Consider a bar of N-type silicon that has a cross-sectional area of $A = 100 \mu m * 100 \mu m$ and concentration of free electrons $n_0 =10^{15} cm^{-3}$.

a) Estimate the number of electrons that pass through a cross-sectional plane in only one direction each second. The temperature is $T= 300 K$ and the effective mass of electrons in silicon is $m^* =0.26*m_0$.

b) Express the flow of electrons calculated in part (a), as an electric current in amperes.

Free electrons from Si are assumed as an “electron gas”. From the kinetic theory of gases ($K$ is Boltzmann constant):

$E_k =(3/2)*KT$

$(1/2)m*V^2 =(3/2)*KT$

Also from kinetic theory all 3 space directions are equivalent so that one can write

$V_x^2 =(1/3)*(V^2)$**The mean square speed in one direction is 1/3 of the total mean square speed.**

Therefore for one direction we have

$mV_x^2 =KT$

$V_x =\sqrt{(KT)/m} =\sqrt{(1.38*10^{-23} *300)/(0.26*9.1*10^{-31})} =1.32*10^5 m/s=$

$=1.32*10^7 cm/s$

(above $m=0.26*m_0$)

The above speed can be taken with +/- sign since it was computed from a $\sqrt{..}$. Half of electrons are traveling back and half are traveling forward.

Area $S =100*100 (\mu m^2) =10^4*(10^{-4})^2 cm^2 = 10^{-4} cm^2$

($1 \mu m =10^{-4} cm$)

Thus the number of free electrons passing in one direction in time unit is ($n_0 =10^{15} (1/cm^3)$)

$N/t =(1/2)*n_0*(V_x*S) =(1/2)*10^{15}* 1.32*10^7* 10^{-4} =6.6*10^{17} (1/s)$

The current is

$I = (e N)/t = 1.6*10^{-19}*6.6*10^{17} =0.1056 A$