>>>>LC circuits scale nicely with frequency. To move from 40 to 30 meters,
>>>>scale the inductance and capacitance by sqrt(30/40).
>>>I would try scaling the inductance and capacitance inversely proportional
>>>to frequency. For example, in going from 7MHz to 10 MHz, a 100 pF
>>>capacitor would be scaled to 70 pF, while the resonating inductor would
>>>scale from 5.17 uH to 3.62 uH. You usually scale all the inductors
>>>and capacitors, including the coupling/shunt elements.
>>>For casual use, this is probably close enough.
>>Probably, but the formula is F=1/(2*pi*sqrt(LC) so scaling by the
>>sqrt of the frequency ratio gives a better approximation. In this
>>case, moving from 40 to 30 meters requires a 0.86 scale factor.
>So, to double F you need to halve sqrt(LC), according to the
Yes, and that requires scaling each of L and C by 1/1.4142135..
Quote:>Or, squaring both sides, F^2 is proportional to 1/LC
Yes, or written another way to express the ratio: F1^2/F2^2=(L2*C2)/(L1*C1)
Quote:>Note that if you make 1/L proportional to F and 1/C proportional
>to F you get an *exact* scaling. Pretty convenient, IMO.
Yes, but the proportionality isn't a linear relationship, as you noted
above, it's actually F^2 that's proportional to 1/LC. Don't confuse
equal reactances with equal inductances and capacitances.
Quote:>It seems that Gary has botched the math. Fortunately, I have
>lots of patience. :-).
Fortunately I have equal patience. I don't think I botched the math,
you botched the concepts by making the leap from F^2 proportional to
1/LC to F being proportional to 1/L and 1/C. That's true, but it's
only true for the combination if you scale *one* of L or C, not when
you scale both at the same time. Then you must consider the squared
terms and the scaling factor becomes F1^2/F2^2. Hmmm, I did botch
the math, F1^2/F2^2 isn't the same as sqrt(F1/F2). But it's not the
same as simply scaling both L and C by F1/F2 either as you propose.
>>>However, stuff like input and output impedances often vary
>>>with frequency unless feedback is used to control them. Thus,
>>>the networks may have to changed to accommodate the new impedances.
>>If you move both reactances equally, the impedance of the network
>>should remain the same at the new frequency as the unscaled network's
>>impedance at the old frequency. If you only scale one of the reactances,
>>the network impedance will change with the frequency shift.
>Sorry about that, I was writing about the impedances of the devices
>being matched by the network, and not the network itself. This is
>the great thing about MMIC or monolithic microwave integrated circuits
>with 50 ohm input and output impedances. Your typical active mixer chip
>isn't as well behaved.
Quote:>I must say that associating terms like "feedback" or "control"
>to passive LC circuits is a bit unusual. You working on something
Not me, you introduced the subject, note the nesting level. I just
commented that you can keep the passive network's impedance transform
constant by scaling both L and C together. If that's not your design
goal, don't do it that way.
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