CHAPTER 1 ELECTRONICS AND SEMICONDUCTORS
Chapter Outline
1.1 Signals
1.2 Frequency Spectrum of Signals
1.3 Analog and Digital Signals
1.4 Amplifiers
1.5 Circuit Models for Amplifiers
1.6 Frequency Response of Amplifiers
1.7 Intrinsic Semiconductors
1.8 Doped Semiconductors
1.9 Current Flow in Semiconductors
1.10 The pn Junction with Open‐Circuit Terminals
1.11 The pn Junction with Applied Voltage
1.12 Capacitive Effects in the pn Junction
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1.1 Signals
Signal processing
Signals can be of a variety of forms in order to carry information from the physical world
It is most convenient to process signals by electronic system, therefore, the signals are first converted into an electric form (voltage or current) by transducers
Input signal
(voice, speed, pressure, etc.)
Signal
Processor
Transducer
v(t)
Electrical Signals
v(t)
Transducer
Output signal
(voice, speed, pressure, etc.)
Electrical Signals
t
t
Signal sources
Thevenin form: (voltage source vs + series resistance Rs)
 Presenting the signal by a voltage form
 Is preferred when Rs is low (Rs can be neglected)
Norton form: (current source is + shunt resistance Rs)
 Presenting the signal by a current form
 Is preferred when Rs is high (Rs can be neglected)
In electronics systems, the signal is taken from one of the two forms for analysis
Two forms are interchangeable with vs(t) = is(t)  Rs
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1.2 Frequency Spectrum of Signals
Sinusoidal signal
A sinusoidal signal is given as: va(t) = |Va|sin(at + a )
Characterized by its amplitude (|Va|), frequency (a) and phase (a )
va(t)
Va
Frequency domain representation
Any time‐domain signal can be expressed by its frequency spectrum
 Periodic signal  Fourier series
 Non‐periodic signal  Fourier transform
t
a
T
Periodic signal
The fundamental frequency of periodic signals is defined as 0 = 2/T.
A periodic signal can be expressed as the sum of sinusoids at harmonic frequencies (n0)
Example: a square‐wave with period T
v(t ) 
Time‐domain representation
4V
1
1
(sin 0t  sin 30t  sin 50t  ...)

3
5
Frequency‐domain representation
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v(t ) 
4V
1
1
(sin 0t  sin 30t  sin 50t  ...)

3
5
Non‐periodic signal
The Fourier transform is applied to a non‐periodic function of time
The spectrum of a non‐periodic signal contains all possible frequencies
Time‐domain representation
Frequency‐domain representation
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1.3 Analog and Digital Signals
Signal classification
Analog signal: signal can take on any value
Digital signal: can only take on finite quantization levels
Continuous‐time signal: defined at any time instant
Discrete‐time signal: defined only at the sampling instants
Sampling: the amplitude is measured at equal time intervals
Quantization: represent the samples by finite values
Quantization error:  Difference between sampled value and quantized value
 Can be reduced by increasing the quantization levels
Continuous‐time analog signal
v(t)
t
Discrete‐time analog signal
Data conversion
t
Analog‐to‐digital converter (ADC):
A/D
converter
...

Analog
vA
input

b0
b1 Digital
output
bN‐1
b
Digital b0
1
input
bN‐1
...
Digital‐to‐analog converter (DAC):
D/A
converter

Analog
vD
output

Sampling
Digital signal
Quantization
3
2
1
0
vD  b0 20  b1 21  ....bN 2 N 1
3,3,3,2,3,3…
t
Quantization error
t
vA = vD + quantization error
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1.4 Amplifiers
Gain of amplifiers
Voltage gain Av  vO / vI
Current gain Ai  iO / iI
Power gain Ap  vO iO / vI iI
Amplifier gains are dimensionless (ratio of similarly dimensioned quantities)
Voltage and current gain can be positive or negative depending on the polarity of the voltage and the current
The gain is frequently expressed in decibels:
 Voltage gain Av (dB)  20 log | Av |
 Current gain Ai (dB)  20 log | Ai |
 Power gain Ap (dB)  10 log | Ap |
 Gain > 0 dB  | A | > 1 (amplification)  Gain < 0 dB  | A | < 1 (attenuation)
 The polarity of the voltage and current is not shown in dB expression
Amplifier power supplies
Amplifiers require dc power supplies
Pdc = VCC ICC + VEE IEE
Pdc + PI = PL + Pdissipated
  (efficiency) = (PL / Pdc )100%
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Transfer characteristics of linear amplifier
The plot of output response vs. input  transfer characteristics
For linear amplifier, the transfer characteristic is a straight line passing the origin with slope = Av
It is desirable to have linear amplifier characteristics for most of the applications
Output waveform is an enlarged copy of the input: vO(t) = AvvI(t)
No higher power terms of vI at the output
Amplifier saturation
Practically, the amplifier transfer characteristic remains linear over only a limited range of input and output voltages
The amplifier can be used as a linear amplifier for input swing:
L/Av  vI  L+/Av  vO = AvvI
For input larger than the swing limitation, the output waveform
will be truncated, resulting in nonlinear distortion
The nonlinearity properties can be expressed as:
vO = a0 + a1vI + a2vI2 + a3vI3 …..
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Nonlinear transfer characteristics and biasing
In practical amplifiers, the transfer characteristics may exhibit nonlinearities of various magnitude
The nonlinearity characteristics will result in signal distortion during amplification
In order to use the circuit as a linear amplifier:
 Use dc bias to operate the circuit near the middle of the transfer curve  quiescent point
 Superimpose the time‐varying (ac) signal on the dc bias at the input
 Be sure that the signal swing is sufficiently small for good linear approximation
 The time‐varying (ac) components at the output is the desired output signal
Slope = Av
vO
vO (t)
Q
VO
VI
vI
vI (t )  VI  vi (t )
vO (t )  VO  vo (t )
vI (t)
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vo (t )  Av vi (t )
dv
Av  O |at Q
dvI
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Symbol convention:
dc quantities: IC, VD
Incremental (ac) quantities: ic(t), vd(t)
Total instantaneous (ac + dc) quantities: iC(t), vD(t)
iC(t) = IC + ic(t)
vD(t) = VD + vd(t)
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1.5 Circuit Models for Amplifiers
Concept of equivalent circuit
Practical amplifier circuit could be rather complex
Use a simplified model to represent the properties and behavior of the amplifier
The analysis results do not change by replacing the original circuit with the equivalent circuit
Voltage amplifiers
A simplified two‐port model is widely used for unilateral voltage amplifiers
Voltage Amplifier
The model is composed of three components:
 Input resistance (Ri): the resistance by looking into the input port
 Output resistance (Ro): the resistance by looking into the output port
 Open‐circuit voltage gain (Avo): the voltage gain (vo/vi) with output open‐circuit
Circuit analysis with signal source and load:
vo
RL
 Avo
vi
RL  Ro
 Overall gain: Gv  vo  Ri Avo RL
vs Ri  Rs
RL  Ro
 Voltage gain: Av 
 Ideal voltage amplifier: Ri =  and Ro = 0
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Circuit parameters in the amplifier model
The model can be used to replace any unilateral amplifier by proper circuit parameters
The parameters can be obtained by circuit analysis or measurement
 Analysis (measurement) of the input resistance:
The resistance by looking into the input port vx
(find ix for a given vx or find vx for a given ix)
 Analysis (measurement) of the output resistance:
Set vi = 0 by input short
The resistance by looking into the output port
(find ix for a given vx or find vx for a given ix)
 Analysis (measurement) of the open‐circuit voltage gain:
Given vx at input
Find open‐circuit output voltage vo
vx
vo is divided by vx
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ix
Ri  vx/ix
ix
vx
Ro  vx/ix
vo
Avo  vo/vx
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Cascade amplifier
Multiple stages of amplifiers may be cascaded to meet the application requirement
The analysis can be performed by replacing each stage with the voltage amplifier model
Buffer amplifier
Impedance mismatch may result in a reduced voltage swing at the load
Buffer amplifier can be used to alleviate the problem
 The gain of the buffer amplifier can be low (~1)
 The buffer amplifier has high input resistance and low output resistance
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Amplifier types
Voltage amplifier: gain of interest is defined by vo/vi (V/V)
Current amplifier: gain of interest is defined by io/ii (A/A)
Transconductane amplifier: gain of interest is defined by io/vi (‐1)
Transimpedance amplifier: gain of interest is defined by vo/ii ()
Amplifier models
Voltage Amplifier
Transconductance Amplifier
Current Amplifier
Transimpedance Amplifier
Unilateral models
The amplifier models considered are unilateral; that is, signal flow only from input to output
The model is simple and easy to use such that analysis can be simplified
Not all amplifiers are unilateral and more complicated models may be needed for the analysis
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Circuit analysis for amplifiers
Voltage Amplifier
Transconductance Amplifier
vo = Avovi RL / (RL+Ro)
vo / vs = Avo[Ri /(Ri+Rs)][RL /(RL+Ro)]
For ideal case (Ri → , Ro → 0): vo / vs = Avo
Current Amplifier
io = Gmvi Ro / (RL+Ro)
io / vs = Gm[Ri /(Ri+Rs)][Ro /(RL+Ro)]
For ideal case (Ri → , Ro → ): io / vs = Gm
Transimpdeance Amplifier
io = Aisii Ro / (RL+Ro)
io / is = AisRsRo / [(RL+Ro)(Ri+Rs)]
For ideal case (Ri → 0, Ro → ): io / is = Ais
vo = Rmii RL / (RL+Ro)
vo / is = RmRsRL / [(RL+Ro)(Ri+Rs)]
For ideal case (Ri → 0, Ro → 0): vo / is = Rm
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Exercise 1.12 (Textbook)
Exercise 1.13 (Textbook)
Exercise 1.14 (Textbook)
Exercise 1: For a voltage amplifier with Ri = 100 k, Ro = 10 k and Avo = 20, find its equivalent models as current, transconductance and transimpedance amplifiers.
Exercise 2: Consider two amplifier stages are cascaded. The model of the first stage is given by Ri = 1 M, Ro = 10 k and Avo = 20, while the model of the second stage is given by Ri = 100 k, Ro = 10  and Avo = ‐2.
(1) Find the voltage amplifier model for the cascade amplifier.
(2) For Rs = 100 k and RL = 100 , find Av and Gv.
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1.6 Frequency Response of Amplifiers
Concept of frequency response
The input signal to an amplifier can be expressed as the sum of sinusoidal signals
Frequency response: the characteristics of an amplifier in terms of its response to input sinusoidals
of different frequencies
Measuring the amplifier frequency response
Applying a sinusoidal signal to a linear amplifier, the output is a sinusoidal at the same frequency
The output sinusoidal will in general have a different amplitude and a shifted phase Transfer function T() is defined as a function of frequency to evaluate the frequency response
 Magnitude of T() is the voltage gain of the amplifier: |T()| = Vo / Vi
 Phase of T() is the phase shift between input and output signals: T() = 
Amplifier bandwidth
The bandwidth is defined within 3dB from the flat gain
For signal containing components outside the bandwidth, the output waveform will be distorted
Flat gain
3dB
vo(t)=Vo sin(t+ +)
vi(t)=Vi sin(t+ )
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Frequency‐dependent components in amplifiers
The frequency response of amplifiers is mainly due to frequency‐dependent components
The most widely used components are capacitors and inductors
Capacitors
d
Current‐voltage relation: iC (t )  C vC (t )
dt
For sinusoidal signals: vC (t )  V0 cos(t   )  iC (t )  CV0 sin(t   )
The ratio of voltage amplitude and current amplitude is proportional to 1/C, and is considered a frequency‐dependent impedance
Phasor analysis: vC (t )  V0 cos(t   )  VC  V0e j
iC (t )  CV0 sin(t   )  I C  CV0e j (  / 2)
Z C  VC / I C  1 / jC
Inductors
d
Current‐voltage relation: vL (t )  L iL (t )
dt
For sinusoidal signals: iL (t )  I 0 cos(t   )  vL (t )  CI 0 sin(t   )
The ratio of voltage amplitude and current amplitude is proportional to L, and is considered a frequency‐dependent impedance
Phasor analysis: iL (t )  I 0 cos(t   )  I L  I 0e j
vL (t )  LI 0 sin(t   )  VL  LI 0e j (  / 2)
Z L  VL / I L  jL
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Deriving the transfer function of amplifiers
Complex frequency (s)
 Treat an inductance L as an impedance sL
 Treat a capacitance C as an impedance 1/sC
 Derive the transfer function with physical frequency T(s) = Vo/Vi
Physical frequency (replace s by j)
 Treat an inductance L as an impedance jL
 Treat a capacitance C as an impedance 1/jC
 Derive the transfer function with physical frequency T(j) = Vo/Vi
Evaluating the frequency response of amplifiers
The transfer function for physical frequency is generally a complex value as a function of 
The magnitude |T()| defines the voltage gain of a sinusoidal at 
The phase T() defines the phase shift of a sinusoidal at 
The magnitude and phase are generally plotted versus frequency to evaluate the frequency response
Frequency‐domain analysis
Use Fourier series/Fourier transform to represent a time‐domain input signal
Use the freq. response to determine the amplitude/phase of the sinusoidal components at the output
The time‐domain output signal can be obtained by adding the output sinusoidal components
Example: vi(t) =  Ansin(nt +n) = A1sin(1t +1) + A2sin(2t +2) + …..
vo(t) = A1 |T(1)| sin(1t +1+ T(1)) + A2 |T(2)| sin(2t +2+ T(2)) + …..
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Single‐time‐constant (STC) networks
The network is generally composed of one reactive component (L or C) and one resistance
Most STC networks can be classified into two categories: low‐pass (LP) and high‐pass (HP)
Low‐pass STC
Bode plot
T ( j ) 
Vo ( j )
1 / jC
1


Vi ( j ) R  1 / jC 1  jRC
General form  T ( j ) 
Magnitude | T(j ) |
K
1  j / 0
K
1  ( / 0 ) 2
Phase T ( j )   tan 1 ( / 0 )
or 180  tan 1 ( / 0 )
Straight‐line approximations:
 Low‐frequency magnitude: |K| in dB
 Corner frequency: 0
 Fast evaluation of gain and phase
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Exercise3: Find the transfer function and plot its frequency response. Is it a STC network?
Exercise 4: Find the transfer function and plot its frequency response. Is it a STC network?
Exercise5: A voltage amplifier is modeled as Ri = 100 k, Ro = 10 k and Avo = 20 V/V. Consider the case where Rs of the signal source is 25 k and the amplifier is loaded with 10 k||100 nF.
(1) Find the transfer function (Vo/Vs) and plot its frequency response.
(2) Given that vs(t) = 10cos(102t)+5cos(104t)+3cos(106t) V, find vo(t).
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1.7 Intrinsic Semiconductors
Covalent bond
Each valence electron of a silicon atom is shared by one of its four nearest neighbors
Electrons served as covalent bonds are tightly bound to the nucleus
Electron‐hole pair
At 0K, no free carriers are available  Si behaves as an insulator
At room temperature, a small amount of covalent bonds will be broken by the thermal energy  electron‐hole pair generation as free carriers
Both electrons and holes are free to move  can contribute to current conduction
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Carrier concentration in intrinsic semiconductor
For intrinsic semiconductors at thermal equilibrium, electron‐hole generation and recommendation rates are equal
The electron and hole concentrations remain unchanged at thermal equilibrium
The conductance of intrinsic semiconductor is proportional to the carrier concentration
The carrier concentration is given by
 n = p = ni (intrinsic carrier concentration)
 np = ni2
 ni2(T) = BT3eEg /kT
 ni increases as temperature increases
 ni decreases as temperature decreases
Intrinsic carrier concentration for Si at room temperature: ni = 1.51010 /cm3
The conductivity is relatively low due to the carrier concentration
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Extrinsic semiconductor
Extrinsic (doped) semiconductor = intrinsic semiconductor + impurities
According to the species of impurities, extrinsic semiconductor can be either n‐type or p‐type
n‐type semiconductor
The donor impurities have 5 valence electrons are added into silicon
P, As, Sb are commonly used as donor
The Si atom is replaced by a donor atom
Donor ions are bounded in the lattice structure and thus donate free electrons without contributing holes
By adding donor atoms into intrinsic semiconductor, the number of electrons increases (n > p)
→ n‐type semiconductor
Majority carrier: electron
Minority carrier: hole
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p‐type semiconductor
The acceptor impurity has 3 valence electron (Boron)
Th Si atom is replaced by an acceptor atom
The boron lacks one valence electron. It leaves a vacancy in the bond structure
This vacancy can accept electron at the expense of creating a new vacancy
Acceptor creates a hole without contributing free electron
By adding acceptor into intrinsic semiconductor,
the number of holes increase (p > n) → p‐type semiconductor
Majority carrier: hole
Minority carrier: electron
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Carrier concentration
Charge neutrality:
 Particles with positive charge:
p: hole concentration (mobile); ND: donor concentration (immobile)
 Particles with negative charge:
n: electron concentration (mobile); NA: acceptor concentration (immobile)
 Charge neutrality (positive charge = negative charge): NA  n = ND  p
Mass‐action law
 np = ni2 for semiconductor under thermal equilibrium
For n‐type semiconductor
n = ND  p
np = ni2
→
2n
ND
[1  1  ( i ) 2 ]
ND
2
2
p  ni / n
n
if ND » ni →
n  ND
p  ni2 / N D
For p‐type semiconductor
p = NA  n
np = ni2
→
2n
NA
[1  1  ( i ) 2 ]
NA
2
2
n  ni / p
p
if NA » ni →
p  NA
n  ni2 / N A
Exercise 1.27 (Textbook)
Exercise 6: What must ND be such that n = 10000p at room temperature (T = 300K)
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1.9 Current Flow in Semiconductors
Free carriers in semiconductors
Mobile particles with positive or negative charges: electrons and holes
The transportation of carriers results in current conduction in semiconductors
Carrier drift
Thermal motion in the absence of electric field:
 The direction of flight being changed at each collision with the heavy, almost stationary ions
 Statistically, a electron has a random thermal motion in the crystal structure
 Net displacement over a long period of time is zero  no net current flow (I = 0)
Thermal motion under electric field E:
 The combined motion of electron under electric field has a random and a drift component
 Still, no net displacement due to random motion component over a long period of time
 The drift component provides the electron a net displacement
Drift is the carrier movement due to the existence of electric field
Electric field E  0
Electric field E = 0
Speed (instantaneous) > 0
No net displacement
Velocity (average) = 0
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Speed (instantaneous) > 0
Drift component due to E
Velocity (average)  0
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Mobility
F = qE
a = F /m* (m* is the effective mass of electron)
Assume the time interval between collision is tcoll and the drift velocity immediately after the collision is 0
Then the average velocity of the electron due to the electric field is:
v d (drift ) 
atcoll
qE

tcoll  E
2
2m*

vd qtcoll

E 2m*
(cm 2 /Vsec)
Mobility () indicates how fast an electron/hole can move under certain electric field intensity
  n is used to specify the mobility of electron
Similarly,  p is used to specify the mobility of hole
In most cases, electron mobility is larger than hole mobility in a semiconductor
Carrier drift in semiconductor
Semiconductor parameters:
 Electron concentration: n (1/cm3)
 Electron mobility: n (cm2/V)
 Hole concentration: p (1/cm3)
 Hole mobility: p (cm2/V)
Dimensions:
 Cross‐section area: A (cm2)
 Length: L (cm)
L
A
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Drift current in semiconductor
Electron current:
 Time interval for electrons flowing across L: T = L/vd = L/nE (sec)
 Total electron charge: Qn = qnAL (Coulomb)
 Electron drift current In,drift = Qn/T = qnAL/T = qnnEA (A)
 Current density Jn,drift = In,drift/A = qnnE (A/cm2)
Hole current:
 Time interval for holes flowing across L: T = L/vd = L/pE (sec)
 Total hole charge: Qp = qpAL (Coulomb)
 Hole drift current Ip,drift = Qp/T = qpAL/T = qppEA (A)
 Current density Jp,drift = Ip,drift /A = qppE (A/cm2)
The electron current and hole current are in the same direction
 Total drift current density: Jdrift= Jn,drift + Jp,drift = (qnn + qpp)E =E
Conductivity  = qnn + qpp (cm)1
Ohm’s Low
I = JA = EA = VA/L = V/R (A)
R = L/A = L/A ()
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Carrier diffusion
Diffusion is a manifestation of the thermal random motion of particles
 Section I: total # = 6 (3 moving to the left and 3 moving to the right)
 Section II: total # = 4 (2 moving to the left and 2 moving to the right)
 Net flux: 1 moving across the interface from section I to section II
I
II
Statistically, a net carrier flow from high to low concentration region in a inhomogeneous material
dp
dx
dn
 qDn
dx
J p ( diff )  qD p
J n ( diff )
Dn: diffusion constant (diffusivity) of e‐
Dp: diffusion constant (diffusivity) of h+
Einstein Relation: Dp/p = Dn/ n = kT/q = VT (thermal voltage)
Total diffusion current density
Both electron and hole diffusion contribute to current conduction
Total diffusion current density:
J diff  J n ( diff )  J p ( diff )  qDn
dn
dp
 qD p
dx
dx
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Graded semiconductor
For a non‐uniform semiconductor, the doping concentration is represented as ND(x)
The mobile carrier will diffuse due to the non‐uniform distribution
The uncompensated space charge builds up a field (potential) for the system to reach equilibrium
No net current flows at any point under equilibrium
The built‐in potential can be derived under thermal equilibrium between points with different doping concentration
Electron diffusion
Electron diffusion
n(x)
Built‐in potential from electron concentration
n(x)
Electron drift
Excess negative
mobile charge
x
x
E
J n  qn n E  qDn
n n E   Dn
dn
dx
VT dn
dV

n dx
dx
dn
dV  VT
n
V21  V2  V1  VT ln
ND(x)
J p  qp p E  qD p
p p E  D p
dp
0
dx
dp
dx
VT dp
dV

p dx
dx
dp
dV  VT
p
E
 n1  n2 e V21 / VT
ND(x)
dn
0
dx
Built‐in potential from hole concentration
E
n2
n1
V21  V2  V1  VT ln
p1
p2
 p1  p2 eV21 / VT
Excess positive
space charge
NTUEE Electronics – L. H. Lu
1‐30
Current in semiconductors
Drift current: Jdrift = qnnE + qppE
Diffusion current: Jdiff = qDn(dn/dx)  qDp(dp/dx)
Total current: J = Jn + Jp = (qnnE + qDn(dn/dx)) + (qppE  qDp(dp/dx))
E
e Jn
h
Jp
Exercise 7: A silicon bar has a cross‐sectional area of 4 cm2 and a length of 10 cm.
(1) For intrinsic Si with n = 1350 cm2/Vs and n = 480 cm2/Vs, find the resistance of the bar at T = 300 K.
(2) For extrinsic Si with NA = 1016 /cm3, n = 1100 cm2/Vs and n = 400 cm2/Vs, find the resistance of the bar at T = 300 K.
Exercise 1.28 (Textbook)
Example 1.9 (Textbook)
Exercise 1.29 (Textbook)
NTUEE Electronics – L. H. Lu
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1.10 The pn Junction with Open‐Circuit Terminals
Physical structure of a pn junction
Close contact of a n‐type semiconductor and a p‐type semiconductor
A two‐terminal device with anode and cathode
pn‐junction in contact
p‐type (NA) n‐type (ND) p‐type: doping concentration: NA
n‐type: doping concentration ND
Majority carriers are crossing the interface by diffusion and recombined in the other side
Leaving uncompensated space charges ND+ and NA  depletion region
In depletion region, electric field (potential) builds up due to the uncompensated space charges
The built‐in potential behaves as an energy barrier, reducing the majority carrier diffusion
This field results in minority carrier drift across the interface in the opposite direction to diffusion
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1‐32
pn‐junction formation (thermal equilibrium)
Depletion region increases due to majority carrier diffusion across the junction
The built‐in potential from uncompensated space charge increases as a barrier for carrier diffusion
Minority carriers are swept across the junction in the presence of the built‐in field  drift current
Equilibrium is reached when Jdiff and Jdrift are equal in magnitude and opposite in direction
No net current flows across the junction
E
p‐type (NA) n‐type (ND) hole diffusion
hole drift
electron diffusion
electron drift
Neutral
Region
p‐type
Jp = 0
Jn = 0
Neutral
Region
Depletion
Region
V0
n‐type
| V0 | VT ln
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p p0
nn 0
N N
 VT ln
 VT ln A 2 D
pn 0
ni
np0
1‐33
The depletion region
Step graded junction (abrupt junction) is used for analysis
Carriers are fully depleted in the depletion region
Neutral region in n‐type and p‐type outside depletion region
Built‐in potential: V0 = VT ln (NAND /ni2)

d 2V
dE

 
Poisson’s equation:
2
dx
dx
 Si
Derivation of pn‐junction at equilibrium:
qN D xn  qN A x p
V    Edx  V0  Emax ( xn  x p ) / 2  VT ln N A N D / n
N D xn  N A x p
2
i

qND
‐xp
xn
x
‐qNA
Electric field (E)
E    v /  Si dx  Emax  qN D xn /  Si  qN Ax p /  Si

Charge density (v)
‐xp

(qN D xn /  Si )( xn  x p ) / 2  VT ln N A N D / ni2

xn
x
‐Emax
Electrostatic potential (V)
W  xn  x p 
2 SiV0 N A
2 SiV0 N D
2 SiV0 N A  N D


qN D ( N A  N D )
qN A ( N A  N D )
q
ND N A
 For NA>>ND: W 
2 SiV0
qN D
 For ND>>NA: W 
2 SiV0
qN A
V0
‐xp
xn
x
Potential of electron
NTUEE Electronics – L. H. Lu
1‐34
Carrier distribution
cm‐3
Neutral n‐type region: pp0
1018
 Majority carrier nn = nn0 = ND
 Minority carrier pn = pn0 = ni2/ND
1014
Neutral p‐type region: 1010
 Majority carrier pp = pp0 = NA
 Minority carrier np = np0 = ni2/NA
106
np0
Depletion region: 102
 n = 0
10‐2
 p = 0
No net current flows across the junction
Built‐in potential across the pn‐junction:
 From hole density: | V0 | VT ln( p p 0 / pn 0 )  VT ln( N A N D / ni2 )
 From electron density: | V0 | VT ln(nn 0 / n p 0 )  VT ln( N A N D / ni2 )
nn0
V0
V0
‐xp
pn0
xn
Example 1.10 (Textbook)
Exercise 1.32 (Textbook)
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1.11 The pn Junction with an Applied Voltage
Depletion region
Forward bias: VF reduces the depletion region and the energy barrier
Reverse bias: VR increases the depletion region and the energy barrier
Forward bias (V = VF)
Charge density ( v)
Reverse bias (V = VR)
Charge density (v)
qND
xp
qND
xn
x
xp
 qNA
x
xn
W
Electric field (E)
x
xp
xn
 Emax
Electrostatic potential (V)
Electrostatic potential (V)
V0+VR
V0VF
xn
x
xp
xn
2 Si (V0  V ) N D  N A
q
ND N A
x
 Emax
xp
Emax  qN D xn /  Si  qN Ax p /  Si
 qNA
Electric field (E)
xp
xn
xn 
NA
2 Si N A (V0  V )
W
N A  ND
qN D ( N A  N D )
xp 
ND
2 Si N D (V0  V )
W
N A  ND
qN A ( N A  N D )
x
NTUEE Electronics – L. H. Lu
1‐36
Minority carrier distribution due to junction bias
Minority carrier distribution is influenced by the junction bias
Diffusion currents exist due to non‐uniform carrier distribution
Junction bias condition:
 Zero bias (equilibrium): V = 0
 Forward bias: V = VF
 Reverse bias: V = VR
Minority carrier distribution for all bias conditions:
( x x ) / L
pn ( x)  pn 0 (eV / V  1)e
 pn 0
n
T
n p ( x )  n p 0 (e
V / VT
 1)e
( x  x p ) / Ln
p
cm‐3
1018
Zero Bias (V = 0)
pp0
nn0
1010
pn0
np0
102
cm‐3
pp0
1018
Forward Bias (V = VF > 0)
pn0eV/VT
nn0
 np0
n (p): excess‐minority‐carrier lifetime  Ln = Dnn (Lp = Dpp ): diffusion length
Boundary condition:  pn(x = xn) = pp0exp[(V0V)/VT] = pn0exp(V/VT)  pn(x = ) = pn0
 np(x = xp) = nn0exp[(V0V)/VT] = np0exp(V/VT)  np(x = ) = np0
1010
pn0
np0
102
cm‐3
1018
pp0
Reverse Bias (V = ‐VR < 0)
nn0
1010
pn0
np0
102
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np0eV/VT
pn0eV/VT
1‐37
Junction current density Assume no carrier generation and recombination within the depletion region:
Jn(xp) = Jn(xn) and Jp(xp) = Jp(xn)
Diffusion currents Jp and Jn at the edge of the depletion region can be obtained by: J p ( xn )  qD p
dpn
dx
J n ( x p )  qDn
dn p
dx
xn
xp

qD p pn 0

Lp
qDn n p 0
Ln
(eV / VT  1)
(eV / VT  1)
Total junction current: J(x) = Jn(x) + Jp(x) = Jn(xp) + Jp(xp) = Jn(xp) + Jp(xn)  Assume Jp and Jn do not change across the depletion region: Jp(xp)= Jp(xn) and Jn(xp)= Jn(xn)
 The total current can be expressed as: J(x) = Jn(x) + Jp(x) = Jn(xp) + Jp(xp) = Jn(xp) + Jp(xn)
 qDn n p 0 qD p pn 0  qV / kT
( e

J  J n (  x p )  J p ( xn )  
 1)  J s (e qV / kT  1)
 L

Lp 
n

The I‐V characteristics of the pn junction
The junction current depends on the junction voltage
The junction current is proportional to the junction area
The junction current is given by I  I s (e qV / kT  1)
D p
 Dp
Dn 
Dn 
Saturation current: I s  qA p n 0  n p 0   qAni2 

 L
L N

Ln 
p

 p D Ln N A 
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Reverse breakdown
Breakdown voltage: a reverse junction bias VR = VZ
A large reverse current flows when reverse bias exceeds VZ
Breakdown region
For breakdown voltage < 5V  Zener breakdown
For breakdown voltage > 5V  avalanche breakdown
Breakdown is nondestructive if the power dissipation is limited
Zener breakdown
Strong electric field in the depletion region breaks covalent bonds, generating electron‐hole pairs
Generated electrons (holes) are swept into n‐type (p‐type region) for a reverse current
Zener breakdown normally takes place for pn junction with high doping concentration
Avalanche breakdown
The minority carriers that cross the depletion region gain sufficient kinetic energy due to the field
The carriers with high kinetic energy break covalent bonds in atoms during collision
More carriers are accelerated by the field for avalanche reaction
Avalanche normally takes place first for pn junction with low doping concentration
Example 1.11 (Textbook)
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Comparison of breakdown mechanism
Zener breakdown Avalanche breakdown
Charge density (v)
Charge density (v)
qND
qND
‐qNA
‐qNA
Electric field (E)
Electric field (E)
‐Emax
‐Emax
Electrostatic potential (V)
Electrostatic potential (V)
Breakdown
voltage
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Breakdown
voltage
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1.12 Capacitive Effects in the pn Junction
Depletion or junction capacitance
Charge density
The depletion width is controlled by the terminal voltage
The change of terminal voltage (dV) will result in dQ at the edge of the depletion region  capacitance
The junction capacitance due to space charge is Cj = dQ/dVR
q Si
ND N A
dQ d qAN D wn 
Cj 

A
dVR
dVR
2(V0  VR ) N D  N A
Cj can also be estimated by a parallel‐plate capacitor:
W
Cj 
p‐type
n‐type
positive space charge
(donors)
negative space charge (acceptors)
2 Si N D  N A
(V0  VR )
q ND N A
 Si A
wdep
C j0  A
A
 Si q N A N D
1
 C j0
2 N A  N D V0  VR
1
VR
V0
 Si q N A N D
1
2 N A  N D V0
Under forward bias conditions, W reduces  larger Cj
Under reverse bias conditions, W increases  smaller Cj
General formula of junction capacitance for arbitrary doping profile:
C j  C j 0 (1 
VR m
)
V0
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Diffusion capacitance
Excess minority carrier stored in neutral regions change with the terminal voltage  capacitance
By integration the excess minority carriers at both sides:
L2p
L2n
Q  Q p  Qn 
Ip 
In   p I p  nIn  T I
Dp
Dn
Small‐signal diffusion capacitance:
Q   T I   T I s e qV / kT   T I s eV / VT
Cd 
dQ  T
 ( )I
dV
VT
Cd is large under forward bias conditions
Cd is neglected under reverse bias conditions
Capacitance of pn‐junction
Operated at forward bias: C = Cj + Cd
Operated at reverse bias: C  Cj
Total capacitance increases as a more positive junction voltage is applied
Exercise 1.40 (Textbook)
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