I think that when the transistor is off and the coil is discharging through the LED to ground, a discharge circuit is created via the supply battery back to the coil, so the battery voltage is added (series aiding) to the discharge which results in more power dissipated in the load.
Hoppy
I think you're on the right track here Hoppy. Yesterday in a chat with ION, I was trying to remember something I was going to point out, then I forgot. What I was going to mention was a "trick" I could use to not only make the efficiency worse, but it serves to help prove the point that it is about potential difference. Imagine adding another battery or variable power supply to the circuit and you discharged through the LED into that rather than GND or Vbat. All are good AC grounds, but the DC level seems to make the difference. If I set my second power supply to say 3V, the power into the LED will be even less than it was when tied to Vbat. It is somewhat analogous to attempting to discharge one capacitor into another, when both are already at the same voltage. It just doesn't happen. The primary inductor becomes a current source when it is discharging its energy, so it doesn't really care where it is terminated, but in order to have more current for longer (and less voltage), it needs to see not only a low impedance, but a potential far away from that at the other end, which is tied to the Vbat. So without our variable voltage supply, the place that allows the greatest potential difference is of course GND. So what would happen if we set our variable voltage supply to say -3V, and discharged through to that? .99
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"Some scientists claim that hydrogen, because it is so plentiful, is the basic building block of the universe. I dispute that. I say there is more stupidity than hydrogen, and that is the basic building block of the universe." Frank Zappa
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