Quote:
Originally Posted by cts_casemod
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Does this sound correct?
Assuming a 50% duty cycle square wave and a battery (Vs) voltage of say 20V and a current of 100A.
SW1 closed
Battery Volts = 20V
Inductor Volts = 20V
Capacitor Volts = 40V
Battery Amps = 100A
Inductor Amps = 100A
Capacitor Amps = 0A
SW1 open
Battery Volts = 20V
Inductor Volts = 20V
Capacitor Volts = 40V
Battery Amps = 100A
Inductor Amps = 100A
Capacitor Amps = 100A
So the capacitor is supplied double the battery voltage for half the time at battery current.
D1 is there to prevent the capacitor being shorted when SW1 closes.
The battery delivers a constant 100A at 20V.
The inductor charges and discharges to 20V at 100A.
As long as the load placed on the capacitor discharges at an average current of half the battery current or less then the capacitor will supply twice the battery voltage.