"Richard Harrison" wrote in message
...
Reg, G4FGQ wrote:
"What allows a class-C amplifier to exceed 50% efficiency is a small
operating angle."
While this is too vague, Richard tries to add detail, but mis-steps just
a bit... and Steve goes into an extended "You ain't quite correct blurb..."
Exactly, and during the majority of the degrees it`s switched completely
off. It draws no current and suffers no "IsquaredR loss" during the
amplifier off-time. Impedance is approximately E/I, but I is the average
I, which is much less than the bursts of I during the conduction angle.
We must be careful with the word "average" here.
First, my "class C" model is a follows:
I liken it to digital or "switched modes". While I have never scoped the
plate to observe this... When the tube is cut off for a large part of the
cycle, there is a high voltage on the tube (I believe it swings higher than
the supply dou to the "ringing" of the plate tuned circuit), but no current.
Hence, ExI=0.
When the tube is on, it is slammed hard on by the "high" grid signal and
there is a high plate current, but the plate voltage is very low (anybody
know how low and if I am all wet? ... tubes aren't quite like transistors
in the digital mode)--therefore ExI=somthing, but since the E is low, it is
lower than in class A during that part of a cycle. There may also be some
effect due to the fact that the plate tank is swinging low allowing the
plate voltage to be even lower.
Did you know that in class A, the plate power dissipated goes DOWN by
the amount that is delivered to the load??? Cool! huh? Isn't physics neat!
Second, it is the RMS current through the tube which will waste power,
so it is what we must be concerned with. Yes, if the tube is off the
current is zero at that time, but the RMS must be considered and it does not
go down as fast you might think. As an example, for the same current
pulses, but spaced out to half the duty cycle, the average drops to half,
but the RMS only drops to .707. There is a square root in there.
[[Anybody see the "AC Watt meter article in QST]] It is an OOPS! Most
power supplies don't draw sine wave current. It is pulses. I have been in
contact with both Bob Shrader (the author) and Stu Cohen (Tech editor) and I
just finished an analysis and am going to make more measurements to verify,
but the numbers Bob published can be as much as 1/3 the true power values
(depending upon the DVM he used and the current waveform of the supplies he
measured.
--
Steve N, K,9;d, c. i My email has no u's.
wa-da ya blokes think.
The switched-off time makes the I in the denominator of E/I very small
indeed and the solution to Ohm`s law is a high impedance without the
dissipation of a resistance that remains in place continuously while
agitating the atoms of a poor conductor to limit current.
Instead, we have a low-resistanc in high conducton for short spurts.
On-time is limited, instead of conduction, to produce a certain
effective resistance.
Another way of saying just wjat I did above, but "effective resistance"
is one way of thinking about it and this resistance must be calculated using
the RMS values.
An automobile Kettering ignition system may use a dwell-meter to
indicate how much of the time the points are closed. An ohmmeter
indicates the resistance between its test prods.
I'd be willing to place a bet (knowing how an analog ohm meter works,
that the *diflection* of the two meter pointers is the same (see below).
Both meters respond to the averacge current through them and both will show
full scale when the points are open (I think thta is the correct polarity).
Here's the "below":
There is, however, the confusion added by the coil/cap waveform for which
the ohm meter is not equiped to limit - whereas, I believe the dwell meter,
if well designed, will have something to limit so as to remofe it as a
complicatin.
--
Steve N, K,9;d, c. i My email has no u's..
The two test circuits
are almost the same although limitation of the deflection of the
dwell-meter is different from limitation of the deflection of the
ohmmeter due to the difference between limited conduction angle ignition
points, and the continuous conduction through a current-limiting
resistor. There`s an analogy between Class C and Class A amplifiers in
there somewhere.
Best regards, Richard Harrison, KB5WZI
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