Chapter 2: Carrier and Transport Transport phenomena Questions provide us the understanding needed to design semiconductor devices If
there is an electric field present in the semiconductor, semiconductor, how do the electrons and holes move?
If there is a concentration gradient in the electron or hole density, density, how do the carriers respond?
Do
electrons in the conduction band fall down into the valence band and recombine with holes?
Is
it possible for electrons in the valence band to jump up into the conduction band? What cause such processes?
1.
CARRIER DRIFT
Objective: An electric field applied to a semiconductor will produce a force on electrons and holes so that they will experience a net acceleration and net movement, provided there are available energy states in the conduction and valence bands. This net movement of charge due to an electric field is called drift . The net drift of charge gives rise to a drift current . What
is drift velocity? ( v d )
Is the velocity component that arises when an electric field, ε is applied to a semiconductor semiconductor.. This electric electric field causes each electron to experience a force - qε due to the field and each electron will be accelerated along the field (in the opposite direction). What
is mobility? ( u )
The mobility is an important parameter of the semiconductor since it describes how well a particle will move due to an electric field.
1. CARR RRIE IER R DRIFT IFT
According to quantum mechanics, electron have a free electron-like behavior and no scattering occurs. In real semiconductors, due to the imperfections, shown in the table, the electrons scatter, which affects their transport ( , , ) properties. Important Sources of Scattering in Semiconductors
Ionized impurities
Due to dopants in the semiconductors
Phonons Alloy Interface roughness Chemical impurities
Due to lattice vibrations at finite temperatures Random potential fluctuations Important in heterostructure Due to unintentional impurities
Results in the Momentum & Energy of electrons will gradually lose coherence with the initial state values. The average time it takes to lose coherence or memory of the initial state properties is called (Mathieson‟s ( Mathieson‟s rule).
1
tot .sc
1
sc1
1
sc 2
...
sc = Scattering time or Mean time between collisions for an electron or Relaxation time or Delay time 3
= d = ave.=
drift velocity = ave. gain velocity (either d n or d p ) e = electron
a) Low Low Elec Electr tric ic Fie Field lds s ( d = immediately before end of collision, after collision the d = 0)
vd
e.F . SC
F = = Electric Field sc = scattering time m* = effective mass (either m n * or m p *) J = Current density = drift current density n = density of charge carrier (either n n = n or n p = p ) = conductivity of material = resistivity of material = 1/ = mobility effects (factor) (either n or p )
m*
J n.e.vd
n.e 2 . SC .F m*
From Ohm‟s Law, J = .F 2
n.e . SC m*
From def. of mobility, - d = - .F (the e move in a direction opposite to the electric field while the holes move in the same direction)
For general :
e. SC m*
For e- :
n
e. .SC mn *
For holes :
p
e. .SC m p *
4
If both e- and holes are present
nn .e. n n p .e. p e(nn . n n p . p ) From Ohm‟s Law, J = .F
J e(nn . n n p . p ).F Example 1 The mobility of e- in pure Si at 300 K is 1500 cm 2/ V.s. Calculate the relaxation time. Given: m* = 0.26 m0
Ans: The time for pure Si:
SC
m *. e
(0.26 x0.91x1030 kg ).(1500 x10 4 m2 / V . s) 1.6 x1019 C
2.2 x1013 s
5
Example 2 The mobility of e- in pure GaAs at 300 K is 8500 cm2/ V.s. Calculate the relaxation time. If the GaAs sample is doped at Nd = 1017 cm-3, the mobility decrease to 5000 cm2/ V.s. Calculate the relaxation time due to ionized impurity scattering. Given: Donor (n-type) doped = m* = 0.067 m 0
(Note:the mobility, decreases with the increase in temp. in order to ionized the dopants because as the temp. rises the atoms in the crystal vibrate with greater amplitude. In other words, the electrons scatter from the dopants from these vibrations and mobility decreases) Ans: The time for pure GaAs 1 SC
m *. e
(0.067 x0.91x1030 kg ).(8500 x10 4 m 2 / V .s) 1.6 x10 19 C
3.24 x1013 s
The time for ionized impurity in GaAs (pure + impurity) 2 SC
m *. e
(0.067 x0.91x10
Mathieson‟s rule:
30
4
1.6 x10
1 2 SC
1 1 SC
2
kg).(5000 x10 m / V .s)
1 imp SC
19
C
1.9 x1013 s
imp SC 4.6x10 13 s 6
Example 3 Consider 2 semiconductor samples, Si and GaAs. Both materials are doped n-type at Nd = 1017 cm-3. Assume 50% of the donors are ionized at 300 K. Calculate the conductivity of the samples. Compare this conductivity to the conductivity of undoped samples. n i = p i = n = p = 1.5 x 1010 cm-3 Given:Pure or undoped density of state (n) for Si Pure or undoped density of state (n) for GaAs n i = p i = n = p = 1.84 x 106 cm-3 = 1000 cm2/ V.s n (Si) = 350 cm2/ V.s p (Si) = 8000 cm 2/ V.s n (GaAs) = 400 cm2/ V.s p (GaAs) Ans: Conductivity for undoped Si:
undoped e( ni .n ni . p )
(1.6 x1019 C )(1.5x1010 cm3 ){(1000 350)}cm 2 / V .s 6 1 x cm 3.24 10 ( ) Conductivity for undoped GaAs:
undoped (1.6 x1019 C)(1.84 x106 cm3 ){(8000 400)}cm2 / V .s
2 47 x109 (cm) 1
7
Cont. Example 3
ndoped (100%) nn N 1017 cm3
For Si & GaAs
ndoped (50%) 1017 x50% 5 x1016 cm3
For Si & GaAs
pdoped pdoped
ni2 ndoped ni2 ndoped
10 2
(1.5 x10 ) 16
5 x10
(1.84 x106 ) 2 5 x1016
4.5 x103 cm3 For Si (very small compared to ndoped ) 6.77 x105 cm3
For GaAs
(very small compared to ndoped )
Conductivity for doped Si:
doped e( nn .n n p . p )
Almost zero
(1.6 x1019 C){(5x1016 )(1000cm1 / V .s) (4.5x103 )(350cm1 / V .s)} 8(cm) 1 Conductivity for doped GaAs:
doped e( nn . n n p . p )
Almost zero
(1.6 x1019 C ){(5 x1016 )(8000cm1 / V .s) (6.77 x10 5 )(400 cm1 / V .s) cm1} 64(cm) 1 8
Example 4 Consider a Si semiconductor at T = 300 K with an impurity doping concentration of N d = 1016 cm-3 and Na = 0. Calculate the drift current density, J, for an applied field, = 35 V/cm. Given: n (Si) = 1350 cm 2/ V.s p (Si) = 480 cm2/ V.s Ans: Since Nd > Na , the semiconductor is n-type at room temperature, we can assume complete ionization: The n = nn ~ Nd = 1016 cm-3 and p = np = 0 2
p
ni
n
10 2
(1.5x10 ) 16
10
2.25x104 cm3
J e( nn . n n p . p ).F e( nn . n ) F
(1.6 x10
19
)(10 )(1350)(35) 75.6 A / cm 16
2 9
Example 5 (Final Sem 1 09/10) An n-type Silicon sample with a conductivity of 0.1 ( cm)-1 at 300 K. Given: a) Calculate the electron and hole carrier density of the material. [4 marks] b) Calculate the Fermi level for n-type and p-type material with the same conductivity using Joyce-Dixon approximation. [4 marks] c) Calculate the intrinsic Fermi level for n-type and p-type of the material.
[4 Marks]
d) Sketch the flat band diagram, indicating clearly the positions of Ec, Ev, EFn, EFp and Ei.
[4 marks]
e) How much is the energy gap has been shifted if compared to the energy gap of 1.1 eV.
[2 marks]
f) Describe why the mobility carrier in an extrinsic semiconductor decreases with the increases of temperature. [2 marks]
10
Solutions: a) n q n n p q p p
b)
n
p
n q n
p q p
0.1(cm) 1 (1.6 x1019 C )(1000cm 2 / Vs) 0.1(cm) 1 (1.6 x1019 C )(300cm2 / Vs)
6.25x1014 cm3 2.08x1015 cm3
n 1 n EFn EC k BT ln N 8 N C C (6.25 x1014 ) 1 (6.25 x1014 ) EFn E C (0.026) ln 19 19 8 (2.78 x10 ) (2.78 x10 ) 6 EFn EC (0.026) ( 10.70) (7.95 x10 ) EFn EC 0.278eV
EFp EFp EFp E
p 1 p EV k BT ln N 8 N V V (2.08 x1015 ) 1 (2.08 x1015 ) E V (0.026) ln 18 18 (9.84 10 ) x 8 (9.84 x10 ) EV (0.026) ( 8.46) (7.47 x105 ) E 0.22eV
11
d) EFn E i exp ni k T B n E E i ln Fn k BT ni n 6.25x1014 EFn Ei k BT ln (0.026) ln 0.276eV 10 1.5x10 ni
c) n
EC 0.278 EFn 0.276 Ei
0.3078 EFp 0.22 EV
E E Fp exp i ni kBT p Ei E Fp ln kBT ni p 2.08x1015 Ei EFp kBT ln (0.026) ln 0.3078eV 10 1.5 10 n x i p
e) From the diagram (d),
EgapBefore = 1.1 eV EgapAfter = 0.276 + 0.3078 = 0.5838 eV E = 1.1 - 0.5838 = 0.5162
12
f) At high temperatures, lattice scattering dominates as the thermal vibrations of lattice atoms increase with T hence increasing the probability of charge carrier-lattice collisions. Hence, the mobility decreases as the sample is heated. or The mobility, decreases with the increase in temperature (in order to ionized the dopants) because as the temperature rises the atoms in the crystal vibrate with greater amplitude. In other words, the electrons scatter from the dopants from these vibrations and mobility decreases.
13
b) Very High Electric Field Transport:Breakdown Phenomena (When > 100 kV/cm, the semiconductor suffers a “breakdown” in which current has “runaway” behavior. The breakdown occurs due to carrier multiplication means the number of electrons and holes that can participate in current fl ow increase. (The total number of electrons conserved))
i) Impact ionization or Avalanche Breakdown ♣
In normal case during transportation, the e-/holes remain in the same band.
♣ At very high this does not hold true. ♣ An e- which is „very hot‟ scatters with an e- in the valence band via coulombic interaction and knocks it into the conduction band as shown in Figure . ♣ Thus the initial e - should have energy slightly larger than the bandgap. ♣ In the final state we have 2 e - in the conduction band 1 hole in the valence band. ♣ Thus the number of current carrying charges have multiplied. The process is called “Avalanching” ♣ The same process could happen to “hot holes”.
Conduction band
-
Initial state has 1 e-
Final state has 2 e+ 1 hole
+ Valence band
Avalanche process 14
Once avalanching starts,
dI ( z ) dz N ( x)
imp I I ( z ) I (O )
exp( imp x)
•I = current • imp = Average rate of ionization per unit distance (coefficient for e-) imp = Average rate of ionization per unit distance (coefficients for hole) N = number of times an initial electron will suffer impact ionization after travelling a distance x
Energy band diagrams under junctionbreakdown conditions-Avalanche multiplication. 15
ii) Band-to-band Tunneling or Zener Tunneling When a strong happens, the e- in the valence band can tunnel into an „unoccupied state‟ in the conduction band or vice versa. As the e- tunnels, the tunneling probability is:
T exp(
4 2m* E g3/ 2 3e F
Electrons in conduction band
) Available empty states (holes) in valence band
F = Electric Field in the semiconductor Example 1 Calculate the band to band tunneling probabilty in GaAs and InAs at an applied Given: m* (GaAs) = 0.065 m 0 m* (InAs) = 0.02 m0 Eg (GaAs) =1.5 eV Eg (InAs) = 0.4eV
Energy band diagrams under junction-breakdown conditions- Tunneling effect
= 2 x 105 v/m.
T GaAs exp(
4 x 2 x0.065x 0.91x1030 kg ) (1.5x1.6x1019 J )3/ 2 3 x(1.6 x10
19
C )(1.05x10
34
7
Js )(2x10 V / m)
)0
T InAs 3.7 x10 6 (means Zener Tunneling is important when ~2 x 105 V/m) 16
b).
HIGH FIELD EFFECTS (Proven thru graph)
• High field transport means Carrier velocity tends to saturate and mobility = v.F starts to decrease (The mobility starts to decrease and becomes independent of the electric field)