Converting a 300 to 75 ohm Balun to 300 to 50 ohms?

[Mikejo]

I would like to know if one could convert one of those 300 ohm to 75 ohm Baluns to a 300 ohm to 50 ohm Balun by adding an additional half of a turn of wire through the ferrite bead?

I read somewhere that the turns ratio is the square root of the higher impedance divided by the lower impedance; (I hope I read that right)!

Example 300/75 = 4
The square root of 4 is 2

Therefore, two turns of wire should go through the ferrite bead to make a 300 to 75 ohm Balun.

I have in the past opened up one of those 300 to 75 ohm Balun's, and I think I did in fact see two turns of wire going through the ferrite bead.

If you use the same formula substituting 50 ohms for 75 ohms you get this:

300/50= 6

The square root of 6 is 2.44

Therefor approximately 2.5 turns of wire should go through the ferrite bead to make it a 300/50 Ohm Balun?

( go easy on me, I'm just trying to learn the stuff) LOL

 

[prcguy]

In general your thinking is correct. If you have an antenna analyzer you can terminate the balun with a 300 ohm non inductive resistor and map it out over your desired frequency range expecting about a 1.5:1 VSWR, then add that half turn and test again. You can also put some capacitance on the low impedance side and pull that down a little. I don't have a formula for that but I've seen some baluns used in the VHF lo range where a capacitor around 25 to 50pf was added to change the ratio slightly. You could tag on maybe 25pf and see what happens on an antenna analyzer, then add more, etc.

If I were doing this I would make changes to two identical baluns, then connect them back to back and measure insertion loss. Divide that by two for the individual balun loss.

I would like to know if one could convert one of those 300 ohm to 75 ohm Baluns to a 300 ohm to 50 ohm Balun by adding an additional half of a turn of wire through the ferrite bead?

I read somewhere that the turns ratio is the square root of the higher impedance divided by the lower impedance; (I hope I read that right)!

Example 300/75 = 4
The square root of 4 is 2

Therefore, two turns of wire should go through the ferrite bead to make a 300 to 75 ohm Balun.

I have in the past opened up one of those 300 to 75 ohm Balun's, and I think I did in fact see two turns of wire going through the ferrite bead.

If you use the same formula substituting 50 ohms for 75 ohms you get this:

300/50= 6

The square root of 6 is 2.44

Therefor approximately 2.5 turns of wire should go through the ferrite bead to make it a 300/50 Ohm Balun?

( go easy on me, I'm just trying to learn the stuff) LOL

 

[Ubbe]

The balun shouldn't be more frequency dependent than the actual ferrite material are. Adding capacitors would make it a tuned LC circuit, something that might not be so bad in some cases. If the ferrite used are specified for 800Mhz it will proably work also at 150Mhz. A 300/75 balun is a fairly non frequency dependent transformer, in this case 1:4 transformer, and its impedance isn't written in stone. If you have an antenna that at a frequency are 200 ohm it will give a 50 ohm impedance at the other side of a 1:4 balun. If you look at this link it seems that loading a properly designed folded dipole with 200 ohm gives a lower SWR compared to 300 ohm. So if you have a folded dipole you want to match to a 50 ohm coax, then that original 1:4 balun might be best to use unmodified. But maybe you only wanted to know the math for the balun calculation as an a exersise and not optimising an actual antenna to a coax. Of course an antenna analyzer would give the actual result from an antenna as it is seldom installed in an enviroment that equals the ideal free space that the antenna developer based his design on.

Folded Dipole

/Ubbe