# Limit current

hello,

I have a question i read several topics but i dont understun why we need limit the input current at 250 mA. Becasue the datasheet of the voltage regulator (AMS1117) give : Current Limit out = 1.1 A therefore Poutmax=1.1A*5V=5.5W and after measure i have : Iin = 0.3A for an input voltage of 9V therefore Pin=2.5w.
Please can you explain me why it a problem ?

Other solution i think design one supply board and powered the arduino mega by the USB Port.

@+
QS

crevette24:
I have a question i read several topics but i dont understun why we need limit the input current at 250 mA. Becasue the datasheet of the voltage regulator (AMS1117) give : Current Limit out = 1.1 A therefore Poutmax=1.1A*5V=5.5W and after measure i have : Iin = 0.3A for an input voltage of 9V therefore Pin=2.5w.
Please can you explain me why it a problem ?

Seems you calculate power in the LOAD, not in the regulator.

For 9volt in and 5volt out, at 300mA, the regulator dissipates (9-5) * 0.3 = 1.2watt.
That is already a lot for the small heatsink area of an Arduino.

Look at the thermal table on page5 of the AMS1117 datasheet.
Say the heatsink area is 225 square mm.
The chip (inside) will rise 65degrees above ambient for every watt.
1.2watt from above will be (1.2 * 65) + ~25 (ambient) = 103 degrees C.
The chip will shut down at 165C.

Maybe there is another way to solve it
Leo..

I don't understand what you mean by 'limit the input current at 250mA' !

where does it come from ?

The limits on arduino boards are on the individual I/O pins, and there is also a limit per Port -

Iin = 0.3A for an input voltage of 9V therefore Pin=2.5w.

Your caclulation is wrong - If Vinput is 9V and Voutput is 5V, and I= 0.3 A then the power in the regulator is (9-5)*0.3 = 1.2W

Other solution i think design one supply board and powered the arduino mega by the USB Port.

BTW, the Mega has a polyfuse on the USB line, which limits it to a max of 500mA

For the moment i dont have a problem but i am worried for the future. In fact i have a touch screen + shield + display + 3*optocoupler.

i think use an external voltage regulator and use a USB port to powered the arduino.

crevette24:
i think use an external voltage regulator and use a USB port to powered the arduino.

why ?

You'd probably better use a power supply to power the mega + the screen and maybe the display

the screen and the display won't be powered by the mega, but by the power source + regulator if needed .

Because for the moment, i use just the USB port (programming and supply) and i don't have a problem. I think my current of my system will be < 500mA .

With this method i don't need modify the shield of my screen.

Ok, but what I mean is that powering via USB connector doesn't allow more current. In fact, it is the opposite.

And if you want to power via USB connector, you need a regulated 5V supply - If it works now with the USB connected to a PC, than it will work if you power it via the power jack (7V - 12V ) or the Vin pin (7V-12V too) without any modification of your shield.
Both will work..... use the way you like, the simpler the better

I agree with you but if i powered by the jack i oblige to use the integrated voltage regulator therefore i can't cooling this device.

With my solution i can add a heat sink on my external regulator (7805) to control the thermal flux.

You can use a 7805 based solution. It can feed both the 5V pin (or USB) and your other other hardware; just split the output.

Note: The 5V pin on the Arduino board is both input and output.