Calculations for the Anode Circuit

The parallel plate output circuit design is directly from the 144 MHz amplifier design notes from the 1999 ARRL Handbook (page 13.23). Here are the design calculations:

Calculate the operating plate load resistance:

 
Rl  =  
Vp   =   3750V   =  2619W  ~  2650W 
AB2Ip (1.7)(0.842A)

Now, calculate the circuit assuming the tube output capacitance plus the stray capacitance of the enclosure results in a capacitance of 24 ohms. Assume an operating tune capacitance of 2 pF, for a total plate output capacitance of 26 pF:

Approximate  Cout  =  26 pF

Calculate the reactance of this value of capacitance, assuming a frequency of 144 MHz:

XCout  =   1   =  42W
 
2pfCout

The circuit Q may be approximated with the ratio of the plate load and the reactance:

 
Ql  =  
Rl   =   2650   =  62
XCout 42

This design is based on using a capacitor in series to match the tube impedance to the 50 ohm coax impedance. The value of this capacitor cannot be calculated directly. However, it is straightforward to calculate the parallel equivalent of the load capacitor using the equation:

 
Xp2  =  
RsRp2   =   50(2650)2   =  367.5W 
Rp  -  Rs 2650  -  50

The actual value of this parallel equivalent may be calculated from the following equation:

Cp  =   1   =  3.0 pF
 
2pfXp

Once the reactance of the parallel equivalent of the load capacitance is known, it is possible to directly calculate the inductance needed if this capacitor were in series:

 
Xp2  =  
Rp2Xp   =   (2650)2(367.5)   =  360.6W 
Rp2  +  Xp2 (2650)2  +  (367.5)2

The reactance of the plate load capacitor (now in series) may now be converted to its capacitance value:

Cs  =   1   =  3.1 pF
 
2pfXs

We must now calculate the value of inductance necessary to form the parallel resonant circuit. We want this inductance to resonate with the value of the parallel capacitance at the design center frequency of 144 MHz. Recall that we had 26 pF of tune + stray capacitance (24 pF tube output + stray capacitance, plus 2 pF of tune capacitance). We now also have the parallel value of the load capacitance that is also in parallel with the tube and tune capacitance:

Cout  =  26  +   3.0  =  29 pF

This translates to a capacitive reactance of:

XCout  =   1   =  38.1W
 
2pfCout

Note that this capacitive reactance is between the plate output line and ground.

The two parallel resonator pipes must exactly offset this value of 38.1 ohms of capacitive reactance to form a resonant parallel circuit at 144 MHz. There are two inductors in parallel used to offset this value, so each of the resonator legs (resonator pipes + doorknob blocking capacitors) must equal 2 * 38.1, or 76.2 ohms. However, we must allow for the four doorknob DC blocking capacitors in the circuit, which will offset some of this inductance.

The two 100 pF doorknob capacitors on each side of the anode clamp add up to 200 pF of capacitance in series with the resonator pipes (dk = doorknob):

XCdk  =   1   =  5.5W
 
2pfCdk

Each resonator pipe/doorknob combination needs to total 76.2 ohms. Therefore, each resonator pipes needs to be 76.2 + 5.5 = 81.7 ohms of inductive reactance. Each one of these inductor pipes then needs to be:

 
Lcoil  =  
Xl   =  0.09mH
2pf

Now, to summarize the values, recall that we used a tune capacitance value of 2 pF. We then calculated a series value for the load capacitor of 3 pF, and a value for each of the resonator pipes of 0.09 uH. The circuit design is complete.

It is necessary to use inductors with a high value of unloaded Q to minimize circuit losses Thus, 1 1/8 OD copper tubing was chosen to build the resonator pipes for the parallel circuit. The fact that there are two inductors in parallel serves to place the resistive loss component in parallel, lessening this effect.

The area of the tune capacitor plates may be calculated using the equation:

 
At  =  
Cd   =   2(0.5)   =  4.4 in2
 
0.2248K 0.2248

or plates 2.125" x 2.125" square. The same equation yields the size of the plate load capacitor:

 
At  =  
Cd   =   (3.0)(0.5)   =  6.6 in2
 
0.2248K 0.2248

or plates roughly 2.5" x 2.5" square. The lengths of the resonator pipes can be calculated using an equation from the ARRL Handbook:

Xin  =  Z0tanl

where:

    Xin = the desired inductive reactance of 82W
    Z0 = the characteristic impedance of the resonator configuration

This characteristic impedance is approximately 130 ohms for a 1.125" pipe centered within the square enclosures at the sides of the amplifier chassis. Solving this equation for l results in a resonator length of 32 degrees, or 7.375.

These sizes and values are very close to the values used inside the actual amplifier. This discussion completes the design summary for the 2M amplifier.


nd2x home back QRO index
go to the top
Best Viewed: MSIE 4.0 & higher, or Netscape 3.02 & higher, and screen set to 800x600 pixels
This Page Last Updated: 9 September 2004
Feedback: Paul S. Goble, III, ND2X
Copyright © 2003 by ND2X and KD5HIO, all rights reserved