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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