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Reverse diode

  1. Dec 13, 2014 #1
    I am not getting why there can't be (or very negligible )current in reverse diode.According to this image
    upload_2014-12-13_15-3-51.png
    positive terminal of battery would suck electrons from n-type and pass them to negative terminal of battery.And Negative terminal would provide electrons to p type and from there we can see in image force on electrons is opposite to electric field so electrons can easily pass the barrier and can go to n-type and hence complete the circuit(i.e initially electrons were sucked from n-type and lastly they go towards n-type)So in this way in reverse diode also there is sufficient (in good amount)current available.
    Where am i going wrong?
     

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  3. Dec 13, 2014 #2

    Drakkith

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    Staff: Mentor

  4. Dec 13, 2014 #3
  5. Dec 13, 2014 #4

    Drakkith

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    Staff: Mentor

    I think you need to look at what happens in the depletion zone. Not sure. (It's very late and I'm tired.)
     
  6. Dec 13, 2014 #5
    I have tried my best to look at all aspects.No ,matter whenever you feel comfortable ,you can answer.Please ..but do explain me this.
     
  7. Dec 13, 2014 #6

    CWatters

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    With no voltage applied there is a thin depletion region at the junction. This has very few charge carriers so it acts like an insulator preventing current flow.

    If you inject more electrons into the p-type you reduce the number of holes even more and the depletion region gets wider. You also get a build up of negative charge near the junction that eventually matches the applied voltage preventing further electrons from flowing into the p-type. In this mode the diode behaves a bit like a capacitor.

    Perhaps see..
    http://www.science-campus.com/engineering/electronics/semiconductor_theory/diode_2.html
     
  8. Dec 13, 2014 #7
    One important thing you must know is we say current to flow in diode iff any charge carriers "cross the junction"between p and n.

    1)ē are majority in N and holes in P.
    2)-ve terminal is connected to P. So it will attract holes from P side . result: holes will move away from junction and get collected in -ve terminal
    3)ē will too move away from junction to the +ve terminal there in N side.
    4)these two will cause widening of depletion region as majority charge carriers move sideways .
    Now ur concept of small current is true . its called reverse leakage current . minority ē in P and minority holes in N will cross junction resulting in current .
     
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