So I am new to the Arduino world and can anybody tell me the optimum input voltage to supply power to the Uno? Everything I plan to connect to the Uno will have it's own power that won't draw (much) from the Arduino.
Thanks in advance.....
It depends on what you have. If you have a 12V battery, then use that. If you have a leftover 5V phone charger, then use that.
"Much" is relative. How much? Some components like LCD screens and WiFi shields draw significant power that you don't expect, like more than 500mA.
Specs advice input voltage of at least 6V, preferably 7V. There's a regulator on board that lowers this to 5V, this regulator needs at least 1V drop to work, hence the 6V minimum. It can take up to 20V.
But beware: if you're working it hard and indeed draw 500 mA, that's (12-5)x0.5 = 3.5W that this little, non-heatsinked chip has to absorb! That may work for a second or even two but not more than that, or it overheats. At 20V it's of course a lot worse. So I wouldn't feed it more than 9V for long term operations.
The above is for the VIN pin.
You also should find a +5V pin on the device, no regulator on this one, so if you have a good 5V supply this works, too. The USB port is another way to provide 5V. Or get a DC-DC buck converter to step down your 12V to 5V at much less loss than the regulator, and supply that to your Uno.
The Arduino is only sourcing enough power to run some relays and soil-moisture sensors (and those only intermittently) right now. I read somewhere that there is a "sweet spot" of input voltage level that doesn't stress the onboard regulator too much. At the moment I have a 12v supply feeding a buck converter to step the DC down to about 9 volts, and I don't remember why that was the voltage I chose. It works fine but, like I say, I try to avoid using the power supply on the Arduino for anything but driving transistors (and I like to use FETs when I can)....
The power-loss (Watts) in a linear regulator is the product of Current x Voltage. The current is the same as the Arduino and the voltage dropped across the regulator is the difference between the supply voltage and 5V. The Arduino's on-board regulator is linear.
A buck regulator is a switching regulator and it can be nearly 100% efficient. (You can get more current out than goes-in, with less voltage-out than going-in.)
So, the most Efficient solution is to bypass the on-board linear regulator and power the Arduino directly with 5V from a switching power supply or a switching regulator. And if you can't do that, 6 of 7V into the regulator is best.
tentoes1962:
It works fine but, like I say, I try to avoid using the power supply on the Arduino for anything but driving transistors (and I like to use FETs when I can)....
So your project is working fine, but you want to change it.
Why ?
So I was simply looking for feedback on the pros and cons of a higher or lower DC voltage on the input and I got what I was looking for: a good discussion. Thanks everyone!
An Uno/Mega has switch-over circuitry to switch between USB supply and external supply.
Switchpoint is set to ~6.6volt.
So anything above 6.6volt on V-in is using that external supply.
There is a reverse protection diode with a ~0.7volt drop between the DC socket and V-in.
If you power the Arduino on the DC socket, the that switch point is 6.6+0.7= ~7.3volt.
So you need 6.8volt minimum on V-in or 7.5volt minimum on the DC socket.
If you trust the DC/DC converter, then you can bypass the 5volt regulator, and inject 5volt on the 5volt pin.
That 'could' be problematic if you also connect USB to your PC.
Leo..
Thank you, Leo for the definitive answer I was looking for!!!!
Wawa:
If you trust the DC/DC converter, then you can bypass the 5volt regulator, and inject 5volt on the 5volt pin.
That 'could' be problematic if you also connect USB to your PC.
Leo..
it may be an old dead thread to you but to me it's what I needed to know. No one actually said that you can completely supply all the arduino's needs by supplying a well regulated 5v to the input pin. I was never clear on this; it 'seemed' like that must be what it means....... I think.........
Now I know what to supply on my end for one of my projects