AVR-Chat

At 11:24 AM 9/24/04 -0400, you wrote:
>It says that it MUST operate down to 4.80V. So, you should
>be safe with a 5V supply. It says that it will work up to
>130% of the rated power dissipation on the coil. Lets go
>through the arithmatic:
>
>Coil is 100 ohms. Rated voltage is 5V. Power is V * I  and
>V = I * R so power can also be written V * V / R. The
>numbers are then 5*5/100 = 25/100 = 0.25W.

We must be reading different data sheets ;)  The rated voltage is 6V and 
the pull-in (must operate) is 75% of that.  I get 4.5V.

Power is nominally 60ma at 6V or 360mW

>130% of this is 0.325W. The coil resistance has not changed
>so we can solve for voltage or 5.7V.

Which gives a max power of 468mW which for 100 Ohm coil gives 6.84V.


>The coil current is one of the reasons your regulator is
>getting warm. If the input voltage is 12 and the output is
>5, the drop is 7V. 7V * 60ma = 420mW. That is almost a half
>watt, just from the relay!

I must have slipped a decimal point or something on this one.  You've got 
this correct (although off of 5V it will be a little less).Somehow I got 
the power loss 3x higher.  Always triple check your calculations Sigh.

Nonetheless I think the basic conclusions are the same.

Robert



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