TEA5757 radio design

Oct 21, 2007 4 Replies

Hello,



I've been "designing" a Philips TEA5757-based radio by fitting readily available components (i.e. from Digikey or Mouser) into the outdated application circuit.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



formatting link
formatting link



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

formatting link

MarkAren snipped-for-privacy@yahoo.com posted to sci.electronics.design:

formatting link

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.

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

Tam/WB2TT t-tammaru@c0mca$t.net posted to sci.electronics.design:

Perhaps stated more strongly than necessary, see my post.

Join the Discussion

Have something to add? Share your thoughts — no account required.

Didn't find your answer?

Ask the community — no account required