fatcat2

modified 5 years ago

Class-E

368

03:51:25

All the class e component values were obtained from a class e design tool. Notice the RLC LPF at the output. This performs two functions; removing the high order harmonics as well as acting as an impedance matching network to convert the 50 ohms load to 25 ohms for a minimum power dissipation. The calculated efficiency was around 90%; is this the actual efficiency of this circuit?

Edit 1:- Thanks @zorgrian! You seem to be right. After meddling with the RF choke, I reached a point when the power dissipation is the least. This works very well, although a lower inductance shifts the wave slightly towards the negative side. 10uH works well but the efficiency is reduced in that case, coz the current transient is steeper. Also, the starting time has been improved. With 10uH, there's enough time to discharge the stored energy so that the output voltage can soar even higher.

Edit 2:- Used trial/error to figure out the right cap value at which the resonant tank, influenced by the low-value inductor, causes a minimum overlap of current and voltage. Thanks for introducing this new idea, @zorgrian.

Edit 3:- So which one actually wins in terms of efficiency? More importantly, I dunno to calculate the efficiency of this amp. 

Edit 4:- Among the last two, the 2nd one is the most efficient one, with an efficiency above 92%. The last one only has an efficiency around 62%, coz the choke was too small to limit the current through it.

published 5 years ago

fatcat2

5 years ago

More filters can be used but this seemed perfectly fine.

jason9

5 years ago

To measure the efficiency accurately you can slow down the simulation by placing a disconnected frequency source at something like 5MHz. I’ve noticed that without such slowing down even 99+% efficiency circuits can be reduced to only 70-80%. To speed up the startup time you can replace the voltage source and choke with a constant current source. That way you don’t need to wait for the inductor to get up to maximum current.

jason9

5 years ago

Also in case you’re interested I made a circuit a long time ago with some filter equations: http://everycircuit.com/circuit/6027199319900160 If you make the filter according to the equations then at the appropriate frequency (the one entered into the equations) the input of the filter will appear to be a purely resistive load with the resistance equal to that of the resistor at the output. This makes it good for putting at the output of a class-E since it’ll appear like it’s not even there from the viewpoint of the class-E amplifier while still filtering out the harmonics. Of course due to the harmonic filtration it’ll have a small effect, but nothing more than simply small. And if the output of the class-E is already fairly pure then adding the filter will have even less effect.

fatcat2

5 years ago

@jason9, the network I used performs two tasks; impedance matching and filtering. The equations for that are pretty different. Anyway, it's at resonance.

zorgrian

5 years ago

The choke L looks too large to me. I guess this was part of the calculation. However, if it is this large then energy is still being pumped out of the choke, for too much time. The result of this is that when the MOSFET is OFF the energy still thumps out, which reduces the efficiency. What we really want here is a hump of energy being taken out of the choke (which must be thought of like a kind of spring and not just a choke) for a short time only.

zorgrian

5 years ago

So, I reduced the choke to 10uH and this seems a bit better.

zorgrian

5 years ago

https://hyse.org/pdf/www.aoc.nrao.edu/~pharden/hobby/Hobby.shtml

zorgrian

5 years ago

The above link is worth checking as it contains Paul Harden's PDF files. Paul's explanations on RF amplifiers are the best. They are awesome!

jason9

5 years ago

That link doesn’t work for me. Also, every class-E design I’ve seen online has a large choke to effectively make a constant current source.

fatcat2

5 years ago

Well, there is a significant difference of nearly 1V between two negative peaks. What's happening? I tried increasing the choke to 47uH (in my new circuit), which resulted in a drop in the overall efficiency along with a much lower "negative ripple magnitude". Dunno why.

fatcat2

5 years ago

In @zorgrian's pdf, it's stated that the RF choke forms a tuned resonant network with the drain shunt capacitance. My first intuition was based on this logic, which eventually collapsed. But now, I think that I'm going back to my previous idea.

zorgrian

5 years ago

@f, this now seems much better. The trial and error method, coupled with some RF knowledge is not as crazy 🥴 as it sounds. Class E necessarily is a tuned system, much like class C but where we deal with hard driven ON OFF states of the MOSFET. In this and most working class E RF amplifier circuits, the choke is as I said, it is like a spring and must be of a value that is 'tuned' to a frequency band and that will work with the Z that is formed by the LPF network. The MOSFET impedance when ON is very very low, so the main Z here is the load. Having too high an inductance in the choke does not help at all here.

jason9

5 years ago

Well, at the very least, it can still work very well (99% efficiency) with a large choke: http://everycircuit.com/circuit/6122692830035968 In fact replacing the choke a voltage source with a current source of 388mA (the average current through the choke) to simulate an infinitely large choke still produces good results that can even be made to perfect results with a bit of re-tuning of LC elements.

fatcat2

5 years ago

Yup; works with a current source. But I don't get how you calculated the efficiency as 99% with a current source; I tried using a 300mA current source but it caused an unmatched waveform (a need for re-tuning the component values). Basically, the MOS charges the RF choke so that a current source is formed. There's no need for giving off all of the energy stored in the choke as it isn't dissipated. Does that explain why two consecutive negative peaks have different peak voltages in my above circuit?

fatcat2

5 years ago

P.S. I don't see any LPF in the actual class-e amp diagram; I've used one here just for transforming the output impedance to a lower impedance, as specified by the online calc.

jason9

5 years ago

There is no LPF in a pure class-E. However, since radio requires a very pure sine-wave so as not to transmit on frequencies at multiples of the base frequency, an LPF is required to purify the output waveform.

jason9

5 years ago

To calculate the efficiency I multiply the average current draw by the input voltage to get the input power and then I assume the output is a pure sine-wave (usually works as long as there isn’t too much distortion) and calculate the output power using the equation V^2/(2R) where V is the peak voltage and R is the load resistance. From there I can calculate the efficiency by dividing the output power by the input power and multiplying by 100 to get a percentage.

jason9

5 years ago

Also, 99% efficiency is only possible if the MOSFET has a very low resistance when on and switches on very close to or at the moment when the voltage hits 0V. Ideally the voltage should be at it’s minimum when it hits 0V so as to provide the longest possible time at 0V for the MOSFET to switch on since real world MOSFETs and MOSFET drivers aren’t instant, but in EC this isn’t much of a concern.

fatcat2

5 years ago

Calculating the input power drawn by circuit no. 3 is a bit tricky as the waveform is sawtooth, along with a DC offset. Thanks anyway. As @2ctiby had told, I maxed the width so that the ON resistance is very less. Edit 4:- All virtual FETs replaced by IRFZ44n (doesn't work IRL). Does the input spike of current reduce the efficiency by a considerable factor?

jason9

5 years ago

You need to slow down the simulation to obtain accurate efficiency results. Without doing so even my 99% efficient class-E is only 70% efficient. To slow down the simulation while still maintaining maximum speed place a disconnected high frequency sine-wave source. Usually about 10 times the frequency of the circuit works well, so in this case 7.5MHz. For the sawtooth it should work pretty well (not necessarily perfectly, but still pretty well) to average the current at the low-peak and high-peak of the sawtooth and multiply the average by the voltage to get the wattage. Also, by input spike do you mean when the simulation starts or when the MOSFET driver turns on the MOSFET creating a spike of current through parasitic resistance?

jason9

5 years ago

If you mean when the simulation starts that doesn’t contribute to efficiency since it’s only when it starts and not during the bulk of the runtime. If you mean when the MOSFET driver turns on the MOSFET, well I did an LTSpice simulation a while ago of one of my class-E amplifiers and it had 90-95% efficiency, and the missing power didn’t add up so I assume it’s due to simulation imperfections and the efficiency is around even as high as 98% despite MOSFET capacitance.

fatcat2

5 years ago

Exactly. These calculations were made by slowing the time to 500ns/s. Are they accurate? Also, I used the RMS value instead of simply averaging the waveform. What MOSFET model did you use in LTSpice? I tried to simulate it but it didn't work because I didn't specify any NMOS model. Did you import any model (say, IRFZ44n)?

jason9

5 years ago

Did you go to 500ns/s by turning the dial? If so, that’s useless because placing a frequency source (square wave, sine-wave, etc.) locks the speed at no greater than 1/(3f) S/S, so the 1MHz source keeps the simulation speed at no faster than 333nS/S and you only experience 1μS/S. If you got to 500nS/S by increasing the 1MHz source to 6MHz to reduce the maximum experienced speed to 500nS/S, that reduces the actual speed to 55.5nS/S, which is pretty good, although I’d personally prefer 7.5MHz rather than 6MHz since 7.5MHz is 10x the operating frequency which is 750kHz.

jason9

5 years ago

I used the FDB33N25 since that’s what was used by this website http://people.physics.anu.edu.au/~dxt103/class-e/ which was the website that inspired me to use LTspice.

fatcat2

5 years ago

I started the simulation, turned the dial to 500ns/s and then again restarted the simulation so that it starts at that speed. Was everything done alright?

fatcat2

5 years ago

I made the circuit in LTspice. Works just fine, except that at 1ns step time, the THD was aroung 2.64%. BTW, the model that I used was BSC118N10NS. The large gate capacitance was taking a toll on the efficiency; there was a significant amount of delay. So I adjusted the shunt cap to 1n and it works just fine. The calculated efficiency was above 99.64% !!! Could've never got such an accurate result in EC. (This time, I didn't take the RMS of the input current)

jason9

5 years ago

Starting at 500nS/S versus turning the dial on the fly doesn’t make a difference. You need to increase the disconnected 1MHz source to 5 or 10MHz to get a properly slow simulation at a reasonable perceived speed (perceived speed being the speed on the dial). The actual speed is always as close to the perceived speed as possible unless kept lower by a high frequency source, so to simulate well with a reasonable perceived speed increase the frequency of the disconnected frequency source. Also, you can safely set LTspice step to 0 and it’ll automatically adjust itself to what it thinks us appropriate. Raising it higher will (as I understand) only make it more inaccurate.

fatcat2

5 years ago

Thanks for that EC tip. Now considering LTspice, it's important to set the step time and turn off the compression. Doing so provides accurate results. The auto-optimization isn't accurate. I tried with a class-A amp and found out that reducing the simulation step to a suitable value gives more accurate results, increased precision and improved THD results. There's a video based on this where @Fez_electronics explains this.

jason9

5 years ago

I see. I’m new to LTspice, so thanks for the tip. Also, what is compression and how is it turned off?

fatcat2

5 years ago

Whenever you simulate something, for example a large circuit, the data for that (including the scope data, fourier data, etc) are stored in the LTC folder location. Without compression, the data file can be very large (up to 150 MB, not considering the schematic data). To avoid this, LTspice compresses the data. But information is lost in this case. So we need to turn off the compression. For that, click the option (in the top panel) with the icon of a hammer and you'll see the control panel. You will also see three tick marks. Un-tick them. That's all. For 1kHz, a step of 250ns will be sufficient. For some cases, the simulation runs so slow that you'd have to increase it to 1us or something.

fatcat2

5 years ago

The format is:- .tran 0 <simulation time> <time to atart saving data> <max step time>

jason9

5 years ago

I see. Thanks.

fatcat2

5 years ago

A while back, you had suggested to take a look at one of your circuits in which you had explained all about di-pole butterworth filters with a DF of approximately 0.7071. The only problem was that there was a significantly large imaginary component along with it. Using another set of equations helps you calculate the values needed for getting a particular input resistance which is lower than the load resistance (here it's 25 ohms). This also let's you calculate the values which will ideally give a zero ohm imaginary component at the resonant frequency. Here you go:- XL = sq.((Ri*Rl)^(2)-(Ri^2)) and Xc = Ri*Rl/XL where XL is the inductive reactance and Ri is the desired input resistance. At the resonant frequency, peaking occurs and the imaginary component is way too small. There will be a phase shift of 45° at the output, however. I got these equations from the net; L-network impedance transformation.

jason9

5 years ago

That’s cool, although the more advanced filter math is still beyond me. I’ll have to play around with it some more to really understand it. And the imaginary component can be canceled out with a parallel inductor or capacitor.

fatcat2

5 years ago

Sorry but the math was wrong; Xl = sq.(Ri*Rl-(Ri^2)) Sorry for that. https://everycircuit.com/circuit/4952540519858176 here's a comparison (sorry for the tonnes of unlisted threads I make). P.S. the mathematical derivations of the filter equations are pretty complicated. I type these formulas in notes and use it whenever I need it.

jason9

5 years ago

That’s interesting. Why doesn’t the top one need an imaginary-canceling cap? Is an impedance reduction of 10 fold some special number that removes such a cap?

fatcat2

5 years ago

Sorry for the delay; I read your comment only a few mins ago. Okay so I did some calculations. Seems that the reactive impedance of the inductor gets cancelled by the parallel configuration coz the cap and the resistor make a phase shift which causes the imaginary part to be equal in magnitude with the inductor's reactance. Okay, sorry for the dumb "explanation". The math is way beyond me dude. Involves a lot of trigonometry; tan, cos and sin where the tan is the most annoying stranger. Lot of variables there. Here's the calc's result: XL = 30 ohms (I had approximated the values while making the circuit and so I'm using the formulae here) and XC = 33.33 ohms. ((ZC)^(-1) +(RL)^(-1))^(-1) = (0.03i + 0.01)^(-1) = 1/(0.0316<71.57°) = 0.0316<71.57° = 10 - 30i. The imaginary part cancels each other. Clearly, the reciprocal of the 0.0316 is itself. But a simple proof (of a single case) doesn't explain why this happens. Unfortunately, I couldn't fins the derivation of these formulae.

fatcat2

5 years ago

This reciprocal thingy occurs in every case.

jason9

5 years ago

jason9

5 years ago

If you’re having trouble with the trig of the vector math, you can stick to just complex numbers. When using complex numbers, all the math works just like Ohm’s law but with capacitors and inductors having imaginary resistances. The resistance of an inductor is always the impedance (voltage divided by current) multiplied by sqrt(-1), and the resistance of a capacitor is always the impedance multiplied by -sqrt(-1). From there you can derive the behavior of more complex systems like filters, all without a single sin/cos/tan.

fatcat2

5 years ago

Oh, thanks but for the resultant impedance, this is necessary. This works just fine here. This means that I shouldn't convert the rectangular form back to the polar form. Thanks.

jason9

5 years ago

No problem.

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