Increasing the resolution of a sensor by using an amplifier

I am not an expert on the subject of power amplifiers, but since there hasn't been an answer, here's my attempt:

1. Defined "good." A straight-forward, inexpensive way would be to measure the output RF with a power meter (directly attached, no antennae) and divide it by the circuit's power consumption. Obviously, that's a full-system efficiency and not specifically the power amplifier.
2. Not especially. Amplifier linearity and power efficiency are inversely proportional. Using a less linear amplifier will be more power efficient, but distorts the signal reducing your SNR and link margin. The theoretical efficiency for a class A amplifier is 50%; in practice, much lower. That isn't to imply this is a "good" amplifier and better options don't exist. Make sure any required output impedance matching is properly designed and populated.
   • In the case of BPSK (or QPSK) you can sacrifice linearity for power gain as the phase of the signal contains the data and not the amplitude. Since you are using BPSK modulation you can safely consider other classes of amplifiers like AB, C, D, etc.
3. There are a couple of work arounds:
   • In mobile technologies, you have things like offset-QPSK and π/4-QPSK to reduce the dynamic range required, allowing a less linear power amplifier.
   • Techniques like pre-distortion are used to correct for amplifier non-linearity, allowing use of less linear amplifiers.
4. Don't use a larger amplifier than you need. Operating a 10W amplifier at 1W may be considerably less efficient than operating closer to 10W. This is a side effect of the bias current in class A amplifiers.

I'm assuming that you are wanting to amplify a 6V ac signal into a 12V ac signal. Here's the problem; using a power op-amp from a 12V supply is not the problem but the amount of output voltage it can produce will be somewhat less than 12Vpeak to peak.

The output transistors will not be able to turn-on hard enough to produce 12Vp-p out. You might find a device that may achieve 10Vp-p however.

Moving on (and assuming you can produce 12Vp-p out of a power op-amp), the next problem is the current - you have calculated that the current will be \$\frac{25W}{12V}\$ and concluded that it will be about 2A - this is correct for DC output voltages but some way off the mark if you want an ac signal.

A 12Vp-p sinewave will have an RMS value of \$\frac{12}{2 . \sqrt{2}} = 4.24V_{RMS}\$

And if you want a power of 25W then the current will be about \$5.9A_{RMS}\$

You may want to rethink your requirements based on the above. Maybe you are more concerned about having a squarewave output and if so then the problem is finding a power op-amp that can produce 12Vp-p from a 12Vdc supply.

One option is to use a H bridge amplifier if the load you want to drive is suitable for this type of output connection. Another option is to use a transformer output to step up by (say) 25% to accomodate the power op-amp's inability to produce 12Vp-p.