Discussion in 'General Electronics Chat' started by samy555, Aug 25, 2013.

1. ### samy555 Thread Starter Active Member

May 24, 2010
116
3
This circuit of the site:

I assume that antenna impedance = 50 ohm
If I ignore the 8-turns ferrite coil, then tried to calculate the antenna impedance that seen looking into point A:
C1 =100pF, C2=150pF and for the inductor L1, I'm supposed 0.5 mm wire diameter according to images, and radius = 5 mm, length = 1 cm, and therefore has inductance approximately = 450 Nano Henry.

The 50 ohm antenna appeared to be only 73.7 ohm!!!
Now I'll turn to the 4:1 turns ratio transformer:

Zs: the total impedance connected to the secondary windings of the transformer
ZP: the total impedance connected to the primary windings of the transformer

My question is: Do I after all these calculations, I can say that the output impedance of the oscillator = 1.8 Kohm?

Thank you for the patience to read all of my long question

2. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
783
I very much doubt your 30cm antenna is anything like a 50 ohm impedance. It will be dominantly capacitive - hence the series inductor loading coil. I would think the effective radiation resistance would be so small even with the loading coil, that the matching network wouldn't accomplish much other than reducing harmonics in the output. One would think an antenna of length ~ quarter wavelength [2.8 metre] would be more effective in producing a higher radiation resistance.

Last edited: Aug 26, 2013