Electric field inside PN junction in equilibrium

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SUMMARY

The electric field inside a PN junction in equilibrium approaches zero as one moves away from the depletion region, particularly for x < -b, due to the cancellation of forces from positive and negative ions. The electric field is not strictly zero at the boundary but diminishes rapidly as the distance from the junction increases. The behavior of the electric field can be likened to that of a dipole, where the electric field strength decreases with the cube of the distance from the junction. Understanding these principles is essential for analyzing semiconductor behavior.

PREREQUISITES
  • Understanding of PN junctions and semiconductor physics
  • Familiarity with electric fields and dipole theory
  • Knowledge of charge distribution in semiconductor materials
  • Basic calculus for understanding field strength equations
NEXT STEPS
  • Study the behavior of electric fields in semiconductor devices
  • Learn about the depletion region and its characteristics in PN junctions
  • Explore the mathematical modeling of electric fields using dipole approximations
  • Investigate the impact of temperature on PN junction electric fields
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Students and professionals in electrical engineering, semiconductor physics, and anyone involved in the design or analysis of electronic components utilizing PN junctions.

anhnha
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I need help to understand this solution:
attachment.php?attachmentid=59361&stc=1&d=1370665607.jpg

attachment.php?attachmentid=59362&stc=1&d=1370665607.jpg

Why electric field is zero at x < - b?
I think it should be non-zero because the electric field from positive and negative ions don't completely cancel each other.
How can you know E rises as x approaches zero?
 

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anhnha said:
Why electric field is zero at x < - b?

I think it should be non-zero because the electric field from positive and negative ions don't completely cancel each other.
It is only a schematic plot. The electric field is not zero just at the boundary of the depletion region. (And the boundary of the depletion region is not sharp, either). But it tends to zero quite fast, as the forces from the positive and negative regions tend to cancel.

anhnha said:
How can you know E rises as x approaches zero?

Imagine you are a positively charged particle. You walk into the depletion region from left, from the n part. When you are inside the depletion region, still in the n part, there are some positive ions both behind you (pushing you forward) and the other positive ions in front of you, pushing you backwards. But all the negative ions on the other side pull you forward. When you are exactly at the interface between the p and n region, (x=0) all the positive ions are behind you, pushing forward and all the negative ions are in front of you, pulling you forward.

ehild
 
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Imagine you are a positively charged particle. You walk into the depletion region from left, from the n part. When you are inside the depletion region, still in the n part, there are some positive ions both behind you (pushing you forward) and the other positive ions in front of you, pushing you backwards. But all the negative ions on the other side pull you forward. When you are exactly at the interface between the p and n region, (x=0) all the positive ions are behind you, pushing forward and all the negative ions are in front of you, pulling you forward.
Thanks, the example is great!
t is only a schematic plot. The electric field is not zero just at the boundary of the depletion region. (And the boundary of the depletion region is not sharp, either). But it tends to zero quite fast, as the forces from the positive and negative regions tend to cancel.

I am not quite understand it. Can you explain it more detail?
 
anhnha said:
Thanks, the example is great!


I am not quite understand it. Can you explain it more detail?

The space charge region around the p-n junction is quite thin, the positive and negative "poles" are close, the electric field is similar to that of a dipole, q and -q charges d distance apart. At far away from the junction, in the line of the dipole at distance x , the electric field is E=kq/(x-d/2)-q(x+d/2)≈2kq/x3.

ehild
 
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Thanks, got it now!
 

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