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Author Topic: Advice for final tank circuit  (Read 38069 times)
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k4kyv
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Don
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« Reply #50 on: January 11, 2010, 08:16:36 PM »


We think of inductive coupling as a ground  isolation trick but in most cases, the link goes to a coax connector mounted on the chassis.  Then a coax to the other chassis and another connector and piece of coax to the next link.  There are all sorts of theories about which end of which coax to ground, adding even more confusion to the subject.


But with link coupling, there still is isolation between tank circuit and coupling coil.  It is not the grounded coax braid that causes the problems, but the rf return path via the chassis that is required for capacity-coupling between stages.

But there is no need to ground the coax braid, or even use coax for that matter. I don't use coax for any of my links. I have had good result with zip cord, sometimes with the conductors separated.  The pair doesn't have to be twisted.  One trick I have used is to strip off the outer jacket and braid from scrap coax and use the inner conductor as the insulated wires for the link.  I just taped them together or used tie wraps, on the assumption that the insulation used to make coax might be a better dielectric than plain old hookup wire.  But in any case, if the stages are only a few inches to a couple of feet apart, almost any kind of insulated wire will produce negligible losses in the link.  I use stranded wire in the case of variable links because it is more flexible.

Quote
If you use a link to twisted pair and a receiving link, there is no ground connection at all and no need to even think about how the driver is tuned.

That's the best way to do it.
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Don, K4KYV                                       AMI#5
Licensed since 1959 and not happy to be back on AM...    Never got off AM in the first place.

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Gito
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« Reply #51 on: January 14, 2010, 03:27:43 AM »

Hi

Because I thing it is important to know ,the drive needed(drive power) for RF tube, I found it on The RCA transmitting Tubes hand book(TT4) page 32 and 33 ,the bottom line is..."  They can be estimated with reasonable accuracy for straight -through amplifiers..At frequency up to about 30 megacycles per second,total tube and circuit LOSSES  are approximately TWICE  the DRIVING POWER..." (data book)

That means the driving power needed = Losses( twice the driving power/data book) + driving power (data book)=  3 times driving power in the technical data book.

Since it is written in The RCA Transmitting tubes(the maker/producer of transmitting tubes.It must be true.

Gito


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K9ACT
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« Reply #52 on: January 14, 2010, 10:48:37 AM »


That means the driving power needed = Losses( twice the driving power/data book) + driving power (data book)=  3 times driving power in the technical data book.


Perhaps a language problem here?  I do not see the jump from 2 to 3 in the text.  The 2X includes all the losses.  

>Since it is written in The RCA Transmitting tubes(the maker/producer of transmitting tubes.It must be true.

I make no such assumption but I am a born-again cynic.

Keep in mind that RCA is/was selling tubes and wants customers to be happy with them and not be plagued with returns and complaints.

It costs THEM nothing to suggest more overhead than is typically needed.  We, on the other hand have to pay for that extra power so we (I) tend to want to know how things work in the real work.

As mentioned, my experience with a number of different power tubes points to the fact that the overhead is excessive.

I would be more interested in knowing what your experience is with all the transmitters you have built.  Put a power meter between your driver and power amp and tell us how this compares with the tube data.

Jack











Gito
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KM1H
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« Reply #53 on: January 14, 2010, 01:14:25 PM »

I think its more than a language problem. Sine the 254 is not even in TT-4 or any other RCA manual why should a generalization even apply?  RCA had their own way of writing specs while others specifically listed that the driving and output power includes expected losses in a properly engineered circuit.

If you look at the actual H&K spec sheet (the developer of the 254) it includes a complete PP 254 circuit with values for ham bands. It states that a driving power of less than 50W is needed for 750W out.

I think we are thru wasting time on the 2X or 3X issue.

Carl
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w1vtp
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« Reply #54 on: January 14, 2010, 01:19:59 PM »

Here are three more hand drawn schematics of proposed fixes to my drive problem.

Ron

I like the option with the link coupling.  Make it swinging and you can adjust the RF drive right where you want it

Al
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Gito
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« Reply #55 on: January 15, 2010, 03:21:55 AM »

hi.

there's no need to put a power meter between the driver and the final,since I put a plate current meter in each of my drivers and You can clearly see  the watt input of the driver(input power) and how much must I load the driver to drive the final to it's rating.and it confirms what the book told us.
Also I have read and analyze find most of the schematic diagram of transmitters confirm with the book.

254W  as Ron wrote  with low voltage on the plate .needs a 25 watt input .
254 W with 400 watt output(high output) with 4000 V (data sheet) on plate it,only needs 12 watt input so 50 watt driver is more than 3 X 12 watt.

Yes the losses are twice the driving power ,if we read it carefully ...that may be lost in tube socket or in component and wiring circuit of the driving circuits.or tube losses due to electron -transit time phenomena,internal lead impedance,or other factors .
so the power  wasted is twice the power driver (needed).
so if we drive with 20 watt ,can the losses be 40 watt?of course it can't
So we must drive with 60 watt power....... 40 watt losses and 20 watt that drives the final tube.
It's my conclusion and confirmed by the ARRL  & RSGB hanbook ---3 to 1 drive power.

Maybe it a waste of time for this Issue.
But it's the basic that at least I must know before building a transmitter.
I don't force anyone to follow what I wrote.
To build  a Transmitter we must have something to learn from,and I believed most of us ,learn from the ARRL handbook.

Gito



 

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K9ACT
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« Reply #56 on: January 15, 2010, 08:53:58 AM »


there's no need to put a power meter between the driver and the final,since I put a plate current meter in each of my drivers and You can clearly see  the watt input of the driver(input power) and how much must I load the driver to drive the final to it's rating.and it confirms what the book told us.

The tube data lists RF power required, not DC input.  For the purpose of this discussion you must know the RF drive.  You have just thrown a possible 50% clunker into the discussion.


>254W  as Ron wrote  with low voltage on the plate .needs a 25 watt input .
254 W with 400 watt output(high output) with 4000 V (data sheet) on plate it,only needs 12 watt input so 50 watt driver is more than 3 X 12 watt.

I don't follow this at all.

>so the power  wasted is twice the power driver (needed).

The power wasted is the drive power (RF) produced minus the power (RF) needed to drive the grid to proper grid current/voltage.


>Maybe it a waste of time for this Issue.

It's an important issue.  The usual applies to those who think it's a waste of time:
read something else.

>To build  a Transmitter we must have something to learn from,and I believed most of us ,learn from the ARRL handbook.

I don't wish to stir the pot on this but I prefer the Radio Handbooks or the so-called West Coast Handbooks.  However, none of them are perfect so we have to use our own brains to sort things out.

Jack




 


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K9ACT
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« Reply #57 on: January 15, 2010, 10:10:40 AM »

As a point of interest, I browsed my 1941 edition of the Radio Handbook and found the attached pertinent paragraph.

I presume "tube data" means data sheets and not extraneous "X factors".

We could argue about the definition of "slightly" but clearly, 2X or 3X would not fit.

Jack


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KM1H
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« Reply #58 on: January 15, 2010, 04:41:01 PM »

I think its safe to say that Gito is completely confused Grin  I just hope others arent.

So lets get back to the original problem....how to get the 6146 driver to transfer sufficient power plus reserve to the 254 grid.

The steps were already covered I believe.

1. Ascertain that the 6AG7 and 6146 can produce sufficient power into a dummy load using a pi-network or other network.

2. Provide the proper LC circuitry to drive the 254. Adding links wastes power on that small chassis. A 2 coil transformer with tuned primary and secondary shouldnt be any harder than an IF transformer....the principle is the same.

3. Something I forgot to mention earlier is to use just the single tuned circuit that started all this and see if tapping down or up doesnt work. The Up means the variable C is tapped down on the coil and the sweet spot on the coil is found for the 254.

4. This isnt rocket science.

Carl
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Patrick J. / KD5OEI
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« Reply #59 on: January 15, 2010, 08:38:14 PM »

As a point of interest, I browsed my 1941 edition of the Radio Handbook and found the attached pertinent paragraph.

I presume "tube data" means data sheets and not extraneous "X factors".

We could argue about the definition of "slightly" but clearly, 2X or 3X would not fit.

Jack

Does the quote mean that the available driving power is to be considered to be what is present at the drive terminals of the driven stage, that is, after the coupling? I tend to think it means that.

Also, for all of this, there is a difference in coupling methods used. parallel circuit, link, pi, etc. and these are not specified within all those statements, nor are their efficiencies. Do the statements refer to AM or FM operation? It does not say but the drive requirement changes with amplitude modulation and not with FM. Could that be the root of the confusing statements?

Just my 2 cents.
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K9ACT
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« Reply #60 on: January 15, 2010, 10:17:13 PM »


Does the quote mean that the available driving power is to be considered to be what is present at the drive terminals of the driven stage, that is, after the coupling? I tend to think it means that.



I tend to think the difference is in the noise.

What is so complicated about measuring the output of a driver/exciter and then adjusting this to the tube data value for drive requirement and then connecting this to the PA grid and see what happens?

It is even easier if you just connect a power meter between driver and grid and determine what power is required for specified output.

Forget how you do this... just DO IT and let's talk about the results, not speculation or parroting books.

I have done it many times and I know what happens or we wouldn't be having this discussion.

js








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Gito
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« Reply #61 on: January 16, 2010, 12:50:29 AM »

Hi.

Jack when We know the input power of the driver,of course we know the output of the driver it.s about 70% right ,I think I don't have to explain it .

Okay I give up,but for me I don't dare to say that I 'have more knowledge than all of You,Especially to the author of the ARRL and RSGB handbook,the RCA transmitting tube handbook.
I got the article from ARRL 1978 hand book ,and the fourth edition RSGB hand book.1969 handbook.


So it's all my opinion bases on the references  come from books that is accepted  in most of the world.Not my personal opinion
And with pracktise ,I found it's confirm what it wrote.

So I don't force anyone to follow/believe what I've wrote

Thanks

Gito,




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Gito
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« Reply #62 on: January 16, 2010, 06:46:38 AM »

Hi

Ron  a suggestion to make a transmitter you want, I changed the schematic from the ARRL handbook you attached ,with a pi section coupling between driver and final Amp.
it's only basic ,you can changed the 6L6 with 6146 ,807  with one or two tube in parallel ,I think 6AG7 have enough  power to drive two 807 or 6146.
Also changing  the HV voltage for the driver tube,Change the final tube HV voltage ,the bias voltage to suit the need of the triode final tube.

You can find the final tube input impedance according the formula that Dan has wrote .
R= (W : I  : I) X 620000
W=drive input power ,I=grid current.

And the driver power is    Plate voltage X Plate current =input power.
in class C the efficiency is about 70 to 75 %  .
The output power of the driver is input power X 70%.

R impedance of the driver is Plate :plate current.
Knowing these two impedance and the frequency of the transmitter, You can design the Pi-Section values.There's also a formula to  calculate in the handbook.


I hoped I'm not wrong

Gito






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W8ACR
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254W


« Reply #63 on: January 16, 2010, 04:21:12 PM »

Thank you Gito, Jack, and all others who weighed in on this topic. My plan at the moment is to use a series fed  plate tank circuit on the 6146 coupled inductively to the grid of the 254W. There will be a balanced grid tank at the grid also, and I will use grid neutralization. I don't know yet what type of plate circuit I will use on the 254W. With grid neutralization, I have the option of either a pi network or a conventional circuit with swinging link output.

A 6146 can run 750V at 150 mA, which should be plenty of power to easily drive the 254W.

If and when this project has a successful conclusion, I will post pictures (hopefully right side up).

Can somebody give me a starting point to try on how many wraps to use on the inductive link both at the plate tank of the 6146 and at the grid tank of the 254W?

Thanks again, Ron W8ACR
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Gito
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« Reply #64 on: January 16, 2010, 05:52:57 PM »

Hi Ron,
my advice ,download the TT4 RCA transmitting tube, It's not only a Tube sheet transmitting tube, in it  You can find all the theory of how to treat a Tube,what is class C,how to find the operating parameter,how to modulate. and in my opinion it's a very good reference book.

On link coupling page 40 it wrote ......in this method of coupling in fig41,substantially IDENTICAL "LINK" windings of A FEW TURNS  each are inductively coupled to the plate- tank coil and to the grid tank coil  of the following stage........

So there's no "rule"  of how many turns You must have....it can be three ,four tuns it doesn't matter,It uses few tutns to get A "very" low impedance connected together  through.a suitable transmission line..... with little  danger of excessive radiation( because very low impedance?) .......... .

And You can read it at the page that I attachced


Gito

modification

Sorry as I read the ARRL handbook there are 2 different link coupling
With flat lines like Co-Ax or parallel feed cable ,we must have the reactance  of the coupling coil as high/the same as the impedance of feed cable at the transmitter frequency.

with unmatched lines,it can be used without much trouble if the line is kept very short comparing  the Frequency of the transmitter (300000 : F =... meter)


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