Good idea about the thread. I dunno if you've found the answer. I think what happends in when the switch (MOSFET or BJT) is closed DC becomes stored in the inductor. When the switch opens, energy leaves the inductor, adds to the supply voltage (this 'steps up' the voltage) and charges the capacitor. it's similar to an AC transformer, but for DC circuit. Pin=Pout. Therefore, as you are stepping up the voltage the current on the output will decrease. Lol thats the best answer I could come up with. Never seen one. Whats your project on?