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I am looking at this circuit:

Common mode current schematic

(taken from here)

In the schematic, the common mode current does not enter into the load - it is bypassed to ground through the parasitic capacitance. I have couple of doubts:

  1. How does this capacitance allow common mode current? It seems like this current should be DC, and capacitors block DC, right?
  2. If there is no current flowing through the load, then how does the load still work?
Greg d'Eon
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Mohammed Azlum
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1 Answers1

2
  1. Why does it seem like this current should be DC? This schematic says "on the mains supply line", so I would guess that it's 50 Hz or 60 Hz AC. (Please correct me if I'm wrong here.) That means the voltage is perfectly happy passing through a capacitor.
  2. That's the point. This current isn't supposed to be the current that operates the load This is a common-mode interference current, which should tell you that it's something you don't want. There is another current loop that is not shown here which would supply the load.
Greg d'Eon
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  • Feel free to accept this answer if you think it answers the question! – Greg d'Eon Apr 25 '15 at 13:06
  • I understood your second explanation.

    I have one more doubt, what is the difference between the common mode noise and differential mode noise ?

     – Mohammed Azlum Apr 25 '15 at 13:13
  • You may want to Google "common mode vs differential mode noise" - there are lots of great explanations online, and they do a far better job than I would. – Greg d'Eon Apr 25 '15 at 13:15