Let's Make Robots!

question about arduino and supply current

hi guys, I just bought a ucam ttl jpeg camera for my arduino and have some very basic questions.

Reading the dataset it says that to power the camera it require 3.0V to 3.6V DC range nominal 3.3V. So I plug it into the 3.3V pin from my arduino mega, setted it up and it worked.

However, im a little confused about the specifications from the datasheet.Namely it says that

Supply Current (ICC) VCC =   3.3V    58(min) 62(typical) 76(max)   mA

But the current draw from the 3.3v pin, according to the website is only 50mA.

So, can someone explain to me why it worked, or am i not understanding things correctly.




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If you leave the camera connected then the 3.3V regulator will overheat and could burn out completely. If your robot is going to be working in sub zero temperatures then you might be able to leave it connected.

The 50 mA is a rating that "should not" be exceeded.  It doesn't mean the supply pin will immediately shut down when you exceed 50 mA.  Kind of like a car engine rated to go 6000 RPM.  You can still get it to go 7000 RPM, but it probably isn't good for it.  The 50 mA rating also takes into account (or should if properly designed) all the other factors (like arduino on-board current, other peripherals, temperature, and other things) that affect the available current, and 50 mA should be what's safely left over after all other worst case situations are accounted for.  It may provide as much as 75 or maybe even 100 mA under good conditions, but it is always best to stay in spec.