🚀 go-pugleaf

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In article <1990Oct6.220317.18337@mthvax.cs.miami.edu>, wb8foz@mthvax.cs.miami.edu (David Lesher) writes...
> 
>{plan to rectify ac-->dc with cap. filter}
> 
>>  This is a BAD idea. If you size the capacitor large enough to get
>>rid of the ripple you will be running the blanket at the AC peak
>>voltage not 110. This will mean it will be running at about twice
>>its rated power. The controller will probably still work, but the 
>>resistance wire in the blankets will get hot too quickly. You are
>>asking for a fire! Please don't do this for safety reasons!!!
> 
>Sorry, I disagree. 
> 
>You only charge the cap up to peak if the load is not there.
>With a load, the cap cannot stay at peak.  What you are implying
>is that you somehow get double the energy from the wall while
>drawing the same current.
> 

    Not at all. Unless there is some current limiting in the
supply, such as winding resitance to limit the current, then
you will get 20 or 30 amps at the peak to charge the cap and
zero the rest of the time. Since he is hooked up to a very
large transformer with likely 100 amp service into the house
he has to draw a lot of current to lower the peak voltage.

   If the cap is near peak sometime during the cycle then there
are 2 possibilites. If it is small compared to the load
and he will get lots of ripple, and a lower average voltage.
If he has lots of ripple then he has a large AC component to
his power, and he hasn't eliminated those 60hz fields he wanted
to get rid of. If on the other hand the cap is large enough
to sustain the voltage with little drop for one half cycle
then it will always be near the peak voltage. The power equation
still works becuase his RMS voltage will be 1.4 times the
aveage line voltage, and thus the RMS current = Vrms/R
will also be 1.4 times as high.

  (1.4 is short hand for square root of 2)

   (sqrt2*V0)*(sqrt(2)*I0)= 2*V0*I0 == twice the power.

					Chuck@mitlns.mit.edu