Why is the balun transformer shorted and not simply removed?

Question

I use a normal balun to match my 300 ohm antenna to my 75 ohm tuner (for "over-the-air" digital TV):

I opened this balun up and found the typical circuit:

Since the second coil from the top is shorted, the top transformer appears to do nothing. So, what would be the harm in removing the top transformer and replacing it with a short? Here is the resulting schematic:

Is that first schematic drawn incorrectly? Or is there some fractional flux linkage between those two magnetic cores that's missing? — glen_geek, Apr 01 '23 at 15:44
@glen_geek Here they show grounding the center point as an option: https://wiki.arising.com.au/wiki/Cost_Effective_Current-Mode_1:4_Balun — GodJihyo, Apr 01 '23 at 17:13
@GodJihyo Seems that my queries were both wrong: apparently, the GND at the source end is distinct from the GND at the load end. OP has drawn the schematic assuming that both GNDs are the same point. — glen_geek, Apr 01 '23 at 17:31
`top transformer appears to do nothing` ... it is still a coil, it does not `do nothing` — jsotola, Apr 01 '23 at 17:35
@jsotola It is still a coil, but it's inductance is effectively zero (since current will flow oppositely in the shorted coil to cancel its magnetic field). — bobuhito, Apr 01 '23 at 18:18
I have very limited RF expertise, but common mode noise is a big thing in the field I am working in and CMRR is highly influenced by the symmetry of impedances (e.g. coil resistance). Is it possible that CMRR is a thing for this application as well? — Christian B., Apr 01 '23 at 19:26
At cable TV frequencies (up to 1GHz now) it is arguably inappropriate to treat this as a lumped-element model. Certainly without including the step impedance changes of the connections to/from the baluns. — TLW, Apr 01 '23 at 20:28

Antonio51 · Answer 1 · 2023-04-01T19:52:24.023

2

This should be ok ...
The point "VII" left "dangling"...

Or not left "dangling".

Your last picture ...

At 300 MHz ...

Other configuration.

edited Apr 01 '23 at 19:52

answered Apr 01 '23 at 16:21

Antonio51

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Can you also simulate the short (the schematic at the end of my question)? – bobuhito Apr 01 '23 at 16:48
It is done in the second picture at point "VII". – Antonio51 Apr 01 '23 at 18:24
No, not a short to ground, but a short across the coil (settling L1=0) as in the schematic at the end of my question. – bobuhito Apr 01 '23 at 18:29
Ah ... ok. Will add in answer ... – Antonio51 Apr 01 '23 at 18:48
By the way, 10uH seems high and 1MHz seems low...wouldn't 0.5uH and 300MHz be more realistic? – bobuhito Apr 01 '23 at 18:53
Sorry for the error. It is 10 MHz. And it is my "L2" shorted. I will add also your last picture. – Antonio51 Apr 01 '23 at 18:59
And 10 uH or 100 nH does not make "more" changes" in the simulation. Even at 300 MHz. – Antonio51 Apr 01 '23 at 19:07
Note that "balun" is used between 3 MHz and 30 MHz ... as pointed out by @Godjihyo link. – Antonio51 Apr 01 '23 at 19:16
In USA, we use this balun for digital TV, for which 300MHz is typical. – bobuhito Apr 01 '23 at 19:18
Ok. it is a "good" use, then. – Antonio51 Apr 01 '23 at 19:27
So it seems that the manufacturers could remove the top transformer to save money, but they don't. So, I'm still hoping you (or anyone) could see a reason why both transformers are being used...? – bobuhito Apr 01 '23 at 19:31
The "two" transformers are "more" symmetric for output as "impedances" are seen. – Antonio51 Apr 01 '23 at 19:35
3

A lumped element circuit does not explain the difference; transmission lines must be used. – Tim Williams Apr 01 '23 at 19:35

score 2 · Answer 2 · answered Apr 01 '23 at 19:36

Transmission line transformers are used, so that the upper transformer, while it may not see any voltage drop across the windings (nominally), it is still used as a delay. It may be wound on a hollow rather than magnetic core, as a consequence. Notice the bottom one must have some voltage drop, so needs a nonzero sized core, and nonzero wire length wrapped around it: therefore there is some delay through the wire pair, which would introduce phase error between the two 300-ohm terminals if wired as the bottom circuit.

Tesla23 · Answer 3 · 2023-04-03T00:25:55.153

This looks like a 4:1 guanella balun, it can be understood as follows:

The voltage V is applied across the input terminals, and the two bifalar wound transformers reproduce V across each of their outputs as shown (think of them as common mode chokes). The bifalar winding acts as a transmission line at high frequencies that tends to extend the high frequency response. As the voltage is doubled as shown, the impedance increases by a factor of 4. To make the output balanced, it is essential that the two transformers are identical.

If the inductance of the transformers is sufficient, then the outputs are floating wrt the input, so we can choose which (if any) output connection to connect to the input ground. For balance, it is common to ground the midpoint (connect A-B), so the output is balanced wrt the input ground. For example, if driving a dipole, the connections are driven +/- balanced around the potential of the outer of the coax feedline.

I'm no expert on the the finer details, but there is plenty on the web.

Just to answer your question, the circuit you have drawn does not have a balanced output.

Why is the balun transformer shorted and not simply removed?

3 Answers3