- #1
shayaan_musta
- 209
- 2
Homework Statement
Give the equilibrium concentration of majority and minority carriers and resistivity for Silicon which is doped with 3x10[itex]^{15}[/itex] boron atoms/cm[itex]^{3}[/itex] at 27°C.
Homework Equations
n[itex]_{o}[/itex] = [itex]\frac{N_{d}-N_{a}}{2}[/itex]+[itex]\sqrt{(\frac{N_{d}-N_{a}}{2})^{2}+(n_{i})^{2}}[/itex]
p[itex]_{o}[/itex] = [itex]\frac{N_{a}-N_{d}}{2}[/itex]+[itex]\sqrt{(\frac{N_{a}-N_{d}}{2})^{2}+(n_{i})^{2}}[/itex]
n[itex]_{o}[/itex]p[itex]_{o}[/itex] = n[itex]_{i}[/itex][itex]^{2}[/itex]
The Attempt at a Solution
DATA
n[itex]_{o}[/itex] (equilibrium concentration of majority carriers) = ?
p[itex]_{o}[/itex] (equilibrium concentration of minority carriers) = ?
[itex]\rho[/itex] (resistivity for Silicon) = ?
N[itex]_{a}[/itex] = 3x10[itex]^{15}[/itex] atoms/cm[itex]^{3}[/itex]
T = 27°C+273 = 273K
n[itex]_{i}[/itex] (for silicon at 300K) = 1.5x10[itex]^{10}[/itex] atoms/cm[itex]^{3}[/itex]
SOLUTION
n[itex]_{o}[/itex] = 0 (I calculated this)
p[itex]_{o}[/itex] = infinity
I used the above given 1st equation to calculate n[itex]_{o}[/itex]. And used 3rd equation to calculate the p[itex]_{o}[/itex].
Actually, I am confused whether I extracted right data or not. And I don't know how to calculate resistivity?
Please tell me where is mistake in the data and Solution.
Thanks.