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Lecture-7
Online-Classes
Corse Title: Solid State Materials and Devices
Course Code: EL-501
Course Teachers: Dr. Sadia Muniza Faraz
Semester: Spring-2020
Offered to: M. Eng. (Electronic Engineering)

Department of Electronic Engineering

NED University of Engineering and Technology Karachi, Pakistan

Chapter-3 Energy Bands and charge Carriers in

Semiconductors

Book: Solid state Electronic Devices

By: Ben G. Streetman & Sanajay
7th edition - (May 11, 2015)
Publisher: Pearson;

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Effect of Temperature on carrier concentration

Exercise
A Si device is to be operated
at 400K and it requires n-
type material
we have two option
Option-1 Doping of Arsenic(As) in Si
with Nd=1.0x1014/cm3
Option-2 Doping of Phosphorous(P) in
Si with Nd=1.0x1016/cm3
Which option is suitable?
Why ?

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Exercise
find the equilibrium electron and hole
concentration for a uniformly doped
sample of Si under following conditions
• (i) Nd=8 x 1015/cm3 T=450K
• (ii)Nd=8 x 1014/cm3 T=650o
Solution:
(i) no  N d  8 1015 / cm3
no  8 1015 / cm3
@ T  450 K
ni  ?( find  value  from  graph)
1000 1000
 2.2 ni  110 / cm
14 3

T 450
ni2
no po  ni2 po  po  ?
no
5

Compensation
• when a semiconductor contains both donors and
acceptors
• donors and acceptors compensate each other
• a p-type semiconductor doped with Na
acceptors can be converted to an n-type
semiconductor by simply adding donors until the
concentration exceeds Na.
• The electron concentration is then given by
no= Nd – Na
• N-type compensated semiconductor
if ( Nd>Na) then no = Nd — Na
• P-type compensated semiconductor
if (Na>Nd) then po = Na — Nd
• Completely compensated semiconductor
(Na=Nd)
• Compensation doping?? 6

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Space Charge Neutrality

• Charge neutrality occurs when all the
charge in a volume adds to zero

Negative charges:
Conduction electrons (density = no)
Ionized acceptor atoms (density = NA)

Positive charges:
Holes (density = po) po - no + ND - NA = 0
Ionized donor atoms (density = ND)

• the sum of the charged carriers and

ions must equal zero po + ND = no +NA
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Carrier Drift
• The process in which charged particles move because of an
electric field is called drift.
• Charged particles within a semiconductor move with an
average velocity proportional to the electric field.
– The proportionality constant is the carrier mobility.
 
Hole velocity
vp   p E
 
Electron velocity vn    n E
qt
Notation: p  *
mp
p  hole mobility (cm2/V·s)
n  electron mobility (cm2/V·s) n 
qt
*
mn 8

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Drift Current
• Drift current is proportional to the carrier
velocity and carrier concentration:
   
vdp   p E vdn   n E
Total current density

J  J p  Jn

J p   q nVdp  q p po E
J n   q nVdn  q n no E

 p  q p po J p   p E  Conductivity
 n  q n no J n   n E  Conductivity

Conductivity and Resistivity

• In a semiconductor, both electrons and holes conduct current:
J p ,drift  qp p E J n ,drift  qn(  n E )
J drift  J p,drift  J n,drift  qp p E  qnn E
J drift  (qp p  qnn ) E  E
• The conductivity of a semiconductor is given by

  qp p  qnn
• The resistivity of a semiconductor is given by
1


10

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Electrical Resistance
I V
+ _

W
t
uniformly doped semiconductor

V L
Resistance R 
I Wt
where  is the resistivity
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Resistivity Dependence on Doping

For n-type material:
1

qno  n

For p-type material:

p-type 1

qpo  p
n-type
Note: This plot (for Si) does
not apply to compensated
material (doped with both
acceptors and donors).

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Problem
The intrinsic resistivity of Ge at 300K is 47 Ω.cm.
(a) What is the intrinsic carrier concentration, given that µn = 3900 cm2
/V.sec and µp = 1900 cm2 /V.sec.
(b) If 5 x 1017 /cm3 donor impurities are added to this sample then find EF -
Ei.

The resistivity is given by

1
 Find  conductivity ( )  ?

The conductivity is given by

  qpo  p  qno  n   qni  p  qni  n

For intrinsic materials   qni (  p   n )
po  no  ni
ni  ? no  ni e( E F  Ei ) / kT

  qpo  p  qno  n
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Problem
An unknown semiconductor was found to have the concentrations of 9 x 1016/
cm3 and 7 x 109/ cm3 for donors and acceptors respectively. The Fermi level is
such that
EC –EF = EF –Ei = 0.275 eV. At room temperature find
1. Intrinsic carrier concentration
2. Band gap (Eg)
3. Intrinsic resistivity

no  ni e( EF  Ei ) / kT
N d  9 10 / cm 16 3

N a  7 109 / cm3 ni  ??
E g  ??
no  N d  N a
  qni ( p  n )
no  ??
given   n  2  p  1400cm 2 / V . sec

1
   ???

14

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Effect of Temperature on Mobility

• Lattice Scattering(phonon Scattering)
• Impurity Scattering

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Mobility Dependence on Doping

Carrier mobilities in Si at 300K

16

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Mobility Dependence on Temperature

3
 impurityT 2

3
 latticeT 2

17

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Example: Resistance Calculation

What is the resistivity of a Si sample doped with 1016/cm3 Boron?
Answer:
1

qn n  qp p
1

qp p


 (1.6 10 19 )(1016 )(450) 1

 1.4   cm

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Example: Dopant Compensation

Consider the same Si sample doped with 1016/cm3 Boron, and
additionally doped with 1017/cm3 Arsenic. What is its resistivity?
Answer:
1

qn n  qp p

1

1.6 10 19 [(1017 )(900)  1016 450]
 0.066  cm

Try  no  N d  N a
  ??

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High field effect

vd = E
• strong electric fields - electron velocity
is no longer proportional to the field
• The heating of free carriers at high
electric fields results in a saturation of
the drift velocity
• Decrease in mobility - originating from
various reasons such as scattering
mechanisms
• the increase in kinetic energy of the
electrons is statistically interpreted as
a raising of the electrons' temperature,
so they are termed 'hot electrons' Hot
carrier effect
• Silicon thermal velocity Vth=1x107
cm/sec
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,

The End

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