Class C amps and saturation (again)
 

wrote:
Thanks to all who responded. I think I'm starting to understand this.
LTSpice helps a lot. I set up a class C amp and looked out power
dissipated in the transistor vs. power dissipated in the load. The big
efficiency gains that come with saturation were very apparent.

To better understand WHY this happens, I set up a spreadsheet that
looked at power dissipated in a variable resistor as it swept from .1
ohms to 10 ohms. It had a fixed 10 ohm resistor in series and 10 volts
DC across both of them. Yes indeed, the power disspated in the
variable resistor drops off dramatically when the resistance (and hence
the voltage across it) gets very low. I guess is why this happens in
the saturating Class C amp, Right?

But I still have some questions. When we design a Class A amp, the
familiar formula Rload = (Vcc-Ve)^2/2Pout allows us to come up with a
load value that will prevent the amplifier from saturating. A load of
this value will cause the voltage across the transistor (collector to
emitter) to vary from zero to twice Vcc. But it won't go into
saturation.

Why then do so many of the books (EMRFD, SSDRA, the W1FB books) call
for the use of essentially the same formula for the load when selecting
a load for Class C amps? We're no longer worried about staying out of
saturation, correct? In fact, we want to saturate. So why the same
formula? In fact, it seems to me that if you have a Class C amplifier
that is designed with this formula and is operating just below
saturation, you can get it to saturate just by increasing the value of
the load presented to the collector. Power out and efficiency
immediately improves. Linearity, of course, does not.

....
In Class A, the active device is conducting the entire cycle, and
conducting in proportion to where you are in the cycle. Because of
that, you can drive a purely resistive load--that is, a non-resonant
load. Class A is used in single-ended broadband linear amplifiers,
such as receiver stages, audio amplifiers, scope vertical amplifiers,
....

In Class B, each active device is conducting for half the cycle. If
you use a single active device, you need a resonant tank to complete
the cycle. You can also have push-pull class B where one device
handles output voltages of one polarity, and the other handles output
voltages of the other polarity. It is usual to bias the active devices
in such a case slightly into class A, so that the nonlinearities right
at the half-cycle point can be smoothed over. But class B in a single
ended RF amplifier, with a resonant tank at the output, is still
capable of linear performance, because you can reduce the conduction
during the active half-cycle and reduce the output amplitude in
proportion.

In both Class A and Class B, it's possible for the output to peak with
the active device having a very low voltage across it, so that the the
output peak-to-peak amplitude at that point is still two times the
supply voltage, or nearly so, assuming it's a good approximation of a
sine wave there. For class A and B amplifiers, though, the output
amplitude may be less than that. The efficiency obviously goes toward
zero as the output amplitude goes toward zero, since some conduction in
the active devices is required to drive any output, and the dissipation
in the active device is proportional essentially just to the current at
low output swings since the voltage on the device is nearly constant;
but the power to the load is proportional to the square of the current.

In Class C, the active device is hard-on for a relatively small
fraction of the cycle. It always drives the output to about two times
the supply voltage, peak to peak, if it's properly designed for low
dissipation in the active device. But a key difference between it and
class A is that the active device is conducting essentially no current
except when there is only a very low voltage across it. In class A,
there's current through the device at all output voltages.

So at full output in all those cases, the peak-to-peak voltage is the
same, and for a given output POWER, that tells you what the effective
resistive load should be at that point.

(In Class D, the voltage at the output of the active devices swings
very quickly between power rails, spending most of the time at one rail
or the other, and the output is varied according to the duty cycle of
the rectangular waveform there. That waveform is filtered to remove
the switching frequency and its harmonics. The result can be a very
efficient amplifier. They are becoming common in audio applications
these days.)

Clearly there's a bit more to it than that simple explanation, but it's
a first approximation that may get you going...

Cheers,
Tom

Class C amps and saturation (again)
 

K7ITM wrote:
. . .
In Class C, the active device is hard-on for a relatively small
fraction of the cycle. It always drives the output to about two times
the supply voltage, peak to peak, if it's properly designed for low
dissipation in the active device. . .

I've designed saturating RF amplifiers of a few watts output in which
the device is on for well over half the cycle, and which have an
efficiency of around 90%. The peak-peak collector voltage is nearly 40
volts when run from a 12 volt supply. The high efficiency implies
relatively low dissipation in the active device. But perhaps this falls
outside some definitions of class C.

Roy Lewallen, W7EL

Class C amps and saturation (again)
 

Roy Lewallen wrote:
K7ITM wrote:
. . .
In Class C, the active device is hard-on for a relatively small
fraction of the cycle. It always drives the output to about two times
the supply voltage, peak to peak, if it's properly designed for low
dissipation in the active device. . .

I've designed saturating RF amplifiers of a few watts output in which
the device is on for well over half the cycle, and which have an
efficiency of around 90%. The peak-peak collector voltage is nearly 40
volts when run from a 12 volt supply. The high efficiency implies
relatively low dissipation in the active device. But perhaps this falls
outside some definitions of class C.

Roy Lewallen, W7EL

A flyback switching RF amplifier, Roy? ;-) One can also use active
devices that pull hard to ground and to a rail, delivering a square
wave output at high efficiency. That square wave can then be filtered
with a network that presents a high impedance for harmonics and passes
the fundamental, and achieve very high efficiency. A square wave is
nice because the even harmonics are theoretically zero, and in practice
can be much lower than the odd, so the filter doesn't have to have
super-steep cutoff. I guess that would be some variation on class D.
It's actually the way the HP8640B generates its signal: divide-by-2
stages from the master oscillator, followed by filters. I suppose it's
really best to describe the operation of any amplifier in some detail,
and not just rely on "Class A" or the like.

Cheers,
Tom


Cheers,
Tom