AVR-Chat

Oooops -

I slipped on the power part. It should have been computed
from 6V, not 5!

On Fri, 24 Sep 2004 12:08:28 -0400
 Robert Adsett <subscriptions@aeolusdevelopment.com> wrote:
> 
> 
> 
> 
> 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
> 
> 
> 
> " 'Freedom' has no meaning of itself.  There
> are always restrictions,
> be they legal, genetic, or physical.  If you don't
> believe me, try to
> chew a radio signal. "
> 
>
                        
> Kelvin Throop, III
> 
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