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Author Topic: Length of coax from xcvr to amp affecting input swr, why ?  (Read 37452 times)
VE7RF
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Posts: 212




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« Reply #30 on: January 03, 2012, 10:28:24 PM »

5. the amplifier input is not presenting a 50 ohm impedance to the radio cable and bridge.

Not only is it something other than 50 ohms, it's not constant over the RF cycle either!

That is very true.
The older radios were better suited to handle this situation.

##  Sue.... take something simple..like a GG SB-220.  With it's oem  5 V zener for bias, the tubes have a 210 deg conduction angle.  That means they don't conduct for the other 150 degs !
A PI net tuned input is used to provide some flywheel action at the cathode. Just put a scope across the cathode on any GG amp, with and without the tuned input in place, and you can see the waveform distortion instantly.   Those peak RF pulses from the tube go around in a big loop..and have to end back up at the cathode.  Those pulses are aprx 3 x the dc plate current for class AB...and aprx  4 x the dc plate current..for Class C.  The tubes can't source anything. They can only sink current.  The return path for those pulses is via the chassis, then up through the C2 cap of any tuned input !  With out the tuned input installed, the path is then via the braid of the 3' of coax to xcvr, then through the C2 load cap of the tube xcvr pi net....then back down the center conductor of the same 3' coax....then back to the cathode. [ that C2 cap in either case has to handle one helluva lot more current than you think. ] 

##  These days, we use SS xcvr's..and each band has it's own high power LP filter.  I firmly believe the Xcvr's  LP filter + coax + the 210 deg conduction angle of the tubes + loaded Q of the amps tuned input all interact.  Change bands, and presto, a different xcvr's LP filter is now being used.  This has nothing to do with defective coax from xcvr to amp, nor defective swr meter inside the yaesu. [ which jives 100% with all my external meters, like coaxial dynamics, bird, array solutions, etc].  This is not transmission line theory either.  On BYPASS, the swr is dead flat into either the ant, or a dummy load. The difference being, we are then driving a 50 ohm, non reactive load...and NOT a the cathodes of pair of tubes..that only conduct 210 degs of each cycle.

## As long as the tuned input in the amp has adjustable C1+C2  caps, they can always be tweaked for a flat swr back to the SS xcvr.  [ and no need to run a sky high loaded tuned input Q]

##  as noted before, I tried loaded Q's of 5-6 on my HB 3CX-3000A7 amp's  tuned input, and with 200w of drive.. I can easily warm up solid 6ga Cu wire used for the 4uh tuned input coil. That tuned input consists of a 4 uh coil, aprx 17 turns of 6 ga wire, wound on a 1.5" ID. The coil is suspended in mid air, between 2 x broadcast caps. each broadcast cap is a 4 x section type..with all 4 x sections strapped in parallel.  Each section is 18-540 pf.  Each broadcast cap is padded with 4 x 500 pf HT-50 type doorknobs on 160M only.  The 4 uh coil is then tapped for each ham band.

##  Using a loaded Q of 5-6 works fine on the low bands, but is flaky on the upper bands.  I tested the assy, with wattmeters on both sides of the HB  tuned input assy....then 50 ohm coax into a dummy load. [the dummy load is not the cathode obviously, but the 3x3 tube is very close to 50 ohms].  With 200w applied, I was only getting 160w into the dummy load on the upper bands. I had to increase the uh on the coil by 1/4 turn on 15m..and retweak both caps..then the PO  shot up to 195 watts.  Now this is with 6 ga solid CU wire used for a tapped 4 uh coil.  Now try the same stunt with the more typ puny 18-20  ga wire, and higher loaded Q on any tuned input, and you will be in for a rude awakening.  And when finally hooked to the cathode of the 3x3 tube, the c1 + c2 values are now off a bit, and in some cases are way off  from the spreadsheets, depending on coax length.

##  My conclusion is,  SS xcvr's with their LP filters for each band, plus  old style bandswitched tuned inputs consisting of 2 x fixed caps  + adjustable slug tuned coil just won't cut it these days.  The short term 'fix' is just to use the auto tuner in the xcvr, if required. The yaesu's will do well over 200w po, and the L4B only requires 110 watts of drive.

## It's a good thing I didn't try and design  9x  bandswitched tuned inputs for the hb 3CX-3000A7 amp. [ that will handle 200-400w each. ] What a pita that would have been.  Loaded Q's of 3-6 is fine, provided you can adjust from the front panel. Otherwise you can't compensate for the narrower swr BW..and cable lengths.  I used a pair of jackson bro's 4" ball drives, with 6:1 vernier's for the C1 + C2 caps on the hb tuned input. they are marked 0-100 across the 180 deg arc. 4" diam = 12" circumference = 6" across the 180 deg.  Then it's a snap to just dial up various pre-sets for each band, and usually several per band.  We used the exact same hb tuned input for the larger YU-148 tubes.  Then coax length is a non issue, and the swr is always 1:1  from 1.8-30 mhz continuously.  As a real test, we stuffed a 800w cxr though it on each band for 10 mins, with no heating issues.

Later... Jim  VE7RF 
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G3UUR
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« Reply #31 on: January 04, 2012, 07:42:05 AM »

##  Sue.... take something simple..like a GG SB-220.  With it's oem  5 V zener for bias, the tubes have a 210 deg conduction angle.  That means they don't conduct for the other 150 degs !
A PI net tuned input is used to provide some flywheel action at the cathode. Just put a scope across the cathode on any GG amp, with and without the tuned input in place, and you can see the waveform distortion instantly.   Those peak RF pulses from the tube go around in a big loop..and have to end back up at the cathode.  Those pulses are aprx 3 x the dc plate current for class AB...and aprx  4 x the dc plate current..for Class C.  The tubes can't source anything. They can only sink current.  The return path for those pulses is via the chassis, then up through the C2 cap of any tuned input !  With out the tuned input installed, the path is then via the braid of the 3' of coax to xcvr, then through the C2 load cap of the tube xcvr pi net....then back down the center conductor of the same 3' coax....then back to the cathode. [ that C2 cap in either case has to handle one helluva lot more current than you think. ]

Jim, the cathode current includes a sizeable grid current, which in the case of a pair of GG 3-500Z tubes means a grid contribution at the peak of up to 1.3A. This makes the peak cathode current 1.5 times the peak plate current.

Your claims of a Q of 5-6 for the tuneable input filter on your HB amplifier don't seem to stack up with your C1 and C2 capacitor values either. With a limit of 2160pF on 75/80m, the circulating current in the input matching network will be around 3 times the output current (Q=3). However, this combined with the increased output capacitance (up to 26pF/meter rather than 14pF/meter in the L4B) would seem to be enough to get a reasonably stable input impedance for any solid state rig to work into.

You'll be able to get Q values of 5-6 easily on the higher bands, of course, because the reactance of your variables will be adequate to achieve that up there. I don't know why you were so surprised that you could get so much loss in the input network at the higher frequencies. Switched inductors are not the best way to achieve high Q values and a few simple calculations of the very low intermediate impedance in the pi-network matching input circuit will soon tell you how high the circulating current will be. XL/Q will give you the loss in the coil and you know the current, so you should be able to estimate the loss prior to building the network. The intermediate impedances can be very low and the currents very high in GG input matching circuits when you're starting from 50 ohms. None of this ought to be a surprise.

Dave.
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W8JI
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« Reply #32 on: January 04, 2012, 05:03:07 PM »

5. the amplifier input is not presenting a 50 ohm impedance to the radio cable and bridge.



Not true.

If line loss was low and the impedance was a stable impedance, and a linear load, SWR would not vary with cable length even if the SWR was not unity. 

SWR would be the same if the line were 3 feet long or six feet long, even if the SWR was 3:1.

73 Tom
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KF7CG
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« Reply #33 on: January 05, 2012, 10:51:55 AM »

What makes the input of an amplifier different from any other load at the end of a coax?

If the coax impedance doesn't match the impedance of the load you will get standing waves in the coax and the attendant impedance transformations that go with feedline transformers.

The well know cases used to be reducing the 200 ohm impedance of a folded dipole driven yagi by using a quarter wave of 75 ohm coax as a transformer.

This may be an extreme in the length case but it shows the concept. 4 to 1 if measured from radio to folded dipole, 1 to 1 if measure at end of coax transformer.

Different frequencies and coax electrical length have different effects on presented impedance and on SWR. Look at the books on transmission line effects and substitute resistive (complex if you want to fight the math) loads/sources at the ends of the line.


David
KF7CG
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VK4TUX
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« Reply #34 on: January 05, 2012, 11:21:24 AM »

What makes the input of an amplifier different from any other load at the end of a coax?
David
KF7CG

The conduction angle of the tube .

Adrian ... vk4tux
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G3UUR
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« Reply #35 on: January 05, 2012, 11:47:35 AM »

What makes the input of an amplifier different from any other load at the end of a coax?

If the coax impedance doesn't match the impedance of the load you will get standing waves in the coax and the attendant impedance transformations that go with feedline transformers.

What makes the input of a grounded grid amplifier different is that the input current is a non-linear function of the drive voltage and that makes the impedance vary over the part of the cycle the tube is conducting. Also, as Adrian has pointed out, the tube is off part of the time and for that time 100% of the drive signal will be reflected back. For the rest of the time when the tube is conducting the reflected power will vary and at a couple of points in the cycle it might be zero fleetingly. This makes a very non-linear load!

One way round this problem is to use an input network with some energy storage to even out the impedance seen by the driver. If the circulating current in the input stage is high enough the flywheel effect will provide a stable load impedance for the driver even though the load on the input matching circuit is varying considerably. The matching circuit only has to have a high enough Q and match to the average value of the varying load. This average value can be found by considering that the power going into the matching network must equal the power going out.

Dave. 
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VE7RF
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« Reply #36 on: January 06, 2012, 05:35:19 AM »

##  Sue.... take something simple..like a GG SB-220.  With it's oem  5 V zener for bias, the tubes have a 210 deg conduction angle.  That means they don't conduct for the other 150 degs !
A PI net tuned input is used to provide some flywheel action at the cathode. Just put a scope across the cathode on any GG amp, with and without the tuned input in place, and you can see the waveform distortion instantly.   Those peak RF pulses from the tube go around in a big loop..and have to end back up at the cathode.  Those pulses are aprx 3 x the dc plate current for class AB...and aprx  4 x the dc plate current..for Class C.  The tubes can't source anything. They can only sink current.  The return path for those pulses is via the chassis, then up through the C2 cap of any tuned input !  With out the tuned input installed, the path is then via the braid of the 3' of coax to xcvr, then through the C2 load cap of the tube xcvr pi net....then back down the center conductor of the same 3' coax....then back to the cathode. [ that C2 cap in either case has to handle one helluva lot more current than you think. ]

Jim, the cathode current includes a sizeable grid current, which in the case of a pair of GG 3-500Z tubes means a grid contribution at the peak of up to 1.3A. This makes the peak cathode current 1.5 times the peak plate current.


###  Right you are Gunga Din !  ur dead on abt the added grid I.

Your claims of a Q of 5-6 for the tuneable input filter on your HB amplifier don't seem to stack up with your C1 and C2 capacitor values either. With a limit of 2160pF on 75/80m, the circulating current in the input matching network will be around 3 times the output current (Q=3). However, this combined with the increased output capacitance (up to 26pF/meter rather than 14pF/meter in the L4B) would seem to be enough to get a reasonably stable input impedance for any solid state rig to work into.


##  when I did that experiment, I used the 160m padder caps on 80m for the extra C.  That gave me an extra 2000pf on each air cap. Now this was for 80m only. So 4160pf max, per cap. 

You'll be able to get Q values of 5-6 easily on the higher bands, of course, because the reactance of your variables will be adequate to achieve that up there. I don't know why you were so surprised that you could get so much loss in the input network at the higher frequencies. Switched inductors are not the best way to achieve high Q values and a few simple calculations of the very low intermediate impedance in the pi-network matching input circuit will soon tell you how high the circulating current will be. XL/Q will give you the loss in the coil and you know the current, so you should be able to estimate the loss prior to building the network. The intermediate impedances can be very low and the currents very high in GG input matching circuits when you're starting from 50 ohms. None of this ought to be a surprise.

##  I never said it was a surprise, just an eye opener, that's all.  On 80m, with high Q, the 6 ga wire is ok. On 40m, it semi ok.  On the upper bands, if u want to run  a  q of 6-10, u better use some 3/8" tubing for the coil.  The L4B tank coil is wound with 1/4" tubing,.and runs HOT on 15M, and real hot on 10M.  that's with 800ma of plate current..and Q's > 10.  That's only 10-12 a of current  through a 1/4" tubing coil.

## Now with 200w  from an xcvr..and a Q of 10 for a tuned input.. = 20A current.  And 20A  will cook  just abt anything on the upper bands.  Anyway, the point being..as long as both C1 and C2  on a pi net for a tuned input is adjustable, then flat swr can always be obtained..with any length of coax. I can calculate current flow..but the question is... what will be the temp rise over ambient?   I hear so much bs.. I thought I  would just run some tests.

Later... Jim  VE7RF

Dave.
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VE7RF
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« Reply #37 on: January 06, 2012, 06:11:21 AM »

What makes the input of an amplifier different from any other load at the end of a coax?

If the coax impedance doesn't match the impedance of the load you will get standing waves in the coax and the attendant impedance transformations that go with feedline transformers.

What makes the input of a grounded grid amplifier different is that the input current is a non-linear function of the drive voltage and that makes the impedance vary over the part of the cycle the tube is conducting. Also, as Adrian has pointed out, the tube is off part of the time and for that time 100% of the drive signal will be reflected back. For the rest of the time when the tube is conducting the reflected power will vary and at a couple of points in the cycle it might be zero fleetingly. This makes a very non-linear load!

One way round this problem is to use an input network with some energy storage to even out the impedance seen by the driver. If the circulating current in the input stage is high enough the flywheel effect will provide a stable load impedance for the driver even though the load on the input matching circuit is varying considerably. The matching circuit only has to have a high enough Q and match to the average value of the varying load. This average value can be found by considering that the power going into the matching network must equal the power going out.

Dave. 

## How do we calculate the average value of the  varying load ??  It's 50 ohms when driven..and sky high during periods of non conduction. [ this is on a 3CX-3000A7].  Ur right, a loaded Q of 3 is more than ample..in some cases. ..and will work, as long as both C1 and C2 caps are adjustable.   In my old  ssb, theory and practise book... it mentions the 20.5' cable that collins used on the 30L1.    " Unless the Z match between xcvr +  output and pa input is reasonably close, the Z variations will be reflected into the driver plate circuit and appear as phase modulation of the SS signal.   The length of the cable  is selected so that when it is added to the lumped constants of the pi net input... the effective length is equal, electrically to  1/2 wave  length..or multiple on each amateur band" .

## same book, but re: to the 30S1.  "to match the two, an ingenious method is used which is a  combo of cable + bandswitched pi nets.  The 20.5'cable  and tuned input are such that  even multiples of 180 deg phase shifts will be provide between driver plate and  pa grid.  Even multiples are needed, since  modulation components  change the amps cathode Z, and this  change is translated into a shift in  reactive  Z  at the driver plate. This shift  phase modulates the driver..and in turn increases overall distortion``

// OK... how do u propose to make high Q ..low esr, coils for the tuned inputs. I`ll be damned if I`m gonna build 9 x pi nets.. and mount it, so all 18 x arco`s can be tweaked.  Just to recap, how high does the loaded Q of the pi net have to be..to match the average load. How do i calculate the average load.   what`s the min loaded Q I can get away with . 

later... Jim  VE7RF
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W8JI
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« Reply #38 on: January 06, 2012, 08:57:57 AM »

Jim,

You certainly have the problem of fractional-cycle loading, anyone knows that.

The cable length affecting SWR reading is a result of the harmonics created by the tube for a fraction of a cycle.

There is considerable harmonic energy at the tube cathode, and regardless of Q, if the harmonics are not bypassed and prevented from reaching the exciter, they will show as reflected power on a directional coupler.

There are ten ways to nit-pick the working of this, and argue how many angels can standing on the head of a pin, but the root cause of the directional coupler seeing an SWR that isn't really there is harmonic content in the waveform.

If you had a Q of 10 in a high-pass to the cathode, you could still have the harmonic energy issue in a grounded grid stage conducting less than 360 degrees.

This is why the network has to be near the cathode, and be a good low pass or band pass, to be reliable. 

73 Tom
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G3UUR
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« Reply #39 on: January 06, 2012, 12:17:46 PM »

## How do we calculate the average value of the  varying load ??  It's 50 ohms when driven..and sky high during periods of non conduction. [ this is on a 3CX-3000A7].  Ur right, a loaded Q of 3 is more than ample..in some cases. ..and will work, as long as both C1 and C2 caps are adjustable.   In my old  ssb, theory and practise book... it mentions the 20.5' cable that collins used on the 30L1.    " Unless the Z match between xcvr +  output and pa input is reasonably close, the Z variations will be reflected into the driver plate circuit and appear as phase modulation of the SS signal.   The length of the cable  is selected so that when it is added to the lumped constants of the pi net input... the effective length is equal, electrically to  1/2 wave  length..or multiple on each amateur band" .

## same book, but re: to the 30S1.  "to match the two, an ingenious method is used which is a  combo of cable + bandswitched pi nets.  The 20.5'cable  and tuned input are such that  even multiples of 180 deg phase shifts will be provide between driver plate and  pa grid.  Even multiples are needed, since  modulation components  change the amps cathode Z, and this  change is translated into a shift in  reactive  Z  at the driver plate. This shift  phase modulates the driver..and in turn increases overall distortion``

// OK... how do u propose to make high Q ..low esr, coils for the tuned inputs. I`ll be damned if I`m gonna build 9 x pi nets.. and mount it, so all 18 x arco`s can be tweaked.  Just to recap, how high does the loaded Q of the pi net have to be..to match the average load. How do i calculate the average load.   what`s the min loaded Q I can get away with . 

Jim,

The input impedance of a pair of 3-500Z tubes in GG, if I remember it correctly from years ago when I calculated it, runs from about 30 ohms at the peak down to about 90 ohms just before they turn off. You can calculate an effective value of input resistance by splitting up the cycle into 10 degree segments and considering the power consumed by each of the various values of load presented to the matching circuit and doing a weighted average of that. The effective value is then the value that would consume the same power if applied continuously. I've done that and it comes out very close to a much easier way, which entails working out the cathode current waveform for an ideal voltage source driver (zero output impedance) and doing a Fourier transform to find out the relationship between the level of the fundamental waveform and the peak current. Both methods give practically the same value, which is around 2.2 times the value at the peak for a pair of 3-500Z tubes, so it's about 66 ohms, theoretically. Bear in mind this is for the ideal situation of a voltage source driving the cathode and in real amplifiers with some source impedance the value is slightly higher, probably 70 to 80 ohms.

When we're considering driving old GG amplifier designs with solid-state rigs there's no real alternative to upping the Q of the input matching network if we want to get away from using magic lengths of inter-connecting cable. The consequence, as you know, is a more limited bandwidth and the need for adjustment to cover the entire amateur band in some cases. Not much can be done about that, other than automating the tuning system, which in an old amplifier wouldn't be worthwhile.

A bit more imagination can be used in the design of matching networks to ease the need for such large values of capacitance, though. Some experimenting with how close to the wind we can sail on Q and pF per meter on the output capacitance will be required, but that should be no problem for some. I don't have the time at the moment because I'm doing up my house to sell. My wife and I are already involved in charity work in the Caribbean and need to downsize so we have somewhere smaller in the UK to use as a base, and hope to be abroad much of the time for the next few years. Much of my gear is packed away or in the process of being packed away.  You have several L4B linears with input SWR problems, so would seem to be in a very good position to conduct a few tests yourself, Jim. Tom should also be in a similar position to do these tests easily. I would think there are many L4B and SB220 owners who could do with some help on this score. I suppose there could be similar problems with some of the Jap GG linears as well. 

Dave.
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VE7RF
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« Reply #40 on: January 08, 2012, 04:12:13 AM »

Jim,

You certainly have the problem of fractional-cycle loading, anyone knows that.

The cable length affecting SWR reading is a result of the harmonics created by the tube for a fraction of a cycle.

There is considerable harmonic energy at the tube cathode, and regardless of Q, if the harmonics are not bypassed and prevented from reaching the exciter, they will show as reflected power on a directional coupler.

There are ten ways to nit-pick the working of this, and argue how many angels can standing on the head of a pin, but the root cause of the directional coupler seeing an SWR that isn't really there is harmonic content in the waveform.

If you had a Q of 10 in a high-pass to the cathode, you could still have the harmonic energy issue in a grounded grid stage conducting less than 360 degrees.

This is why the network has to be near the cathode, and be a good low pass or band pass, to be reliable. 

73 Tom

###  I just looked at the schematic of the AL-1500... and it depicts pi nets for each band, consisting of fixed caps, and a slug tuned adjustable coil.   The C1 + C2 values  for each band of the AL-1500  are no where near as high as the L4B tuned input values [80-10m, the L4B has no 160m].

##  OK, I just looked at the tuned inputs on the AL-82..which uses 2 x 3-500Z... and it too, has lower values  than the L4B.  In both the AL-1500 and also AL-82... C5 + C6 are the tuned input caps..with C5 closest to the cathode.  The AL-82 also uses fixed caps and a slug tuned coil.

##  I will have to do some tweaking and experimenting on one of the 4 x L4B's.  It's a bit awkward, since the caps and slugs are below the chassis..and with bottom lid off, the air is lost to the tubes [blower + chimneys].

Later... Jim  VE7RF
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VK4TUX
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« Reply #41 on: January 08, 2012, 04:31:50 AM »

This guy (W2AJI)

http://home.earthlink.net/~cherokeehillfarm/id2.html

has an interesting take on the subject;

"Why does this work as it does? 

In cases involving RF signals, some time will pass during the 'round trip of the reflected energy and the phase of the reflection will also depend upon this length of time. Imagine that a resistor in a black box is at the end of a length of cable. From the outside world this length of cable will give the reflection from the resistor a phase shift since the signal must make a round trip through the length. If a 100 ohm resistor has an SWR of 2, a cable long enough to invert the signal after the round trip will make it look like a 25 ohm resistor, also with an SWR of 2 but with inversion (a cable with a multiple of 1/4 wavelength would do the trick). Since the impedance looking into this black box is a function of the SWR and the cable length, it can be seen that intentionally mismatched lines can be used to transform one impedance into another. Notice that the 1/4 wave cable inverts the impedance and preserves the SWR. This impedance inversion may be used to match two impedances at a particular frequency by connecting them with a 1/4 wave cable with an impedance equal to the geometric mean of the two impedances. (The geometric mean is the square-root of their product.) A 50 ohm, 1/4 wave cable will match a 25 ohm source to a 100 ohm load : sqrt(25 x 100) = 50. Of course, it is not always easy to find the desired impedance cable!

Multiples of 1/2 wavelength will give enough delay that the reflection is not inverted and the impedance will be the same as the load. Such cables may be used to transfer the load impedance to a remote location without changing its value (at one frequency).

Other cable lengths will transform an impedance which differs from the cable's impedance with a reactive component. If the load is a lower impedance than the cable, a length below 1/4 wave will have an inductive component and above 1/4 (but below 1/2) wave a capacitive component. If the load is a higher impedance than the cable, the reverse is true. Above 1/2 wavelength, the reactance will alternately look capacitive and inductive in 1/4 wave multiples. This reactance will combine with the load's reactance and offers the possibility of resonating the reactive component of the load. Therefore, a cable with the "right" length and impedance can match a source and load with different resistance and reactance values. Obviously, these calculations can become quite involved and most engineers resort to a Smith chart, a computer program or perhaps the most common method, trial and error with a SWR meter or return loss bridge!

Our trial and error experiments with an external or internal transceiver SWR meter, suggest that the solution length is usually between 20 and 22 feet.  Therefore, considering for example that on 20 meters where this length is greater than a 1/4 wavelength, your input SWR is too high and your amplifier is presenting a load impedance of less than 50 ohms, this solution length will have a capacitive component. This would be true for 15 meters as well.

It is also possible that the longer length of coax acts as a "stub" and strips shield currents from the interconnecting cable which are fooling the internal SWR bridge, and therefore it eliminates power foldback.  In this case, a line isolator installed between the driver and the amplifier may also be of major benefit in eliminating the power foldback.

This solution is simple to achieve, and inexpensive as well.  Just a little patience is required when trimming the cable.  In the event that your transceiver does not directly read SWR, you may install a SWR bridge coupler onto the output connector of your transceiver to obtain the solution length of cable.

If your linear amplifier input circuits are tunable, you can finish up with adjusting the input coils for the absolute lowest input SWR on each band.  You will be pleased to see that after following this solution, your transceiver will be able to operate at an even lower level of drive to produce full amplifier output. "
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« Reply #42 on: January 09, 2012, 03:20:41 AM »

Jim,

You certainly have the problem of fractional-cycle loading, anyone knows that.

Apparently not everyone!
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« Reply #43 on: January 09, 2012, 04:18:49 AM »

There is considerable harmonic energy at the tube cathode, and regardless of Q, if the harmonics are not bypassed and prevented from reaching the exciter, they will show as reflected power on a directional coupler.

There is but you seem to only half understand the problem. Considerable fundamental reflected power occurs until the Q is high enough for the flywheel effect to stabilize the load to the driver. A high enough Q in this context is still very low, perhaps somewhere between 1 and 3. Nobody really knows yet because the designers of 1980's and onwards GG linear amplifiers didn't understand enough to ask the question and did things by trial and error rather than measuring what input network Q was required to just be able to get over the problem.

Once you've decided on a configuration to use for the matching network, the attenuation of the harmonics and Q are inextricably linked. If you increase the capacitor values of a CLC pi-network input matching circuit to increase the harmonic attenuation, you inevitably increase the Q. There's no way round that and if you know your theory you just accept that both are being increased if you're talking about increasing the Q of the network. The real question, which you "experts" still haven't answered is how little Q can we get away with and how does that differ for the various input matching circuits?

Quote
If you had a Q of 10 in a high-pass to the cathode, you could still have the harmonic energy issue in a grounded grid stage conducting less than 360 degrees.

Trust you to think of an absurd comparison, Tom!  No one would use this configuration, so why raise the issue. The rest of us are not talking about using a Q as high as 10 in any configuration, though we know it would provide the necessary stabilization of the load for the driver, because the operational bandwidth would be too narrow. In case you hadn't noticed, we're discussing where between 2 and 5 we'd find the best compormise for use with a solid-state driver. Above a Q of 2 the bandwidth between the 1.5:1 SWR frequencies no longer allows operation over the whole band on 75/80m, but there is still harmonic contamination in the reflected signal. How little can we get away with above this?

The second harmonic is probably only in the region of 7dB down at the cathode in the ideal case of the driver looking like a perfect voltage source. A CLC pi-network input circuit with a Q2 of 2 might give you 17dB attenuation in the ideal case, but would probably be lower by a couple of dB because the driver is not conjugately matched. You'd see a 1.2:1 SWR in this perfect case and much worse in practice.

A simple fix would be to use yet another low-pass filter bank between the amplifier input and the driver transceiver, but that would not address the problem of IMD impairment if that's happening as a result of increased non-linearity at the cathode/filament due to a higher than desirable source impedance. However, these filters would only need to be third order but their phase delay would need to be included in the number of 180 degree phase delays between the solid-state PA in the driver and the tube cathode to optimise the source impedance.

This should all have been addressed thirty years ago and shouldn't have been left until now. Someone has been derelict in their engineering duty and we know where to point the finger, Tom. It's all very well you criticizing other designers like Warren Bruene but he had very good theoretical knowledge for his day, unlike some of you modern designers. He also had to make do with limited test equipment compared with todays designers.

Jim, the figure given for the input impedance of a pair of 3-500Z tubes by Bill Orr is around 60 ohms, so it's not as high as the optimum output inpedance of the L4B matching networks would suggest. You've also got to be very careful comparing the input matching circuits of the different amplifier designs because of the reactance of the filament chokes and other circuitry around the input. Very often this reactance is a bit low at the low frequency end of the range and has to be tuned out by some of the output capacitance of the input network. So, unless you know the exact impedance looking into the cathode/filament on the various bands with the tubes off, you have to rely on the input values and the inductor values to give you a hint of what the output capacitance should be solely for matching to the tube input impedance.

Dave.
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« Reply #44 on: January 09, 2012, 04:25:44 AM »

But the Q needed to have the harmonic current low enough for one transceiver not to object may well not be enough for another. So to cover all transceivers now and in the future may not be possible if a relaively wide band (eg 75/80) is to be covered.
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