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DynamoBen
- 15th September 2006, 22:20
I'm a hobbyist with a fairly basic question. Its something I've never had to deal with.

In many of my projects, I have installed a 7805 by default knowing that I wouldn't ever need more than 1A of power. However I'm at the point were I would like to switch to a 78L05 to save on space. How does one go about calculating whether the 100mA the 78L05 will be enough?

precision
- 16th September 2006, 05:40
You can use dc-dc converter as small as 2mm X 2mm

http://www.linear.com/ltmagazine/LTMag_V16N2_Jul06.pdf

DynamoBen
- 16th September 2006, 05:44
You didn't understand the question.

I'm trying to figure out my current draw based on the my schematic.

Jerson
- 16th September 2006, 06:28
The method I would choose is to be absolutely sure. What you do is to measure the current with the worst case load scenario and build your power supply for this. The other technique, which I think you are asking about, is to check the datasheets for the parts involved and check the Icc requirement of each. The total current required will tell you if the 78L05 can handle this.

Jerson

Melanie
- 16th September 2006, 10:21
The easy way...

Cut the +5v Supply lead, insert a DVM to measure your current consumption. Write some code to switch-ON (or work) every possible peripheral that draws current. Add a little as a 'safety margin'. Personally I never run any 78L05 over 50mA which allows for a 100% safety margin. They do get hot when you run them near their maximum, and I don't like anything getting hot in designs...