if duty cycle is 50%, it means that only for 50% of time input
> voltage is connected to the load. So average output voltage in
> this case is Vcc/2 and power dissipated on the load is 4 times
> less then if there was no switching.
Power depends on the RMS voltage, not average voltage, so it's only
2 times less. (Another way to look at it - you're switching the load off for 50% of the time, so the power dissipated will be halved.)
I did not bother reading much of the original article because it probably covers the topic well and straightforwardly.
What Andrew Gabriel does not realize is that this kind of application most likely uses an inductive input ripple filter. This filter smooths the pulses into a dc output with low ripple. sufficient load current flowing so that current in the inductor never goes to zero. Look up "inductor input ripple filter critical current."
Under such circumstances, power during the too high pulse voltage is stored in the inductor. when the pulse is turned off, current continues to flow through the inductor through a free-wheeling diode (or bridge rectifier). That allows the energy stored in the inductor to provide power to the load while the pulse is off. The result is that except for power lost in the semiconductors from switching transients and diode drops, the efficiency is going to be close to unity. There are a few other losses from residual resistance, or magnetic losses.
The higher frequency you can chop, the smaller you can make the filter components, although other losses may go up.
To sum up, dc to dc converters of this nature can be made highly efficient. The efficiency will be limited by how large you are willing to make the components.
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