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Author Topic: Tip for gaining 10%-20% in Amp. Efficiency  (Read 1377 times)
VE7ALQ
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Posts: 349




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« on: August 09, 2005, 12:49:03 PM »

As is well known, a class AB2 amplifier generates harmonics.  Of these harmonics, the third is the highest in energy.  According to "Power Vacuum Tubes Handbook" by Jerry C. Whitaker, CRC Press, 1999, inserting a parallel "trap" circuit tuned to the third harmonic in series with the plate lead(s) will stop the tube(s) shunting the third harmonic to ground, and thereby wasting energy as unnecessary plate dissipation.  Apparently a 10% - 20% improvement in efficiency can be obtained by this simple technique.
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WB2WIK
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« Reply #1 on: August 09, 2005, 02:14:25 PM »

Two problems with that theory:

1.  To achieve a 20% efficiency improvement implies that the third harmonic energy was consuming 20% of the available power.  This is 20% distortion, which is horrendous and no linear amplifier that's not being overdriven into saturation would have this much distortion.

2.  You'd need to bandswitch that trap, since a single trap could only work for a single band.  If you had a 3rd harmonic trap in line on 40 meters, your amplifier wouldn't have any output on 15 meters (for example).  You'd need a new trap for each band operated.  

WB2WIK/6
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N0TONE
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Posts: 173




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« Reply #2 on: August 09, 2005, 02:50:23 PM »

The pi network presents a very low impedance to that third harmonic, which is what a trap would do, so we already have that feature built into ham amps.

AM
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VE7ALQ
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Posts: 349




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« Reply #3 on: August 09, 2005, 02:56:30 PM »

The output current, as measured at the plates of the amplifier tubes, is anything but sinusoidal.  It consists of portions of a sinusoid, goes to zero during the portion of the cycle that the tube is biased to cutoff.  If you want to know how much power is in the fundamental, how much is in the harmonics, you have to do a Fourier Transform of the current pulse.  Only the fundamental frequency is of interest, the harmonics are shunted to ground (at least with the common "PI"-net and "PI-L"-net) output circuits.  Having the tube produce a harmonic component is normal, but this content represents wasted power, as it is shunted to ground by the output circuit.  Since the third harmonic is the primary unwanted current, a parallel trap tuned to the frequency of the third harmonic increases efficiency from 10% - 20%.

Yes, I agree that this trap has to be tuned for each band, and probably isn't practical for us amateurs unless we are building monoband linears.
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VE7ALQ
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« Reply #4 on: August 09, 2005, 03:11:19 PM »

I agree that the "PI" and the "PI-L" network present a very low impedance to harmonics.  However the plate current rich in harmonics is being collected at the plates.  If these harmonic currents are not blocked, they cause the plate(s) of the amplifier tube(s) to heat.  A parallel trap has HIGHEST impedance at resonance, and it is such a parallel trap I saw in series with the plate lead in a circuit in "Power Vacuum Tubes Handbook",Jerry C. Whitaker, 1999, CRC publishing company.
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K0IZ
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« Reply #5 on: August 09, 2005, 06:13:00 PM »

20% of 2400 watts (or so, input) is a bunch of watts.  Would take a big trap to dissipate that much power!

I think there is something wrong with your math.  For a reasonably linear amp, the third harmonic should be at least 10db down (not very linear), more likely 20db or so.  This is before the beneficial impact of the plate circuit.  At 10db, this would be 10%, at 20 db 1%.  A 20% third harmonic would be about 7db down, which would make for a nice, oldfashioned tripler stage, not a linear amp.

John.
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VE7ALQ
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« Reply #6 on: August 09, 2005, 06:36:32 PM »

The trick is, you BLOCK the third harmonic current from flowing with a parallel LC trap which goes in series with the plate lead.  Since there is no longer a third harmonic component to the plate current, the tube plate no longer has to dissipate third harmonic energy.
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K7KBN
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« Reply #7 on: August 09, 2005, 08:46:49 PM »

Build it.  Let us know how it works.
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73
Pat K7KBN
CWO4 USNR Ret.
VE7ALQ
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Posts: 349




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« Reply #8 on: August 09, 2005, 08:56:59 PM »

Please give me some time to go through the 750 page plus "Power Vacuum Tubes Handbook",1999,Jerry C. Whitaker, published by CRC.  The circuit for the amp with third harmonic suppression, is in there.  I have seen it but  just don't know exactly where the blazes it is in the book.
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WA9SVD
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« Reply #9 on: August 10, 2005, 11:00:16 PM »

Getting rid of the third harmonic by a trap won't increase the output at the fundamental frequency.  Whether it's 20%, 2% or 0.2%, or any value, the harmonic power will be dissipated in the trap, so you will never "recover" that power in the output.
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N0TONE
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Posts: 173




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« Reply #10 on: August 11, 2005, 02:09:30 PM »

The third harmonic in a normal amp is not generating any appreciable power.

Power requires in-phase volts times amps.  Even if there is a 3rd harmonic current (amps) at the anode, the narrowband pi network ensures that there is not a third harmonic voltage.  

The third-harmonic parallel "trap" which he refers to turns a class C amp into a class F amp, in the solid state world.  It's a way to make the collector waveform more closely resemble a square wave.  In the solid-state world, we think of it as a way to squeeze an extra 10-15% out of an already-optimized class C amp.

Its effectiveness is dependent on the amp already operating close to saturation; a class AB amplifier (such as is required for SSB) won't enjoy as many benefits, and it doesn't do a thing for an amp that's operating substantially below saturation.  Also, because it will have an impact on gain only near saturation, it would lead to increased IMD.

Most efficiency-enhancement techniques require the amp to be operating in saturation to begin with.  We hams are the odd lot, because we are the sole survivors who don't use Class C amps.  The entire commercial world operates with AM (plate modulated class C) or FM (class C).

AM
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WA0LYK
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Posts: 85




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« Reply #11 on: August 20, 2005, 06:46:23 AM »

A little late to the discussion so may not be noticed.

I agree with the last poster.  There are two components to IMD, the exiciter and the amp.  A trap could help remove the IMD component contributed from the exciter.  It won't do anything for the IMD generated in the amp.  The amp generates the IMD in the tube itself during the amplification process.  As a result, you may help eliminate effects outside the tube, but it won't do anything for the efficiency of the amp!

Here is the real rub.  IMD products are generated at the audio frequencies.  In other words, your IMD products are close to the fundamental frequency in terms of audio frequencies, i.e. 3, 4, ..., 8, 10 kHz depending on what audio range you are using.  You WON'T build a trap that can eliminate the 3rd order IMD without also eliminating the fundamental frequency!

Jim
WA0LYK
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W0FEN
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« Reply #12 on: August 22, 2005, 06:21:02 AM »

There is a way to eliminate ENTIRELY the harmonics in the amplifier.  Try rebiasing to class A this will produce an exact copy of the input signal.  You will not be happy with the overall effeciancy of the amplifier after that.
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VE7ALQ
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Posts: 349




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« Reply #13 on: August 22, 2005, 08:12:25 AM »

I agree that placing a third harmonic trap across the plate leads (like the RF Parasitic Chokes) will not increase your output.  It should, correct me if I'm wrong, reduce the plate current as the path for the third harmonic signal would be blocked in the parallel trap.  As there will be no third harmonic energy to speak of reaching the plates of the amplifier tubes, then the tubes should run a lot cooler, and draw less power than a conventional design with no parallel third harmonic trap.
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