Ken Smith wrote:


The strongest argument for dropping the impedance matching concept is PA
efficiency, and therefore maximum signal swing. Obtaining maximum swing is a
load line issue.

What do you mean by "maximum signal swing" in this context. I can get a
bigger swing by leaving the output completely unloaded and hence causing
the actual efficiency to be zero.

LOL. Sure, the purpose of a power amp is to actually extract power. This is a
good start.

Perhaps a simplistic (and of course idealized) class A example would help. And
I want to remind that this is a simplification of the first order design cut.

The first assumption/idealization for the class A example would be to demarcate
between strong and weak non-linearity. This demarcation is basically the
boundary of clipping, both positive and negative. That is, we want our class A
amp to swing to the rails but not go beyond. In our class A design we will
accept weak non-linearity but not strong non-linearity.

Lets say we have selected a class A device/amplifier for which we can statically
dissipate 10W. The drain DC circuit is simply an RF choke (see, it is a simple
example!). Let's say the quiescent values are a 10 V supply and 1 A of current
(10 W). The question is: how do we load the device to extract maximum power
given the "no clipping" (strong non-linearity) constraint?

Say we load it with 20 ohms, what happens? The max positive swing before
clipping is Id*rL = 1*20 = 20 V. The max negative swing is, of course, Vd = 10
V. Since the 10 V is the lesser of the two swings, our non-clipping design
constraint limits us to 20 Vp-p. So we can deliver (Vp)^2/(2*rL) =
(10)^2/(2*20) = 2.5 W.

Say we load it with 5 ohms, what happens? The max positive swing before
clipping is Id*rL = 1*5 = 5 V. The max negative swing is, of course, Vd = 10
V. Since the 5 V is the lesser of the two swings, our non-clipping design
constraint limits us to 10 Vp-p. So we can deliver (Vp)^2/(2*rL) = (5)^2/(2*5)
= 2.5 W.

Say we load it with 10 ohms, what happens? The max positive swing before
clipping is Id*rL = 1*10 = 10 V. The max negative swing is, of course, Vd = 10
V. Since they are equal, we get 20 V of p-p swing. So we can deliver
(Vp)^2/(2*rL) = (10)^2/(2*10) = 5 W.

Our circuit loaded with 10 ohms delivers twice as much power as with the lesser
5 ohms or greater 20 ohms. That is, extracted output power is peaking at some
finite non-zero value. This is also easily seen to be most efficient point for
this simplistic example.

In no way was the ouput-Z of the amplifier considered in deciding how to load it
for the purpose of extracting maximum power from the circuit. The output-R is
completely irrelevent.

This example is intended to be illustrative rather than exact.

The reactive component issue is still there too. Reactive loads cause
increased currents in the output stage without delivering any power to the
load so they still need to be reduced as much as practical.

Yes, I already noted that for that portion of the impedance, it should be tuned
out *as best* possible.

"...(to be fair, the time-averaged reactive output component is tuned out as
best possible)."