Topband: impedance transformer windings
amdx at att.net
amdx at att.net
Tue Oct 28 01:07:46 EST 2003
Hi Steve,
Someone once decreed, "the reactance of the windings should be at least
four times the impedance the winding is designed to look into." So if you
have a 950 impedance, 4x950=3800 ohms of reactance. Using 1.8Mhz as your
minimum frequency, the inductance would need to be L = 3800 / 2 x pi x 1.8Mhz.
So L= .000336 henrys or .336mh.
To find the number of turns the formula is N=1000(sqrt(L in mh/A sub L))
Or N=1000(sqrt(.336/8500))= 6.28 turns on the primary.
To find the number of turns on the secondary use Np/Ns = sqrt(Zp/Zs)
or Np/Ns = sqrt(950/50) = 4.36, this is the ratio of primary to secondary
turns. So Np/Ns = 4.36 or Ns = 6.28/4.36 = 1.44 turns on the secondary.
(NOTE) Np=number of turns on the primary, Ns= number of turns on the secondary
Zp=impedance of primary, Zs= impedance of secondary
So now we have 6.28 turn pri and 1.44 turn secondary, not good,
I would try a 9 turn pri and 2 turn secondary this is a ratio of 4.5 close to
the 4.36 we calculated.
I have used a 13 turn primary and 3 turn secondary with a gapped pot core,
I gapped the pot core to give me the desired primary reactance. This was used
mostly in the MW BCB.
Mike
> I've been trying to put together an impedance transformer on the Amidon
> BM-73-202 binocular core. It is to be used on a flag RX antenna for 160m so I
> need about a 950:50 (19:1) transformation. The published AL is 8500 mH/1000
> turns. This translates to 8.5 mH per turn. From a Handbook formula, I
> calculated I need 1.4365 mH on the HI Z side and 0.0765 mH on the Lo - Z side.
> As you can see, even one turn thru the core exceeds these by a large margin. Do
> I need a core with a lower AL? Seems like many folks recommend the BN-73-202
> though with a 3:12 (W7IUV) or 1:4 (W8JI) turns. But I can't arrive at these
> numbers by equation. I'd like to learn how this works rather than just build it
> without learning.
>
> Thanks all. Steve KK7UV.
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