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I am trying to reverse engineer this video balun. It is meant to convert a 100~120 balanced video signal on twisted pair to a 75ohm unbalanced signal. I've removed it from it's housing, but normally there are a pair of screw terminals on the skinny end and a BNC connector on the fat end (square pads are positive):

enter image description here

It has a trifilar winding composed of L1a, L1b, & L1c on a toroidal core. All three windings are the same (~30 turns). I have no data on the core itself, but I would assume we are getting proper transformer action. There is a 4V bidirectional zener diode in parallel with the balanced input and a parallel RC circuit in series with two of the windings. I am familiar with Ruthroff and Guanella baluns, but I am having a bit of a time understanding how this one works. I've reverse-engineered the PCB to get the following schematic (note the dots on the windings):

Schematic diagram of Balun circut

I know that the zener is just there for transient protection and is not part of the actual balun circuitry. I suspect that the RC circuit is there to provide gain that increases with frequency (impedance drops from 1k at DC to 256 ohms at 6MHz), and perhaps does a dirty 100ohm to 75ohm conversion via loss?

Any concise explanation of the way the coils achieve the balun operation and the role of the RC circuit would be greatly appreciated. These devices are ubiquitous in the CCTV industry and I've never been able to suss out how they work in a way that satisfies me...

Thanks!

EDIT: in an earlier version I erroneously stated that the current in each winding would be the same due to proper transformer action. This clearly is not the case; I'm not sure why my mind assumed that...

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  • \$\begingroup\$ Have you done any research about BALUNs, on HAM radio forums/web sites? I guess it's just an ordinary BALUN, no need to reverse engineer and to reinvent warm water. \$\endgroup\$ – Marko Buršič Mar 26 at 19:34
  • \$\begingroup\$ This is a good suggestion, and I actually am an amateur radio operator and have worked with baluns a lot. However, this is not one of the standard topologies (Ruthroff or Guanella, or their derivatives) as I can see it. It appears to be based on auto-transformer action. I know that it works, but I would like to understand how for my own benefit. \$\endgroup\$ – Chris Mar 26 at 21:35
  • \$\begingroup\$ Are you sure that the Unbalanced- net is connected to the bottom side of the capacitor as opposed to the top side of it? \$\endgroup\$ – joribama Mar 27 at 1:16
  • \$\begingroup\$ The only thing between the balanced -ve and unbalanced -ve terminals is the transformer coil L1c. \$\endgroup\$ – Chris Mar 27 at 19:52
  • \$\begingroup\$ joribama, after looking at the PCB again do double-check, and in light of Jasen's answer below, I agree that the unbalanced -ve should connect to the top side of the RC circuit to completely block DC from the balanced side, but it doesn't. The circuit is as shown above... \$\endgroup\$ – Chris Mar 27 at 23:24
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It's not doing impdance transformation. 75 ohms is close enough to 110 ohms that it doesn't matter.

For now ignore the capacitor and treat it as a short circuit. you can see that the signal comes in on the lower two transformer segments goes out on the upper two segments it's acting as an autotransformer. shifting the signal up by half its voltage converting balanced to unbalanced. (this is the Ruthroff topology mentioned in comments)

So what about the capacitor?

Video is a baseband signal the balun need to pass DC so what the designwer did is insert that capacuitor into the autotransformer between balaced+ and ground (unbalanced - is signal ground)

So DC signals put in on balanced + and balanced - will come out unaltered between unbalanced + and ground.

the resistor is probably there to stop charges from building up on the capacitor.

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  • \$\begingroup\$ Thank you, this is a very good explanation. \$\endgroup\$ – Chris Mar 27 at 18:36
  • \$\begingroup\$ One thing that I will add is that baseband video does not extend all the way down to DC, because it uses a vestigial lower sideband for the Luminance part of the signal. the cutoff is somewhere around 400kHz IIRC. I ended up doing some simulations in LTSpice (assuming perfect coupling in the transformer). The capacitor appears to do most of the work on common-mode rejection, but it is not perfect and varies with frequency, but at least it's something. Again, thank you very much for your answer! \$\endgroup\$ – Chris Mar 27 at 20:03
  • \$\begingroup\$ a baseband video signal carries information at frequencies down to the vertical rate of 50 or 60 Hz it may be possible to recover the low frequencies by detecting the blanking pulses and adjusting the compensation but I wouldn't want to bet on it. \$\endgroup\$ – Jasen Mar 27 at 21:45

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