Reg Edwards wrote:
=======================================
All power amplifiers have a tuned circuit in the plate. It is
essential to reduce output power contained in the harmonics.

In any case, power in the harmonics is wasted power.

With a tuned circuit in the plate it is impossible to achieve a
rectangular voltage output waveform. It is always a sinewaveform.

A rectangular plate current in conjunction with a tuned load always
causes harmonic power to be wasted at the plate.

So one might just as well use a sinusoidal driving waveform, Class-C
or not. It's easier. It also avoids generating and wasting harmonic
power in the driver.


Not true.

A network with an inductive input will allow a square waveform at the
device output but not waste significant energy in harmonics. I've done
that in designs.

In the 1950's RCA had an AM BC transmitter that drove a tube with a
near square wave, and had a near square wave. The RCA transmitter used
a low-mu triode that had parallel tuned circuits in the grid and anode
set at the third harmonic. It had conventional networks feeding the
grid and to the antenna from the plate resonantor. That transmitter
made over 95% anode efficiency.

73 Tom

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?

Saturation is itself a somewhat mushy point. There's a V_sat specified
on the datasheets but the actual definition of saturated is entirely
application-sensitive. As a practical matter as you add more base
current you will go further into saturation (up until you melt the
base-emitter junction and then all sorts of wacky things ensue).

Choice of drive level and output level and load impedance in a Class C
amplifier certainly will in many cases put the device into saturation.

Tim.

What says the group? Do we saturate in Class C or not?
-----------------------

In order to get any semblence of efficiency, the drive waveform should drive
the stage from near-cut-off to near-saturation, somewhat like a switch.
However, depending on how hard the sinusoidal input signal drives the "Clacc
C" stage, the conduction-angle and thus the duty-cycle of the output (before
filtering) will increase or decrease, resulting in variable output power in
the load.

Joe
W3JDR


"Tim Shoppa" wrote in message
oups.com...
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?

Saturation is itself a somewhat mushy point. There's a V_sat specified
on the datasheets but the actual definition of saturated is entirely
application-sensitive. As a practical matter as you add more base
current you will go further into saturation (up until you melt the
base-emitter junction and then all sorts of wacky things ensue).

Choice of drive level and output level and load impedance in a Class C
amplifier certainly will in many cases put the device into saturation.

Tim.

W3JDR wrote:

In order to get any semblence of efficiency, the drive waveform should drive
the stage from near-cut-off to near-saturation, somewhat like a switch.
However, depending on how hard the sinusoidal input signal drives the "Clacc
C" stage, the conduction-angle and thus the duty-cycle of the output (before
filtering) will increase or decrease, resulting in variable output power in
the load.

Joe
W3JDR


Who said the drive had to be sinusoidal? If the final can run almost square
wave, why can't the driver?

Yes, exactly. We just assume sinusoidal drive because that's the most common
case

Joe
W3JDR


"John - KD5YI" wrote in message
newszDtg.17708$Wh7.10655@trnddc07...
W3JDR wrote:

In order to get any semblence of efficiency, the drive waveform should
drive the stage from near-cut-off to near-saturation, somewhat like a
switch. However, depending on how hard the sinusoidal input signal drives
the "Clacc C" stage, the conduction-angle and thus the duty-cycle of the
output (before filtering) will increase or decrease, resulting in
variable output power in the load.

Joe
W3JDR


Who said the drive had to be sinusoidal? If the final can run almost
square wave, why can't the driver?