ELEC2104 – Week 13
Review
Pure silicon
• Intrinsic/silicon semiconductors – poor conductor, high resistance
2
= 2 = 3− −6
Review Exercise
1. N-type silicon is obtained by doping silicon with
(a) Germanium
(b)Aluminium
(c) Boron
(d)Phosphorus
2. The concentration of minority carriers in a doped semiconductor under
equilibrium is
(a) Directly proportional to the doping concentration
(b) Inversely proportional to the doping concentration
(c) Directly proportional to the intrinsic concentration
(d) Inversely proportional to the intrinsic concentration
3
Doped silicon
• Extrinsic/doped semiconductors -
4
pn = ni 2N-type P-type
≈ , ≈

2

≈ , n ≈ 2
Mechanisms leading to current flow
• An electric field or a concentration gradient leads to the movement of these charge carriers
• Drift current
• Diffusion current
• Total current
5
Fermi level and Carrier concentration






Many
electrons
Few holes
Few
electrons
Many holes

≈ = − / ≈ = − /
6
Metal-semiconductor contact (Schottky)
Au Si
Vacuum




+

−
Net charge

ℰ
=
ℰ electric field strength
permittivity of semiconductor
ℰ = − = −
= − 12ℰ = 12 2

(Gauss’)


= − − −
total transferred charge
depletion region length
electron density in Si
voltage drop over depletion
metal work function
electron affinity
Will have an example
calculation of in tutorial
Positive
number
7
PN junctions
• Putting a p-type semiconductor to a n-type semiconductor, we can create a PN junction
8
= 2
How to calculate ? Fermi level have to LEVEL
pn junction
− , = ln, − = ln





n-siliconp-silicon

,
,
= 1 , − , = ln2 = 2 1 + 1
9
Review Exercise
• The depletion region of a PN junction has
(a) Only free mobile holes
(b)Only free mobile electrons
(c) Both free mobile holes as well as electrons
(d)Neither free mobile electrons nor holes
• If positive terminal of a battery is connected to the anode of the diode, it is
(a) Forward biased
(b)Reverse biased
(c) Equilibrium
(d)Saturated
10
“panic”
Biasing PN Junction
11
• When a PN Junction is forward biased
(a) The depletion region decreases
(b)The barrier reduces
(c) Holes and electrons drift away from junction
(d)None of the above
• When a PN Junction is reverse biased
(a) Holes and electrons concentrate towards junction
(b)The barrier reduces
(c) Holes and electrons move away from junction
(d)None of the above
p n- - - - + + + +
Forward bias
Reverse bias
PN Junction as Diodes
• Junction equation
• where
• ID = diode current
• VD = diode voltage
12
if exp(/) ≫ 1
if exp(/) ≪ 1
“leakage”
current
= , = , =
Diode Models
• To calculate the Q-point
• Load line analysis
• Mathematical iterative method – Newton Rhapson
• Ideal diode model Constant voltage diode (CVD) model
13
Thevenin equivalent of a
more complicated circuit
Review Exercise
• Two diodes with a constant forward voltage drop of 0.7 V are
used in the circuit below. The range of input voltage Vi which
the output voltage VO = Vi is
(a) -0.3 V < Vi < 1.3 V
(b) -0.3 V < Vi < 2 V
(c) -1.0 V < Vi < 2.0 V
(d) -1.7 V < Vi < 2.7 V 14
Rectifiers
• A resistor in parallel allows a path for the capacitor to discharge
• During the first quarter cycle, the diode is on and the capacitor charges up to the
peak value.
• At the peak input, the diode cuts off and the capacitor discharges exponentially
through R.
15
≈ −


1 − Δ

≈ −


Δ ≈
12 2 −
: peak
: diode turn-on
: ripple
Δ: charging time
Full-Wave Bridge Rectifier
16
Diode applications
17
BJT
18
E C
B
Base-Emitter
Junction Base-Collector Junction
Reverse Bias Forward Bias
Forward Bias Forward Active Region Good amplifier
Saturation Region
Closed switch
Reverse Bias
Cut-off Region
Open switch Reverse Active Region Poor Amplifier
Heavily doped Emitter
Thin Base
BJT
• NPN
• PNP
19
BJT – Operating conditions
20
Review Exercise
• Consider a NPN transistor, VBE = 0.7 V and VCB = 0.2 V then the
transistor is operating in the
(a)Forward active mode
(b)Saturation mode
(c)Reverse active mode
(d)Cut-off mode
21
BJT – Q Point
• Forward – active region
22


=


= ∝
= ( + 1)
= ∝1−∝
≈ exp
= + 1
BJT model – Large signal analysis
Q-point: (IC, VCE)
BJT – Small signal analysis
• Hybrid π small signal model
• Early effect
23
= = =
=
Review Exercises
• The “Early Effect” in a BJT refers to the reduction of the effective
base-width caused by
(a)Electron-hole recombination at the base
(b)Reverse-biasing of the base-collector junction
(c)Forward-biasing of the emitter-base junction
(d)Early removal of stored base charge during saturation to cut-off switching
24
BJT Amplifier Topologies
25
Common-Emitter Common-Base Emitter-Follower
=
= Input grounded, =
MOSFET
NMOS PMOS
• In a MOSFET, the polarity of the inversion layer is the same as
(a) Charge on the gate electrode
(b)Minority carriers in the drain
(c) Majority carriers in the substrate
(d)Majority carriers in the source
26
G
D
S
MOSFET
• Different regions of operation
27
Review Exercise
• The effective channel length of a MOSFET operating in saturation
decreases with increase in
(a)Gate voltage (c) Source voltage
(b)Drain voltage (d) Body voltage
28
MOSFET Large-Signal Models (VGS > VTH)
VDS << 2(VGS-VTH)
)(
1
THGSoxn
ON
VV
L
WC
R
−
=
µ
DS
DS
THGSoxntriD V
VVV
L
WCI 


 −−=
2
)(, µ
VDS >> (VGS-VTH)
Deep linear/triode Linear/triode Saturation
VDS << (VGS-VTH)
Q-point: (ID, VD)
• MOSFET model – Large signal analysis
29
Review Exercises
• An N-type MOSFET is biased at VGS > VTH and VDS > (VGS – VTH).
Considering channel length modulation effect to be significant, the
MOSFET behaves as a
(a)Voltage source with zero output impedance
(b)Voltage source with non-zero output impedance
(c)Current source with finite output impedance
(d)Current source with infinite output impedance
30
MOSFET Small Signal Model
31
≈
1
λ
( )THGSoxnm VVL
WCg −= µ
Doxnm IL
WCg µ2=
THGS
D
m VV
Ig
−
=
2
approximate
Approximate
(most used since
don’t need VGS)
exact
Gate has no
current
MOSFET Amplifier Topologies
32
Common-Source
Source-Follower
Op-Amps
• Non-inverting op-amps
• Inverting op-amps
33


= 1 + 2
1


= −2
1
Z2
Z1
Z2
Z1
No current through + and –
Current unknown at output
With feedback, + and – has equal voltage
Filters
34
Passive Active
Bode plot for a constant gain
• Assume the system provides a constant
gain to all frequencies
• If K is minus value
= Magnitude: 20 log10 ( = ) = 2010Phase: ∠ = 0°
Magnitude: 20 log10 ( = ) = 2010
Phase: ∠ = 180°
H = 1 + 1
1 + 1
35
(example,
let’s not
cancel )
Bode plot for zero at the origin
• Zero at the origin: =
∠ = −1 0 = 2
20 log = 20log
ω

20 dB/decade
ω
∠
90°
H = 1 + 1
1 + 1
1
36
20 log 1/ = −20log
∠ = −1 −1/0 = −2
ω

-20 dB/decade
ω
∠
−90°
H = 1 + 1
1 + 1
1
Bode plot for pole at the origin
• Pole at the origin: = 1/
37
20 log 1 +
1
= � 20log1 = 0, → 020log
1
, → ∞
∠ = −1
1
= � 0, = 045°, = 190°, → ∞
ω
∠
ω

20 dB/decade

Frequency 1 where the two lines meet is
called corner or break frequency
90°45°
.
xIm
Re
= tan−1
H = 1 + 1
1 + 1
Bode plot for simple zero
• Simple zero: = 1 +
1
38
• Simple pole: = 1/(1 +
1
)Bode plot for simple pole
20 log 1/(1 +
1
) = � 20log1 = 0, → 0
−20log
1
, → ∞
∠ = −1 −
1
= � 0, = 0−45°, = 1
−90°, → ∞
ω

-20 dB/decade
ω
∠
−90°

Frequency 1 where the two lines meet is
called corner or break frequency
−45°
.
x
Im Re
= tan−1
H = 1 + 1
1 + 1
39
Bode plot summary
• To sketch the Bode plots for a function of H(ω)
40
Four Key Steps
1. Rewrite transfer function in proper form
2. Separate the transfer function into its constituent parts and record all
the corner/break frequencies
3. Sketch the factors one at a time
4. Combine additively the graphs of the factors
H = 1 + 1
1 + 1
Bode plot summary
41
Bode plots
42
ω1
A1
|A| dB
-20dB/dec
concave
downward
ω1
∠A
-45°/dec
0°
-45°
-90°
0.1ω1 10ω1
1
11 + ω1
flat gain
flat gain
normal pole
Bode plots
43
ω1
A1
|A| dB
20dB/dec
concave
upward
ω1
∠A
45°/dec0°
45°
90°
0.1ω1 10ω1
1 1 + ω1
flat gain
normal zero
flat gain
Active Filters: Inverting Op-Amp
• Active High Pass Filters
44
= −2
1
11 + where = 11 = 2 -3-dB cut-off frequency
Gain set by ratio of resistors R2 and R1
Filters that employ gain elements are called active filters


Bode Plots
• Terms with 1 +
ω
correspond to frequencies to the right of the flat gain.
45
ω1 ω2 ω3ωaωbωc
Am
|A
|
= 1 + ω1 1 + ω21 + ω3
“normal” poles and zeros
Bode Plots
• Swapping numerator and denominator vertically inverts the magnitude graph.
46
ω1 ω2 ω3ωaωbωc
Am
|A
|
= 1 + ω31 + ω1 1 + ω2
“normal” poles and zeros
Bode Plots
• Terms with 1 + ω

correspond to frequencies to the left of the flat gain.
47
ω1 ω2 ω3ωaωbωc
Am
|A
|
“inverted” poles and zeros
= 1 + ω1 + ω 1 + ω
Bode Plots
• Again, swapping numerator and denominator vertically inverts the magnitude graph.
48
ω1 ω2 ω3ωaωbωc
Am
|A
|
“inverted” poles and zeros
= 1 + ω 1 + ω1 + ω
Bode Plots
• If there is no “flat gain”, use a reference value:
49
90°
45°
0°0°
-45°
-90°
= 4 ω4
ω3
3
= 3 1
ω3
= 3 ω3
ω4
4
|A|
∠ A
Bode Plots
• Exercise: Express the gains in factored pole-zero form
50
ω1
1
ω2 ω1
1
ω2 ω1
ω2
Bode Plots
• Exercise: Express the gains in factored pole-zero form
51
ω1
1
ω2 ω1
1
ω2 ω1
ω2
= 1 11 + 1 1 + 2 = 1 1 + 11 + 2 = 1 1 1 + 21 + 1
1
Front cover of final exams
52
Exam Preparation
• Note: some questions may depend on the answer from previous parts
• Keep answers to as many significant numbers as possible
• Write codes. Save them.
• Prepare your own formula sheet
• Express them in different forms e.g.
• = exp or = ln
• IC, IB, IE in terms of beta
• ID and VGS
53
GOOD LUCK!!



























































































































































































































































































































































































































































































































































































































学霸联盟