What is the doping and hole concentration?

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SUMMARY

The discussion focuses on the relationship between doping concentration and hole concentration in semiconductors, specifically using the law of mass action. The derived equations show that the electron concentration (n) and hole concentration (p) can be calculated using the doping concentration (D) and intrinsic carrier concentration (I) at different temperatures, notably at 750K and 300K. The final calculations yield a doping concentration of approximately 1.7 x 1023 cm-3 at 750K and a hole concentration of 6 x 1014 cm-3 at 300K, confirming the relationship between these variables.

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  • Knowledge of intrinsic carrier concentration calculations
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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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