one more thing, in most cases, "wire size" should always be the largest you
can get away with (have handy, etc)--not only is it more efficient (lower
ohmic loss of power) it usually holds form better... I would not be all
that concerned about capacitance between turns--unless for very high freq
use...

Regards,
John

"John Smith" wrote in message
...
| Errr, impedance of the winding should be "inductive reactance"--excuse me,
| these "unreal" things are confusing to me... grin
|
| here is an piece I got somewhere I have been using with tollerable
| results...
| CALCULATING THE TURNS COUNT
|
|
|
| Lets calculate the turns count for impedance matching a beverage antenna
| with an impedance of 450 ohms to 50 ohm coax. Because this is a step-down
| transformer, the primary (attached to the antenna) will be the larger
| winding and we'll deal with that first.
|
|
|
| The first formula to use will give us the desired inductance of the
primary
| winding:
|
|
|
| desired L of winding = XL/2p¦
|
|
|
| where L= Inductance in millihenries XL=Reactance in ohms ¦=Lowest
| frequency of operation in kHz
|
|
|
| XL may be found by multiplying the impedance of the antenna to be matched
by
| a factor of 4. This XL would be
|
| 4 x 450 ohms or 1800 ohms. To make things easy, lets use 500 kHz. as our
| lowest frequency of operation.
|
|
|
| So, L of the primary winding = 1800/2 x3.1416 x 500 or .573 mH
|
| Now that we know the inductance (L) needed for the primary winding, we can
| apply the following formula to determine the number of turns needed for
the
| primary winding.
|
|
|
|
| N = 1000 ÖL/AL
|
|
|
| In narrative, this formula should be read: Number of turns required (N) is
| equal to 1000 times the
|
| square root (Ö) of the Inductance (L) divided by the constant AL.
|
|
|
| The constant AL is determined from the Amidon technical literature and
takes
| into account the RF qualities and the size of a Type 43 toroid that is
1.14
| inches in diameter. The AL for the FT-114-43 is 603.
|
|
|
| So, working the formula above, N = 1000 Ö.573/603 = 1000 x .030825 = 30.8
| turns, use 31
|
|
| DISCLAIMER:
| If your radio blows up from the use of my advice I can ONLY be held to
feel
| sorry...
|
| Warmest regards,
| John
| "John Smith" wrote in message
| ...
||I think, others will correct me if I am wrong--the "optimal" number of
| turns
|| would would present an impedance of 4x (four times) the impedence
|| (resistance) of the load/feed, at the lowest freq of operation...
| sometimes
|| this cannot be met, and other values must be used... you should be able
to
|| compute this with the "Al" value of the toroid--with data from the
|| manufacturer...
|| Remember, I am a "newbie", greater authorites will provide finer
| details...
||
|| Warmest regards,
|| John
||
|| "Fred W4JLE" wrote in message
|| ...
|||
||| Using a T200A-2 for a 1:1 balun, what is the optimum number of turns and
||| best wire size? 200 watts max power.
|||
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|
|