SaM ~ Ex. Electro-Neutrality of an n-type Doped Silicon

Index

• Calculation of the Electro-Neutrality
• ~Ex. n-type Doped Silicon

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Calculation of the Electro-Neutrality

Like we have done for intrisic Silicon, when we calculated the Density of Carriers, we will use the same formula, however in this case we have a different formula for the Electron and Hole Density of States Functions, since we are talking about extrinsic or doped Silicon.

So starting with the new “Electron and Hole Density of States Functions”:$N_C^{\prime }(E)=N_C(E)+N_A\cdot \delta (E-E_A)$And:$N_V^{\prime }(E)=N_V(E)+N_D\cdot \delta (E-E_D)$Where:

• $N_A\cdot \delta (E-E_A)$ and $N_D\cdot \delta (E-E_D)$ represents the single extrinsic energy level added by doping the material.
  $\delta (E)$ in this case is the Dirac’s Delta.

So given this changed Density of States, we continue with the calculation of the density carriers, the density of electrons in the Conduction Band will be:$n={\int }_{E_C}^{\infty }{N}_C^{\prime }(E)f_e(E)dE$And, the density of holes in the Valence Band will be:$p={\int }_{-\infty }^{E_V}{N}_V^{\prime }(E)\left(1-f_e(E)\right)dE$Where:

• $\left(1-f_e(E)\right)$: is the probability of finding a hole at a certain energy $E$.

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~Ex.: n-type Doped Silicon

Let’s take a look at the n-type doped Silicon, so: $N_A\text{or}{N}_A^{-}=0$, and:$n=p+N_D^{+}$ The only thing to remember is that when calculating the integral for the dopant, we need to use the “Degeneracy Fermi-Dirac Function” ($f_D(E$): $N_D^{+}={\int }_{\infty }^{\infty }{N}_D\cdot \delta (E-E_D)(1-f_D(E))dE$Where:$f_D=\frac{1}{1+\frac{1}{g_D}\cdot e^{\frac{E-E_F}{kT}}}$And:$1-f_D=\frac{1}{1+g_D\cdot e^{\frac{E-E_F}{kT}}}$Where:

• $g_D$ is the “degeneracy factor”.

So, at the end we will obtain:$n=N_C\cdot e^{-\frac{E_C-E_F}{kT}}$(Same formula as in the intrisic case)

But also:$n=N_V\cdot e^{-\frac{E_F-E_V}{kT}}+N_D^{+}$We can also calculate:$N_D^{+}=\frac{N_D}{1+g_D\cdot e^{\frac{E-E_F}{kT}}}$Where:

• Like in the intrinsic case, $N_C$ and $N_V$ do not depend on $E$, we can think of them as constants depending on the material and temperature.

==I can see $n$ as the number of electron which you have in the conduction band. But also I see $n$ as the sum of the hole that left behind this electron here, which says that this electron here comes either from the valence band or from the donors==.

By doping I transform material which is approximately an insulator at room temperature, into a conductor (even a good conductor).

Remeber that the Mass Action Law needs to be respected, so by doping we have “transformed” some holes in electrons or vice-versa.

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Complete Notes