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Topic: Maximum current from the 5V pin? (Read 51201 times) previous topic - next topic

bilinsky

Hi, currently I'm using the usb cable as power and use the 5V pin to power some sensors, however I think I need more power because I get communication errors sometimes. If I power the arduino by say 12 volt and 2A, is it possible to draw the equivalent power (~ 5A) (except for the power loss for the arduino itself) from the 5V pin? I didn't find specs regarding the max current from the 5V pin.

cerocca

Hi,
checking from the Duemilanove schematic seems that the voltage regulator used by the arduino is the MC332690-5.0

I searched the web for a datasheet and once found, it says:

Quote

800 mA, Adjustable Output, Low Dropout Voltage Regulator


so regarding your question the answer is NO, you cannot draw 5A from the 5V pin (but neither 1A... and anyway after can come the heatsink problems)

hope this helps

C

Grumpy_Mike

You can't get more than 500mA from the USB connector and sometimes your hardware might limit that down to 100mA.

Have a look at this for a discussion on power supplies:-
http://www.thebox.myzen.co.uk/Tutorial/Power_Supplies.html

bilinsky

#3
Mar 19, 2009, 01:17 pm Last Edit: Mar 19, 2009, 01:17 pm by bilinsky Reason: 1
Ok, thanx, good to know that 800 mA is the max.  The power supply I bought today is then somewhat of an overkill ;), but I can power my sensors directly from that one instead of using the pin on the arduino.  

Grumpy_Mike

Quote
good to know that 800 mA is the max.


That is only the current rating of the regulator. You can't actually get anywhere close to this current from it due to the maximum power dissipation.
See:-
http://www.thebox.myzen.co.uk/Tutorial/Power_Examples.html

spiffed

Quote
If I power the arduino by say 12 volt and 2A, is it possible to draw the equivalent power (~ 5A) (except for the power loss for the arduino itself) from the 5V pin?

The short answer is no, the long answer is maybe, but with a different setup.

The onboard regulator is linear so theoretically, ignoring any chip-specific limitations, you can't draw more current than you put in. 12V@2A in == 5V@2A out, the rest of the power 14watts is burned as heat in the linear regulator! This said, the regulators specced limit is 800mA, but thermal dissipation is going to keep you down in the 400-500mA range.

If you want to convert 12V@2A to 5V@4.5A or so, you'll need a 'switching buck regulator', and a beefy one at that. You're likely further ahead finding a cheap bench supply or repurposing a PC power supply.

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