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Author Topic: Class-C Tube Amplifier Circuits  (Read 4861 times)
KB1WSY
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« on: May 05, 2013, 01:42:58 PM »

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
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KB1WSY
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« Reply #1 on: May 05, 2013, 01:47:42 PM »

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
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WB6BYU
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« Reply #2 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.

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.
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G3RZP
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« Reply #3 on: May 06, 2013, 01:28:15 AM »

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....
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KB1WSY
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« Reply #4 on: May 06, 2013, 06:12:58 AM »

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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AA4PB
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« Reply #5 on: May 06, 2013, 10:00:36 AM »

Its because a class C amp is used only to amplify a constant amplitude sine wave and the plate circuit includes a L/C resonant circuit. The plate current pulses excite the resonant circuit which completes the sine wave.
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KB1WSY
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« Reply #6 on: May 06, 2013, 10:12:34 AM »

Its because a class C amp is used only to amplify a constant amplitude sine wave and the plate circuit includes a L/C resonant circuit. The plate current pulses excite the resonant circuit which completes the sine wave.

Brilliant, thanks!!! A good example of something that seems obvious once you know it, which explains why my textbooks don't bother to explain it.

73 de Martin, KB1WSY
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QRP4U2
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« Reply #7 on: May 06, 2013, 06:57:03 PM »

Quote
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?

You have to take the amplifying device into consideration.

In class A, the device is dissipating about 70% of the input power over the full 360 degree conduction cycle.

In class B, the device is dissipating about 50% of the input power over the 180 degree conduction cycle.

In class C, the device is dissipating about 25% of the input power over the 90 degree conduction cycle.

The shorter the time the amplifying device is conducting, the more power that is available to the output circuit, because the amplifying device is wasting less power during this shorter time. Hence, the output power to input power efficiency rises.

In class C, the energy during that shorter time period is being pumped into the LC cicruit, in this case, a resonant circuit which recreates the full sine wave.
« Last Edit: May 06, 2013, 07:33:21 PM by QRP4U2 » Logged

AC0OB - A Place Where Thermionic Emitters Rule!
KC2ZFA
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Posts: 15




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« Reply #8 on: May 06, 2013, 07:33:50 PM »

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.

I like this description...thanks !

And that is why Class F (with parallel L/C traps tuned to 3f at the cathode and plate) gives the 90% efficiency it is famed for.
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QRP4U2
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« Reply #9 on: May 06, 2013, 07:48:18 PM »

The OP is trying to understand basic C class theory so discussion of other classes of operation should be relegated to a later thread.

Quote
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??

A tube is biased below cutoff so that only a short time period of the driving signal triggers the tube to conduct. So if we're feeding the grid of a class C tube biased negative below cutoff, we only allow a small positive portion of that driving sine wave to force the tube to conduct for a short period of time.

Phil - AC0OB

« Last Edit: May 06, 2013, 07:56:05 PM by QRP4U2 » Logged

AC0OB - A Place Where Thermionic Emitters Rule!
QRP4U2
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« Reply #10 on: May 07, 2013, 05:45:32 PM »

Martin, glad to see you are a Homebrewer.

Since I can't seem to upload pdf files or schematics, here is the next best thing:

http://www.clarisonus.com/Archives/TubeTheory/ClassC.pdf

Look at page 3, right side for the waveforms.

Consider a Class C broadcast transmitter transmitting on 1 MhZ. The input waveform to the grid, an RF sine wave, will have a total period of 1 microseconds. The positive portion of the grid waveform will be going from zero voltage to maximum voltage back down to zero voltage in 500 nanoseconds.

1. The grid drive will be a sine wave of 200 Volts Peak-to-Peak and we are only interested in the positive portion or the 100 volt positive segment

2. (From bottom to top) the DC grid bias supply is set to -90 volts, which is connected to a 10K resistor which is connected to a 2.5 mH choke and then to the grid,

3. Plate voltage will be 3000 Volts with a maximum current of 0.5 Amps

4. The grid voltage rises from zero to +10 volts at the 50 nanosecond time click, grid and plate current starts to flow, but not much

5. At 100 nanoseconds the grid voltage is at 58 volts and plate current is significant

6. At 250 nanoseconds, the grid voltage is 100 volts maximum and the plate current is maximum, input power at this time is 1,500 Watts

7. From 255 nanoseconds on, the grid voltage starts to fall, the plate current starts to fall

8. At 450 nanoseconds, the grid voltage starts to fall below + 10 Volts and the plate and grid current tends toward zero,

9. At 500 nanoseconds the grid voltage finally goes to zero, plate current is zero


So the time between about 100 nanoseconds and 400 nanoseconds is where the majority of energy is dumped into the resonant circuit, usually a Pi-Network or Pi-L network circuit which not only stores energy, but provides an impedance match between the tube plate and the output circuit.

I hope this helps

Phil – AC0OB
« Last Edit: May 07, 2013, 05:52:02 PM by QRP4U2 » Logged

AC0OB - A Place Where Thermionic Emitters Rule!
KB1WSY
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Posts: 806




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« Reply #11 on: May 08, 2013, 04:59:21 AM »

To everyone: thank you for your detailed answers, which I am still digesting. My understanding of C-class amplification is now about 80 percent better! In the end there's no substitute for hands-on experience. I'm looking forward to restoring my ancient oscilloscope, then testing a tube configured for various amplification classes. I'd like to build a basic one-tube amplifier circuit and mess around with grid bias and other settings, feeding the tube from my vintage RF signal generator and passing the output to a tank circuit then viewing the output on the 'scope. I'm also looking forward to getting a detailed appreciation of the tube types (diode, triode, tetrode, pentode). I can build a crude "breadboard" configured for tubes -- I have seen various solutions for that but in this case it should be very simple, just a small metal chassis with a single tube socket and a couple of terminal strips I can solder and re-solder to. The other format that I have seen is a piece of flat aluminum mounted vertically, with the tube(s) on the backside and the wiring easily accessible on the front.

73 de Martin, KB1WSY
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