2E26 Transmitter

[AB2EZ]

Carl

I agree that I used the term "light loading" incorrectly, and that Gito has stated it correctly.
My descriptions of what is actually happening in the transmitter are, however, correct.

I would speculate that the reason that "light loading" was defined (by the industry and the hobbyist community) this way is as follows:

The manufacturers of radio equipment (in the earliest days) wanted to ensure that when their customers tuned up their rigs they would start with a high r.f. load impedance (at resonance) on the tube. This would ensure that when one "quickly dips the plate current"... the plate current, "at dip" would be low. As a result, the period during which high plate current would flow (before "dipping the final") would be as brief as possible.

Then, as per the instructions in the manual, the user would "increase the loading", [i.e. reduce the r.f. load impedance on the tube (at resonance)] to slowly increase the plate current (and "re-dip the final")... until the plate current "at dip" was the correct value for maximum power output.

By describing the process this way, and defining the initial dial setting of the loading capacitor (in the case of a pi-network) as "zero", the manufacturers avoided the situation where the "dip" at resonance would be too shallow... and therefore avoiding damage to the equipment by allowing the plate current to be too high for too long a period of time.

Again: "Set the loading knob to zero, and quickly adjust the tuning knob to dip the plate current"

To make this easier for a non-expert to understand ... they referred to the initial setting (high r.f. load impedance on the tube, at resonance) as "light loading"

With a pi network... "light loading" corresponds to the maximum capacitance setting of the loading capacitor. Therefore, (for example) in a Johnson Ranger... when you turn the loading control knob to "zero" on the front panel dial, the loading capacitor is fully meshed. When you set the auxiliary loading switch to "zero", all of the extra capacitors are placed in parallel with the variable capacitor.

The fact that the industry and the user community gives this definition for the expression "light loading" is a matter of record. I stand corrected. [If I were on the committee that standardized this definition... I would have objected, and proposed something like "safe loading"]

However, what is important (in my opinion) is understanding what is actually happening in the transmitter.

Because "light loading" was defined this way, I believe that more than half of the people who understand enough about their equipment to care about this sort of thing incorrectly believe that if the plate current "at dip" is too low ("light loading") that they need more loading capacitance (when actually they need less loading capacitance)

Best regards
Stu

[w1vtp]

You guys are making this thread like going to "Tuning up Your Transmitter Properly, 101" or "Whats Going on Behind the Panel While Tuning up, 101" class.  I mean that in a good way.  What I have been doing for decades intuitively is being explained clearly (what's going on behind the panel).

Excellent.  I'll be most interested in what the final diagnosis is for this particular xmtr finally is.

Al

[Gito]

Hi

Stu,what You wrote is correct ,but we must have enough knowledge to visualize what You had written .
Actually what I wrote came from my experience(and theory),
I build a small BC station,with 813 final,with fixed screen voltage,fixed min bias.

On switching on(first switch) ,only the filaments,and the oscillator ,and driver is on,tuning the driver to get the right grid current(813) and than switch on The the HV and Screen voltage.
Since I used Meters on the grid ,screen grid ,plate to measured their currents.
I can clearly see the influence ,when the grid drive is smaller an needed,or higher than needed,sometimes  we need a higher grid drive because the Final tubes got weak.
That's why I encourage myself to wrote before /from my experience .
So there's a correlation between Grid drive/grid current ,screen current ,plate current and loading of the final Rf tube.Especially when We used automatic bias for the control grid,screen voltage with a dropping resistor.

Carl wrote on taping the coil with a coupling C he manage to load on 75 meter(the highest output),It's actually like using a link coupling with a higher  winding  turn to couple  to the antenna.
Looking at the old ARRL books that used the "same" circuit it used a 20 Kohm  to drop resistor from the B+ to get it's screen voltage.
In your Circuit It used a 33Kohm dropping resistor.
So changed Your dropping resistor to 20Kohm (V=I X R)
Hoping the screen voltage get higher /with higher screen voltage we can get Higher  plate current/Screen voltage has a great Influence on the Tubes output

Gito

[KM1H]

Stu, the amplifier manufacturers make it even harder to know what you are doing unless you take the cover off.

They pretty much all use a 0-100 scale on the panel or vernier drive scale. Some are set so that 0 = minimum C, and others are just the opposite.

I built my first KW amp in 1957 and followed by others and blindly followed the ancient dip and load bit; I sure learned what white plates were and Im still amazed those 250TH's didnt melt down. In the Navy they used RF ampmeters which made more sense, just keep twiddling until you got maximum.

Then in 63 when I went to work at National Radio I discovered the screen current meter and later the grid meter for GG triodes. These days I almost completely ignore the plate meter and just watch the grid current and the Bird meter for linear or Class C amps.......so much easier! And less destructive to sensitive tubes.

Carl
KM1H

[WU2D]

Let's get back to the transmitter - is it working yet? It would make the basis for a 25W input power rig. Charles Caringella's book, Practical Ham Radio Projects shows such a 5763-2E26 setup.

Here is a ham that built the mobile one like yours: www.qsl.net/k3hln/10m_mobl.htm

The typical lineup with the 2E26 was to use a 6CL6 or a 5763 in a Colpitts oscillator to drive it.

I have attached the data sheet for the 2E26.

Mike WU2D

[Carl WA1KPD]

Hi all
As I said that due to business and personal travel it would be a while before I got back to it. I finally had a chance this weekend.

Using an online calculator I was able to use existing parts to set up a PI Network for 50 ohms using existing parts and my junque box.
The good news is that I am getting about 6 watts out. The bad news is that I cannot get the loading cap to interact at all. The setting makes no difference on the dip.
My calculations are based on 320 V at 30 MA

Plate cap 0-75 mmfd
Inductor 30 mh
Loading cap 0-1000 mmfd


The thing I am wondering is since the coil is still vertical on its side with the loading side near the chassis is there capacitive coupling from the coil to the chassis?

I am around for several week so I hope to spend some time on it. Thanks for the help

So we are getting there but I wanted to let you all know where I am on the project.    

Carl
/KPD
    

[N2DTS]

Pie net, right?
So you have a plate tuning cap to ground, then the coil, then the 1000 pf loading cap to ground and you adjust the loading cap and see no change in final plate current or power output?

That makes no sense...

Brett

[AB2EZ]

Carl

What is the frequency on which you are operating? What is the load (antenna) impedance into which the transmitter is looking?

If there is a dip in the plate current, but the plate current is too low (at dip), then the value of the inductor is too large for that frequency of operation. If that is the case, reduce the size of the inductor.

On the other hand, if the dip in the plate current is shallow (too much plate current at dip), then you need more loading capacitance. The loading capacitor's reactance must be less than the load (antenna) resistance (e.g. 50 ohms) to have much of an effect.

A 1000 pF capacitor has a reactance of approximately 80 ohms on 160 meters and 40 ohms on 75 meters.

If you actually expect the tube to be running (at carrier) at 320 volts B+ and .030 amps of average plate current... then the input power at carrier will be 9.6 watts. In Class C operation, you might expect (realistically) 75% efficiency... so you should be looking for around 7.2 watts of output power.

If you are operating the tube at 320 volts B+ and .030 amps; and you are operating it in Class C, then the optimal load impedance on the tube (for maximum output) at the operating frequency = 0.5 x 320 / 0.03 ohms = 5333 ohms. This is a fairly high value of load impedance... but that's what you need if you are really running the tube at 320 volts and 0.03 amps.

If you are operating this rig on 160 meters, then 30uH is ok for the inductor, but 1000pF is too low for the loading capacitor (you should use around 3000 pF, and you may need more).

If you are operating this rig on 75 meters, then 30uH is too high for the inductor (you should use around 15 uH), and 1000pF is marginal for the loading capacitor (2000 pF would be better)


Stu

[KM1H]

That Tune cap is way small also, I suggest around 250-400 for testing and with a 12-18 uH coil. Clip in 500pf fixed for the load if required but I doubt it as most of the kit and HB rigs used a 2 section BC variable that worked fine on 80.

Getting the coil away from metal is always a good idea. At that power level an inch at least, preferably 2"

Once you establish the correct values then you can finalize the actual components

Carl
KM1H

[AB2EZ]

Yes, the tuning capacitor is too small (once you adjust the value of the inductor)... but you don't need quite that much:

At 3.885 MHz, you will need a 112 pF tuning capacitor (some of which is the 2E26 output capacitance) to resonate with a 15 uH inductor.

Stu

[KM1H]

A 250pf is commonly used in kits and HB and should be used as a starting point. A 400 or anything in between is fine also, especially while testing.

Carl
KM1H

[Carl WA1KPD]

Carl

What is the frequency on which you are operating? What is the load (antenna) impedance into which the transmitter is looking?


Stu

Hi Stu
I am running it at 3.885 mhz into a 50 ohm dummy load.. Using the PI network calculator at http://www.qsl.net/wa2whv/radiocalcs.shtml with 8000 ohms  (which I figured the impedance would ennd up with 400V @ .050 amps) It calculates:
 
Plate C at 67 PF
L at 25
Load C at 268

My L is  on the high side but I figured that there was enough play in the C to be OK. Based on the comments you and Carl made clearly the L has to come down substantially. It is acting just as you described with a very deep dip and no ability to alter it with loading C.

Let me try to change it tonight and I will let you know the results.

Thanks to all

Carl

/KPD

[KC4VWU]

Which brings up another question I've been intending to ask. Without using specific values as an example, what difference does tank coil diameter make? If you come up with the same value whether you use a 3/4" or a 3" dia form? I can understand the mechanical aspect of it having to use larger material for power handling capability and skin effect would necessitate a larger radius. Available space considerations also. Is that about it?

Phil   

[w1vtp]

Which brings up another question I've been intending to ask. Without using specific values as an example, what difference does tank coil diameter make? If you come up with the same value whether you use a 3/4" or a 3" dia form? I can understand the mechanical aspect of it having to use larger material for power handling capability and skin effect would necessitate a larger radius. Available space considerations also. Is that about it?

Phil    

Play with this calculator and see

http://www.daycounter.com/Calculators/Air-Core-Inductor-Calculator.phtml

Al

Or use the attachment (input only the yellow fields)

[KM1H]

There are calculators on line to determine the best wire size, coil diameter, and length ratios when going for the highest unloaded Q. For low power 1 to 1 1/2" diameter seems to be the most used with plug in coils, miniductor, and wound ceramic forms.

Carl
KM1H

[AB2EZ]

Carl

To determine the proper rf load impedance, for a Class C amplifier, the formula to use is:

0.5 x B+ / average plate current (not B+ / average plate current... which is the formula for the modulation resistance).

If the B+ = 400 volts and the average plate current is 50 mA (at carrier).... then the rf load impedance should be 4000 ohms.

Best regards
Stu

P.S. The reason for the above is as follows

1. You want the rf load impedance (at the fundamental frequency of operation) x the amplitude of the component of the plate current at the fundamental frequency of operation to equal the B+

2. But, in a Class C amplifier... the amplitude of the component of the plate current at the fundamental frequency of operation is 2x the average plate current. [This follows from expanding the actual Class C amplifier plate current waveform in a Fourier series  ]

3. Thus, you want the rf load impedance (at the fundamental frequency of operation) x 2 x the average plate current to be equal to the B+

For other classes of operation, the factor: 0.5 is replaced by a different factor. In a Class A amplifier, the factor is 1. In a Class B amplifier, the factor is x 2/pi = 0.637

[Carl WA1KPD]

Thank you Stu-

That clears up a lot for me. I have been calculating the modulation impedance not the rf load impedance. Now the calculations from the web site agree and make sense.

Interestingly I can think of one other transmitter project I gave up on because I could not get it to load and I was making the same error in the claculations. Finally I gave up in frustration.

I did at least get the coil to the right value tonight before family obligations set in. I'll give it another try tomorrow night.

Tomorrow I'll add in some small (or large ) fixed caps to see if it will start to behave

73, and thank you very much

Carl

[w1vtp]

Aint this fun?   I'll be doing these craculations pretty soon for the 813 xmtr soon

Al

[Carl WA1KPD]

I added in the fixed cap and now the rig seems to work just fine. Good dip on the shallow dip on the final and I am getting 18 watts out (using a Bird into a 50 ohm dummy load). This is with about 450 V power supply and 50 ma of plate current- assuming the shunts are still good.

The plate cap is at about 105 pf
The L is 15 .7 uh

The only thing that is odd is there is virtually no loading capacitance needed.

Adding a large 1000 pf variable shows that adding even a slight amount of capacitance starts to bring the power down.

I'd be interested in opinion  but unless that reaction (pardon the pun) indicates some other problem, I'm perfectly happy with it the way it is.
Thanks and 73
Carl
/KPD

[Carl WA1KPD]

I added in the fixed cap and now the rig seems to work just fine. Good dip on the shallow dip on the final and I am getting 18 watts out (using a Bird into a 50 ohm dummy load). This is with about 450 V power supply and 50 ma of plate current- assuming the shunts are still good.

The plate cap is at about 105 pf
The L is 15 .7 uh

The only thing that is odd is there is virtually no loading capacitance needed.

Adding a large 1000 pf variable shows that adding even a slight amount of capacitance starts to bring the power down.

I'd be interested in opinions  but unless that reaction (pardon the pun) indicates some other problem, I'm perfectly happy with it the way it is.
Thanks and 73
Carl
/KPD

[AB2EZ]

Carl


Great!

The fact that almost no loading capacitance is needed means that the optimal r.f. load on the 2E26 is occurring when the output-side resistance presented to the pi-network is 50 ohms (the dummy load's value). Putting capacitance (loading capacitance) in parallel with the 50 ohm load reduces the effective load resistance that is being presented to the pi network. For example: a 50 ohm resistor (load) in parallel with a loading capacitor that has 70 ohms of capacitive reactance... appears to the pi network as a resistive load whose value is 50 / [1 + (50/70 x (50/70)]~ 50 / [1.51] ohms = 33 ohms. The more loading capacitance you place in parallel with the 50 ohm load (the dummy load), the lower is the effective resistance seen by the pi network. Right now (for the value of inductor you are using, and the frequency you are using) the optimal value of effective resistance presented to the pi network is 50 ohms... so adding loading capacitance makes the output power decrease.

Although non-intuitive: when you increase the loading capacitance... and, therefore, you lower the effective load resistance seen by the pi-network (as stated above)... you increase the r.f. load seen by the tube at resonance. [The r.f. load seen by the tube at resonance = Q*Q*effective load resistance. But: Q= the inductive reactance of the inductor / the effective load resistance. Therefore the r.f. load seen by the tube = (the inductive reactance of the inductor) x (the inductive reactance of the inductor ) / the effective load resistance.]

If you made the inductor even smaller... for example, you reduced it to 90% of the current value of 15.7 uH... then you would need loading capacitance to obtain the optimal output power... because the new pi network (10% less inductance) would present the optimal r.f. load to the tube when the effective load on the pi network is about 81% of the current value of 50 ohms. If the inductance is multiplied by a factor of k, then the required effective load on the pi network (for maximum output power) will be multiplied by k x k. The square of 0.9 (90%) is .81 (81%)

Also, with the inductor you are using now:  if your antenna system (rather than the dummy load) presents a resistance of more than 50 ohms... you will need to use the loading capacitor  to tune the rig to maximum output. [You can demonstrate this by putting some additional resistance in series with your dummy load]. If your antenna presents a resistance of less than 50 ohms, then you will not be able to tune your rig to full output (because the loading capacitor can make the effective resistance smaller than the resistance of the antenna; but it cannot make it larger]. Therefore, you might want to make the inductor somewhat smaller (say 90% of its current value) in order to match a wider range of antenna resistances (by adjusting the loading capacitor).



Congratulations on the new (working) rig!

Stu

[Carl WA1KPD]

I appreciate all the help and education Stu and others. Now that it is working I can clean it up and make it presentable for use in the truck.


Carl
WA1KPD

[KM1H]

Translating to English and a lot less words:  Reduce the L until you can make full use of the Load cap.

Otherwise you will have a problem loading in the real world under certain VSWR conditions.

Carl
KM1H

[AB2EZ]

From "The 2000 Year Old Man",  recorded by Mel Brooks and Carl Reiner:

Reiner: What language did you speak 2000 years ago?

Brooks: We spoke "basic rock"

Reiner: Can you give us (the audience) an example of something said in "basic rock"

Brooks: Okay... let's see... "Hey, what are you doing with that rock? Hey, put down that rock. Hey, don't throw that rock at me"