Print Page - Current dip at resonance

Again, I am going to bare my ignorance on a topic but must ask.

Exactly why does the plate current (indicated on the b+ leg of the power supply) "dip" at resonance? I have been dipping and loading transmitters for years without this understanding.

Now that I am building from scratch a small transmitter and trying to figure out some of the parts parameters the question keeps ringing in my ears.

I am not trained in electrical engineering so I was hoping someone might have a "lay" explanation.

Tom

Tom

When the tank circuit is tuned to resonance (using the tuning capacitor), the plate voltage swing (+ and -) at the fundamental RF frequency is maximized*.

Separately, as you load the rig up (adjusting the loading capacitor), the impedance of the tank circuit, at resonance, as seen by the output tube, gets larger and larger, and the associated plate voltage swing (+ and -) at resonance gets larger and larger.

As you approach optimum loading, the amplitude of the plate voltage swing will approach the value of the B+.

Therefore, at or near optimal loading, the plate voltage will swing up to 2x the B+ on positive RF excursions, and down to 0 volts on negative RF excursions. The negative RF excursions will cut off the instantaneous plate current, and will, as a consequence, cause the average plate current to go down... therefore leading to a plate current "dip" when the tank circuit is tuned to resonance... provided the loading is sufficient.

*The time-varying part of the plate current flows to ground through the tank circuit. The tank circuit, typically having a Q of greater than 10, presents a high impedance to the plate current at the fundamental frequency... when tuned to resonance at the fundamental frequency... and a much lower impedance to the plate current at harmonics of the fundamental frequency.

Stu

Stu and Phil, that was very helpful and I had a glimmer of this sequence of events but not firm understanding. Unfortunately, as I grew up on the fringe of radio experimentation and having no formal training I just seem to have missed some of the fundamentals. I think if I had had more test equipment and time I might have had better opportunities to learn. Now semi retired and wanting to expand my hobby, more and more things are being dredged up that I need to understand.

Thanks again, 73, Tom

Well, while sitting here today watching the "super bowl" (and I don't even like football) I tried to meld this more technical explantion of the dip in current in the B+ supply with my older and now somewhat defunct understanding.

I think that my personally derived explantion doesn't quite hold water as well as it used to for me. In the past (distant) my big worry was to get the plate circuit into resonance before I melted the plate of the output tube. So, I was intuitively thinking in more of a "power" explanation. I.e. if the amount of power was fixed from the supply where P=IE then the reason I needed to "quickly" get the circuit into a resonance condition was to try to share the "power" dissipation between the output tube's plates and the antenna (via the quickly tuned pi network) therefore reducing the seemingly designed overload of the tube.

This didn't account for the dip in any way that I could guess as I=P/E dipping was still confusing or at least unexpected. I tried to explain it with simple DC equations. A poor choice of simplifying it for sure.

This is why I wish I had gone into electrical engineering... but, you can't do everything.

I will work more on this tonight.

Thanks again, Tom

That analogy helps a lot as well. I guess I hadn't even considered what the result would be of not having a pi network or antenna attached to the circuit at all.

One might expect a runaway for sure of current through the tube as it oscillates.

Fifty years ago an "elmer" of mine helped me with the design of a very large tesla coil system and I remember we didn't even supply the large triode with DC, just AC from my HV transformer. Who really knows what the frequency of the oscillation was but he cautioned me to only run it for short periods of time so as to not be "locate-able". The load of the primary coil and its tuning did seem to keep the plates from getting past "orange" in color and the 12" corona was pretty exciting. Sometimes wires would "burn through" and provide somewhat of an open condition for the tank circuit and I had to keep my hand on the HV power plug to get it turned off before the tube burned up.... mostly I didn't have to worry about the "locate-ability" as the thing would have fairly regular circuit failures. The worst being that in my home basement the overhead was lath and plaster which provided a great "route" for the corona to tickle the ceiling... looked impressive until smoke was noticed in the overhead.... had to bring in a hose and put a lot of water and hope into putting that fire out. The folks never found out or the rest of my hobby time might have expired.

Tom

Quote from: W7TFO on February 02, 2014, 08:50:13 PM

Think of it as the balance between throttle and clutch in your truck.

You step on the gas (transmitter out of tune), and the RPM's (current) goes up.

You let out the clutch (achieve resonance), and the RPM's settle down into the load (you're moving).

I tried to make it simple.

73DG

D

You've been spending too much time breathing in desert dust from all that high RPM wheel spinning.

Fred

Hi

to extend a little bit more in this topic, i still don't understand this following phenomenon:

- assumed the output network is pi section meaning C plate - L - C load
- above network design surely to a certain power output to load the plate to 50 ohm

Question is: Why when we open C load (decreasing capacitance) the plate current is increasing? and the power also increase

My understanding when changing the value in the network will change the conversion impedance no longer to 50 ohm. Meaning that looking to feeder line still 50 ohm but reverse looking into the c load is changing, so standing wave will occur.

Is my understanding correct?

Best regards

yb0djh, 73s.

The Pi network (as well as T networks, etc) have to be looked upon in their dual and simultaneous roles:

It acts as a Parallel Resonant Circuit,

It is an impedance transforming device.

If you look at the Pi network as two mirror image L networks, the impedance transformations make a lot of sense:

http://books.google.com/books?id=9X5gy_mDamsC&pg=PA15&lpg=PA15&dq=Pi+network+as+double+L&source=bl&ots=6oPMNW5R53&sig=CDCFkZmOt1fq2XilaxxHg1ENWxc&hl=en&sa=X&ei=8ebvUtvME6amygGa44CQDQ&ved=0CCYQ6AEwAA#v=onepage&q=Pi%20network%20as%20double%20L&f=false

Phil - AC0OB

Without reference to the two previous posts I have another question.

A couple of years ago I built a little QRP 40 meter CW transmitter using a reference that stated that the 6CL6 as used was "electron coupled". I "kind" of understand that concept but an interesting thing came out of my experimentation with the final product.

The Plate current, measured between the power supply and a series RF choke on the B+ line will NOT show a dip at resonance. I carefully looked at the B+ line current and it peaked at maximum output.... no dip.

In the final iteration of the transmitter I finally put a little panel lightbulb as a tunable output indicator and it correlated exactly with maximum brilliance and maximum output to a dummy load.

Tom

Tom

This indicates that the "loading" was not optimized*. I.e., you could be getting more RF output power. As a result, even at resonance, the amplitude of the + and - swing of the plate voltage at the RF frequency was significantly less than the B+; and, therefore, the minimum plate voltage on each RF cycle was significantly higher than zero volts. For a tetrode or a pentode (including the electron coupled oscillator application), the average plate current is (almost entirely) controlled by the screen voltage until the plate voltage begins to swing down close to zero on each RF cycle.

You need more loading capacitance or a larger pi network inductor (more inductance). Putting a light bulb in series with the dummy load will make the loading worse (further from optimal).

Separately... make sure that the screen of the tube is properly bypassed to ground with a 0.002uF (or larger) capacitor.

*The impedance to ground, at the resonant frequency, looking from the plate of the tube into the pi network is proportional to L x L / Reff, where L is the inductance of the pi network coil, and Reff is the effective resistance associated with the actual output load (50 ohms or whatever it is) in parallel with the "loading" capacitor. When you increase the loading capacitor's value, the effective resistance of the loading capacitor in parallel with the actual load gets smaller. I.e. more loading capacitance => a smaller value of Reff. If L is too small, or Reff is too large, then the impedance to ground at the resonant frequency is too low. As a result, the + and - swings of the plate voltage at fundamental frequency of the transmitter will be less than they could be. I.e. the RF output power will be less than it could be. With proper loading, the impedance to ground, at the resonant frequency, looking from the plate of the tube into the pi network will be just high enough to produce a voltage swing whose amplitude is equal to the B+ on the plate. When this happens, there will be a dip in the average plate current due to the instantaneous current being pinched off on the negative peak of the plate voltage on each RF cycle.

Stu

Stu, I am goin to pull that little transmitter off the shelf again and see what I can do... it really only put out about 1.8 watts and that was less than advertised by the circuit description for the unit.

I do remember noticing that the tuning capacitor was less critical than I expected no matter where I tapped the coil....

The little tune-up light is automatically removed after loading the antenna or dummy. When tuned to maximum brilliance and then switched out of the circuit and the antenna switched in, the output seemed very close to maximum shown with the light.

I also included an antenna relay that worked with the CT of the power supply... that part works fine.

I will drag it out and post a photo.

I wonder if I could get away with using the Tek 465 I have to look at the voltage swing at the plate?

Tom

The scope could be used if you have a 10x probe with a voltage rating higher than 2x the B+. The probe will add some capacitance in parallel with the tuning capacitor. So the setting of the tuning capacitor for resonance will be different with and without the probe.

I suggest that you not use the light bulb in series with the dummy load because the added resistance of the bulb will add confusion with respect to getting the pi network to properly load the transmitter.

If you want to observe the output power, use the other channel of the scope to observe the voltage across the dummy load. For a 50 ohm dummy load, the power going into it is V x V / (2 x 50 ohms); where V is the amplitude of the sine wave voltage across the dummy load (in volts).

Stu

OK Stu, this will take a few days as I will want to refresh my knowledge of the plate voltage to be sure I won't go over the scope limits.

The light bulb is only in if I hold a momentary slide switch and so it won't be a problem of confusion... the dummy can be on and used seperately.

I only used the little transmitter for a few months and then shelved it for a Viking II project that I acquired... too many projects and I am still only semi retired but .... it is what it is.

Thanks for your time, Tom

In my 400 lbs of Nothing project, the BC-653A had a conventional output network, basically a parallel tuned circuit. What was odd was the way they coupled to the antenna which was a 15 ft whip on the half-track or whatever. This being a very short whip for 2.5 - 4.5 MHz, they knew it would be a very low radiation resistance and capacitive. It is typically about 1 Ohm with 50 pF of capacitance. So they simply came off the top of the tank with a variable inductor which served to cancel out the capacitance and resonate the whip. Now yours truly wanted to feed a conventional 50 Ohm system, a coax fed antenna. So I simply used a series 50 pF cap from the antenna post to the coax center, then a big triple section variable to ground (and shield) off the same coax hot. Observing the output power, I moved the tap until I got maximum power out into a 50 Ohm dummy load. The best loading was at around 500 pF. So with the tuning maxed, tap set for max and loading cap at 500 pF, I was getting max power out. Only now did I go back and see what the plate current was doing. And wouldn't you know, you could detect a very slight dip exactly at the point of maximum power out.

Ok, I think I see the problem...s. I posted a picture of my little homebrew one tube transmitter that I built from junk in the shop. It appears the coil may well be a few turns to few. I put the o'scope on the plate and watched the voltage (rf) nearly double at resonance. Just as I now understand it should. But I also played with my tap on on the coil you can see and it turns out that I have virtually no more turns as I approach the highest output. I simply may not have enough turns. The tuning capacitor is about mid-range at resonance as you can see in the photo and the loading cap seems to work as it should.

And I see I DO have a very slight plate current dip at peak output. It is just so slight that I missed it when I first looked for it.

When I built this little transmitter I only had about 275 volts for the plate and I fixed the screen at 150 with the OD3. I know that with the 6CL6 I could have gone to higher screen and plate voltage if I had had a different transformer. But it was what I had.

The (uncalibrated wattmeter) shows about 2.5 watts output into the dummy load. Maybe the bright side is that the reports I get from contacts all have heard a clean chirp free signal.

Thanks for the insights, I learned a lot and appreciate the help. 73, Tom

Quote from: k7mdo on February 05, 2014, 10:31:55 PM

Tom, to my eyes you have plenty of turns for 40M in fact maybe too many! When I look at the coil you have it looks like it would work fine on 80M. Now the problem becomes, if you reduce the turns you will need more "C" for the tuning capacitor. Decreasing the L and increasing C will raise the Q and your dip should be sharper. In addition if you get a shallow dip with two much plate current you will have to raise the C on your loading cap. So play around with tapping down on the coil while adding some more C which will be required from your tuning cap. You can just pad it but be aware that point is one of the highest voltage points in your circuit as you correctly observed on your scope.

Have fun!

Joe, W3GMS

Yes, I thought the rig was for 80 meters till I read the xtal.

But little tubes could run an odd plate impedance and might need more L.

Not having a good L/C meter, I just guess at coils by looks, then try different settings to see where I get a nice sharp dip at a good power output and call it good.

For 40 meters, the plate tuning cap looks small and the coil large, for 80 meters, the coil looks fine and the cap looks way too small. The loading cap looks small for any band.
The loading cap can be padded with a fixed cap if need be.

Tom

To get more output, and to make the Q of the output pi network higher, you primarily need to add more loading capacitance. You should not decrease L*

The Q of the output pi network is proportional to L/Reff; where

L is the inductance of the inductor, and
Reff is the resistive part of the impedance of the (parallel) combination of the 50 ohm load and the loading capacitor

To make Reff smaller, and therefore to increase Q, and also to increase the impedance seen by the tube looking into the pi network, you have to:

Make sure that the magnitude of the impedance of the loading capacitor can be set to values that are significantly less than 50 ohms. I.e. |impedance of the capacitor| must be significantly less than 50 ohms

For a 0-500pF loading capacitor (as an example), the minimum impedance at 7 MHz is -j/[2 x pi x 7,000,000 Hz x .000000000500 farads] = -j45 ohms. The magnitude of the impedance of the capacitor is 45 ohms. This is too large an impedance, and therefore too small a value of the loading capacitance.

I can't tell for sure, from the picture, but I suspect that your loading capacitor has less than 500pF maximum capacitance. Let's assume that its maximum capacitance is Cmax pf.

You should add around 500pF of fixed capacitance in parallel with the loading capacitor, to obtain a loading capacitance range of roughly 500pF to (500 + Cmax) pF. If that isn't enough, add another 500pF in parallel with the loading capacitor for a total of 1000pF of fixed capacitance + 0pf to Cmax pF of variable capacitance.

*As an aside, decreasing the value of L will decease the Q of the pi network, not increase it.

Stu

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.

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.

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

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.

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

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

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

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

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

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

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

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

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

I am wondering if I could use my "antenna analyzer" to look "into" the tuned pi network with the dummy in place? I can access the pi network with everything up to it disconnected prior. Set it at the crystal frequency and see what the impedance is looking that direction... what the tube is seeing... thoughts on that?

Yes, I expected to see a different L possible with the change in plate current... however, the change was pretty small, only a few milliamps... hard to tell if small L changes actually show any significant differences.

Well, I am going to have to get new batteries for the analyzer to do that test and it is too late tonight.

73, Tom

The AM Forum

THE AM BULLETIN BOARD => Technical Forum => Topic started by: k7mdo on February 02, 2014, 03:57:26 PM