>
>On Sat, 15 Jul 2000 12:54:17 -0500 "Phil Clements" <philk5pc@tyler.net>
>writes:
>.
>>Much has been written for the proper size conductor
>>to be used in the construction of tank coils, but very little
>>data is available for the proper construction of the "L" coil in=20
>>a Pi-L network, other than its value. Is there a source for
>>this ?
>
> Since the Q is (typically) rather low compared to the PI application,
>the circulating current is also reasonable, requiring just a small
>increase in wire size.
>
As I see it, the load is in series with the L - so the current is exactly
equal. .
>>The torroid is very popular for this purpose, as it is more
>>compact and self-shielding, allowing close-to -chassis mounting.
>>Has anyone run any tests on the maximum power handling capability
>>of such a coil?
>
> That depends mostly on the core and is easily calculated from the
>numbers in a Micrometals (or Amidon) catalog. If you need more power
>capability you just stack cores.
>
The question was about conductor size. .
>>There is probably a formula for a simple "L" coil; given an image
>>inductance of 300 ohms and 50 ohms output, but I have never seen
>>it.
>
> Yes, there is ...presuming you meant to type impedance immediately
>above...
>
> I think you can find it in the Orr Handbook... and probably the older
>ARRL books... don't know about the new ones. :-)
>
> Q = Square Root (Rin/Rout -1)
>
> L = Rin / (Q * 2 * Pi * F)
>
>>I guess my bottom-line question is: what would a 9.5 uh 20 amp rated
>>"L" coil look like?
>
> Just guessing that this is for 160M... #10 wire should suffice, either
>a section of Air-Dux or wound on a toroid.
>
>
>73,
> Marv WC6W
>
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>*
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>
- Rich..., 805.386.3734, www.vcnet.com/measures.
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