[Amps] MOSFET amp for 80 m

[Angel Vilaseca]

Manfed,

Thank you very much for taking the time to write this extremely detailed 
answer. I will study it carefully.

Vy 73

Angel HB9SLV

Le 21.07.2018 à 19:40, Manfred Mornhinweg a écrit :
> Angel,
>
>> I built your ESR meter last year, and right away, it allowed me to 
>> repair several SMPS I had collected over the years!
>
> Good! It's a really useful gadget.
>
>> OK, since it is a push-pull, the calculation must be done with twice 
>> the supply voltage. I did not tnink of that.
>
> Yes, basically that's it.
>
>> But could it also be said, as you wrote on this page of your website:
>>  https://ludens.cl/Electron/mosfetamps/amps.html
>
>> "When Q1 conducts at the peak of its semicycle, it will have 1.3V on
>> its drain, making 12.5V between Vcc and its drain. That places 12.5V
>> on one half of T2. Due to the tight coupling, that forces 25V to
>> appear across the entire T2, making Q2's drain go up to 26.3V. The
>> primaries of T1 will see a total of 25V."
>
>> Or are these two ways to express in fact the same phenomenon?
>
> Yes, it's the same thing. The 1.3V in that example depends on the 
> tarnsistrs chosen, and the operating conditions, such as current and 
> frequency. It also depends on how much compression (distortion) you 
> are willing to accept. So that number is not fixed at all. But 
> whatever it is, you can plug it into the equation and then calculate 
> the power output you should get.
>
> >Also, the other thing I could not understand was the high output
> >impedance that the author of the project claims: 42 ohms.
>
> He says 48 ohm, but that's not entirely correct. The impedance at 
> which the transistors are loaded is simply the impedance of the load 
> (antenna  or whatever), transformed by the lowpass filter (ideally the 
> filter should reflect the same impedance). So, if the filter is 
> optimal, and you are transmitting into a 50 ohm dummy load, those same 
> 50 ohm appear as the load between drains, given teh fact that there is 
> just a 1:1 balun in between.
>
> >As Bill Turner writes in another post: "The output impedance of the amp
> >should be very small, close to zero and the load impedance around 50 
> >ohms. "
>
> Bill is right.
>
> >That is exactly what I thought too!
> >Instead, this amp uses no output transformer, only a 1:1 balun. How can
> >this possibly work?
>
> I see that you still don't understand the difference between output 
> impedance and load impedance. So here is the explanation:
>
> Output impedance is truly the impedance you would measure if you 
> replaced the load by an impedance meter and measured into the 
> amplifier's output. Instead load impedance is the impedance of the 
> load, and load impedance to the transistors is the load you and the 
> circuit apply to the transistors. The two don't need to match, and in 
> fact in power amplifiers they SHOULD NOT match, because if they match 
> then the efficiency will be terrible!
>
> When the output and load impedances match, you get the highest 
> possible power from a circuit. But nothing guarantees that the power 
> source (amplifier) will not blow up under those conditions! Usually it 
> would.
>
> In the interest of efficiency, power amplifiers are normally designed 
> in such a way that the load impedance applied to them is much higher 
> than their internal output impedance, and the transistors and 
> everything else are dimensioned to handle the actual currents and 
> powers that appear under those operating conditions.
>
> In pure theory, if you have an amplifier with a certain output 
> impedance, the behavior at different load impedances is:
>
> Infinite load impedance: No power output, zero effficency.
>
> High load impedance, compared to output impedance: Modest output 
> power, good efficiency, tending to approach the theoretical efficiency 
> of the amplifier class.
>
> Load impedance equal to output impedance: Highest power output, 
> efficiency is only 50% of the theoretical efficiency of the operating 
> class.
>
> Low load impedance compared to output impedance: Modest output power, 
> very poor efficiency.
>
> Zero load impedance: No output power, zero efficiency.
>
>
> Now let's put some numbers to that small 2*IRF510 amplifier: For 
> simplest analysis let's assume square wave, fully saturated operation, 
> at a frequency so low that the FETs switch very cleanly. The RDSon of 
> that FET is given as 0.54 ohm. So during one half cycle you will have 
> one FET pulling its drain to ground via 0.54 ohm, and the other FET 
> open, and during the other half cycle the roles will reverse.
>
> You have a 50 ohm dummy load, and let's assume that the lowpass filter 
> isn't there, or the frequency is so low that the filter passes the 
> strong harmonics. In that case the unaltered 50 ohm load appears 
> between drains. One drain is in open circuit, but that point is 
> connected to the bifiliar feed choke, which is really a 1:4 
> autotransformer. It will transform the 50 ohm load down to 12.5 ohm 
> and apply that to the conducting FET. Since the FET has 0.54 ohm of 
> internal resistance, the total resistance from the power supply to 
> ground is 13.04 ohm. If the supply voltage is 13.8V, the supply 
> current (and drain current of the conducting FET) will be 1.058 
> amperes, and the input power will be 14.6W. Of this input power, 0.57W 
> are dissipated in the FET, while the bulk of the power, 14.03W, go to 
> the load. Thus the efficiency is a very high 96%, thanks to the load 
> impedance being much higher than the output impedance.
>
> By the way, the output impedance is the 0.54 ohm of the conducting 
> FET, up-transformed in a 1:4 ratio by the bifiliar choke. So it's 2.16 
> ohm - very low, as Bill said.
>
> Now let's get a little more real: We are (hopefully) dealing with sine 
> waves. Saturation of a FET happens at best during the peak of the 
> waveform. This allows placing a lowpass filter of the proper band in 
> the circuit, but also makes the efficiency much lower. How much lower? 
> Well, instead of calculating impedances it's easier to calculate 
> voltages and currents. What was the effective voltage in the square 
> wave example becomes the peak voltage in a sine wave. That makes the 
> RMS voltage decrease to 0.71 of what it was before, and makes the 
> output power one half of what it was before. Thus we get 7W output, 
> not 14W.
>
> The RMS current in the sine wave is also 0.71 times that of the square 
> wave, so it's 0.751 A, but this is NOT the supply current! That's 
> because RMS does not equal average. There is roughly a 0.9 factor 
> between them. So the actual supply current is 0.676A, and at 13.8V, 
> the input power is 9.33W. With 7W output, the efficiency is now 75%. 
> The maximum theoretical efficiency of a class B amplifier is 78.5%, 
> and in our example it has dropped to 75% due to the non-zero output 
> impedance of the amplifier.
>
> Now let's get another bit more real: MOSFETs unfortunately are not 
> linear. They are in fact extremely nonlinear! And switching MOSFETs 
> like the IRF510 are far more nonlinear than MOSFETs optimized for use 
> in linear amplifiers. To more or less linearize an amplifier using 
> switchmode MOSFETs, we need to do two things: Bias them for a strong 
> idling current, moving the operating point deep into class AB; and 
> applying strong negative feedback. The high idling current directly 
> reduces the efficiency (but not much the output power), while the 
> negative feedback eats a big chunk of the gain that would otherwise be 
> available, and consumes power in its resistors, thus also reducing the 
> efficiency. These are the main reasons why such amplifiers in practice 
> only get to roughly 50% efficiency. There is a trade-off to be made 
> between efficiency, linearity and gain, by juggling with the amount of 
> idling current and feedback.
>
> And to get fully real, we need to consider that every component in the 
> circuit has loss, every conductor has inductance, the FETs have some 
> losses related to their limited speed, and all these things cost us 
> both output power and efficiency. Also throughout this analysis I have 
> thought of "impedance" as "resistance", but of course in RF circuits 
> impedances almost always include reactances, which make the detailed 
> analysis a bit more complicated. For example, waveforms at different 
> places of the circuit are no longer exactly in phase with each other, 
> and this effect increases as we raise the frequency.
>
> A good question is what's the actual value of the amplifiers output 
> impedance is, when operating in its reasonably linear range rather 
> than in saturation. If the FETs were perfect, they would act as 
> voltage-controlled current sources. Thus, lacking any feedback, the 
> output impedance would be infinite, given that a change in drain 
> voltage would not cause any change in drain current. But FETs are not 
> perfect, and so they do have a finite impedance, which varies 
> constantly along the waveform.
>
> If the amplifier had extremely strong negative feedback, then this 
> feedback network would basically stabilize the output voltage, making 
> it insensitive to output current changes. This would cause an output 
> impedance of zero! But of course, we cannot have infinite negative 
> feedback. The feedback can only be as large as the excess gain we have 
> available. The result is that negative feedback produces a lowish, but 
> definitely nonzero output impedance. Maybe the author stated the 48 
> ohm output impedance based on the FET characteristics and the amount 
> of negative feedback he used, but such a calculation is imprecise and 
> unreliable, because FET transfer characteristics are only loosely 
> specified, and vary hugely with instantaneous drain current, frequency 
> and temperature. Even if the amplifier should have a (dynamic) output 
> impedance of 48 ohm under some specific conditions, and the load is 50 
> ohm, this is quite irrelevant regarding efficiency and power output. 
> It just means that the negative feedback is roughly 3dB under those 
> conditions!
>
> Since the linear mode output impedance is not useful in calculating 
> the efficiency, better use the method outlined above, and use the 
> actual currents and voltages rather than those calculated for a 
> simplistic approximation. But it's still useful to make such 
> simplistic approximations, to understand the basic situation.
>
> Manfred
>
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