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I am having a tough time wrapping my head around this question.

I am well aware of the benefits of using a twisted pair for signal wires: the cancellation of common mode noise coupled onto the wires, and the cancellation of radiated noise from the loops when there are changing currents through the wires.

Now, we have a power supply with a fast rise time and long cables. This is causing concern about generating a voltage spike at the load from the wire's inductance and the load capacitance.

Someone recommended we twist the power supply cables to help with this... does this actually help? Thank you!

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  • \$\begingroup\$ Why does your power supply need to have such a fast rise (turn on) time? Is this something you need, or is it just a characteristic of the PS? \$\endgroup\$
    – SteveSh
    Dec 19, 2019 at 20:56
  • \$\begingroup\$ A fast rise time at turn on will not cause a voltage spike at the load. The distributed capacitance along with the cable inductance will slow down the flow of initial current to your load, resulting in a nice, RC-like rise in voltage across your load. \$\endgroup\$
    – SteveSh
    Dec 19, 2019 at 20:59
  • \$\begingroup\$ Hi Steve, thanks for commenting. That fast rise time is not needed for this test, but as you mentioned just the nature of the supply being used. My fear was that the current flowing to charge up the load capacitance would store energy in the wire's parasitic inductance. Once the load capacitance is fully charged to nominal, the stored energy in the parasitic inductance would actually charge it a bit more past the nominal voltage of the power supply. Not saying this is true, but it was something I wanted to double check. \$\endgroup\$
    – TI_Lover
    Dec 20, 2019 at 0:39

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I assume you want the fast rise time to appear at the load on the end of the long cables. Or you have a high di/dt load and you want to minimize the voltage dip at the load. In that case twisting the cables may help. The magnetic fields of close conductors carrying opposite currents will tend to cancel, reducing the inductance associated with the cabling.

The bigger the loop area between the two conductors the higher the inductance. And since V=L*di/dt for a fast di/dt you want a small value of L (large V/L).

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  • \$\begingroup\$ Thanks John! It seems this statement is what I was overlooking: "The magnetic fields of close conductors carrying opposite currents will tend to cancel, reducing the inductance associated with the cabling." I figured the inductance of a wire was purely a function of it's dimensions. However, it looks like placing it in a twisted pair actually can reduce its inductance. Neat :) This web page agrees with you: sites.google.com/site/markgurries/home/… \$\endgroup\$
    – TI_Lover
    Dec 20, 2019 at 0:43
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Cable impedance depends on the sq.rt. of L/C ratio while C depends on gap between conductors and increases about 1pF per twist. But Inductance increases at a lower rate so impedance only drops 50% or so unless they were very loose to begin with.

If the load has any capacitance then the step voltage load impedance becomes the very low ESR of the caps. The relatively high twisted cable impedance differences have no effect on the terminal voltage rise but the radiated current B field in A/m due to A/us rise time affects the radiated noise if that is of any concern.

Cable inductance per meter is the more significant for effect of ramped current rise, which is typically 1uH/m and dI/dt=V/L

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