How much power (current) is available from the 5V regulator on board the Arduino UNO? I don't see it specified on the website.
The onboard 5V regulator is rated at max 500mA.
Any idea how much of that 500mA the Arduino UNO board is using?
30mA or so (with an onboard LED on).
How much current you can use is dependant on the input voltage you are using.
The linear regulator dissipates excess voltage from the input so that the output is constant.
It depends on the regulator spec based on size and heatsinks used but that regulator is likely a LM1117 or similar model that is actually rated at 800mA but you almost certainly would never be able to use it as such.
For my example and likely equal to what's on the arduino, the spec is 79C/W. And has a 125C max rating.
So on a 12V input. (12-5V) = 7V
150C / (79C/W) = 1.58W
1.58W / 7V = 226mA current rating.
So if you use 12V input you can only use 226mA or risk killing the regulator. Not likely a problem unless you use more then 5 LED's (@20mA) or are powering something else.
If you want that 500mA rating. 1.58W / 0.5A = 3.16V
5V + 3.16V = 8.16V input max.
If you want what I'm fairly sure is a 800mA rated regulator. 1.58W / 0.8 = 1.975V
1.975+5 = ~7V input.
If you want to do high voltages and use high currents well then you need a switching regulator to do the regulation to 5V and then connect it to the 5V line directly. cough
harddrive123:
How much current you can use is dependant on the input voltage you are using.The linear regulator dissipates excess voltage from the input so that the output is constant.
It depends on the regulator spec based on size and heatsinks used but that regulator is likely a LM1117 or similar model that is actually rated at 800mA but you almost certainly would never be able to use it as such.
For my example and likely equal to what's on the arduino, the spec is 79C/W. And has a 125C max rating.
So on a 12V input. (12-5V) = 7V
150C / (79C/W) = 1.58W
1.58W / 7V = 226mA current rating.
So if you use 12V input you can only use 226mA or risk killing the regulator. Not likely a problem unless you use more then 5 LED's (@20mA) or are powering something else.
If you want that 500mA rating. 1.58W / 0.5A = 3.16V
5V + 3.16V = 8.16V input max.
0.8A =If you want what I'm fairly sure is a 800mA rated regulator. 1.58W / 0.8 = 1.975V
1.975+5 = ~7V input.If you want to do high voltages and use high currents well then you need a switching regulator to do the regulation to 5V and then connect it to the 5V line directly. cough
Based on the instructions for powering the Arduino, supplying 5V to the 5V pin is a bad thing and I agree. Applying 5V to the 5V pin is placing a voltage on the output of the regulator. Not a good thing.
The two best design options (IMHO) are to either use a 5V/12V power supply, so the 12V runs the Arduino and the 5V runs the "accessories". Or use a 12V supply, split it, and use one half to run the Arduino and the other to run an auxiliary 5V regulator (perhaps something in a TO-220 package with a heatsink) that then runs the accessories. Either way the best design is to feed the arduino the 7-12V it wants and use another power supply to run the accessories.
I am looking at running the arduino itself, a couple of opamps, an LCD, and an LED display. The biggest culprit is the LED display which uses 60mA. It must be LED because it needs to be big and bright. The LCDs don't fit that requirement.
LED display which uses 60mA. It must be LED because it needs to be big and bright
60mA is the equivalent of 3 LEDs. Not big, nor bright.
Based on the instructions for powering the Arduino, supplying 5V to the 5V pin is a bad thing and I agree. Applying 5V to the 5V pin is placing a voltage on the output of the regulator. Not a good thing.
Where did you read those instructions?
florinc:
Where did you read those instructions?
https://www.arduino.cc/en/Main/ArduinoBoardUno
5V.This pin outputs a regulated 5V from the regulator on the board. The board can be supplied with power either from the DC power jack (7 - 12V), the USB connector (5V), or the VIN pin of the board (7-12V). Supplying voltage via the 5V or 3.3V pins bypasses the regulator, and can damage your board. We don't advise it.
The onboard regulator is good for 800mA, but you'll likely only see that with 7.5V on the barrel jack.
Powering via the 5V header pin will also work, add a 1N4001 type diode from 5V (anode) to Vin (cathode) to avoid reverse driving the regulator.
Splitting the 12V into an Arduino feed and a parellel feed to other higher current stuff is a good idea.
florinc:
60mA is the equivalent of 3 LEDs. Not big, nor bright.
Compared to the tiny digits on the LCD, the 3-digit 1.2" tall red LED on black display is neon bright.
CrossRoads:
Powering via the 5V header pin will also work, add a 1N4001 type diode from 5V (anode) to Vin (cathode) to avoid reverse driving the regulator.
I'm not following this plan. The 5V pin on the header is the output of the 5V regulator (per the schematic). Applying 5V to the 5V pin with a diode from 5V to Vin doesn't do much. The diode is forward biased from 5V to Vin but incurs a diode drop (0.7V). The regulator is well below the minimum input voltage and cannot boost. At best, the regulator output is 4.3V. Probably less. The regulator will still be reversed biased.
See page 10 "Protection Diode".
Other 5V regulators (non 1117 type) did not have the reverse driven problem, like the MC33269 used on the Duemilanove.
3-digit 1.2" tall red LED
If this is a 7-segment LED display, a digit is 7 LEDs. To display an 8, it takes about 140mA (7 x 20mA).
florinc:
If this is a 7-segment LED display, a digit is 7 LEDs. To display an 8, it takes about 140mA (7 x 20mA).
Well the spec sheet says 60mA max (which took me a while to find on the workbench). Before I found the spec sheet, I measured it at 47mA with three 8s displayed. Not all, especially red, LEDs do not take the same current. I have an LED bargraph that is eye damaging bright at 2mA, it is spec'ed at 1mA max.
CrossRoads:
See page 10 "Protection Diode".
Linear & low-dropout (LDO) regulators | TI.comOther 5V regulators (non 1117 type) did not have the reverse driven problem, like the MC33269 used on the Duemilanove.
I read the datasheet and the reverse protection diode speaks to the input being grounded not open and floating (with or without the diode). From the block diagram on page 2, i would worry about the 4.3V going through the chip directly to ground. It won't go through Vout since Vout>Vin. I would need to contact the manufacturer to verify the situation of Vout=5V and Vin floating, with or without the diode.