Class-E Amplifier

I tuned the transformer to not have the nasty 100mOhm at primary side and got about 84% efficiency http://everycircuit.com/circuit/6529658261864448

You linked me the wrong link.

http://everycircuit.com/circuit/6339603274661888

Wait, I’m confused. How does increasing the transformer inductance from 10uH to 360uH change its resistance? And also, even then, isn’t 84% a bit bad for a class-E? The site I got this from said that it had an efficiency above 90%.

If I add negative resistance resistors to the transformer to null any DC resistance, and I make the MOSFET perfect, it still only gets 85% efficiency, and then if I increase the transformer’s inductance to 360uH in addition it still gets 85% efficiency. Where is the loss?

Actually if you look closer to the transformer, I didn't changed anything. I just fliped primary to secondary and adapted the inductivity. Nothing else.

Oh, I see. But still, why is it so inefficient even when all resistance has been removed?

All resistors are not removed. At least there is the 100mOhm at output. But right its not consumming to much power. I guess we do not calculate correctly the input power and also the output. Cuz, the out is not a pure sinwave and the input we (at least I guess you did the same) averaged the current which is alao not correct. Another thing is the simulation speed, maybe its much to fast. Also there is a lot reactance which is difficult to handel. More reasons?

Huh, I removed the transformer resistance with negative resistors, and averaging the input current shouldn’t cause a very noticeable amount of deviation from the correct value, and the output is very very close to a pure sine wave, and the simulation speed shouldn’t be too fast with the square wave source holding it back, and I don’t imagine that reactance can be much of a problem, but it might actually be a problem because I once made a circuit where energy seemed to be disappearing into this air when it should have a 100% efficiency, so actually reactance might be the problem.

I think is by far not a pure sinwave check this we loss a lot energy in harmonics http://everycircuit.com/circuit/6339603274661888

Ok, I decreased the simulation speed to help handle the reactance and took away the resistance and got 99.3% efficiency. I’ll check the efficiency with the normal resistance.

With normal resistance, it gets 75% efficiency, and then with negative resistors used to decrease the transformer’s DC resistance to 10mOhms I got 90% efficiency. I say problem solved.

Ok, with 0ohms transformer resistance, just MOSFET resistance, it gets a nice and even 90% efficiency.

Oh, sorry, I didn’t see your comment until just now. I don’t think that’s actually a problem because I measured 99.3% efficiency even with the impure sine wave, so I suppose it doesn’t matter that much.

Ok, but if you take away the harmonics from output power it will drop again a little

Oh, I suppose that’s right, but the measurements should be close enough anyway.

Class E is the most efficient PA output system in RF. It doesn't need the transformer and could be much simplified.

http://everycircuit.com/circuit/6134050686500864

Check the above circuit

Your circuit could be much simplified. I just put the transformer there for impedance matching and the extra components are a low pass filter to filter out any harmonics. Actually, I think I’m going to lengthen the low pass filter to better filter the harmonics.

Turns out I didn’t need to lengthen the filter.

@hurz, those harmonics were caused by clipping of the wave by the OP-amp.

Right, shit clipping at +-15V 😵, anyway I see about 3% of harmonics you loss for the efficiency.

3% is not that big a deal. If it’s 50W out, then 3% equates to 1-2W extra dissipation which is practically nothing.

I never said its a big deal. But you have to take this into account when calculating efficiency, cuz the harmonics are part of the output which yourself said you see as perfect sinwaves and calculate the rms from it by dividing peak / 1.41. So the out looks a little better/higher then it actually is. Nothing else.

Actually, to calculate power I use the I^2*R method, and to calculate “I” instead of (peakCurrent*sqrt(1/2))*(peakCurrent*sqrt(1/2)) I do (peakCurrent^2)*(sqrt(1/2)^2) which simplifies to (peakCurrent^2)/2 which removes the possibility of error from using 0.707 instead of the actual square root of 1/2, but none of that really matters. Anyway, harmonics, depending on their order (2nd, 3rd, 4th, 5th, etc) and phase they can either increase the peaks like in a triangle wave, or decrease them like in a square wave, so it depends on the type of harmonics whether you end up overestimating or underestimating the efficiency. Anyway, 3% isn’t that much, so it shouldn’t matter so much. It’ll only matter when you can actually see the distortion on the wave itself.

I use an UPN calculator and do not have such problems. And 3% are 3%. Its not much but its not nothing. And funny how you explain harmonics.