In article , Cecil Moore wrote:
Ken Smith wrote:
RF transmitter power amps are certainly "impedance matched" to the
intended load. Take a look in the ARRL "The radio amateur's handbook".
If you have the 1944 addition, you will need to start reading at page 96
in the lower right column. If you don't have that, try Motorola's AN-721.

A CMOS Class-E amp is in full saturation (0.5v at 2a)
for 10% of a cycle and off (12v at 0a) for the other
90% of a cycle. The tank circuit changes the digital
energy to analog energy by filtering out everything
except the fundamental frequency component. How
in the world does one determine the steady-state
impedance of the CMOS source? Isn't the best one
can do with a digital switch is to keep it within
specified parameters? The CMOS device dissipates 2
watts for 10% of the time - therefore 0.2 watts
steady-state.

For what you say here really to be true the transistors must switch very
fast. About 25pS switching speed is needed at about 400KHz. If we take
that to be the case however, I think you will see why matching still
applies.

Lets take the reactive component first. If there is a reactive component
to the loading, the current in the switch will have a higher RMS value
without that increase in RMS increasing the radiated power of the system.
So the reactive component of the matching is fairly obvious.

Imagine that you have a well designed Class-E circuit loaded with the load
the designer optimized it for.

Now imagine that you slightly increase the resistance slightly. When you
do so, the current into the load will decrease but the voltage will not
increase enough to compensate for this.

Now lets assume that you slightly decrease the resistance. Since we are
assuming that this is a well designed case, we can assume that the
designer took steps to ensure that the output devices would be protected
from excess currents. This could be done by reducing the operating
voltage of the output section, for example. In any case, the voltage on
the load will decrease by a larger factor than the current will increase.

So it is obvious that the reactive part is matched and the resistive part
is matched just as it would be in a non-class-E output section.


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