MOSFET RF power amplifier design

[spacecase0]

I am not sure of the calculations to figure out the idle current,
because each half of the sine wave is created by separate mosfets (or sets of them), you have to set the idle current so that the waveform matches up at the zero voltage point.
each mosfet is slightly different, so you need adjustment there anyway...
and you sure don't want to much idle current as that just makes waste heat.

any more questions, or is this all set now and it time to bring up how you match impedance to what ever you are running the output power into ?

[Buddhafollower]

Matching impedances isn't hard at all actually. Just the square root of the primary impedance divided by the secondary impedance and this gives the ratio of the transformer.
There are cases, as I have seen, when you have to match impedances for the input of the amplifier. At low frequencies the impedance of the mosfets is higher than the impedance of the driver, which in my case will be 50Ohm. So in such a case I invert the quantities, meaning that I divide the input impedance of the transistors by the 50Ohm input impedance of the amplifier. So I do the calculation this way. It seems the correct thing to do because I really see no other way.

[spacecase0]

there are ways to match impedance very closely,
this is just one page that seems to talk of popular ways
https://www.hamradiosecrets.com/home...nna-tuner.html

if you have impedance that changes all the time, there are ways to deal with that as well, but hopefully you don't have that going on.

seems like you are mostly set to do what you want to do.
good luck and I will answer more questions if you have any.

[Buddhafollower]

You see in this image there are two capacitors at the input of the amplifier. Two capacitors of 100nF after the secondary of the signal transformer. How do I calculate their values? If i follow the example you gave me with the input impedance, there will be two capacitors of the mosfets in series with these two capacitors of 100nF. Is it accurate?
In this second image that I uploaded i specified how I see things.

[spacecase0]

your second image seems to be correct to me.
the 10 ohm resistor in there is going to set most of the input impedance
and with that 5:2 transformer... (2.5 (turns ratio) squared gets you impedance ratio, so 10ohms times 6.25)
is that 62.5 ohms in for just the resistor and transformer,
and it is lowered by the 4.72nF to ground, but that will change with frequency,
so, my guess is they are trying for about 50ohms in with that circuit (as that is a very common input number in radio electronics.
and likely your signal generator is 50 ohms out as well (all of mine are).

[Buddhafollower]

I was thinking that they put they two 100nF capacitors there in order to slightly decrease the gate capacitance. That's why I had drawn them like that in the second image. This value of capacitance is appropriate for the frequency they use also decrease the input impedance of the mosfets a little. That's what I think. That's how I see the problem. They chose a value that will have a low reactance at certain frequencies.

[spacecase0]

the capacitor is in there because you need to isolate the DC for the bias voltage,
they chose a value that is 25 times the gate capacitance, that is large enough not to loose much.
you could choose larger value, but there is no point, and you just risk DC leakage.

[Buddhafollower]

In the above image of the amplifier I got an input impedance of the mosfets of 31.84Ohm( at the frequency of 1MHz). The ratio of the transformer will be 1.253 from my calculation. What I don't understand is where did they get the 5:2 turns? You said the ratio is 2.5 but I still don't understand... I don't understand what is the purpose of that resistor parallelled with the secondary of the input RF trafo and how exactly is that going to set the input impedance since I done the math and got a totally different thing?

[Buddhafollower]

Yes, well... I guess you are right. I did the calculation for the 30m band, meaning 10.15MHz, and indeed the ratio is 2.5. But still I know not how they picked the 5:2 number of turns. If I divide 50Ohm by 2.5 I get 20. So I guess that will be the ''2''. Or you round it up to the nearest full number. And then you have 2.5*2= 5 turns. So we will have in the primary 50Ohm, 5 turns and in the secondary, 314.46Ohm, 2 turns.
You have to know that I divided the bigger impedance by the smaller (50Ohm). I had no other choice.
Please correct me if I'am wrong.

.........
And still I do not understand. From where is this 5:2 number of turns? So 4 divided by 2.5 is 1.6 and it doesn't work well enough. Then 5:2.5= 2. So 5:2.
But then another question arises in my unripened mind, which is why is the secondary of 'two turns' supplying the gates of the mosfets with signal? I don't get it.
I guess that you will feed a smaller inductance to a higher impedance.

[spacecase0]

5:2 refers to 5 turns on the primary and 2 turns on the secondary.
the ratio of 5/2 is 2.5, but that is for voltage and not impedance
impedance ratio = turns ratio squared.
and that would be 6.25 for the 5:2 transformer
and looking at your math, you are dividing the wrong direction,
the lower number of turns will have the lower impedance.
so the 2 turn side is 8 ohms if the 5 turn side is 50 ohms.
the 2 turn side is run it into a 10 ohm resistor, so that gets it all pretty close without the mosfets even being connected.
the mosfets are an additional load, so it lowers that 10 ohm number a bit (the target being 8 ohms)

the reason why they do all this is that radio waves in a circuit will bounce back if they hit an impedance that is not the same as where it came from.
the more the impedance mismatch the more of the radio wave is bounced back.
radio waves coming back into the input can do bad things
so, you have to set up the input of the amplifier to match your source.
and transformers can't easily be made with partial turns...
you also can't have to many turns or you can over magnetize the core...
so you might not have ratios like 12:3 (that might be a closer number match for this circuit)
so there are limits.
the impedance of the mosfet input is reasonably high,
so they needed additional load (the resistor)
and they needed the voltage in to the gate of the mosfet to be in the correct range, and this also sets the turns ratio of the transformer.
so whoever designed this circuit likely choose values that are close enough for a wide range of conditions
designing is all about tradeoffs, and this circuit is not bad at all.

make sense now ?

[Buddhafollower]

Is it all just a matter of voltage value to the gates of the mosfets? That's why this number of turns in the secondary? Let us forget about matching impedances a little. So how exactly I can determine the voltage that will go to the gates? This is of course an AC signal so it is stepped down by the transformer. But why is so and how can I calculate the voltage values? I guess that depends greatly on the output of the driver. In certain cases can someone design a step up transformer or are thay all step down ones?
Thank you,

[Buddhafollower]

As for the division, you know very well I had no choice. It does work though.

[spacecase0]

yes, mosfets are voltage devices for the input, and the output is equivalent resistance ( and is clearly set by that voltage on the input).
how you get the required voltage in is that you look at the gain required and calculate from the data sheet on the part.
the gate-source voltage absolute limit is 20V ( + or - )
gate threshold voltage for that mosfet is listed as Min. 2V Typ. 2.9V Max. 4V
and that is why you need an adjustment for the bias voltage for each mosfet in the circuit, none of them are quite the same.
as far as how much voltage turns on the mosfet past that point,
there is going to be a saturation voltage that turns it on and more voltage does not help,
by the graph it looks like that is somewhere near 8V depending on your current through the device
so about the most AC signal you want in would be 4V peak (not RMS) (4V bias plus the 4V peak in get you 8V, and you are not going to need more than that for this part)
so knowing that you need less than 4V PTP in, how much less ?
there is the actual gain, they call it Forward Transconductance in the specifications... and that number is 35 with the min. being 25, no max value is listed...
the units are "s" and I just have no idea about it, I did read it all at one point, but it did not make sense to me,
I guess that figuring out a unit for a voltage in that relates to an equivalent resistance out is not going to end up with units that make sense.
so I just ignore the value and go look at the typical Characteristic graphs in the specification shet as that has the information I am looking for.
and testing with a signal generator gets you your number quite fast.

remember that bipolar transistors are not the same at all, they are current in that is setting current out with a gain amount. easier to figure out the math, but more annoying to design circuits for.

[Buddhafollower]

The ''S'' I believe stands for ''Siemens'' and it is a measure of Admittance. Here Dr. Dollard puts it like this.

[Buddhafollower]

Please look at this datasheet of the bipolar transistor 2SC3264. Do you think that in a class B configuration this will be able to achieve at least 7MHz? 10.15 MHz?
It is strange because it has a very high transition frequency...