think it sonsumes about 40mA in idle),
a Adafruit Bluetooth LE breakout board (not sure how much this takes up)
6 Standard IR Leds (each LED has a 33 Ohm resistor, so i guess for all 6 around 800mA)
In total i would guesstimate around 1A peak for all the components hooked to the Due.
Julia
Reading the official Arduino Due page:
External (non-USB) power can come either from an AC-to-DC adapter (wall-wart) or battery. The adapter can be connected by plugging a 2.1mm center-positive plug into the board's power jack. Leads from a battery can be inserted in the Gnd and Vin pin headers of the POWER connector.
The board can operate on an external supply of 6 to 20 volts. If supplied with less than 7V, however, the 5V pin may supply less than five volts and the board may be unstable. If using more than 12V, the voltage regulator may overheat and damage the board. The recommended range is 7 to 12 volts.
In other words using Vin pin (and GND) or the barrel jack makes absolutely no difference: in both cases your input is going through the on-board voltage regulator. Another bit of info you can determine by the above quote is my original suggestion to use something in the 6-9V range was erroneous, 7-12V is the suggested range and you probably should go for 7.5V or thereabout.
With regards to your current requirements, if you are in the 1A region than you should go for nothing less than that and probably aim at 2A.
I believe there is a mistake somewhere in your calculation anyway, as the Arduino Due isn't capable to regulate 1A, so either you are powering your LEDs directly from your wall wart or you are damaging your Due:
Digital I/O: pins from 0 to 53
Each of the 54 digital pins on the Due can be used as an input or output, using pinMode(), digitalWrite(), and digitalRead() functions. They operate at 3.3 volts. Each pin can provide (source) a current of 3 mA or 15 mA, depending on the pin, or receive (sink) a current of 6 mA or 9 mA, depending on the pin. They also have an internal pull-up resistor (disconnected by default) of 100 KOhm.
A 33Ohm resistor used to limit current from a 5V source fed into an IR LED mean about 110mA, an order of magnitude greater than what I/O pins on the Due can provide... So, either you are controlling those LEDs through transistors or.... 
Hi,
thanks for the additional info. Concerning the voltage for the Arduino Due, i will go for 7.5 V. i had the same amount on my Mega and it worked very good.
2 A sounds like a good safe value for the step-down converter, i will use one in this range.
I am powering the LEDs with transistors, according to the schematic attached. This worked well and i didnt have any problems with this setup on my Arduino Mega so far. Also take into account, that the LEDs are IR and only used to transmit short bursts, so the max current is only drawn for a very short time.
Its a little off-topic, but since we are already talking about damaging the Due, i have another question:
Can i use the 5 V output of the Due, power some 5 V device and put the anode to the Due's GND? Will 5V at the Due GND damage it or are 5 V just probematic at I/O ports?
Cheers,
Julia
Looking at your schematic Julia, do I interpret it correctly by saying VCC_1 is the Due 5V pin? If it is and you have 6 of those IR LEDs you are using 660mA out of the maximum 800mA the Arduino due voltage regulator can output. Please don't think the 3.3V and 5V current limit can add up: the 3.3V output is a further regulation out of the 5V regulation (according to the schematics on the Arduino website) which means 800mA is total maximum limit.
On top of that, even if you are sending just short energy bursts to those LEDs I bet you are waving in almost 3 times their maximum current rating (unless you are using power LEDs, they are almost always rated at 40mA) which means you are stressing them quite hardly for no benefit: you should just replace the series resistors so to limit the current to 40mA (or whatever is those LEDs current rating if you have the datasheet): that will put the LEDs within their specs and reduce the load on your Due voltage regulator.
When picking the step down buck regulator don't forget the input voltage range: most regulators out there on eBay aren't rated for 40V input.
Now, regarding your GND pin question, you can safely connect the 5V pin to GND through something which limits the current flow within the voltage regulator limits (like a resistor or an LED with a series resistor and so forth) not just because the 3.3V limit does apply to I/O pins only, but because the limit is posed by the voltage regulator and not the MCU. If, by accident, you short out the 5V or 3.3V pin with a GND pin (like by putting a power LED anode and cathode in there without a series resistor) what you will fry is the voltage regulator, the MCU will still be intact.
As stated before, please note I do not own a Due and I'm talking only based on what I can see on paper with my limited electronics knowledge 
rlogiacco:
Looking at your schematic Julia, do I interpret it correctly by saying VCC_1 is the Due 5V pin? If it is and you have 6 of those IR LEDs you are using 660mA out of the maximum 800mA the Arduino due voltage regulator can output. Please don't think the 3.3V and 5V current limit can add up: the 3.3V output is a further regulation out of the 5V regulation (according to the schematics on the Arduino website) which means 800mA is total maximum limit.On top of that, even if you are sending just short energy bursts to those LEDs I bet you are waving in almost 3 times their maximum current rating (unless you are using power LEDs, they are almost always rated at 40mA) which means you are stressing them quite hardly for no benefit: you should just replace the series resistors so to limit the current to 40mA (or whatever is those LEDs current rating if you have the datasheet): that will put the LEDs within their specs and reduce the load on your Due voltage regulator.
When picking the step down buck regulator don't forget the input voltage range: most regulators out there on eBay aren't rated for 40V input.
Now, regarding your GND pin question, you can safely connect the 5V pin to GND through something which limits the current flow within the voltage regulator limits (like a resistor or an LED with a series resistor and so forth) not just because the 3.3V limit does apply to I/O pins only, but because the limit is posed by the voltage regulator and not the MCU. If, by accident, you short out the 5V or 3.3V pin with a GND pin (like by putting a power LED anode and cathode in there without a series resistor) what you will fry is the voltage regulator, the MCU will still be intact.
As stated before, please note I do not own a Due and I'm talking only based on what I can see on paper with my limited electronics knowledge
Hi,
thank you for the useful information. So at 5 V, if i want 40 mA per LED i should switch to 125 Ohm per LED, correct?
Greetings,
Julia
IR led forward voltage should be about 1.5V (refer to your LED datasheet if you have one), which means 3.5V need to be dropped by the resistor (5 - 1.5) because of Kirchhoff's voltage law.
Ohm's law states R = V / I, where V = 3.5 and I = 0.04 which gives 87.5R... for simplicity round it up to 100R to get 35mA of current through your IR LED.
Please note the same 100R resistor would limit a generic blue LED current to 17mA: different LEDs have different forward voltages and the LED datasheet is your master on this, but in case of generic, cheap, unknown LEDs you can refer to tables like this one.
