I need to estimate the signal delay through the 1:4 Guanella balun shown here: enter image description here

Is the delay simply the delay through the winding on CORE1 or CORE2?

For example, assume each winding comprises 0.1-m of coax with velocity factor of 0.78, is the delay: $$delay = \frac{0.1 \text{-m}}{0.78 \;\text x \;3\text x 10^8 \text{-m/s}} = 0.43 \text{ns}$$

  • $\begingroup$ How are you winding this balun with coax? $\endgroup$ Dec 6, 2019 at 19:47
  • $\begingroup$ @PhilFrost-W8II "For example, assume each winding comprises 0.1-m of coax with velocity factor of 0.78..." $\endgroup$
    – Brian K1LI
    Dec 7, 2019 at 14:04
  • $\begingroup$ I see you wrote that, but I'm not sure a balun like this can be fabricated with coax. And from the schematic, it's not really clear specifically how you've connected the coax. Unless maybe you're driving the coax with only a common-mode current, in which case the velocity factor of the coax would be irrelevant since it's basically just fat, expensive wire. $\endgroup$ Dec 8, 2019 at 5:39

1 Answer 1


A colleague used a 2-port vector network analyzer (VNWA) to perform a useful measurement.

After calibrating the instrument to set the phase of S21 to zero at 20-MHz - no delay - he measured the phase added by about a meter of small diameter coax as 28.49 degrees. He then wrapped the entire length of the same coax on a large toroidal ferrite core, FT240-43 (2.4-in O.D., 43-mix ferrite) and measured the phase as 28.43 degrees, indicating that "ferrite loading" of the coax added no discernible delay. This is not surprising, in hindsight, but the experimental verification is worthwhile.

I modeled a Guanella balun using SimSmith. As shown, I used the resistance and reactance specified for a binocular type-43 ferrite core to establish the impedance to currents on the outside of the coax shield at 2-MHz:

SimSmith RUSE block for Guanella balun

As shown by SimSmith's Wave function:

SimSmith balun delay measurement

the delay from the input to the output of the balun is:

17.97ns - 17.54ns = 0.43ns

which is the delay expected from the length of coax used to wind each of T1 and T2:

$\text{delay} = \frac{\text{4-in x .0254-m/in}}{\text{0.78 x 3x}10^8 \text{m/sec}} = \text{0.43-ns}$


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