> -----Original Message-----
> From: Roman Black [mailto:fastvid@EZY.NET.AU]

> Hey, this is a good argument! It would be cool if some
> others added their 2c worth??
> -Roman

well, ok. Here's my 2c, but please remember to adjust for inflation:

> I think once the inductor is "charged",
> ie, it's field is established, the current fed in is
> transferred to the output. So each time Q1 turns on,
> the current is fed straight to the output

But when Q1 is off, where does the energy go? Does it not get wasted as heat
in the diode and feedback windings? When Q2 is off, it certainly does not
get transferred to the output. Slow decay or not, the inductor is going to
discharge through the diode. You charge the inductor, but then don't do
anything with the energy you stored there.

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