What is the doping and hole concentration?

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Homework Help Overview

The discussion revolves around the concepts of doping and hole concentration in semiconductor physics. The original poster presents a problem involving the law of mass action and calculations related to dopant and hole concentrations at different temperatures.

Discussion Character

  • Exploratory, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to derive expressions for the electron and hole concentrations based on doping concentration and intrinsic carrier concentration. They raise questions about the validity of their calculations and estimations at different temperatures.

Discussion Status

The discussion appears to be ongoing, with the original poster seeking confirmation on their calculations. Several participants have responded with brief posts, indicating a lack of detailed engagement or feedback on the original poster's approach.

Contextual Notes

The original poster mentions specific temperatures (750K and 300K) and provides numerical estimates for intrinsic carrier concentration, which may influence the discussion. There is an emphasis on the relationship between doping concentration and hole concentration, but no explicit consensus or resolution has been reached.

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Homework Statement



(a) Show law of mass action.
(b) Find dopant concentration and hole concentration.[/B]
2012_B6_Q3.png


Homework Equations

The Attempt at a Solution



Part(a)[/B]
Bookwork.

Part(b)

Letting the doping concentration be ##D##, we have:
D = n - p
For ##I = n_{intrinsic} = p_{intrinsic}##, we have:
D^2 + 4I^2 = (n+p)^2 = (n-p)^2 + 4np

Which gives us the result
n = \frac{1}{2}\left( \sqrt{D^2 + 4I^2} + D \right)
p = \frac{1}{2}\left( \sqrt{D^2 + 4I^2} - D \right)

For the doping to work at 750K, I estimate the doping concentration to be ##D \approx n_{intrinsic} = \sqrt{np}##.
At 750 K,
D = \sqrt{np} = 1.7 \times 10^{23}
At 300K,
I = \sqrt{np} = 1 \times 10^{19}

Since ##D >> I##, we have
p \approx \frac{I^2}{D} = 6 \times 10^{14}

Is this right?
 
bumpp
 
bumppp
 
bumpp dope
 
bumping on dope
 
bumping dope concentration?
 
bump on dopant concentration
 
bumpp
 
  • #10
BUMPP
 

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