[Top] [All Lists]

Re: [Amps] MOSFET amp for 80 m

Subject: Re: [Amps] MOSFET amp for 80 m
From: Angel Vilaseca <>
Date: Thu, 26 Jul 2018 14:46:58 +0200
List-post: <>

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 :

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:

"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.


Visit my hobby homepage!
Amps mailing list

Amps mailing list
<Prev in Thread] Current Thread [Next in Thread>