Olin

Let me clarify my position on the cap charge question.
Neglecting pawer factor and efficiency the power source
can deliver 12 watts.  If 12 V is converted to 300 V, the
original stated voltage, the maximum available current is
limited to 40 mA.  300 V X 40 mA = 12 watts.
The current, I = dQ/ dt.  Q = CE so I = C dE/ dt and
I dt = C dE.  Integrating both sides,  It = CE and t = CE/I
The time t = 160uF X 300 V / .04 A = 1.2 Seconds.
This time was confirmed on Ansoft's program "Serenade"
and Electronic Workbench version 5.1  They show a dV/dt of
250 V/s.  1.2 S X 250 V/S = 300 V.  Of course with lower
duty cycle we may be able to use 80. mA and reduce the
charge time to 0.6 seconds.

Charlie W8AWS

Midland Electronics
     Specialties

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