Discussion in 'Electronic Design' started by [email protected], Oct 21, 2007.

1. ### Guest

Hello,

available components (i.e. from Digikey or Mouser) into the outdated
application circuit. I'm currently stuck on the FM front-end tank
circuit; I'm not quite sure I understand how it's done in the
application diagram (page 27 of the TEA5757 datasheet). By my
calculations, it shouldn't work.

The tank consists of a dual varactor (BB804), a 10 pF trimmer, and a
RF coil that I cannot find the specs for (Toko MC117 E523FN-2000242).
The schematic says the coil has 38 pF capacitance, and from comparison
with current Toko coils and googling I'm guessing that it's an
unshielded coil with a Q > 100 and an inductance < 100 nH. From the
BB804 datasheet, each individual varactor has an effective range of
20-60 pF (generous assumption given 12V supply), so the series
combination results in 10-30 pF.

Altogether, the capacitance range is 48 to 68 pF, and 68/48 = 1.42. We
need (108/88)^2 = 1.5 to tune the FM radio band. Stray capacitance and
the trimmer don't help. I doubt Philips would provide a bum
application diagram, so I must be missing something.

Thanks,
Mike

2. ### MarkArenGuest

Hi Mike,

Digging for data on the Toko MC117 series... http://81.149.89.17/Pages/MOLD/page91.htm

The E523FN-2000242 isn't shown, but there does appear to be a
relationship between PN and inductance (2xxxxxx indicates 2.5 turns)
which seems to be in the 55-66nH range.

Hope that helps a little bit.

Regards,

Mark

3. ### JosephKKGuest

MarkAren posted to sci.electronics.design:

Just a wild guess, but the 38 pF may be the capacitance required to be
resonant at 100 MHz. Inductors below 1 uH rarely have parallel
capacitances above 1 pF. And remember that 10 pF is a variable
trimmer.

Put this into your assumptions and see how it works.

4. ### Tam/WB2TTGuest

For L in uH and C in PF, the LC ratio at 88 MHz is 3.27. The LC ratio at 108
is 2.17. If Cmax - Cmin is 10 PF, then L= (3.27 - 2.17)/10 = .11 uh. That
means Cmax is 3.27/.11=29.7 PF and Cmin=2.17/.11=19.7PF. Sanity check:
Cmax - Cmin = 29.7 - 19.7 = 10 PF; qed.

If they say a 100 nH inductor has a capacitance of 38 PF, that is a garbage
statement.

Tam

5. ### JosephKKGuest

Tam/WB2TT [email protected]\$t.net posted to sci.electronics.design:
Perhaps stated more strongly than necessary, see my post.