To: | JMLTINC@aol.com |
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Subject: | Re: [Amps] Line Sections As Plate Lines |
From: | jeff millar <wa1hco@adelphia.net> |
Date: | Fri, 24 Sep 2004 19:30:17 -0400 |
List-post: | <mailto:amps@contesting.com> |
The line is not physically 1/2 wave long because it's loaded at both
ends by capacitance. I use the RF Dude's smith chart program (or the
Linux one by Lapo Pieri, IK5NAX) and enter some starting values. Then
just tune the values with the keys until you get a match, change the
length of the line, readjust tune and load C and repeat. A few minutes
of playing will show how to move the values to get closer to the 50 Ohm
center point. You stated 74 Ohms of stray C, which corresponds to 15 pF at 144 MHz, allocate 9 pF to the tube and 6 pF to the other end to keep the tune and load flapper small. 144.2 MHz Frequency 100000 Ohm open circuit impedance (dummy hi-z starting point) Parallel Plate impedance 2000 Ohms ( 3600V / (1.8 * 1 Amp)) Parallel Plate output C 9 pF (GS-35 has 4.4 pf Cout plus 4.6 pF stray) Series Transmision line of 87 Ohms and 114 degrees long Parallel tune C of 4.3 pF Series load C of 3.4 pF This produces a nice match to 50 Ohms The length of the line is computed directly from length in degrees, assuming velocity factor of 1 for air coax, for example (113 degrees /360) * 300 M meters_per_sec / 144.2 MHz = 66 cm = 26 inches. Other combinations work as well 75 Ohm line Tline of 75 Ohms, 116 degrees long Tune C of 6 pF Load C of 3.6 pF 100 Ohm line T line of 100 Ohms, 92 degrees Tune C of 8.4 pF Load C of 4.2 pF More Tune C, shorter line T line of 50 Ohms, 105 degrees Tune C of 24 pF Load C of 5.5 pF Less tune C, longer line T line of 50 Ohms, 140 degrees Tune C of 3.7 pF Load C of 3.4 pF This calculation can be done with transmission line formulas and math, but its complex numbers with real and imaginary impedances...lots of opportunity for silly math errors. The smith chart program lets you enter approximate components and then tune all the values with up/down keys on the computer until the end point of the arcs lands on the 50 Ohm point in the middle. The length of the anode can get modeled as either a lumped capacitance of about 5 pF or as a short section of lower impedance transmission line. The impedance of the transmission line depends on the diameter of center conductor and the dimensions of the box. I don't have the formulas handy. The process basically is to decide how big you want the outer box and how small you're willing to go for the center without skin effect losses becoming too much. Using high impedance lines make for a shorter overall design but risk heating and possible detuning due to thermal effects. The web designs you pointed to look like about 50 Ohms. Knowing the impedance helps to ensure the model works correctly. hope this helps, jeff, wa1hco JMLTINC@aol.com wrote: Hi again guys-
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