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Author Topic: Current dip at resonance  (Read 27463 times)
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k7mdo
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« on: February 02, 2014, 03:57:26 PM »

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
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« Reply #1 on: February 02, 2014, 04:16:44 PM »

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  
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« Reply #2 on: February 02, 2014, 04:42:59 PM »

In addition to what Stu said and going back to basics,

Consider the series resonant circuit with L-C in series: At resonance, the Current through the circuit is maximum.

For the parallel resonant circuit with L-C in parallel (which is essentially the pi network), the current at resonance is minimal while the voltage across it is maximum.

http://www.allaboutcircuits.com/vol_2/chpt_6/5.html

Phil - AC0OB

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k7mdo
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« Reply #3 on: February 02, 2014, 07:35:10 PM »

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
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k7mdo
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« Reply #4 on: February 02, 2014, 08:42:12 PM »

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
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« Reply #5 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
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k7mdo
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« Reply #6 on: February 02, 2014, 10:33:28 PM »

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

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« Reply #7 on: February 02, 2014, 10:49:01 PM »

The way I always looked at it is the power has to go someplace, it can go into the antenna, or into the tube plates.

Dipping and loading matches the power to the antenna, like a transmission matches the motor power to the wheels.
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« Reply #8 on: February 03, 2014, 03:09:30 AM »

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
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« Reply #9 on: February 03, 2014, 08:46:01 AM »

Come-on? really who dips any more? Who ever dipped? Tune for max! - da dipp da dip dip...
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« Reply #10 on: February 03, 2014, 10:33:27 AM »

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.
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« Reply #11 on: February 03, 2014, 02:05:14 PM »

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
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« Reply #12 on: February 03, 2014, 03:16:58 PM »

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
 
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AB2EZ
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« Reply #13 on: February 03, 2014, 04:42:29 PM »

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
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« Reply #14 on: February 03, 2014, 06:04:14 PM »

Hi Phil


Thanks for explanation and for the link.

Now i can imagine what was happening. Meaning that the interim impedance moving up and down. Because load impedance is fix, conversion impedance reacting back to plate tube load either decreasing or increasing.


Agus, yb0djh. 73s
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k7mdo
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« Reply #15 on: February 03, 2014, 07:30:38 PM »

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?
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« Reply #16 on: February 03, 2014, 08:17:08 PM »

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
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« Reply #17 on: February 03, 2014, 09:44:29 PM »

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
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« Reply #18 on: February 03, 2014, 10:06:01 PM »

Tom

I'm glad to be of assistance. Keep us all updated on your findings.

Best regards
Stu
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« Reply #19 on: February 04, 2014, 07:13:25 PM »

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.
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« Reply #20 on: February 05, 2014, 10:31:55 PM »

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





 


* P1000910.jpg (272.49 KB, 1191x670 - viewed 422 times.)
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« Reply #21 on: February 06, 2014, 08:07:42 AM »

Nice looking transmitter.
To get more voltage, you could always go cap input.

The coil LOOKS like it should be fine for 40 meters.
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« Reply #22 on: February 06, 2014, 09:21:23 AM »

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 


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
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« Reply #23 on: February 06, 2014, 10:06:56 AM »

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.
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« Reply #24 on: February 06, 2014, 11:32:59 AM »

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
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