Class-C Tube Amplifier Circuits

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MARTIN MARRIS:
I've been trying to assimilate the basics of tube amplifier theory. I'm a bit confused about Class C, specifically, this passage in ARRL's "Understanding Amateur Radio" from the 1960s:

"Here the [grid] bias is set well beyond the cutoff point -- up to twice the bias required for cutoff, usually. A large grid driving voltage is used -- so large that making it larger wouldn't cause any further increase in plate current. This is called driving the tube to saturation. Class C operation gives the highest plate efficiency that can be obtained. It is useful in driving r.f. power amplifiers that are to be used for code work, or are to be modulated [later in the amplification chain] for phone."

The book goes on to point out that (as I already knew) the hitch is that you cannot use Class C amplifiers on an already modulated signal.

The bit that I find confusing is the notion that driving the tube to saturation would be the most efficient scenario. Surely once you are at saturation you are stuck with a certain power ratio between the grid input and plate output, and it cannot be further increased. I assume that I should understand it as saying, "driving the tube to saturation and not beyond."

Correct?

73 de Martin, KB1WSY

MARTIN MARRIS:
And another thing I find a bit confusing is that, as I understand it, in Class C the grid is operating below cutoff for some of the time, typically half of the time (180 degrees) on the fundamental and even less if the tube is used as a frequency multiplier, not just as an amplifier. How can that be efficient, in terms of power ratio?

(Please forgive my elementary questions, I'm trying to get it straight!!)

KB1WSY

Dale Hunt:
Efficiency is NOT related to the power ratio:  that is the stage power gain.  You can have
amplifiers with a very high impedance that require input voltage, but little input power
(because little or no current is required), but that is not efficiency.

Efficiency is the ratio of the RF output power to the DC input power drawn by the stage.
So if the stage draws 150mA at 300V (45 watts) from the plate supply and delivers 30W RF
to the load, it is 30/45 = 67% efficient, regardless of the driving power required.

A tube that acts as a pure switch is most efficient because in neither state (on or off) does
it dissipate power:  when off there is no current flowing, and when on there is no voltage
drop across it.  It is those intermediate times when it is neither fully off nor fully on that cause
the efficiency to drop.

Peter Chadwick:
Class C as an amplifier is best at about 140 degrees. Shorter pulses give lower power, while longer ones mean more dissipation.

The tube is driven to saturation by the peak of the RF input cycle.

You can get higher efficiency than Class C by running in a switching mode. Back in the 1960s, Marconi's did some high efficiency  broadcast transmitters driven with third harmonic peaking and trapping the third harmonic energy: because this made the PA tubes more of a switch, they could get 85 to 90% efficiency in the BC band at 1 MHz. That was with three tubes producing 750kW, and a big electricity bill....

These days, a lot of guys are using Class E amps with FETs to get high efficiency.

My father told me that at one stage in WW2, the RAF had a tx with a 8018 tube, which was a selected 807 for better performance at 120MHz. That was sort of Class C (probably nearer Class B) and the previous stage was modulated. As they never appeared on the surplus market, I assume they weren't a success....

MARTIN MARRIS:
Quote from: WB6BYU on May 05, 2013, 02:23:57 PM

Efficiency is NOT related to the power ratio:  that is the stage power gain.  You can have
amplifiers with a very high impedance that require input voltage, but little input power
(because little or no current is required), but that is not efficiency.


Yes I misspoke, and should have used clearer terminology. As I understand it, amplification in a tube is when a change in the grid *voltage* causes a change in the plate *current* -- and if the grid is never driven positive, this can happen without taking any current (and thus no power) from the grid circuit. Under those circumstances, one measure of the gain of the tube is to measure the change in grid voltage and the resultant change in plate current. Having got those figures, you remove the grid signal and then find out how much of an increase in plate *voltage* would cause the same, previously recorded change in plate current. The ratio of the change in the grid voltage to the change in plate voltage (that will produce the same current change) is a measure of the gain of the tube.

You'll have to forgive my blatant mistakes. I've been dabbling with tube gear for just over a year but without a deep understanding of what is going on; but now I do want to "get it" and this will involve asking some outstandingly dumb questions from time to time. I have plenty of textbooks but sometimes there's no substitute for actual advice from helpful people like yourselves. Followed by some very basic hands-on experiments on a single-tube basis. (I still lack the time to build complete radios, but that will come!!!) My dream is to build something stage by stage and actually understand the purpose of every component and how it all hangs together, i.e. to go beyond the block diagrams.

I wonder if someone can reformulate for me why it is that driving a tube way below cutoff, so that only a portion of the waveform is amplified, is actually more efficient than otherwise?? This statement is made baldly in the textbooks but I am still having some trouble getting my head around it. Naively, I would think that amplifying the whole waveform would be more efficient (with or without actually driving the grid positive). That's what I'm trying to get at! Is it because you can drive the tube so much harder, that it more than compensates for the partial absence of waveform?

73 de Martin, KB1WSY

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