In HMC453ST89's datasheet, it gives the electrical specifications when Vs = +5V:
The max. output power is around 32.5dBm, about 1.6W. But for a 50 Ohm load, 1.6W means the peak voltage is about 12V. How is it possible with Vs = 5V?
1.6W mean's the peak voltage is about 12V. How is it possible with Vs = 5V?
Look at the data sheet application circuit diagram below. I've added a red box and a blue box: -
The red box is around L1 (an inductor) and when you have a collector of a transistor being pulled up to Vs, the peak-to-peak voltage that can be attained is nearly twice Vs. So, if Vs is 5 volts then you can probably achieve something like 9 volts p-p at that node.
But to achieve 25 volts p-p (about 1.6 watts into 50 ohms) you need some help from the tuned circuit inside the blue box. You need some form of resonance that lifts the 9 volt p-p to 25 volts p-p; an increase of about 9 dB: -
The interactive 2nd order low-pass filter tool above is loaded with 5.1 nH, 50 ohm and 22 pF and this resonates at around 400 MHz but, importantly it lifts the voltage amplitude by over 10 dB to give the required signal level into a 50 ohm load to achieve 1.6 watts.
So, it's a combination of class B/C output stage with an inductor and using a filter to resonate the signal to a higher amplitude.
This power is never achieved.
The 1db compression point is the theoretical point where the input/output curve of your amplifier deviates from a perfect linear amplifier by 1dB.
You can just make an exponential curve fit at much smaller differences, and then find the point where that extrapolated curve deviates by 1dB.
Having the 1dB compression point significantly above the maximum output power means that you're pretty linear for all actually achievable output powers.