If Impedance is given as following,
Center Frequency : 455KHz
Pri = 60KOhm
Sec = 600Ohm
Inductance = 680uH
Tuning Capacitane = 180pF + 5pF(ext.)
What do they mean?
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Do they mean impedance at 455KHz?
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Then, Shouldn't primary impedance depend on the inductance? which is 680uH which gives 2*pi*455KHz * 680uH = 2KOhm?
Best Answer
The inductance of 680 µH would refer to the "Pri" (primary) winding. This inductance resonates at 455 kHz with 180 pF in parallel. The "Sec" (secondary) winding of 600 Ohms would be an untuned winding.
Turns ratio is the square root of impedance ratio:
$$ \sqrt{60000 \over 600} = 10:1 $$
Yes, these impedances are at 455 kHz, and with this design impedance, you get a loaded Q suitable for AM radio use. A primary impedance of 60 kOhms would yield a bandwidth of about 14 kHz. Loaded Q would be about 30.
The unloaded Q of the primary winding is considerably higher than 30. Normally, a transistor collector would drive this winding, and would present about 60 kOhm impedance (1/hoe) to the primary winding. I1, combined with R1 simulate the output port of that driving transistor. The secondary winding would drive the base of the next transistor, and present a reasonable match to its base-to-emitter input impedance. I would guess that you'll find a tiny 180 pF polystyrene capacitor buried in the plastic base. However, some I.F. transformers expect you to supply this capacitor externally.
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