in article DM8Ka.13887$ snipped-for-privacy@rwcrnsc51.ops.asp.att.net, Peter at snipped-for-privacy@aol.com wrote on 6/24/03 8:36 PM:
I have a few questions about this LC circuit.
>
> It's supposed to drive a 50 ohm impedance coax cable. When I calculate Xc
> and XL I get about 170 for both. But how do these two together match the > 50ohms?
>
> My other question is, how are L and C values calculated? If you do the
> resonant frequency formula (1/ 2pi (sq rt of LC) and solve for LC, you will
> get one number, you can make L and C any value to satisfy the formula. I can
> make the 5.6uH 11uH and make the 180pF 90pF. So where do these values come > from?
>
> To give an example, if I want to make a 5MHz filter, and solve for LC, I
> will get a value of some number, say 100. that means if I make L * C 100, I
> will be all set. But how do I know what to make L and C to make up the 100?
> I can make L 90 and C 10, vice versa, 50 and 50, so on and so on. >
> Any help would be appreciated.
>
> Thanks
>
> I presume that R1 is irrevocably part of the source. You also did not specify
the bandwidth requirements for the match.
Obviously, when L1 and C1 resonate, the combination looks like a short circuit. That is what seems to be selected.
There must be more to the story, because why not just connect R1 and R2 directly instead of through an LC network? The combination does give some selectivity. Because the resistors are somehow specified, the L/C will be related to the required bandwidth or damping.
Bikll