Why Isn't There More Current in Reverse Diode?

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Discussion Overview

The discussion centers around the behavior of current in a reverse-biased diode, specifically why there is negligible current flow under these conditions. Participants explore the underlying mechanisms, including the role of the depletion region and charge carrier dynamics.

Discussion Character

  • Technical explanation
  • Debate/contested

Main Points Raised

  • One participant questions the absence of significant current in a reverse-biased diode, suggesting that electrons should flow from the n-type to the p-type region when a positive voltage is applied.
  • Another participant points out the existence of a small current known as reverse leakage current in reverse-biased diodes, referencing external sources for clarification.
  • A participant expresses confusion and requests further explanation on where their understanding might be incorrect.
  • Several participants mention the importance of the depletion zone, indicating that it acts as an insulator due to the lack of charge carriers, which prevents current flow.
  • One participant explains that injecting electrons into the p-type region increases the depletion region's width, thus hindering current flow and likening the diode's behavior to that of a capacitor.
  • Another participant clarifies that current in a diode is defined by charge carriers crossing the junction, noting the movement of holes and electrons in response to applied voltage.

Areas of Agreement / Disagreement

Participants express differing views on the mechanisms of current flow in reverse-biased diodes. While some acknowledge the existence of reverse leakage current, others debate the implications of charge carrier movement and the role of the depletion region, indicating that the discussion remains unresolved.

Contextual Notes

Participants reference external sources and concepts such as the depletion region and reverse leakage current, but there is no consensus on the interpretation of these phenomena or their implications for current flow in reverse-biased diodes.

gracy
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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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I think you need to look at what happens in the depletion zone. Not sure. (It's very late and I'm tired.)
 
Drakkith said:
I think you need to look at what happens in the depletion zone. Not sure. (It's very late and I'm tired.)
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.
 
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.

...Negative terminal would provide electrons to p type

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
 
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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