Silicon and Germanium conductivities

Q1) Find the current flows through a silicon bar of $20 \mu m$ length having a cross-section of $10 \mu m^2$ and having free-electron and hole-concentrations of $10^{18} cm^{-3}$ and $10^5 cm^{-3}$, respectively, when a $5 V$ is applied across the silicon-bar. Use $\mu_n =1200 cm^2/(V*s)$ and $\mu_p= 400 cm^2/(V*s)$

Q2) The intrinsic concentration of  charge carriers in a semiconductor is $n_i=4*10^{-19} m^3$. Being given the mobilities $\mu_n =0.38 m^2/(V*s)$ and $\mu_p =0.18 m^2/(V*s)$ find the concentration of acceptors needed to be introduced so that the electronic conductivity will be equal to the hole conductivity.



$\sigma = (1/\rho) = e*n*\mu_n +e*p*\mu_p =1.6*10^{-19}* (10^{18}*1200 +10^5*400) =$

$=192 [1/(\Omega*cm)]$

electrical resistance is

$\rho =5.2*10^{-3} (\Omega*cm)$

For Silicon bar the length is $L =20 \mu m =20*10^{-4} cm =2*10^{-3} cm$

and $S =10 \mu m^2 =10*(10^{-4})^2 cm^2 =10^{-7} (cm^2)$

Resistance is

$R = \rho* (L/S) =5.2*10^{-3} * (2*10^{-3})/10^{-7} = 104 \Omega$

Current is

$I =U/R =5/104 =0.048 A =48 mA$


We know the dependence between the electron and hole concentrations after doping with acceptors $N_a$:

$n*p =n_i^2$     and $p =N_a +n$
The electronic and hole conductivities are therefore

$\sigma_n = \frac{e\mu_n*N_a}{2}\left[\sqrt{1+\left ( \frac{2n_i}{N_a} \right )^2}-1\right]$

$\sigma_p = \frac{e\mu_p*N_a}{2}\left[\sqrt{1+\left ( \frac{2n_i}{N_a} \right )^2}+1\right]$

But since we have the condition of equal mobilities $\sigma_n =\sigma_p$  then

$N_a =n_i*\left(\sqrt{\frac{\mu_n}{\mu_p}}-\sqrt{\frac{\mu_p}{\mu_n}}\right)=3.06*10^{19} m^{-3}$