Designing Pi networks

[M1ECY]

I am having some difficulty with the design of a suitable pi network - this is for a small 160m HB rig

As I understand it for class C the formula plate impedance calculation

Plate volts / 2Plate Current

Using that formula I get 2500r for the 6V6.

Now, the rest of the formulae have got me flummoxed!

I have tried using various online calculators, strange as it is, they all give differing results!

Now, not being fully versed in whats best, an I aiming for small L and big C, or vice versa?

At present the PA efficiency is poor - 19w dc in for 5w rf out.....

This is with a 35uh coil, and 150pf (ish) tune cap - about 750 pf on the loading.

Anyone able to walk me through this minefield?

Cheers
Sean

[G3UUR]

Sean,

You need a Q of at least 10 in the tank circuit to convert the current pulse from the 6V6 to a reasonable sine wave. With your present values you're way short of that.

2500/10 = 250.  Thats the reactance of your tuning capacitor and for 1.8Mhz you'll need greater than 350pF.

The tank inductor comprises two parts. The first part is defined by the tank Q1 (the main Q) and has a reactance the same as the tuning capacitor, ie 250 ohms. That makes it 250/(2pi x 1,900,000) or about 21uH. Looking at the tuning capacitor and the first part of the tank inductor as an 'L' network, the impedance of 2500 is transformed down by (Q squared +1), which makes it 24.75 ohms at the intermediate point. This then has to be transformed by the second part of the pi network up to 50 ohms. This time there is a different Q squared + 1 step-up with a much lower value of Q2. You have to work backwards from 50 ohms for this one. 50/24.75 = 2.02. Subtracting the 1 gives 1.02 and the square root is 1.01, so your output capacitor needs to have a reactance of very close to 50 ohms and the extra bit of inductor needs to be very close to 24.75 ohms reactive. That makes to extra bit of L about 2.1uH at 1.9MHz.

So you need a total tank inductance of 23.1uH, a tuning capacitor of 400pF maximum and a loading capacitor of around 1800pF maximum (your present variable with an extra 1000pF fixed across it would probably do).

Hope you managed to follow that.

73, Dave.

[M1ECY]

Well Dave,

That makes more sense than my continual reading of different books - it's been a long time since school and Maths!

I know what you have written was in at least 3 or 4 texts I have been looking at, just not explained that way.

Great Stuff - can now go rewind the tank coil to something sensible, and work a bit more C into the parts that need it.

Cheers
Sean

[w3jn]

Dave explained it very well.  Once you know that a pi-network is actually TWO L-C "L" networks in series the thing is pretty easy to figure out.  The first LC consists of the plate tune cap and most of the tank inductor - this transforms the impedance DOWN.  Then the next part is a LC consisting of a small inductance (the rest of the tank inductor) and the loading cap which transforms that low impedance back UP to your antenna impedance. 

The two inductors in series, which we consider in the calculations, are generally only one coil when you're building the thing.

[DMOD]

Hi Sean,

If you have Matlab I can send you a listling that calculates the values (send me a PM),
or you can use one of these online calcs such as:

http://www.raltron.com/cust/tools/network_impedance_matching.asp


Phil - AC0OB

[Detroit47]

Dave you did a great job on that. How about going into Q with an explanation. As to how and why.

John N8QPC

[k4kyv]

Dave explained it very well.  Once you know that a pi-network is actually TWO L-C "L" networks in series the thing is pretty easy to figure out.  The first LC consists of the plate tune cap and most of the tank inductor - this transforms the impedance DOWN.  Then the next part is a LC consisting of a small inductance (the rest of the tank inductor) and the loading cap which transforms that low impedance back UP to your antenna impedance.  

The two inductors in series, which we consider in the calculations, are generally only one coil when you're building the thing.

Basically that is how it functions, but one complication I see that might muddle the calculations is that the tank inductor in the pi-network is not two isolated sections; it is one continuous coil, with mutual coupling between the part that serves as the inductor in the first L-network and the rest of the coil that serves as inductor the second L-network.

Another way to look at the pi-network (probably nothing more than a matter of semantics) is to think of it as an unbalanced L-C tank circuit, identical to what one would use with a coupling link at the grounded end of the coil. Instead of the coupling link, one could simply tap the load onto the bottom of the main coil, a few turns up from the grounded end. Or, one could obtain identical results by using a CAPACITIVE voltage divider, by inserting a large value of capacitance in series with the main tuning capacitor to ground, and feeding the load at the point where the two capacitors are wired in series.  Instead of moving the tap on the coil around as explained previously, loading would be adjusted by varying the ratio of capacitance between the main tuning capacitor and the one at the bottom end of the voltage divider.  Now, take it one step further: reverse the ground and output leads, grounding the common point between the two capacitors, and feed the load from the point where the bottom capacitor is connected to the coil.  Bingo! You have a pi-network.  The main tuning capacitor is the plate tuning cap, and the larger capacitor at the bottom is the loading cap.

[G3UUR]

John, I'm not sure what you mean. Are you asking why the Q needs to be as high as 10 to produce a reasonable sine wave at the output?

Don, the poor guy only wants to know what values he needs for his 6V6 pi-network tank circuit. BTW, the mutual coupling is allowed for if the total inductance is correct. And, there is a subtle but important difference between the pi-network circuit and a capacitive divider in the tuning capacitor leg of a parallel tank circuit, which has to do with the way the two circuits are driven.

That reminds me: Sean, many designs for 160m transmitters in British magazines use small close-wound PA coils. The PA efficiency usually suffers as a result. It's best to space the turns slightly, even though that may mean using thinner wire. That may sound counter-intuitive, but because of the proximity effect the increase in HF resistance due to the thinner wire is less than that caused by the magnetic effect of close adjacent turns on their neighbours restricting where current can flow even more than the skin effect alone.

73,

Dave. 

[M1ECY]

Hi Dave,

Ah, yes, had come across this now.

It was a PW design by G3OGR, I started to build it about 5 years ago, but it never worked then - I rebuilt it last week, using Octals (dont like B9a.....)

Now it all works VFO and buffer wise, and the PA works a lot more efficiently with the correct LC ratios.

I think the inductor has too much self capacitance though - using a nice big ceramic former and 2mm wire efficiency almost doubles!

So, off to have another go at the tank coil, but with spaced turns....

It's great to find so many helpful types here!

Cheers
Sean

[k4kyv]

Don, the poor guy only wants to know what values he needs for his 6V6 pi-network tank circuit.

Well, remember that we AMers are renown for our old buzzard transmissions. 

And, there is a subtle but important difference between the pi-network circuit and a capacitive divider in the tuning capacitor leg of a parallel tank circuit, which has to do with the way the two circuits are driven.

Hmmm. I can see that with the pi-network, drive from the PA tube appears only across the portion of capacitive divider formed by the plate tank tuning capacitor, while the the loading capacitor and load remain isolated from the direct excitation from the tube. With the capacitive divider in the tuning capacitor leg of a parallel tank circuit, the PA driving voltage appears across both the plate tuning capacitor and the bottom end of the capacitive divider, so that the load is included as the part of the circuit directly excited by the tube. Is that the difference in the way the circuits are driven, that you were referring to?

[W4AMV]

Dave, I think John is asking about the (Q^2+1) in the math application and where it comes from. However, I maybe wrong. In any case, the technique of Z-matching as you outlined is so key and fundamental that it is worth dragging in the details (AGAIN) as to where it comes. In addition, the idea of transforming a network from its series to parallel form and visa-versa. A worthwhile post (in my opinion).

[M1ECY]

Well, it has helped me!

TB rig now producing about 9W RF for 16W input - this is much better.

Onto the modulator now.....

Thanks
Sean

[IN3IEX]

Take also into account the inductor for the anode V+ supply.
It happened to me. I built a linear amp that worked perfectly except in the "middle" of 20m band, where the output was 50%.
It worked well at 14.0 and 14.5....
That RF blocking solenoid had a series resonance at 14.25 MHz....while experimenting a big spark developed between a point along the coil and a nearby shield, it was very revealing.  I resolved by inserting a piece of a broken Amidon core inside the glass tube on which the solenoid was wound.

Giorgio

[G3UUR]

Sean, glad to hear you've got things percolating better now.

Don, you've got it in one. That's exactly what I meant.

Giorgio, the plate choke matters as far as efficiency is concerned, but that is covered very badly by the various handbooks. It's a relatively simple matter to develop a procedure to deal with the consequences of using low inductance for the choke, though, and if any of the current gurus understood amplifier operation as well as George Grammer it would have been done decades ago. The big problem is getting the right balance of inductance for low frequency operation and freedom from series resonances in the upper frequency region.

John, if as suggested, you wanted to know where Q^2+1 comes from, it's to do with the way 'L' networks convert from the parallel version of the two components to the series version. The Q must be the same in both versions and so must the phase angle for them to be equivalent. To achieve this the resistance Rp in the parallel combination of Rp and Cp must be Q^2+1 times the resistance Rs in the series equivalent of Rs and Cs. I don't know whether I've explained that well enough for you, but let me know if I haven't and I'll try another way.

73, Dave. 

[G3UUR]

Sorry, I should have said the magnitude and the phase angle need to be the same in both parallel and series versions, then, of course, the Q comes out to be the same anyway.

My apologies: I was in too much of a hurry.

Dave.