If a RF transmitter has an output impedance of 50 ohms, doesn't that mean that you're wasting half your power?

I've always wondered the same thing and the answers below don't quite convince me either way.

Firstly, an observation: HF rigs tend to only be <50% efficient (eg: a typical 100W rig will draw ~22A giving about 30% efficiency. This is a bit low as some of that 22A goes into auxiliary circuits such as the front panel, CPU, etc but most of it goes to the PA). No doubt FM only rigs with non-linear power amps will be more efficient, but I haven't measured one..

Secondly, some circuit theory: maximum power is drawn from a source if the load impedance is the complex conjugate of the source impedance (or load/source resistance are equal if the reactive component is zero, which we tend to aim for). This is irrelevant of any transmission line effects involving reflections/SWR/etc. It applies at DC just as it does at RF.

However, maximum power is very different from maximum efficiency, which tends to require a low source impedance relative to load (+/- any quiescent power dissipated in the source, which complicates things beyond "make the source impedance zero", but I won't go into it here).

So, what's the deal? I suspect (let me stress: this is only a hypothesis) that the linear power amps in HF rigs are designed with an output impedance which is optimised for the transistors. ie: the biasing resistors/chokes/etc are chosen considering efficiency, power, linearity, heat, etc. If this is done a matching circuit could then be used to match the amp's output impedance to the 50 ohm output. An output LPF is always needed to kill harmonics and it's easy to make a LPF which also happens to be a matching circuit. Mutli-band rigs always have a bank of switched filters, again a different match can be built into each filter.

I'm thinking that the above design is necessary given that 100W into 50 ohms is a peak voltage of 100V (~70V RMS). Without some kind of transformer (which would transform impedance just like it transform voltage/current, they're all interrelated) this is impossible with a 13.8V supply rail on the power amp. A matching circuit will also change the voltage/current ratio, it's required as part of the impedance transform, except unlike a transformer (which can be very wideband, several octaves if not decades) a matching circuit is quite narrow. The bandwidth of a matching circuit depends on the ratio of the impedance transform: large changes in impedance result in a narrow band matching circuit.

My impression is that power amps use a combination of transformers (typically baluns used to drive a push/pull topology built from a pair of N-channel parts, P-channel parts are more expensive and less efficient, in general) and matching circuits. But this is by no means my field of engineering expertise.

tl;dr I'm not sure, but hopefully the above technical ramble will inspire someone who does know to comment. Worst case scenario I'll reverse-engineer the FT-857 schematic.

/r/amateurradio Thread