[SOLVED] About amplifiers and amplifications (vswr judgement + ntone amplif. pow. supply) (?)

[LandLack]

Dear community,
I know that to someone these questions may seem very simple, but I have some doubts and need clarifications.

1-I have an amplifier (CMA-63 from minicircuits) which has to amplify N (as much as possible) tones with input power amplitude of -12dBm each;the gain is about 20dB, so the outputs should be around 8dBm (6.4mW) each. The power source needed for this device is 5V (0.5W). Let`s just ignore the intermodulation issues, I want to know if increasing the number of tones (so there are Nx6.4 mW total output, I think) may change the behavior of the amplifier, with a physical limit near N=79 (where the supplied power is about equal to the output power); or am I doing some error at this point, and forgetting something important? (Someone I know told me that I could use a greater number of tones if I ignored the intermodulation, but I don`t understand why).
2- I want to confront (with numbers) the VSWR of two amplifiers; of course I know that the input VSWR should be the lowest as possible, but is output VSWR better the highest as possible (because the power doesn`t enter back in the device)?​

Thank you for responding anyone of these two questions.

Regards,
John

 

[chuckey]

Suppose the output device in your amplifier can use all of the 5V, then with one carrier of 6.4mW, which has a amplitude of Z volts at the output device, adding more carriers means that the output device must be able to handle 2Z or 3Z or 4Z volts (depending on the number of carriers). Its in the design of the amplifier which determines how close to 5V N X Z is. If it was designed as a CW amplifier then Z may be 4.5 V so its one carrier only. Another point is that for one carrier the Po = 6.4 mW and the peak power = 6.4 mW, for two carriers Po = 12.8 mW but the peak power is now 25.6 mW (twice the voltage, twice the current = 4 X the power), for three carriers the peak power is 9 X 6.4 mW. I think you are being optimistic for the amplifier to handle 79 carriers, I would think 10 would be the maximum ignoring i/p problems.
Frank

- - - Updated - - -

For most power amplifiers the output VSWR is most importent, because the reflected power is of a level which can do real harm. In real life any solid state amplifier with an output power > 5 W has its output connector coloured red, warning people to disconnect the input before removing this connector. High power transmitters have a VSWR trip of lower then 1.05, because the reflected power can cause KVs of RF volts to appear across circuits and cause flash overs.
Frank

 

[LandLack]

Suppose the output device in your amplifier can use all of the 5V, then with one carrier of 6.4mW, which has a amplitude of Z volts at the output device, adding more carriers means that the output device must be able to handle 2Z or 3Z or 4Z volts (depending on the number of carriers). Its in the design of the amplifier which determines how close to 5V N X Z is. If it was designed as a CW amplifier then Z may be 4.5 V so its one carrier only. Another point is that for one carrier the Po = 6.4 mW and the peak power = 6.4 mW, for two carriers Po = 12.8 mW but the peak power is now 25.6 mW (twice the voltage, twice the current = 4 X the power), for three carriers the peak power is 9 X 6.4 mW. I think you are being optimistic for the amplifier to handle 79 carriers, I would think 10 would be the maximum ignoring i/p problems.
Frank

- - - Updated - - -

For most power amplifiers the output VSWR is most importent, because the reflected power is of a level which can do real harm. In real life any solid state amplifier with an output power > 5 W has its output connector coloured red, warning people to disconnect the input before removing this connector. High power transmitters have a VSWR trip of lower then 1.05, because the reflected power can cause KVs of RF volts to appear across circuits and cause flash overs.
Frank

Thank you very much for your response, I've understood most of it; I have a 8 tone signal to amplify, so the possible peak power may be 64 times greater than the power of every tone, right? If this is true, this may be a bad thing since it is 409 mW = 26 dBm, while my 1dB compression point is 18dBm.
But this (the constructive interference of the voltages and currents) should happen if the tones had carrier frequencies which were one the multiple of the other, right? I'm telling this because I've made a transient simulation with my 8 tones, and the voltage value of the output did not have too high values (lower than 0.7V), so I suppose that the voltages did not add too much with each other; this means that I'm safe or has the simulation some inaccurate value?
Again, thank you for helping and sharing experience.
Best Regards,
John

- - - Updated - - -

No, wait, my reading was completely wrong. Although the signals don't have frequencies which are multiply with each other, there is one peak in which I have 3.2V, which means around 200 mW. You were right. This response saved me from producing a very big error. Have a nice day,

Best Regards,
John