Joel Kolstad wrote:
Hi Tom,

Thanks for the information; someone else pointed me yesterday to the idea that
you'd generally match the input for optimum noise performance rather than
optimum power transfer. The data sheet claims that the optimum noise figure
is 0.8dB, but the part itself is marketed as "typical 1.8dB NF" -- which they
obtained from their own eval board that, as far as I can tell, was matched
pretty close to optimum power transfer. In any case, at present mine is also
matched at the input for optimum power transfer, and I'll measure its noise
figure and see whether or not I'm around 1.8dB or better... if so I don't
think I'm going to worry about it too much.

Output matching will transfer the greatest power to the load.

Yeah, but it's not practical, is it? I'm looking back at (approximately) a
current source, and my load is dictated as (approximately) 50 ohms, so the
load itself is what's driving the power transferred, no?

Well, no... Clearly it's not exactly a current source. Perhaps it's
1000 ohms, or 10000 ohms, (plus some reactance, of course) which is
high but not a pure current source. Then a network that matches 1000
ohms, or 10000 ohms, to 50 ohms (and cancels the reactance), will give
you the most power output. That is a linear, small-signal model, but
that should be a pretty good approximation for any application where
you need a low-noise preamp.

You do need to consider losses in whatever matching network you use;
and many matching networks will be highly resonant to transform between
impedances that are in a large ratio. Realize that the match at the
operating frequency may give you a load at some other frequency which
causes instability...

If the source (the output impedance) was really a current source, you
could get near-infinite power gain (again, assuming that S12 is zero)
by transforming the load to as high an impedance as possible. Consider
the small-signal low-frequency model of a bipolar transistor where the
output is a current source shunted by a resistance. Assume that
(internal effective) resistance is infinite. Now the current from the
current source all flows into the load. Since power is i^2*R,
increasing R increases the power without bound. A matching network
just transforms your practical load (e.g. 50 ohms) to the desired load
R.

What's practical depends on how wide a bandwidth you want, and how good
you are at designing and building impedance matching networks. (I can
personally come up with lots of IMpractical networks! ;-)

Cheers,
Tom