In message <19980510030240.AAB548@[205.231.11.82]>, Rich Measures
<measures@vc.net> writes
>
>You have not explained how lowering R is a parallel L/R suppressor
lowers
>VHF-Q. By your logic, the lowest Q would result when R = 0-ohms
Yes, in a parallel circuit, Q is maximum when R=0. Here are the
equations:
In a parallel circuit, Q = X/R. X, of course, is frequency dependent,
and if R is made of a highly conductive material (nichrome, copper,
silver, etc) then R is also frequency-dependent, predictable by the skin
effect equations. But the frequency dependence of X will dominate.
In a series circuit Q = R/X
Look at the parallel circuit this way: If I make a (practically
impossible) perfect inductor of 1uH, then at 14 MHz, it has a reactance
of 87.9 ohms with nothing in parallel. Nothing in parallel means
infinite resistance in parallel. This has an infinite Q. If I put an
87.9 ohm resistor in parallel (assuming a perfect resistor and perfect
inductor), then the Q is equal to 1. If the resistor goes down further,
the Q increases.
One thing that we MUST bear in mind: when we use the algebraic
manipulations to transform a series RLC circuit into a parallel RLC
circuit, the values we arrive at are valid for one frequency only. The
equations are not intended to imply that we can actually build a series
RL circuit that behaves identically to a parallel RL circuit.
When you first start using the duality relationships, it's easy to be
surprised - a series circuit with a huge resistor compared with the
reactor (capacitor or inductor) transforms to a parallel circuit with a
very small resistor compared to the reactor. After a while, you just
get used to it, then you assume it, then you assume everybody else knows
it and always did. That's how we absorb knowledge and accept it. But
these relationships do seem strange at first.
Arlen
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