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.
|
|
|
|
|
|

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.
||
||
||
||
||
||
|
|

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.
|||
|||
|||
|||
|||
|||
||
||
|
|

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, have you ever tried winding the largest possible wire around a
ferrite toroid ring?
----
Reg.

Reg:

Yes, I have, started winding 'em with too BIG of wire to, had to drop a
couple sizes... which is why I said:
"...should always be the largest you can get away with...", as to "wire
size."
A 2" core is pretty big though, that is what he has got-if I know those
numbers correctly, and unless he is going to the basement (160M) or a KW
(probably could get by with a KW on a 200A too, but I'd go 2.5" core) he has
some room...
It is a 200A, they have a bit more inductance per turn too (bit fewer turns
for same mh/uh), than just a 200, if I remember datasheets correctly...

Warmest regards,
John

"Reg Edwards" wrote in message
...
|
| 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, have you ever tried winding the largest possible wire around a
| ferrite toroid ring?
| ----
| Reg.
|
|