Single-ended to differential filter mapping

A balun after the unbalanced filter.

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

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

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

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|  James E.Thompson, P.E.                           |    mens     |
|  Analog Innovations, Inc.                         |     et      |
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[...] [ 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

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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

[CUT]

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

Dario