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I Hate Old Radios!




 
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Author Topic: I Hate Old Radios!  (Read 8697 times)
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ashart
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« on: October 21, 2013, 04:06:29 PM »

Ok, here's the deal.  I've finally got BigRig   ( see www.w8vr.org ) on the air, and I've been having a great time on 40 AM.  In fact, I've done more hamming on 40 AM in the past couple of months than I've done otherwise over the past 4 years!  What fun!!  But now I've got to get down to brass tacks and finish some of the fine points of the rig.

I'm trying to finalize my 810 triode final amp operating points, and have encountered the following Catch 22:

1.   The RCA Transmitting Tube Manual, #TT-4 on Page 44, states:   Experience has shown that the most satisfactory relation between power output and power gain in straight-through class C amplifier service is achieved at a conduction angle of about 140 degrees.  The use of larger conduction angles reduces driving-power requirements, but results in substantially reduced plate-circuit efficiency.  On Page 47, it is stated that this recommendation applies to Class C Telegraphy Service. 

FB for CW work.  But for phone, the same manual states on Page 49:

2.    [In plate-modulated Class C Telephony Service] It is also usually desirable to employ a conduction angle smaller than that used in telegraphy service to assist in obtaining linear modulation, as discussed previously.

Yet, no matter how many times I read that manual, either manually(!) or by computer search, I just can't seem to find any reference to using different conduction angles for CW and for phone.

A man could get a headache fiddling around with these stupid radios!

If you can direct me to an authoritative discussion of how different conduction angles are optimized for plate-modulated Class C triode amps, and for CW Class C triode amps, I would certainly appreciate it very much.

But please, I'm a crotchety old man, grouchy and cantankerous, and easily frustrated wading through a bunch of old wives' tales about how this rig or that rig always gets 5 and 9 from Indiana and works lots of DX while never even considering conduction angles, or how if I were smart, I'd just use 27 volts of bias, or how I'd be better off if only I bought a Model XX transmitter instead.  In my stubborn ignorance, I crave equations and technical rationale!  Please help me find some!

73.

-al hart
al @ w8vr . org

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« Reply #1 on: October 21, 2013, 06:12:51 PM »

Al .... stick to yer guns !   

this world is full of political correctness and devoid of critical thinking  Tongue  Embarrassed  Lips sealed
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« Reply #2 on: October 21, 2013, 06:55:28 PM »

I don't know if this will answer all your questions but the Terman and Roake  paper on Class C Design (circa 1936) has a some design data on page 627 in which he states:

"..A value of 120 degrees represents a reasonable compromise
between high efficiency and large output and will be tentatively selected..."


Phil - AC0OB

* Class C Design by Terman.pdf (1438.59 KB - downloaded 300 times.)
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« Reply #3 on: October 21, 2013, 10:47:53 PM »

Hi Al,

I think your problem is that you are asking the wrong question. If I understand correctly, you seem to be looking for the answer to the following question:

"What is the best conduction angle for my class C plate modulated final?"

I think the proper question is a bit more general. Something along the lines of: "What are the optimum operating parameters for my final amplifier tube?"

In the 1947 West Coast Radio Handbook there is a semi-scholarly discussion regarding operating conditions of class C amplifiers on pages 58-62. Lots of math, you'll love it. Anyway, They start out by asking three questions:

1. What is my desired power output?
2. What is my desired plate voltage?
3. What is my desired plate efficiency?

Conduction angle is discussed and is calculated as part of the process, but the text does not stress that you should be shooting for a particular value.

I respect your quest for purity of knowledge, but I think in the case of class C operating conditions, using published tube data and recommendations from the manufacturers is all you need to do for excellent results.

I have had some modest success in building triode RF decks. I usually look at the recommended grid bias value, and increase it 20-25%. This usually works well if you have the grid drive available. For example, my 822 final is recommended to be operated at -250VDC grid bias. I am running it at -300VDC. I have no idea what my conduction angle is, but the final amp tunes properly and easily on 160/75/40 meters and I get good reports on the audio. I think the bottom line is that the actual conduction angle is not critical as long as you are in the ballpark for the type of service you are using.

May the EE's have mercy.

Ron
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« Reply #4 on: October 21, 2013, 11:19:54 PM »

Hi Al,

Take a look here:

http://n4trb.com/AmateurRadio/RCA_Ham_Tips/issues/rcahamtips0103.pdf

73DG
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« Reply #5 on: October 21, 2013, 11:32:16 PM »

I just found another file in my stash that might help as well:


* ClassC Tube Operation.pdf (618.76 KB - downloaded 302 times.)
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« Reply #6 on: October 22, 2013, 12:41:51 AM »

A good running, clean class C plate modulated final is a compromise of many things.  There is the theoretical, which is good, but then there comes a time when we need to tune it in the real world.

I feel the best way to optimize the system is by using a spectrum analyser and O'scope looking for the cleanest THD/IMD vs: best audio peaks vs: best efficiency.  All homebrew rigs are one-of-a-kind, have unique parts and require trying various voltages, various tank Qs, loading, drive and all combinations of grid, screen and plate parameters. There usually is a sweet spot where we find the best way to run a rig. I have usually been able to squeeze an additional -10dB of THD cleanliness out of a rig by playing around. If we are running a big rig, that extra -10dB can make a huge difference up the band for our neighbors.

In the case of conduction angle, just adjust drive, grid leak and fixed bias to see where things run the best along with the other desired goals mentioned.  I cannot stress enough the need to run a THD and IMD test to make sure the rig is performing at its peak under various modulation levels - and clean at the same time. A $25 SoftRock board hooked to a computer is all we need for the spectrum analyser using audio tones from ~30 HZ to 8 Khz.   Both sine and triangles are helpful.


T
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« Reply #7 on: October 22, 2013, 10:16:49 AM »

Well, when I was building class C triode finals (I built a lot of them), I found in every instance that running as small conduction angle as possible (and subsequently higher drive and bias) resulted in the highest plate efficiency along with the best modulation linearity.  Furthermore, 100% grid leak bias was also better from a modulation linearity standpoint.

The plate efficiency was very important at the time because power was measured in terms of power input to the final RF amplifier and of course good modulation linearity particularly at high percentages of positive modulation is always important.

The parameter to watch out for is the grid dissipation.  I used to run right to the edge on grid dissipation, but never over.

Anyway, just my experience from the past.
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« Reply #8 on: October 22, 2013, 11:58:55 AM »

Steve:  Yes, pushing the final deep into class C seems like magic. In fact, I use about 60 watts to drive my 4X1's in class C. This is the result of very heavy grid leak bias and grid current that is a little above recommended.  I found the positive peaks were higher and the THD was better.


Speaking of specifics and tuning....

Jeff / W2NBC and I were talking about THD and tuning his class C final.  He used a spec analyzer and optimized his 4-400A plate modulated rig.  I was impressed by his results. He said he was able to get the rig working much cleaner.  His spectrum THD pictures were at least -45dB in some cases.

Here's an excerpt of what he found. I have also experienced similar tuning techniques that helped THD and IMD:

"the biggest discernable change in both transmitters running Class C was screen voltage.. Increasing V was the key.. just a small increase in the CT. KW made -8 db difference .. in the "Gates 500" made a -6 db improvement.. loading also made a difference.. lots of folks lightly load to get the headroom for positive peaks. WRONG choice for THD .. It's true that positive peaks will improve in most cases, ..BUT..the caveat is that THD really starts to suffer.. "


T
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Nothing like a new homebrew rig. Come into the shack, flip on the switches and everything works perfectly.

And, nothing like an old dog.
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« Reply #9 on: October 22, 2013, 07:52:12 PM »

now this has been some critical thinking .... I would not argue with the results but instead wonder if there is a possibilty of getting some tablurized data to see if a trend can be spotted

thanks for sharing ....
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« Reply #10 on: October 22, 2013, 11:57:23 PM »

I don't know my conduction angle but I run a 4-1000 class C well beyond cutoff, about -200V on the grid, 200 on screen @ carrier, and 2500 on plate. I run as much bias as I have to keep the angle small and also drive it hard to get as much current in the plate current pulse as possible.

Can't compare modulating that to an 810 but it's certain the narrow angle has improved my efficiency greatly, maybe 80% or better, but it's getting into the region of metering where meter errors add up to a percentage of the total, making errors.

I run a low screen volts (instead of 450) and low plate volts (instead of 3500) because I am keeping it throttled way back to stay legal. For an 810, no need to throttle back and no screen hassles.

One issue I ran into is drive. At some point, too much is as bad as just a bit too little. I do the drive adjustment with 100% modulation of the plate supply voltage with a sine, and adjust drive power so the peaks are best.

These are not tales, but facts of how this one runs and that the smaller the angle the better the efficiency seems as long as there is enough emission for the high peak currents. At some point too short an angle with too high currents (same power input) will be inefficient due the resistances in the tuned circuit.

Class C final can be roughly compared to the high efficiency horizontal output circuit on a TV set, except for the relationship between the scan repetition frequency and the actual tuned frequency.
The power tube conducts for no more than about 12% of the total scan time. But the tuned frequency of that circuit is about 3x the scan rate (these would be triplers if the 'damper' diode did not kill the wave after the first half cycle, so the actual angle is 36%, or 129 degrees.
TVs were made for efficiency so they could use the cheapest parts. Just something to think about. other applications can shed light on concepts.
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« Reply #11 on: October 23, 2013, 03:26:17 PM »

Adding to DMOD's pdf:

Quoting "Electronic Designers Handbook," Landee, Davis & Albrecht, McGraw-Hill, for modulated class C power amplifiers, "In practice the plate efficiency is not constant throughout the modulation cycle.  The efficiency is a maximum near the zero modulation and decreases somewhat during both the crest and trough of the modulation cycle."

They go on to recommend a combination of fixed and grid leak bias, approximately 50/50, the fixed for lower cycle plate voltage linearity and safety, the grid leak for the higher cycle plate voltage region and linearity.  

When only fixed bias is used, linearity begins to suffer at an increasing (constant) rate about half way through the cycle and towards the higher plate voltage region.

Proper Drive, constant (and adjustable for practical purposes) screen voltage and applied grid voltage resulting from a combination of initial fixed and variable voltage through the cycle dependent on drive are all necessary for a reasonably linearly modulated signal.  There is no cook-book figure that fits all tubes.  Several steps are required with trial values (called iteration) with tube load line/characteristic curve charts and some formula juggling to get even a paper solution.  This was how it looked in the pre-general purpose digital computer era.  

In all commercial cases, final linearity is determined looking at the Lissajous trap figure comparison of modulated carrier (vertical scope deflection plates) to modulation signal (horizontal deflection)  If two slopes on the 'trap' are discovered, circuit constants and voltages are juggled to arrive as closely as possible to one slope.  The difference in slope at the peak plate voltage (say the right side of trap) compared to the initial cycle voltage (left side of trap) allows a modulation factor to be determined.

So for bias, most hams simply go the 50/50 route, e.g., using an 813 tube, fixed bias is -75 volts and the series grid resistor then determined for an additional -75 volts for the approximate -150 volts required using typical grid current as given in the tube tables.  This resistor would be halved for two 813's in parallel.

Then the screen voltage is determined by RCA's charts, say 350 to 400 volts for the same 813.  The ham may make this variable so that he gets the specified voltage under loaded conditions.  He will also insure that the screen has a bleed or other means to dump reverse current from the screen.  

So the conduction angle difference of class C CW vs. cyclical AM modulation as well as decent linearity will automatically follow if the above conditions are met.  The angle difference alone for efficiency's sake is not worth worrying about for amateur power ratings.  
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« Reply #12 on: October 23, 2013, 04:07:42 PM »

Quote
The parameter to watch out for is the grid dissipation.  I used to run right to the edge on grid dissipation, but never over.

How do you measure that Steve?

In a BC-610 the grid voltage stated by Eimac is -160 volts and the grid drive is 60 ma. for a 250TH tube with 2000 volts on the plate.Plate current is .250 ma.  Grid dissipation is 40 watts max.

Not an engineer here, just a hamboner Grin Grin
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« Reply #13 on: October 23, 2013, 04:22:21 PM »

At least be thankful Big Rig is on the air. I hope to work you soon!

Philip
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« Reply #14 on: October 23, 2013, 05:09:47 PM »

Hey Al,

What happened to your owl? I thought that was a pretty cool avatar. Cool

Has anyone here given you satisfaction or is there still NO JOY in Farmington? I'm afeared that you might just need to mosey over to the MSU dept of electrical engineering and spend some time in the library. Tongue

Ron
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« Reply #15 on: October 24, 2013, 01:59:57 AM »

He did ask for formulas. I'd like to see them posted..
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« Reply #16 on: October 24, 2013, 09:20:39 AM »

Quote
The parameter to watch out for is the grid dissipation.  I used to run right to the edge on grid dissipation, but never over.

How do you measure that Steve?

In a BC-610 the grid voltage stated by Eimac is -160 volts and the grid drive is 60 ma. for a 250TH tube with 2000 volts on the plate.Plate current is .250 ma.  Grid dissipation is 40 watts max.

Not an engineer here, just a hamboner Grin Grin

Usually for control grids you are safe using the maximum grid current as your guide, as the control grid dissipation is sometimes not specified in the tube data sheet.  There is usually a maximum grid voltage specified as well.

If you did want to measure the grid dissipation, you would need to know the actual voltage from grid to cathode when the grid is conducting, along with the grid current.  You could measure the grid voltage using an oscilloscope.  Since the grid acts like a diode, the grid dissipation (and current) is not linear with respect to grid voltage.  The peak grid dissipation and peak grid current will be significantly higher than the average power and grid current.  For this reason, you will often see the grid dissipation calculations specified using the peak RF grid voltage because this is where the greatest grid heating will occur.

Even this is somewhat inexact because the waveform itself comes into play.  A square wave will generate more grid heating than a sine wave of the same peak voltage.  In my own calculations, I do not use the peak grid voltage unless it is a square wave or something close to a square wave.

I have designed a number of pulse width modulators using vacuum tubes, and in this type of operation there was no grid leak resistor, and the grid was driven hard using a square wave from a low impedance source. The grid dissipation in this case was vital to know, and easy to calculate because the peak voltage was a well know value.  Under full-on operation, where the driving waveform is 95% on and 5% off, the grid dissipation would be right at the edge to achieve maximum modulator efficiency.

Again, for normal class C operation using a reasonable grid leak resistor, you can use the grid current as the important parameter.

I don't know if this made things clearer or not!!
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« Reply #17 on: October 24, 2013, 03:08:49 PM »

Thanks Steve, for the answer. I copied this from an old data sheet and it has a formula to calculate the power in the grid circuit but they used a volt meter in those days as a measurement device. I would think that what ever is used to measure, would have to be shielded very well to work  in most environments.

* GRID DISSIPATION.doc (66.5 KB - downloaded 106 times.)
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« Reply #18 on: October 24, 2013, 03:58:39 PM »

"  I Hate Old Radios!  "
I take it this is in the realm of, "no lids, no kids, no space cadets.... "  Grin

and,

"He did ask for formulas. I'd like to see them posted.."

I think several have tried to explain that there is no simple cook book answer. One must overlay the characteristic curves previously and courteously  determined by the RCA's and Eimacs of the world for each tube desired along with a set of desired operating voltages with load lines for Class C CW vs. Class C modulated.  It is an iterative process

So where does one start with the 'formulae?'  (Please consult any decent introduction to electronics for nomenclature of identities in the following formulae.  I believe a gentle hint to hit the books has already been mentioned.)

May we start with:
Maxwell-Boltzmann energy distribution (gasses) vs. Fermi-Dirac energy distribution (metals)?

Work function of escaping electrons ; Boltzmann factor, e ^-E/kT?

Emission current proportional to flux of escaping electrons...
Emission current, "I" from an area approx. equal to triple integral minus infinity to plus infinity u*e ^ -E/kT du dv dw?    ....boiling down(pun intended) to J = A0 T^2*e^-W/kT, Richardson's equation where J is the current density or current per unit surface area.   This assumes you remember from physics that KE = 1/2m v^2 + v1t so that
 E = 1/2m(u^2+v^2+w^2)

Thence to practical emitters, e.g., tungsten, thoriated tungsten, oxides..
along with their work functions.  A nice 3x3 table might be shown here with amp cm^-2(K)^-2, volts and T, deg. K (white, bright red, dull red).
Hey, didn't I just describe this in a previous thread?
..well, moving along.

Next diodes and space charge where we see a set of characteristic curves for the first time, of Ip vs. Ep  where a series of constant currents are described by Richardson's equation.  All curves at low Ip and Ep rise on same universal curve that does not depend on cathode temp. as long as it is high enough (this is the region of space-charge limited current, the 3/2 power law) but then diverge into almost constant Ip vs. Ep at selected temperature points, eg. the 2400 deg. flattened curve is below the 2600 deg. curve.

..and on and on through triodes, tetrodes.

Many years and much experimental and industrial thought went into describing characteristic curves for thousands of tube types.  Get out those curves obtainable on the net, plot your load lines as instructed, find those creating the least distortion.  For the 810 in this case, Apply voltages, particularly constant screen and quiescent constant grid as instructed then use old fashioned Lissajous curves or modern spectral analysis to tweak it.  You'll find a beautiful combination of  grid drive, plate tune an load through, say an injected sine wave analysis, pretty neat stuff.

 

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« Reply #19 on: October 24, 2013, 04:57:13 PM »

Rick:

The 810 is just a triode.  No screaming screen present.

73DG
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« Reply #20 on: October 25, 2013, 10:08:30 AM »

Well darn, I knew that. Makes it even simpler.  Grin

The whole house of cards felled by one little piece of feather hanging out of the fox's jaws, heh, heh.
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« Reply #21 on: November 08, 2013, 08:03:20 AM »

Hi Al,

I'm not sure that anyone here really gave you a satisfactory answer. I have another suggestion. There are several commercial broadcast transmitters that use 810's as the final amplifier. Perhaps you could research the operating parameters used by Collins, RCA, Gates, etc., in their designs. I would think that you would do well to emulate those designs. I know - more empiric evidence, but useful nonetheless.

Ron
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« Reply #22 on: November 08, 2013, 10:55:05 AM »

To all respondents:

Thank you all for your widely-based responses to this query.  I've finally satisfied myself on the matter by running through the detailed analysis offered by the RCA Transmitting Tube Manual and by backing that up using characteristic curves and other data from GE and Penta.  Some of the non-RCA curves were easier to use for finely-resolved graph points because the available graphical data from the RCA manual was not as highly-resolved as that from the other sources. The RCA manual's graphs looked a bit like those cartoons that made the office rounds after too many passes through the Xerox machine!  

It seems that deciding on mode-certain angles of conduction and proceeding from there to determine other details of the operating point was not the best solution.  I eventually put together the operating-point analysis based on numerical and graphical data available from the various sources for the different phone and CW operating conditions, and from there was able to calculate the conduction angle.

It is indeed different for the two modes, but not greatly so - only a few degrees - thus to some extent supporting the occasional opinion offered here that it wasn't extremely important.  Nonetheless, it was satisfying to know that the operative arithmetic supports that position and I was able to establish a seemingly proper bias for 'phone and for CW because my rig has mode-switched bias voltages.  


73.

-al hart, w8vr
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« Reply #23 on: November 13, 2013, 08:13:10 AM »

I'm following this thread with interest as I just discovered a 810 and a matching socket in my junkbox and thus, I began thinking about what to do with it.
Thanks to this thread and forum the result is likely a tube AM transmitter for 80m or 40m.
The real adventure will be doing it on a collage student budget Smiley
Should I make a separate thread for the project?
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