HCMUT – Falculty of Electrical and Electronics Engineering
Course: Solid State Electronic Devices (EE2411) – Instructor: Hồ Trung Mỹ
Solved problems for Chapter 4 – Excess carries in Semiconductors
Notes:
· For an electron: m = 9.1 x 10 –31 kg and q = 1.6 x 10–19 Coulomb, h = 6.62 x 10–34 Joule‐sec.
· For silicon at T = 300 K, ni = 1.5 × 1010 cm–3. The Boltzmann constant kB = k = 8.61 × 10−5 eV/K.
Silicon bandgap energy Eg = 1.12 eV.
· For problems about carrier concentrations, we use the aproximation X >> Y when X/Y ³ 100.
· Commonly accepted values of ni at T = 300 K
Ø Silicon (Si) ni = 1.5 x 1010 cm–3
Ø Gallium arsenide (GaAs) ni = 1.8 x 106 cm–3
Ø Germanium (Ge) ni = 2.4 x 1013 cm–3
1. Excess electrons are generated in a semiconductor to a concentration of dn(0) = 1016 cm–3. The excess
carrier lifetime in the semiconductor is 5 x 10–6 s. The source generating the excess carriers is switched off
at t = 0. Calculate the excess electron concentration at t = 1 ms.
Solution
The excess electron concentration at any time t is given by
dn(t) = dn(0) exp(–t/tn)
Putting the given values of dn(0) and tn in the above equation, we get
dn(1 ms) = 1016 x exp(–1/5) = 8.19 x 1015 cm–3
2. For the situation presented in solved numerical 1., calculate the recombination rate of excess electrons
for t = 5 ms.
Solution
The excess electron concentration dn(t) at t = 5 ms can be evaluated using the Eqn,
dn(t) = dn(0) exp(–t/tn)
(1)
Putting the given values in Eqn (1), we get
dn(5 ms) = 1016 e–1 = 3.68 x 1015
(2)
The recombination rate Rn for excess electrons is given by
(3)
Rn = dn(t)/tn
Putting dn(5 ms) from Eqn (2) and the given value of tn in Eqn (3) yields
Rn = 3.68 x 1015 / 5x10–6 = 0.74 x 1021 cm–3 s–1
3. An intrinsic Si sample is doped with donors from one side such that ND = N0 exp(–ax).
a) Find an expression for the built-in field E(x) at equilibrium over the range for which Nd >> ni
b) Evaluate E(x) when a = 1 mm–1 at T = 300 K.
Solution
a) Find an expression for the built-in field E(x) at equilibrium
dn
At equilibrium Jn = 0 and J n = qnmn E (x) + qDn
dx
Dn 1 dn
kT 1 dn
=(4.1)
E (x) = q n dx
mn n dx
where kT/q = 0.026 V @ 300 K, and
dn d (N0 exp(- ax))
=
= - aN 0 exp(- ax) = - an
dx
dx
Therefore
kT
q
b) E(x) = 104 cm–1 x 0.026 V = 260 V/cm
E (x) = a
4. A particular sample of an n-type semiconductor at T = 300 K has a carrier concentration n0 = 5 x 1015
cm–3, ni = 1010 cm–3, and p0 = 2 x 104 cm–3. A source creates excess carriers in the semiconductor such that
SSED – Solved problems For Chapter 4 – page 1/3
dn = dp = 5 x 1013 cm–3. Calculate the position of the Fermi level for thermal equilibrium and positions of
quasi-Fermi levels for electrons and holes at non-equilibrium conditions with respect to the intrinsic Fermi
level. (Note: EFi º Ei)
Solution
The position of the Fermi level at thermal equilibrium is given by
(1)
Putting kT = 0.026 eV and the given values of n0 and ni in Eqn (1) leads to
This yields
EF – EFi = 0.3412 eV
The quasi-Fermi level for electrons in non-equilibrium is given by
(2)
where dn = 5 x 1013 cm–3. Thus, Eqn (2) results in
which implies
EFn – EFi = 0.3414 eV
One can notice that EFn is not much different from EF. This is because dn does not change the total
electron concentration significantly.
The quasi-Fermi level for holes in non-equilibrium is given by
(3)
with dp = 5 x 1013 cm–3
Equation (3) thus yields
which implies
EFi – EFp = 0.2214 eV
EFp is thus signifi cantly different from EFi.
5. A Si sample with 1015 cm–3 donors is uniformly optically excited at room temperature such that 2x1019
cm–3 electron-hole pairs are generated per second. Find the separation of the quasi-Fermi levels and the
change of conductivity upon shining the light. Electron and hole lifetimes are both 12 ms. Dp = 14 cm2/s.
Solution
The light induced electron-hole pair concentration is determined by
dn = dp = gopt = 2x1019 x 12x10–6 = 2.4 x 1014
Þ n = n0 + dn = 1015 + 2.4 x 1014 = 1.24 x 1015
Þ p = p0 + dp = (1.5 x 1010)2/1015 + 2.4 x 1014 » 2.4 x 1014
Ø Separation of the quasi-Fermi levels
EFn – EFp = kT x ln(np/ni2) = 0.026 x ln(1.24x1015 x 2.4x1014 / (1.5 x 1010)2) = 0.5461 eV
Ø The change of conductivity upon shining the light
Ds = q(mndn + mpdp) = qdn(mn + mp)
where mp = Dp/VT = 14/0.026 » 539 cm2/Vs and mn = 1350 cm2/Vs
Ds = 1.6 x10–19 x 2.4x1014 x (1350 + 539) = 0.0725 S/cm
6. [Neamen] Determine the induced electric field in a semiconductor in thermal equilibrium, given a
linear variation in doping concentration.
Assume that the donor concentration in an n-type semiconductor at T = 300 K is given by
SSED – Solved problems For Chapter 4 – page 2/3
Nd(x) = 1016 – 1019 x (cm–3)
where x is given in cm and ranges between 0 £ x £ 1 mm
Determine the induced electric field in the semiconductor.
Solution
Taking the derivative of the donor concentration, we have
dNd(x)/dx = –1019 (cm–4)
The electric field is given by Equation (4.1), so we have
0.026 ´1019
E (x) = 16
V/cm
10 - 1019 x
At x = 0, for example, we find
E(0) = 26 V/cm
7. [Neamen] Assume the donor concentration in an n-type semiconductor at T = 300 K is given by
Nd(x) = 1016 exp(–x/L) where L = 2 x 10–2 cm.
Determine the induced electric field in the semiconductor at (a) x = 0 and (b) x = 10–4 cm.
Solution
Taking the derivative of the donor concentration, we have
dNd(x)/dx = –1016 /L x exp(–x/L) (cm–4)
The electric field is given by Equation (4.1), so we have
kT 1 dn kT 1
E (x) = =
q n dx q L
E(x) = 0.026 /(2 x 10–2) = 1.3 V/cm = const
Therefore, E = 1.3 V/cm at x = 0 and x = 10–4 cm.
SSED – Solved problems For Chapter 4 – page 3/3