0 0 votes The intrinsic carrier density at $300 \mathrm{~K}$ is $1.5 \times 10^{10}$ / $\mathrm{cm}^{3}$, in silicon. For $n$-type silicon doped to $2.25 \times$ $10^{15}$ atoms $/ \mathrm{cm}^{3}$, the equilibrium electron and hole densities are $n=1.5 \times 10^{15} / \mathrm{cm}^{3}, p=1.5 \times 10^{10} / \mathrm{cm}^{3}$ $n=1.5 \times 10^{10} / \mathrm{cm}^{3}, p=2.25 \times 10^{15} / \mathrm{cm}^{3}$ $n=2.25 \times 10^{15} / \mathrm{cm}^{3}, p=1.0 \times 10^{5} / \mathrm{cm}^{3}$ $n=1.5 \times 10^{10} / \mathrm{cm}^{3}, p=1.5 \times 10^{10} / \mathrm{cm}^{3}$ Carrier Transport gate1997-ec carrier-transport silicon electronic-devices numerical-answers + – admin 424 views answer comment Share Follow Add Sync Questions Print 0 reply Please log in or register to add a comment.
0 0 votes at eqillibrim n= nd n=2.25*10^15 np=ni*ni p=ni*ni/n=2.25*10^20/2.25*10^15=1* 10*5 p=1* 10*5 rashmi8303 answered Feb 4 rashmi8303 comment Share Follow 0 reply Please log in or register to add a comment.
