flowing current wastes power

Hi, I don't quite understand why a Class-E Power Amplifier is efficient. See the attached Class-E diagram.

My understanding is that C1/L1 resonate at the switching frequency, so that the transistor drain is at zero volts when current is flowing, and C2/L2 filter out unwanted harmonics.

But for C1/L1, when current is flowing (with the drain at zero volts), current is flowing from the supply to ground. No power is dissipated in the transistor due to the zero-volt drain, but this energy still comes from the supply and doesn't go to the load - so surely this is wasted energy, and so lowers efficiency? I believe that the Class-E can be 100% efficient, but I guess I don't understand something because I don't see why...

Ansu

Actually for Class-E operation C2 should be in series with L2.

Read the article written by Nat Sokal WA1HQC which is the inventor of the Class-E PA, and you will understand the behavior of the circuit.

http://www.arrl.org/tis/info/pdf/010102qex009.pdf

Hi, thanks for the reply. I still don't quite understand though, sorry!

I've seen this paper and unfortunately it hasn't helped clear things up. In his words: "Efficiency is maximized by minimizing power dissipation, while providing a desired output power.".

My understanding of this is that he's saying that you want to transfer power from the supply to the load without dissipating any power in the process. But there are three terminals in the system: supply, output and ground. If 45% of the system power is transferred to ground, and 45% of system power is transferred to the output, and only 10% of system power is dissipated in the process, then according to (my interpretation of) the statement above, the system would be 82% efficient (0.45/(0.45+0.1)).

But surely you also care about all the power that you're (very efficiently) dumping to ground during your zero voltage switching moments?

Ansu

The basic confusion is in your assumption of creating losses when "system power is transferred to ground". While the switch is closed, the current rises and energy is transferred to L1.

The real class-E amplifier is however creating some losses due to finite switching times. Also discharging C1 to the transistor is necessarily involving losses. But these are minor contributions, so the overall efficience around 80-90 % can be achieved.

Right, I think I understand now - current is sort of the effect of moving energy, so by moving energy from the supply into the inductor, a current flows through the switch. Is it sort of like a capacitor with the bottom plate attached to ground? ie. if you add a positive charge to the top plate, the bottom plate attracts a negative charge, which looks like a positive charge being repelled into ground. But the charge we added didn't go to ground (the energy wasn't wasted).

I've simmed the Class-E using the design equations in that paper - it does give very high efficiency, just about 100% (with ideal components), so it does seem to be my understanding of how the inductor energy transfer behaves that's the problem.