Thanks to all who replied to my question about converting Heathkit
amps to six meters. It seems the conversion is not that difficult, so
I think I'll try to convert my own SB-1000. There is one thing I
don't quite understand, though, and that is the use of an "L" network
in the plate circuit. Perhaps someone could help me out.
The existing tune capacitor apparently has too much minimum
capacitance to use it as-is on six, since the tank Q would be
excessive. Many conversions replace it with a vacuum variable but
I'd like to keep the original if possible.
What I'd like to do is use a coil from the plate of the 3-500Z to the
tune capacitor (through the DC blocking cap) to step down the 3500 ohm
plate impedance of the tube to something around 1000 ohms, which would
be within the range of the tune cap. As I understand it, this coil
forms a simple L network with the output capacitance of the tube
(including stray capacitance). So far, so good. The problem is the
numbers don't come out right. Perhaps I have made a math error, but I
can't see what it is.
The ARRL handbook, 74th edition, has an example of how to do this on
page 13.21, using an 8877 as an example operating on 29.7 MHz. They
make the following assumptions:
1. Tube output capacitance, including stray, is 15 pF.
2. Tube output impedance is 2200 ohms.
3. The "step down" impedance which the pi-net sees is 1205 ohms.
Using the classic formulas for an L network, matching 2200 ohms to
1205 ohms, requires a capacitive reactance of 2421 ohms paralleled
with the 2200 ohm resistance, and an inductive reactance of 1095 ohms
in the series arm. At 29.7 MHz, these values are 2.2 pF and 5.9 uH.
Obviously, the 2.2 pF value can not be achieved, since the parallel
capacitance already present is 15 pF. This is where I get lost.
In the ARRL's example, they show how to convert the parallel circuit
to an equivalent series circuit, insert an appropriate coil, and reach
a solution. My question is why are their numbers so different from
what an L network requires? Apparently there is more going on here
than simple L-type impedance transformation.
And that's where I need some enlightenment. All comments will be
greatly appreciated.
73, Bill W7TI
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