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Author Topic: Pi Network Calculators differ  (Read 8536 times)
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W4RFM
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« on: November 21, 2014, 03:25:10 PM »

Using two different on line pi network value calculators, I have come up with two answers.  For a plate load impedance of 6,250 ohms and a Q of 12, I have received C1=76.7pf, L=22 uH. C2=260 pf.  The second calculator gave me C1=78pf, L=23, C2=883pf.  First calculator was the Jim Hawkins page, and the second was the Raltron page.  Anybody have an idea about the C2 difference?
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AB2EZ
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« Reply #1 on: November 21, 2014, 04:11:44 PM »

If the RF load impedance seen by the tube (looking into the pi network) is 6250 ohms, and if the Q is 12, then the load across the output of the pi network has to be:

6250 / [(12 x 12) +1] = 43.1 ohms

The question is:

What value of loading capacitor, C2, when placed in parallel with a 50 ohm antenna load, will produce a combined impedance of 43.1 ohms + jX at the operating frequency. Note that it doesn't matter what the reactive component, X, is.

From the values of C1 and L, it appears that the operating frequency is around 3875kHz.

As a consistency check:

The impedance of the inductor at the resonant frequency is j x 22uH x 2 x pi x 3875kHz = j535 ohms. This is 12 x j x 44.6 ohms.

Therefore, for a Q of 12,  the resistive part of the load across the pi network must be in the ballpark of 44.6 ohms... which is reasonably consistent with the 43.1 ohm value calculated above.

The required value of C2 is around 329pF

I.e. at 3875kHz, a 329pF capacitor in parallel with a 50 ohm resistor is equivalent to a 43.1 ohm resistor in series with 17.2 ohms of capacitive reactance. The load impedance across the output of the pi network is 43.1 ohms - j17.2 ohms

This is in the ballpark of the value of C2 produced by the first calculator.

The value of C2 produced by the second calculator would result in a Q of roughly 23, and an RF load impedance on the tube of roughly 12.4k ohms.

Stu


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W3RSW
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« Reply #2 on: November 22, 2014, 08:52:25 AM »

You used pure Resistive values in your first calculation without any orthogonal component. Pure R's leading to pure Q's.

Personally, I used to hand calculate PI nets from the ARRL handbook's derived equations.

Nowdays something like this that Bob mentioned is handy. You need ActiveX along with Java though.  Build them then just tune out the difference  Grin  Both close enough at HF for the real world.
http://www.qsl.net/wa2whv/radiocalcs.shtml
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« Reply #3 on: November 22, 2014, 12:47:26 PM »

Rick

I think you are correct:

If one defines the Q as the ratio 2 x pi x f x L / R, where f is the resonant frequency, and R is the resistive part of the load across the output of the pi-network (including C2)... then my calculation for the required value of the loading capacitor, C2, is correct.

However, if one defines Q in terms of the rolloff (output voltage v. frequency) at the output of the loaded pi network assuming that the RF source impedance is 6250 ohms.... then my calculation for C2 is not correct... and I defer to the PIEL application (Jim Hawkins' application).

260pF (PIEL) v. 329pF (my calculation). The effect on the behavior of the actual circuit is that using C2=329pF would produce a pi network (as per the PIEL application) with a Q of 12.35 instead of 12.

Note:

Even PIEL has a conceptual issue with regard to the calculation of Q. The PIEL application assumes that the RF source impedance is 6250 ohms. But, for a typical tetrode, the RF source is actually a current source... and what one is trying to do is to place an RF load across the output of the current source (tetrode) whose value is 6250 ohms at the operating frequency. I.e. one is not trying to "impedance match" a 6250 ohm RF source. Rather, one is trying to place a load across the source that is just large enough to cause the amplitude of the fundamental frequency component of the voltage across the output of the source to equal the value of the B+.

I suspect that whoever created the 2nd pi-network calculator application that Bob is referring to (http://www.raltron.com/cust/tools/network_impedance_matching.asp) was making yet a different assumption about what the 6250 ohms refers to. For example, if he modeled the RF source (incorrectly for a tetrode) as an ideal current source in parallel with a 12500 ohm output impedance... and, if the impedance looking into the pi-network, at resonance, is 12500 ohms... then the parallel load across the ideal current source (when it is connected to the pi-network) would be 6250 ohms. This may explain why the second calculator application specified C2 to be 883pF. Doing so would produce a pi-network whose input impedance is around 12500 ohms... although that is not consistent with the schematic shown on the Raltron web site, and it is not how one should design a pi network whose input impedance (with a 50 ohm output load) is 6250 ohms.

Separately:

I just now looked on Google... and I found this third pi-network calculator:

http://www.eeweb.com/toolbox/pi-match

It also assumes a source impedance of 6250 ohms. It calculates C2 to be 329pF.

Stu
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« Reply #4 on: November 23, 2014, 10:05:39 PM »

The source frequency I had in mind was our beloved 3880 KHz, I wish I was as good at math as you guys.
Thanks.
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« Reply #5 on: November 28, 2014, 03:44:56 PM »

one item that was not mentioned... using the correct plate load impedance multiplier for the class of service.
Many of the old handbooks have charts assuming class C.
I am sure you already are well past that.
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WB2CAU
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« Reply #6 on: December 02, 2014, 04:31:45 PM »

Since a PI-network is usually adjustable over a range, are the values from the calculation the absolute value for resonance?  Or is it the suggested value of the variable capacitor?

Wouldn't it be prudent to choose a variable capacitor where the mid point adjustment would provide the value so there is adjustment range?  I would think you would not want to be all the way over to one side of the adjustment for resonance.

For example, if it calculates that you need 100 pf for the loading capacitance (C2), wouldn't you be better off using a 200 pf variable capacitor that has a mid-point capacitance close to 100 pf?  Or is that assumed to be obvious?

I've never seen that explained.  Am I missing something?

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« Reply #7 on: December 02, 2014, 05:39:06 PM »

You are correct for a single band transmitter.
Think about it some more.  Your almost there.

Look at a typical multi-band transmitter's tune capacitor; you'll probably find a couple of sections, one about twice or three times the capacity of the other at any point of rotation and with cap. sections switched in as required.

Notice also the load cap. has perhaps additional fixed capacitors in parallel for the lower two freq's, say 80 and 160.

Oh, ok, won't make you work for it....

Notice that as you go up in freq., the tune cap. approaches less mesh for each higher freq. segment. But you can get a decent design by having only one variable capacitor for several different, widely differing bands, by designing so that each band is happy with the spread allotted.  Note that the higher the freq., generally the less delta capacitance is required to tune from one end of a band to another. Perhaps 1/4 turn of the var. cap. is required on 80, but only 1/8 on 40, etc.  -Somewhat oversimplifying but you get the idea.

Notice also the band switching and how it ties to the various combinations of tune and load capacitance.  

So, you've guessed it; for even more variability in a PI net, even the coil has been split into several sections, those switched too, and can even have a rotary/variable coil in the more versatile lash-ups.

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« Reply #8 on: December 04, 2014, 08:54:30 PM »

I am building a one band, almost one frequency AM rig. 3880-3885. 2x 4-125A's modulated by 2x 811a's.
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« Reply #9 on: December 04, 2014, 11:13:27 PM »

Bob

In your first post... you asked why two different on-line calculators gave you different results for the design of a pi network with an input impedance of 6250 ohms, and a Q of 12.

However, in your latest post. you have indicated that you plan to use a pair of 4-125 tubes in the modulated RF output stage... I think you are using the wrong value for the required impedance looking into the pi network.

With 2000V DC for the B+ on the plates, at carrier; and a total of 300mA of plate current, at carrier (2 tubes); the required RF load (looking into the pi network), for class C operation, is approximately:

0.5 x 2000V / 0.3A = 3333 ohms.

I.e. In class C operation,  the optimal RF load impedance (at the carrier frequency) is approximately: 0.5 x the modulation resistance.

As an aside:

In ideal class A operation, where the amplitude of the RF portion of the plate current is equal to the average plate current, the optimal RF load impedance is: V(B+) / I(average) = the same as the modulation resistance.

In ideal class B operation, the optimal RF load impedance is: (2/pi) x [V(B+) / I(average)] = 0.637 x the modulation resistance

In class C operation, the optimal RF load impedance approaches: 0.5 x [V(B+) / I(average)] = 0.5 x the modulation resistance... as one biases the amplifier further away from class B (further into class C).


Stu
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