Why Does the Depletion Layer Affect Acceptor Atom Concentration in PN Junctions?

Click For Summary
SUMMARY

The discussion centers on the behavior of acceptor and donor atoms in PN junctions, specifically how the depletion layer affects their concentration. At 300K, ionization of acceptor atoms (Na) on the P side and donor atoms (Nd) on the N side leads to a high concentration of holes and electrons, respectively. As diffusion occurs, holes migrate to the N side and electrons to the P side, creating a depletion layer where the fixed charges of Na and Nd are revealed. The concentration of acceptor atoms (Na) in the neutral region of the P side remains constant, as it is defined as charge density per cubic centimeter, unaffected by the depletion layer.

PREREQUISITES
  • Understanding of solid state electronics principles
  • Knowledge of PN junction behavior at thermal equilibrium
  • Familiarity with charge density concepts in semiconductor physics
  • Basic grasp of diffusion processes in semiconductors
NEXT STEPS
  • Study the effects of temperature on PN junction characteristics
  • Learn about the mathematical modeling of charge carrier concentrations in semiconductors
  • Explore the role of the depletion region in diode operation
  • Investigate the impact of doping concentration on semiconductor behavior
USEFUL FOR

Students and professionals in solid state electronics, semiconductor physicists, and anyone seeking to deepen their understanding of PN junctions and charge carrier dynamics.

CoolDude420
Messages
199
Reaction score
9

Homework Statement


I'm doing a course in solid state electronics and we just started PN junctions etc. I'm a bit confused.
So here is what I know so far:

Initially the P and N regions are separated by a a PN junction. At 300K as the acceptor atoms on the P side and donor atoms on the N side are ionised, there is a very high concentration of holes on the P side and electrons on the N side.
Diffusion occurs. Holes go to N side. Electrons go to P side. As the holes go to the N side they leave behind negatively charged acceptor atoms. And the electrons from the N side leave behind positvely charged donor atoms. So at this point there is a large concentration of negatively charged acceptor atoms near the junction and a large conc. of positvely charged donor atoms near the junction. This area is known as the depletion layer.

Now here is my question relating to this diagram:

1706855efd.jpg


In my notes it says that concentration of acceptor atoms in the neutral region of the P side(i.e excl the depletion layer) is Na. However that's clearly a contradiction because some of those Na acceptor atoms are now in the depletion layer. So shouldn't Na be the total number of acceptor atoms in the entire P region including the part of the depletion layer which is in the P side. Same for N side.

Homework Equations

The Attempt at a Solution

 
Physics news on Phys.org
The donor and acceptor atoms don't move. They are fixed in the lattice. Only the mobile charges, the holes and electrons, can move. Also note that the Na and Nd quantities are charge densities - i.e, so many atoms per cubic centimeter. So there are Na acceptor atoms per cubic centimeter everywhere on the P side, but in the neutral region they are neutralized by Na holes per cubic centimeter, so that region is charge neutral. In the depletion region, the acceptor charge is "uncovered" because the holes have moved out, so all that remains is the fixed lattice charge of Na acceptor atoms per cubic centimeter. Does this help?
 
  • Like
Likes   Reactions: CoolDude420
phyzguy said:
The donor and acceptor atoms don't move. They are fixed in the lattice. Only the mobile charges, the holes and electrons, can move. Also note that the Na and Nd quantities are charge densities - i.e, so many atoms per cubic centimeter. So there are Na acceptor atoms per cubic centimeter everywhere on the P side, but in the neutral region they are neutralized by Na holes per cubic centimeter, so that region is charge neutral. In the depletion region, the acceptor charge is "uncovered" because the holes have moved out, so all that remains is the fixed lattice charge of Na acceptor atoms per cubic centimeter. Does this help?

Amazing. Thank you for the brilliant explanation!
 
Glad I could help.
 

Similar threads

Replies
3
Views
2K
  • · Replies 12 ·
Replies
12
Views
3K
  • · Replies 1 ·
Replies
1
Views
10K
  • · Replies 2 ·
Replies
2
Views
11K
  • · Replies 30 ·
2
Replies
30
Views
7K
  • · Replies 2 ·
Replies
2
Views
3K
Replies
3
Views
2K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 1 ·
Replies
1
Views
3K