Connect with us

TEA5757 radio design

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

Scroll to continue with content
  1. Guest


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    available components (i.e. from Digikey or Mouser) into the outdated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

  2. MarkAren

    MarkAren Guest

    Hi Mike,

    Digging for data on the Toko MC117 series...

    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.


  3. JosephKK

    JosephKK Guest

    MarkAren posted to

    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

    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

  5. JosephKK

    JosephKK Guest

    Tam/WB2TT [email protected]$ posted to
    Perhaps stated more strongly than necessary, see my post.
Ask a Question
Want to reply to this thread or ask your own question?
You'll need to choose a username for the site, which only take a couple of moments (here). After that, you can post your question and our members will help you out.
Similar Threads
There are no similar threads yet.
Electronics Point Logo
Continue to site
Quote of the day