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Author Topic: Current dip at resonance  (Read 27460 times)
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N2DTS
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« Reply #25 on: February 06, 2014, 01:13:38 PM »

Don't know, but that looks like a lot of coil for 40 meters, the turns are close spaced...
The loading cap looks like a dual 365pf for a total of 730pf.

I would want 1200 to 1500 but 730 might work.

I would guess the tuning cap is 50 to 75 pf, might be fine on 40 with the right coil.


Very high Q starts wasting power in the coil I think, 12 to 15 is the usual Q wanted from memory.

If it dips and loads to the rated current, its fine, more voltage would give more power out.

2.5 watts will make a lot of contacts on 40 though.
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k7mdo
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« Reply #26 on: February 06, 2014, 02:01:10 PM »

I will try answer the last two posts with what I have found.  The plate tuning cap originally had more plates but I removed some as the insertion of the variable set was only on a slight overlap when the output power was maximum on the power meter.  I do find that the coil is "one" turn from the maximum number of turns from the bottom... you may notice that I have a slider so I can move the tap anywhere up or down the coil.  If I go to the end (the one more turn) the output goes down about 1/2 watt when I dip and load.  The loading capacitor is a 365 pf broadcast unit and when loading into the 50 ohm dummy it is about 1/3 engaged.  The two caps are in their "best" loaded output position as shown.  I did try 80 meter xtal many times and found very low output with maximum number of turns being used on the L.  I finally gave up and went with 40 meters.  I can vouch for the 80 meter crystals as they are used often in other equipment I have. I have a 20 meter crystal and it does provide about 1/2 the output but I don't remember how many turns I had to back up to to even get that. 

In the picture you will note my homebrew line voltage reducer I was using with the little rig so in order to check for what happens with higher plate voltage I just plugged into the wall socket at 125 VAC (home level) and I got a little highter output but the filament voltage went up to 6.7 VAC and I feared for my little 6CL6 so I scrounged around and found a 1 ohm resistor at about 1 watt and put it in series with the filament supply... reduced it to 6.1 VAC and the tube is probably happier.

Topic creep is probably happening but I sure have learned a bunch from the posts.


* LineReducer.JPG (111.34 KB, 1191x670 - viewed 348 times.)
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« Reply #27 on: February 06, 2014, 02:13:48 PM »

Eyeballing the inductor... based on the photo:

75 mm long
30 turns (10 turns per 25mm)
37.5 mm in diameter

From an on-line calculator: http://hamwaves.com/antennas/inductance.html

The inductance of the inductor is about 13.5uH

The impedance at 7MHz about j600 ohms.

This is a little high for 40 meters. You want the magnitude of the impedance of the inductor to be around 10 x Reff (i.e. Q=10), and you would want the value of Reff to be around 50/2 = 25 ohms.

But the inductor is adjustable.

A better value of impedance of the inductor at 7MHz is around j250 ohms.... which implies (from the on-line calculator) using about 50% of the turns (i.e. 15 turns) of the physical inductor

I suggest using the actual physical dimensions in conjunction with the on-line calculator to determine how many turns correspond to around j250 ohms of impedance at 7MHz.

To resonate the inductor at 7MHz, you will need -j250 ohms of impedance from the tuning capacitance, including the output capacitance of the tube, and various sources of stray capacitance. The tuning capacitor should, therefore, have a maximum capacitance of at least 90pF [i.e. 1/(2 x pi x 7,000,000 Hz x 250 ohms) = 90pF].

With the above, and assuming that the loading capacitor is large enough to produce Reff = 25 ohms [i.e. the loading capacitor's capacitive reactance = 50 ohms => C(loading) = at least  450pF, preferably more*]; then the impedance seen by the tube, looking into the pi network will be Q x Q x Reff = 10 x 10 x 25 ohms = 2500 ohms.

This should be about right.

*Reff is the resistive part of the impedance of: the loading capacitor in parallel with the 50 ohm dummy load. Call the value of the loading capacitance C farads

The impedance of the loading capacitor in parallel with the 50 ohm dummy load is = 1/[(1/50) + (j x 2 x pi x 7,000,000 Hz x C)]] = 50 ohms / [1 + (j x K)] = (with a bit of algebra): [50 - j x 50 x K] ohms / [1 + (K x K)], where K = (2 x pi x 7,000,000Hz x C x 50)

If were choose C so that K=1, then Reff = 50 ohms/[1 + (1x1)] = 50 ohms / 2 = 25 ohms
To make K=1, we need to have the magnitude of the impedance of C equal to 50 ohms. I.e. we must have -j/[2 x pi x 7,000,000 Hz x C] = -j50 ohms



Stu  
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« Reply #28 on: February 06, 2014, 04:19:53 PM »

Fun topic.

What is the power input and output?
70% should be easy if the values are in the ball park.
At 275 volts on the plate, and 2.5 watts out, if you have 70% that is about 3.25 watts in and about 11 ma.

You are good if running 275 volts and about 11ma of plate current.
You should be able to load it up past what the tube is rated for.
6CL6, 7.5 watts of plate disipation.
275 volts, 80ma, 22 watts in, 15.4 watts out, 7 watts plate disipation out of 7.5.
That is what the tube can do.

What is the dip in plate current like, broad and shallow, or sharp and deep?
If it is very broad, the Q is low, if its very sharp, the Q is high.
 
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« Reply #29 on: February 06, 2014, 05:05:45 PM »

Since this is an electron-coupled oscillator... I'm pretty sure that the cathode current will be the sum of a DC + a sine wave whose amplitude is equal to the DC level (maybe a little flattened at the top and/or the bottom).

Therefore, the plate current will be approximately a DC + a sine wave whose amplitude is equal to the DC level.

Therefore the tube will be operating in the equivalent of class A with a 100% sinusoidal RF current swing.

This implies that:

a) The efficiency will be somewhat less than 50%

b) The optimal impedance, at resonance, looking from the tube into the pi network (for maximum power output) will be: the B+ / the average plate current. I.e. for 50mA of plate current (at resonance) and 275V of plate voltage, the optimal impedance, looking from the tube into the pi network will be 5500 ohms. NOT 2500 ohms, as I assumed in the post above.

[Note that the optimal impedance, at resonance, looking from the tube into the pi network is about 0.5 x the B+/the average plate current for class C operation; and about (2/pi) x the B+/ the average plate current for class B operation.]

Therefore, with Reff = to around 25 ohms, we want to set the Q of the pi network to around 14.8. This means that we want the impedance of the inductor to be about 14.8 x 25 ohms = 370 ohms (not 250 ohms as I calculated in the post, above)

Likewise, we will need around 60pf of tuning capacitance to achieve a resonating impedance of -j370 ohms.

With 275V B+ and 0.05A of average plate current at resonance, and with the tube fully loaded to achieve 50% output efficiency... the plate will be dissipating about 50% of 13.75W = 6.875W on key down. The output power will also be about 6.875W on key down.

One can compare all of this to the Ameco AC-1, which is a very similar ECO... except that it uses a 6V6 instead of a 6CL6. The 6CL6 should be able to put out a bit more power than a 6V6.

Ref: www.tuberig.com

Stu

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« Reply #30 on: February 07, 2014, 08:33:46 AM »

That looks like a 3 to 4 MHz command set VFO inductor which is an excellent coil. But you will need to use less of it as discussed. The 6CL6 should easily do 5 Watts as an oscillator. So If you can get 5 to 6 Watts out of an ECO with a 6CL6 tube, you are about maxed out without cracking the crystal.
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k7mdo
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« Reply #31 on: February 07, 2014, 11:32:01 AM »

This weekend I will see if I can get some more data on the coil and cap... I do have an LCR meter that will verify the current parameters.  I just have to unsolder a couple of joints to isolate the components.

I will also wire in a milliamp meter and check the "sharpness" of the dip.  

The only thing that I have reported that is probably suspect is that the wattmeter I have is not calibrated in anyway and could well be off by 100%.

Snow here this morning and 20 degrees F.  This is not Oregon weather... we hate it.

Tom
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« Reply #32 on: February 07, 2014, 02:07:20 PM »

That is why they have ham radio, when its nasty out, you get to play radio in the nice warm house, or, if in some places in PA, you get to check out the disaster radio setup running off batteries and generators.

My first home brew rig was the one tube eco in the 1969 handbook, it worked great, 30 watts out I think.
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k7mdo
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« Reply #33 on: February 08, 2014, 08:59:18 PM »

OK, I spent a couple of hours tearing into my little transmitter to verify some of the component values and the plate and cathode currents.

!. The tank coil as in use on 40 meter CW is 1.375" in Dia. with a coil height of 1.60" and I use the entire coil of 28 turns for a measured 10.4 microH.

2. Plate tuning cap range Open 17.4 pF and meshed 60 pF (max output occurs at about 40% closure.

3. Loading cap is open 32.8 pF and meshed 557 pF (seems to function OK on antenna and dummy load at 1/3 of range)

4. Cathode current measured at key is 20 mA at max output of near 3 watts on meter.

5. Plate current is 14 mA at max output and at dip.

Dip is sharp from one direction from about 17 mA down to the 14 mA but as you rotate the cap past the dip it only rises just very slightly then drops away. That is probably why I initially missed the dip as I wasn't looking as closely as one should for that tiny dip from one direction of rotation.  But it is there.

Anyway, great time for shack work as the snow is now turning to freezing rain.

73, Tom
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« Reply #34 on: February 08, 2014, 09:10:56 PM »

Sounds like its working well then.

I think the dip should be deep though, a little less coil is more Q and a sharper dip.
That likely won't improve anything though.
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« Reply #35 on: February 08, 2014, 10:34:39 PM »

Tom

It's working... but still not working well.

Please refresh my memory:

What is the plate voltage?
What is the screen voltage?
Is the screen bypassed to ground with at least 1000pF of capacitance?
What is the size of the grid-to-ground resistor?

Stu

P.S. Making the inductance less will not increase Q. Doing so will decrease Q.

Q = 2 x pi x f x L / Reff
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« Reply #36 on: February 08, 2014, 10:58:31 PM »

275 volts on the plate, 150 on the screen, regulated.
275 volts, 14 ma,  3.85 watts in for 3 watts out.

Did I remember wrong? For some reason I thought more coil was less Q.

Anyway, 275 volts and .014ma is 19.6K ohms plate impiedance!
That will call for little caps and a big coil for a Q of 12!
for 40 meters, something like 25pf, 24UH, 150 pf... 
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« Reply #37 on: February 08, 2014, 11:38:13 PM »

The plate current is much lower than it should be in this application. Something is not right. It might be too large a grid-to-ground resistor.

Stu
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« Reply #38 on: February 09, 2014, 04:12:19 AM »

Grid to gnd is 68k ohms.

Screen bypass cap is .005 microF.



Tom

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« Reply #39 on: February 09, 2014, 10:40:31 AM »

Tom

Try reducing grid-to-ground resistance. Tack another 68k resistor across the existing grid to ground resistor. [The average grid current flows through the grid-to-ground resistor, and produces a negative bias (to ground) on the grid (as it is supposed to do). If this resistor has too high a value, the average plate current and the average cathode current will be too low for this application]

If this does what I think it will do, the average plate current and the cathode current will increase noticeably... and the available output power will also increase.

The behavior of the tuning and loading will be different. You should start with the loading capacitor set to maximum capacitance. Then tune to resonance. Then reduce the loading capacitance, and re-tune to resonance. After a few iterations, you should be at maximum power out (more than before), you should see a noticeable dip when you tune to resonance, and the plate current at resonance should be considerably higher (30mA or more). [Also note that, at around 10uH, the inductance of the coil is still okay for this application... once the average plate current is increased to around 30mA or more]

If the plate current remains low after reducing the grid-to-ground resistance, try the following:

Go to key up (no current). Remove the crystal. Very briefly go to key down to measure the cathode current. With 275V on the plate, 150V on the screen, and approximately 0V on the grid (i.e. no crystal, no oscillation, grid connected to ground via the grid-to-ground resistor acting as a grid leak) the cathode current should be around 87mA (i.e. 70mA of plate current and 17mA of screen current).

http://www.mif.pg.gda.pl/homepages/frank/sheets/093/6/6CL6.pdf  (see page 4)

Stu
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« Reply #40 on: February 09, 2014, 04:34:41 PM »

Stu, I will give this a try tonight or tomorrow night...  let you know the result.  thanks for your interest, Tom
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« Reply #41 on: February 09, 2014, 06:16:38 PM »

Tom

Please also measure the DC resistance of the path between the cathode and ground (i.e. the resistance between cathode and ground with the key closed and the power off, including the DC resistance of the cathode choke).

This resistance should be less than 10 ohms.

If this resistance is too large, it will produce a positive bias voltage between the cathode and ground... which will limit the sum of the plate and screen current to too low a level.


Stu
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« Reply #42 on: February 09, 2014, 07:30:06 PM »

Will do Stu....  I will be interested to see what happens with the grid leak halved.... T
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« Reply #43 on: February 10, 2014, 12:46:51 PM »

Stu:
I paralleled another 68k on the grid resistor and the result was: cathode current went up to 24.3 mA at apparent resonance and the wattmeter showed 3+ watts out which is up from the roughly 2.5 watts with the "stock" 68k only in the circuit.
Cathode resistance to ground is 10.6 ohms.
Big change was to the loading capacitor settings (why? I can't imagine) but now, when I "dip and load" in sequence, the loading cap intially shows increases of power out but eventually makes no appreciable difference all the way to it's minimum of about 32 pF.  Before the grid leak change there was a reasonable maximum of about 1/3 of travel of the loading cap into the dummy?  The plate cap also ended up in a "less meshed" condition at resonance with the lower grid leak but not a significant change. It got to be "Antique Roadshow" time so I had to quit for the evening.

Tom
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« Reply #44 on: February 10, 2014, 01:10:04 PM »

Tom

okay...

The behavior of the rig suggests the following:

The loading cap is set to minimum capacitance for maximum output => the impedance of the loading cap is much less than -j50 ohms when adjusted for maximum output => the loading cap is having no effect, because the impedance looking into the pi network is too high even with 0 loading capacitance => the inductor value is too large

Solution:

Reduce the inductance, in steps of around 10%, until the optimal setting of the loading capacitor is not the minimum capacitance.

If the pi network is working properly (not too much inductance), then the maximum power output should occur with the loading capacitance set to produce an impedance of about -j50 ohms at 7 MHz (i.e. in the ballpark of this value). This implies that the loading capacitance should be about 450pF (variable part + any added fixed capacitance)

Stu
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« Reply #45 on: February 10, 2014, 01:39:36 PM »

Stu, is there a "best" grid leak value I should aim for?  Or do you think I should drop it to the 34k rather than the 68k?

The loading cap I have is 34pF to 500pF roughtly so your email is a little confusing to me as you mention a max of about 450 or so and at the same time mention reducing it's total ....  not sure what you mean there.. 

You know the set works really well as originally made with the 68k grid leak and the current plate and loading caps... it just seems a little weak at 275 volts on the plate...  I now wish I had looked for a bigger transformer when I started the construction.  I may build another one (breadboard it) with a higher voltage plate supply to see just how far I can push the 6CL6. The current one would end up a hack job if I tried to change it.. and it makes contacts "as is".

73, Tom
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« Reply #46 on: February 10, 2014, 02:12:56 PM »

Tom

275V should be enough plate voltage for a 6CL6.

The issue is that now that you have more plate current (so far, not a lot more, but more) you need to adjust the pi network to properly match the new optimal output load impedance. For a fixed value of B+ (275V), more average plate current implies a lower value for the optimal load impedance.

The impedance, looking into the pi network, at resonance, is (2 x pi x f x L) x (2 x pi x f x L) / Reff

where f is the frequency, and Reff is the real (resistive) part of the impedance of the 50 ohm load in parallel with the impedance of the loading capacitor (having value C farads)

The impedance of the loading capacitor is -j / (2 x pi x f x C)

Reff = 50 ohms / [1 + (K x K)], where K = 50 x C x 2 x pi x f

The largest value that Reff can take on occurs when K<<1, and this largest value of Reff equals 50 ohms. This occurs when 2 x pi x f x C is small compared to 1/50. For example, if C is only 32pF (the minimum value of the capacitance of the variable capacitor), then 2 x pi x f x C (at f= 7,000,000 Hz) = 0.0014 siemens, and K= 0.07.

Right now, having increased the plate current (somewhat), it is apparent that your coil has too high a value of inductance. The symptom is that maximum power output now occurs when C=the minimum value it can be adjusted to (32pF). I.e. the maximum power output occurs when K is much less than 1, and Reff, therefore, is approximately 50 ohms.

The net result is that the impedance looking from the tube into the pi network, at resonance, is too high.

If you reduce L, then the optimal value of Reff will become something less than 50 ohms... the associated optimal value of C (not the maximum value of C for the variable capacitor, but the setting/value that produces maximum output power) will be in the ballpark of 450 pF, and the output power will be considerably higher.

Separately, a grid-to-ground resistor value of 34k ohms is better... because the higher value for this resistor results in a larger circulating current through the crystal (not obvious... but higher R increases the Q of the input tuned circuit) which puts more stress on the crystal... possibly damaging the crystal.

Stu

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« Reply #47 on: February 10, 2014, 02:54:07 PM »

Remember I have an L that is variable with moveable tap... makes me wonder if a little experimentation with reducing the L to fewer turns might not be in order....  I am at the maximum of 28 turns now and early on there was some discussion that this was too high of L for 40 meters...  I am going to experiment with that as it is easy to do... Tom
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« Reply #48 on: February 10, 2014, 03:33:06 PM »

No joy on moving the tap to fewer turns at least for any different output... it actually looks like a few more turns might make a difference but I don't have any easy way to experiment with that in the existing layout. 

T
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« Reply #49 on: February 10, 2014, 03:50:21 PM »

Thats odd, more current lowers the plate impedance which should call for less coil.
The original setup had a VERY high impedance.
Its lower, but its still very high.
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