wrote:
Back in July I wrote:

I'm reading David Rutledge's excellent "The Electronics of Radio."
In Chapter 10 -- Power Amplifiers, he discusses Class C amps and says,
"In addition, if we drive the transistor clear to saturation, using the

transistor as a switch, the dissipated power can be greatly reduced,
because the saturation voltage is low. This is Class C
amplification..."
I'd always throught that in Class C, while you'd operate the device so
that it was cutoff during most of the cycle, but not saturated.
Is this just a different definition of Class C?
I checked back with SSDRA and EMRFD, and didn't see anything about
driving Class C amps into saturation?
What says the group? Do we saturate in Class C or not?
--------------------------------------------------------
I've been thinking about this some more. The 1980 ARRL handbook points
out that "Solid State power amplifiers should be operated just below
their saturation points for best efficiency and stability." Also, the
formula that we use to determine load resistance (Rl=Vcc^2/2Po) implies
that we are looking for a combination of Vcc, Load resistance and power
out that will prevent saturation.
And wouldn't we end up with far lower harmonic content if we only clip
one side of the wave form (at cutoff) instead of both sides (cutoff and
saturation)?

I know there are more exotic modes beyond C, but for plain old ordinary
ham radio applications, don't we normally avoid saturation in Class C
amps?

Also, what about this business of having the efficiency improve through
saturation "because the saturation voltage is low" Could that be
right? If you put a voltage across a conductor and generate a large
current, you can't sit back and say "Great! Power consumption across
the conductor is low because the voltage drop across it is now
minimal!"

73 from London
Bill M0HBR N2CQR CU2JL
http://www.gadgeteer.us

The definition I've always heard for class C is that conduction of the
active device is for less than half of each cycle. Nothing is said
about saturation.

Of course, if the active device conducts when there is appreciable
voltage across it, it will dissipate more power than if it were
conducting the same current at lower voltage.

Cheers,
Tom