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TEA5757 radio design

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

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  1. Guest

    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    available components (i.e. from Digikey or Mouser) into the outdated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

    http://www.nxp.com/acrobat_download/datasheets/TEA5757_5759_3.pdf
    http://www.nxp.com/acrobat_download/datasheets/BB804_3.pdf
     
  2. MarkAren

    MarkAren Guest

    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. JosephKK

    JosephKK Guest

    MarkAren posted to sci.electronics.design:
    http://www.nxp.com/acrobat_download...p.com/acrobat_download/datasheets/BB804_3.pdf

    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/WB2TT

    Tam/WB2TT Guest

    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. JosephKK

    JosephKK Guest

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