question about RF push pull amplifier !

[bobytronic]

Hi,

In the classical push pull power amplifier with 2 N-Chanel MOSFET and a wideband transformer 1:1 , How can i determine the GAIN with this kind of amplifier ?


Best Regards

 

[StoppTidigare]

Hi, a very rough approximation, would be that the outputpower is

RL*I^2, where I=75V/10.

I've roughly put the coupling factor k between the transformers equal to 1, but it should rather be somwere 0.5<k<1.

Output impedance of transistor (for BJT at least it is always lesser than 50 ohm ) is always low I guess. I hope It is the same as for MOSFET. I don't have my reference beside me...

As far as for the input power, you can do a rough estimate that your source impedance is perfectly matched to the PA input impedance.(R_source=R_in)
So it should become So if you have U_source your U_in then becomes U_source/2 and your I_in is U_source/2R, so your P_in will become
U_source^2/4*R.

But you asked for the gain : P_out/P_in, so you must decide how high input signal is ENOUGH.

Could somebody pick up the ball ?.....
regards,
StoppTidigare

 

[bobytronic]

Hi, thanks for your reply !!

I don't understand all that you say... why 75/10 ???? why 10 ?

I wish to have a sinusoidal 70V RMS voltage on a 50 ohms Load(RL) . So , at full power (98W) i have a current load of 1.4 A.(RMS). that's right ?

Now , the only solution i have found to design an
/high voltage/
/High power/
/input signal between 1Mhz & 5 Mhz/
is this kind of topology: this is the classical broadband power amplifier for radio antenna.

I know that the power supply in center dot of the output transformer (ferrite core) is the limit of the high voltage i can deliver to the load !
I know calculate the best reactance of this transformer ( Xp at fmin = 5*RL) ....L = Xp/(2*pi*Fmin).... N = 1000 sqr(AL (mH) / L (mH/1000 turns)) etc ....

But the problem is that i dont know how calculate the VOLTAGE gain.
if i have 5V Rms at the input ... what i have on the load ?

I just have the transconductance gm of the MOSFET !!


Best Regards

 

[StoppTidigare]

Hi sorry about the "75/10" without any comment. I just did a rough estimation, thinking that output impedance of BJT is lower than 50 usually.

Then this output impedance is parallell with your when you have the load attached to the collector and is therefore dominating. So thats why I estimated output_impedance of transistor parallell with load =10 Ohm.

How you should calculate the gain ? Calculate the gain for one only, they work one at a time.

gm is derived as a small signal quantity, you shouldn't consider it I believe...
Could somebody continue ?

You work with large signals. hmm......... gm times the input voltage gives you the output current in the small signal regime. This current into the load would give you V_out and then you would have Vout/Vin.

Regards,
StoppTidigare

Be aware of though that you want to transfer maximum power to the load, that means that you want your output voltage and current to be in phase.
You want to maximize the product U*I
Thats why people usally look for P_out/P_in insted of voltage amplification in these matters.