Selected Problems Tutorial # 3
1.
The concentration of conduction electrons in pure
Si at 300K is 1016 m-3. Assume that by doping the
lattice with phosphorus, you want to increase this
number by a factor of a million (106). What fraction
of the silicon atoms must you replace with
phosphorus atoms?
(At room temperature, the thermal agitation is
effective enough so that essentially every phosphorus
atom donates its “extra” electron to the conduction
band).
Answer:
concentration of phosphorus atoms
(1016 m-3) x (106) = 1022 m-3 = np
concentration of silicon atoms in a pure silicon lattice
NA
nSi 
A
NA = Avogado’s number 6.02 x 1023 mol-1
A = molar mass of silicon 28.1g/mol
 = density of silicon = 2330 kg/m3
(6.02  1023 mol 1 )(2330kg / m3 )
nSi 
0.0281kg / mol
= 5 x 1028 m-3
The ratio of the concentration of Si atoms and the
concentration of P atoms is the fraction of Si atoms
that need to be replaced with P atoms
nSi 5  10 m
6

 5  10
22
-3
nP
10 m
28
-3
This says that if only one silicon atom in five
million is replaced by a phosphorus atom, the
number of electrons in the conduction band will
be increased by a factor of 106.
2.
A Si sample contains 1016 cm-3 In acceptor atoms.
The In acceptor level is 0.16 eV above Ev, and EF
is 0.26 eV above Ev at 300 K. How many (cm-3) In
atoms are unionized (i.e. neutral)?
Solution
Na = 1016 cm-3
Not all of them are ionized. The concentration of ionized atoms is
given by Naf(Ea), where f(Ea) is the probability of electrons being
found in energy state Ea.
f(E a ) 
1
1 e
(E a  E F )/ kT

1
1 e
(0.16-0.26)/0.0259
 0.9793
The concentration of unionized atoms is
1016 – 0.9793 x 1016 = 2.1 x 1014 cm-3