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)."

In article , gwhite wrote:
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.

No, the purpose of the power amp is to deliver power, not extract it.


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.

Don't bother with the over simplified Class A case. RF power
amplification is rarely done class and and it is a digression from the
actual topic.

[...]

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.

At some point as you decrease the resistance, the output will drop to zero
as the amplifier fails or it will start to decrease in some more
controlled manner as the protection circuits take control. If we assume
the latter case, it is easy to see that the power reaches a maximum value
and then decreases as the resistance is lowered. The point at which the
power is at the maximum is the point at which the load is matched. If you
make a small change in the load and observe the voltage and current when
that small change is made, you will see that that is indeed the output
impedance of the amplifier. I think this is the part you are not
grasping.


--
--
forging knowledge

Ken Smith wrote:

In article , gwhite wrote:
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.

No, the purpose of the power amp is to deliver power, not extract it.

Well really, once the device and supply have been determined, we can indeed view
it as extraction. We'll load it to extract the most. If you want to mince
words and call it "deliver," that is fine.

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.

Don't bother with the over simplified Class A case. RF power
amplification is rarely done class and and it is a digression from the
actual topic.

Well, class A is certainly done. Two cases are where the extra little bit of
linearity is desired and at high frequencies, were PAE starts to take a bite as
the gain drops below 10 dB.

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.

At some point as you decrease the resistance, the output will drop to zero
as the amplifier fails or it will start to decrease in some more
controlled manner as the protection circuits take control. If we assume
the latter case, it is easy to see that the power reaches a maximum value
and then decreases as the resistance is lowered. The point at which the
power is at the maximum is the point at which the load is matched. If you
make a small change in the load and observe the voltage and current when
that small change is made, you will see that that is indeed the output
impedance of the amplifier. I think this is the part you are not
grasping.

No, this is exactly where I'm saying you are incorrect. You are not getting the
practical limitations and are mistakenly applying linear concepts. It doesn't
work if you want to extract maximum power from the DC supply through a real
device, converting the DC power into RF power.

G. White wrote:
"This example is intended to be illustrative rather than exact."

Nobody contradicts that maximum power from an amplifier requires an
impedance match, so far.

Nobody contradicts that efficiency in an ideal Class A amplifier can
never exceed 50%, so far.

Some texts show waveform distortion typical of a single-ended Class A
amplifier and note that "maximum undistorted power" requires a load
impedance 2 to 3 times that of the amplifying device.

G. White gives a peak to peak voltage of 20 and a power out of 5 watts.
RMS volts are 7.07 in this case. The load then is the square of the
volts divided by the power. This is 50/5 or 10 ohms for the load. If
this matches his amplifier, 5 watts is the maximum output.

G. White said that he did not consider the output-Z of the amplifier in
its loading. Ignoring an amplifier`s impedance does not make it go away.
Ignoring an amplifier`s impedance does not revoke the maximum power
transfer theorem, either.

Best regards, Richard Harrison, KB5WZI