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Single-ended to differential filter mapping

Discussion in 'Electronic Design' started by Dario, Oct 13, 2006.

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  1. Dario

    Dario Guest

    Hi everybody!
    I'm trying to transform an unbalanced low pass filter to the
    differential version of this. I can't find serious documentation on the
    internet, so I hope somebody in this newsgroup tell me some references.
    Any suggestions are welcome.

    The particular filter is a elliptic form, fifth order and low pass. Are
    standard ways to resolve my problem?

    Thank you very much
    Dario
     
  2. Tam/WB2TT

    Tam/WB2TT Guest

    A balun after the unbalanced filter.
     
  3. Dario

    Dario Guest

    The particular filter is a elliptic form, fifth order and low pass. Are
    Thank you for your answer, but I can't use balun because the system
    works at low frequencies (about 60 MHz) and a balun is too big for my
    purposes.

    I know that, after designing an unbalanced filter, we can transform
    this filter in a balanced one... There are some rules that I don't know
    and that I can't find on the internet. Have you any other suggestions?

    Thank you very much again
    Dario
     
  4. Tam/WB2TT

    Tam/WB2TT Guest

    The size of the balun is going to depend on the amount of power , and the
    lowest frequency you want to pass. At low power, a 1:1 balun could just be a
    6" piece of RG174, or the like, with a bunch of ferrite beads on it. So long
    as you don't need to have it work below about 5 MHz, you won't need too many
    beads. You could also use a transmission line transformer; TV antenna
    transformers are 4:1, but the same basic windings can also be connected as a
    1:1.

    Tam
     
  5. Jim Thompson

    Jim Thompson Guest

    Does the unbalanced filter have a terminal common to both input and
    output?

    If so...

    Draw the filter.

    Below that drawing, draw the filter, but flipped over the x-Axis.

    Erase the common terminal/line.

    Combine _vertical_ L's and R's by doubling

    Combine _vertical_ C's by halving.

    ...Jim Thompson
     
  6. john jardine

    john jardine Guest

    [ Occasional ones I've seen, do the above mirror but halve the vertical
    L's C's and R's and for each horizontal series trap, halve the L and double
    the C ].
    john
     
  7. Dario

    Dario Guest

    Does the unbalanced filter have a terminal common to both input and
    Thank you very much for your answer, I have tried to follow your
    suggestions, but the result is not very good. Maybe I've not
    understoodthe rules at all.

    My unbalanced filter is this (ASCII coding):
    0-----L1-----L2-----L3------0
    | |
    | |
    | |
    L4 L5
    IN | | OUT
    | |
    | |
    C1 C2
    | |
    | |
    0---------------------------0
    where L1 L2 L3 L4 L5 are inductors and C1 and C2 are capacitors. The
    filter was designed for 50ohm impedance matching both for input port
    and output port.

    The result of your suggestions is this:

    0-----L1-----L2-----L3------0
    | |
    | |
    | |
    L4*2 L5*2
    INdiff | | OUTdiff
    | |
    | |
    C1/2 C2/2
    | |
    | |
    0-----L1-----L2-----L3------0
    where vertical lines were simplyfied halving capacitors and doubling
    inductors. Using Ansoft design, with two transformers at input and
    output ports, the resulting transfer function is not the same of
    unbalanced filter (evaluated to differential ports).

    Is my elaboration correct? Or have I made some mistakes?
    Thank you very much again.

    Dario
     
  8. Dario

    Dario Guest

    [CUT]

    I'm sorry, your answer is correct, I have found the problem in the
    measurement system. Thank you very much again!

    Dario
     
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