Transcript 10. Diodes – Basic Diode Concepts

Chapter 10
Diodes
1. Understand diode operation and select diodes for various
applications.
2. Analyze nonlinear circuits using the graphical load-line
technique.
3. Analyze and design simple voltage-regulator circuits.
4. Solve circuits using the ideal-diode model and
piecewise-linear models.
5. Understand various rectifier and wave-shaping circuits.
6. Understand small-signal equivalent circuits.
10. Diodes – Basic Diode Concepts
10.1 Basic Diode Concepts
10.1.1 Intrinsic Semiconductors
* Energy Diagrams – Insulator, Semiconductor, and Conductor
the energy diagram for the three types of solids
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10. Diodes – Basic Diode Concepts
10.1.1 Intrinsic Semiconductors
* Intrinsic (pure) Si Semiconductor:
Thermal Excitation, Electron-Hole Pair, Recombination,
and Equilibrium
When equilibriu m between
excitation and recombination
is reached :
electron density  hole density
ni  pi  1.5  10 10 cm -3
for intrinsic Si crystal at 300 K
( Note : Si crystal atom density
is ~ 5  10 22 cm -3 )
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10. Diodes – Basic Diode Concepts
10.1.1 Intrinsic Semiconductors
*Apply a voltage across
a piece of Si:
electron current
and hole current
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10. Diodes – Basic Diode Concepts
10.1.2 N- and P- Type Semiconductors
* Doping: adding of impurities (i.e., dopants) to the intrinsic semiconductor material.
* N-type: adding Group V dopant (or donor) such as As, P, Sb,…
n  p  constant for a semiconductor
For Si at 300K

n  p  ni2  pi2  1.5  10 10

2
In n - type material
n  N d the donor conceration
n  N d  ni , p  pi
We call
electron the major charge carrier
hole the minor cahage carrier
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10. Diodes – Basic Diode Concepts
10.1.2 N- and P- Type Semiconductors
* Doping: adding of impurities (i.e., dopants) to the intrinsic semiconductor material.
* P-type: adding Group III dopant (or acceptor) such as Al, B, Ga,…
n  p  constant for a semiconductor
For Si at 300K

n  p  n  p  1.5  10
2
i
2
i

10 2
In p - type material
p  N a the acceptor conceration
p  N a  pi , n  ni
We call
hole the major charge carrier
electron the minor cahage carrier
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10. Diodes – Basic Diode Concepts
10.1.3 The PN-Junction
* The interface in-between p-type and n-type material is called a
pn-junction.
The barrier potential VB  0.6  0.7V for Si and 0.3V for Ge
at 300K : as T ,VB  .
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10. Diodes – Basic Diode Concepts
10.1.4 Biasing the PN-Junction
* There is no movement of charge
through a pn-junction at
equilibrium.
* The pn-junction form a diode
which allows current in only one
direction and prevent the
current in the other direction as
determined by the bias.
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10. Diodes – Basic Diode Concepts
10.1.4 Biasing the PN-Junction
*Forward Bias: dc voltage positive terminal connected to the p
region and negative to the n region. It is the condition that
permits current through the pn-junction of a diode.
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10.1.4 Biasing the PN-Junction
*Forward Bias: dc voltage positive terminal connected to the p
region and negative to the n region. It is the condition that
permits current through the pn-junction of a diode.
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10. Diodes – Basic Diode Concepts
10.1.4 Biasing the PN-Junction
*Forward Bias:
Current
Voltage
Forward direction
Diode as an insulator
Diode as a conductor
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10. Diodes – Basic Diode Concepts
*Reverse Bias: dc voltage negative terminal connected to the p
region and positive to the n region. Depletion region widens
until its potential difference equals the bias voltage, majoritycarrier current ceases.
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10. Diodes – Basic Diode Concepts
*Reverse Bias:
majority-carrier current ceases.
* However, there is still a very
small current produced by
minority carriers.
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10. Diodes – Basic Diode Concepts
10.1.4 Biasing the PN-Junction
* Reverse Breakdown: As reverse voltage reach certain value,
avalanche occurs and generates large current.
Diode is acting as a
Conductor after the
Breakdown voltage
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10. Diodes – Basic Diode Concepts
10.1.5 The Diode Characteristic I-V Curve
Switch ON
region
Switch OFF
region
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10. Diodes – Basic Diode Concepts
10.1.6 Shockley Equation
* The Shockley equation is a theoretical result
under certain simplification:
  vD  
  1
i D  I s exp
  n VT  
where I s  10 -14 A at 300K is the (reverse) saturation
current, n  1 to 2 is the emission coefficient,
kT
VT 
 0.026V at 300K is the thermal voltage
q
k is the Boltzman' s constant, q  1.60  10 -19 C
 vD 

when v D  0.1V, i D  I s exp
 n VT 
This equation is not applicable when v D  0
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10. Diodes – Load-Line Analysis of Diode Circuits
10.2 Load-Line Analysis of Diode Circuit
dv
di
We can use v  iR, i  C
, v  L ,...
dt
dt
  vD  
  1
but when there is a diode : i D  I s exp
  n VT  
It is difficult to write KCL or KVL equations.
For the circuit shown,
KVL gives :
VSS  R i D  v D
If the I - V curve of
the diode is given,
we can perform the
" Load - Line Analysis"
iD 0
vD 0
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10. Diodes – Load-Line Analysis of Diode Circuits
Example 10.1- Load-Line Analysis
For the circuit shown,
Given : VSS  2V, R  1kΩ ,
the I - V curve of the diode
Find : the diode current and voltage
at the operating point (Q - point)
VSS  R i D  v D , i.e.,
2  1000 i D  v D
 perform load - line analysis
 at the operating point
VDQ  0.70 V, i DQ  1.3 mA
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10. Diodes – Load-Line Analysis of Diode Circuits
Example 10.2 - Load-Line Analysis
For the circuit shown,
Given : Vss  10 V, R  10 k ,
the I - V curve of the diode
Find : the diode current and voltage
at the operating point
VSS  R i D  v D , i.e.,
10  10k i D  v D
 perform load - line analysis
 at the operating point
VDQ  0.68 V, i DQ  0.93 mA
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10. Diodes – Zener Diode Voltage-Regulator Circuits
10.3 Zener-Diode Voltage-Regulator Circuits
10.3.1 The Zener Diode
* Zener diode is designed for operation in the reverse-breakdown
region.
* The breakdown voltage is controlled by the doping level (-1.8 V to
-200 V).
* The major application of Zener diode is to provide an output
reference that is stable despite changes in input voltage – power
supplies, voltmeter,…
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10. Diodes – Zener-Diode Voltage-Regulator Circuits
10.3.2 Zener-Diode Voltage-Regulator Circuits
* Sometimes, a circuit that produces constant output voltage while
operating from a variable supply voltage is needed. Such circuits
are called voltage regulator.
* The Zener diode has a breakdown voltage equal to the desired
output voltage.
* The resistor limits the diode current to a safe value so that Zener
diode does not overheat.
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10. Diodes – Zener-Diode Voltage-Regulator Circuits
Example 10.3 – Zener-Diode VoltageRegulator Circuits
Given : the Zener diode I - V curve, R  1k
Find : the output voltage for VSS  15 V and
VSS  20 V
KVL gives the load line :
VSS  R i D  v D  0
From the Q - point we have :
vo  10.0 V for VSS  15 V
vo  10.5 V for VSS  20 V
5V change in input
 0.5V change in vo
Actual Zener diode
performs much better!
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10. Diodes – Zener-Diode Voltage-Regulator Circuits
10.3.3 Load-Line Analysis of Complex Circuits
* Use the Thevenin Equivalent
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10. Diodes – Zener-Diode Voltage-Regulator Circuits
Example 10.4 – Zener-Diode Voltage-Regulator with a Load
Given : Zener diode I - V curve, VSS  24V, R  1.2k , RL  6k
Find : the load voltage v L and source currents I S
Applying Thevenin Equivalent  VT  VSS
RL
R RL
 20V , RT 
 1k 
R  RL
R  RL
 VT  RT i D  v D  0
 v L  -v D  10.0 V
I S  (V SS - v L )/R  11.67 mA
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10. Diodes – Zener-Diode Voltage-Regulator Circuits
Quiz – Exercise 10.5
Given : the circuit and the Zener diode I - V curve as shown.
Find : the output vol tage v o for i L  0, i L  20mA, and i L  100mA
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10. Diodes – Ideal-Diode Model
10.4 Ideal-Diode Model
* Graphical load-line analysis is too cumbersome for complex circuits,
* We may apply “Ideal-Diode Model” to simplify the analysis:
(1) in forward direction: short-circuit assumption, zero voltage drop;
(2) in reverse direction: open-circuit assumption.
* The ideal-diode model can be used when the forward voltage drop and
reverse currents are negligible.
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10. Diodes – Ideal-Diode Model
10.4 Ideal-Diode Model
* In analysis of a circuit containing diodes, we may not know in
advance which diodes are on and which are off.
* What we do is first to make a guess on the state of the diodes in
the circuit:
(1)For " assumed on diodes" : check if i D is positive;
(2) For " assumed off diodes" : check if v D is negative
 ALL YES  BINGO!
 not ALL YES  make another guess....
iterates until " ALL YES"
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10. Diodes – Ideal-Diode Model
Example 10.5 – Analysis by Assumed Diode States
Analysis the circuit by assuming D1is off and D2 on
+10V
+7V
(1) assume
D1 off, D2 on 

(2) assume
D1 on, D2 off
+6V
R1
+3V
Vd1 != +7V, because it should be in forward bias!
i D2

 0.5mA OK!
v D1  7V
not OK!
+3V
 i D1  1 mA OK!
v D2  -3 V
OK!
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10. Diodes – Ideal-Diode Model
Quiz – Exercise 10.8c
* Find the diode states by using ideal-diode model. Starting by
assuming both diodes are on.
(1) assume
D3 on 
D4 on
(2) assume D3 off and D4 on
iD 3


 -1.7 mA, not OK
i D 4  6.7 mA, OK
 i D 4  5 mA, OK
v D 3  -5 V,
OK
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10. Diodes – Piecewise-Linear Diode Models
10.5 Piecewise-Linear Diode Models
10.5.1 Modified Ideal-Diode Model
* This modified ideal-diode model is usually accurate enough in
most of the circuit analysis.
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10. Diodes – Piecewise-Linear Diode Models
10.5.2 Piecewise-Linear Diode Models
v  Ra i  Va
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10. Diodes – Rectifier Circuits
10.6 Rectifier Circuits
* Rectifiers convert ac power to dc power.
* Rectifiers form the basis for electronic power suppliers and battery
charging circuits.
10.6.1 Half-Wave Rectifier
Rewerse
Forward
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10. Diodes – Rectifier Circuits
* Battery-Charging Circuit
No current flow
In this half period.
Current Flow
On the battery
* The current flows only in the direction that charges the battery.
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10. Diodes – Rectifier Circuits
* Half-Wave Rectifier with Smoothing Capacitor
* To place a large capacitance across the output terminals:
Capacitor is
discharging
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10. Diodes – Rectifier Circuits
10.6.2 Full-Wave Rectifier Circuits
Diode B
OFF
Diode A
OFF
* Center-Tapped Full-Wave Rectifier – two half-wave rectifier with out-ofphase source voltages and a common ground.
* When upper source supplies “+” voltage to diode A,
the lower source supplies “-” voltage to diode B;
and vice versa.
* We can also smooth the output by using a large capacitance.
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10. Diodes – Rectifier Circuits
10.6.2 Full-Wave Rectifier Circuits
* The Diode-Bridge Full-Wave Rectifier:
A,B
C,D
Without Capacitor
With Capacitor
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10. Diodes – Wave-Shaping Circuits
10.7 Wave-Shaping Circuits
10.7.1 Clipper Circuits
* A portion of an input signal waveform is “clipped” off.
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10. Diodes – Wave-Shaping Circuits
10.7 Wave-Shaping
Circuits
10.7.2 Clamper Circuits
* Clamp circuits are used to
add a dc component to an
ac input waveform so that
the positive (or negative)
peaks are “clamped” to a
specified voltage value.
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10. Diodes – Linear Small-Signal Equivalent Circuits
10.8 Linear Small-Signal Equivalent Circuits
* In most of the electronic circuits, dc supply voltages are used to
bias a nonlinear device at an operating point and a small signal
is injected into the circuits.
* We often split the analysis of such circuit into two parts:
(1) Analyze the dc circuit to find operating point,
(2) Analyze the small signal ( by using the “linear smallsignal equivalent circuit”.)
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10. Diodes – Linear Small-Signal Equivalent Circuits
10.8 Linear Small-Signal Equivalent
Circuits
* A diode in linear small-signal equivalent
circuit is simplified to a resistor.
* We first determine the operating point
(or the “quiescent point” or Q point) by
dc bias.
* When small ac signal injects, it swings
the Q point slightly up and down.
* If the signal is small enough, the
characteristic is straight.
 d iD
i D  
 d vD
 d iD

 d vD

 v D
Q
i D is the small change in diode current
v D is the small change in diode voltage
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

Q
10. Diodes – Linear Small-Signal Equivalent Circuits
10.8 Linear Small-Signal Equivalent
Circuits
Define the dynamic resistance of the diode as :
 d i
rd   D
 d v D
 di
i D   D
 d vD
 
 
 Q 
 d iD

 d vD
1
We will have :

v D
 v D  i D 
rd
Q
Replace i D and v D by id and v d denoting
small changes, we have for ac signals :
vd
id 
rd
Furthermore, by applying the Shockley equation,
n VT
we have : rd 
I DQ
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

Q
10. Diodes – Linear Small-Signal Equivalent Circuits
10.8 Linear Small-Signal
Equivalent Circuits
vd
n VT
id 
, rd 
rd
I DQ
* By using these two
equations, we can treat
diode simply as a linear
resistor in small ac signal
analysis.
* Note: An ac voltage of fixed
amplitude produces
different ac current change
at different Q point.
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10. Diodes – Linear Small-Signal Equivalent Circuits
10.8 Linear Small-Signal Equivalent Circuits
i D  I DQ  i d
v D  V DQ  v d
vd
n VT
id 
, rd 
rd
I DQ
(1) V DQ and I DQ represent the dc signals
at the Q point.
(2) v d and i d represent the small sc signals.
(3) v D and i D represent the total
instantaneous diode voltage and current.
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10. Diodes – Linear Small-Signal Equivalent Circuits
Voltage-Controlled Attenuator
* The function of this circuit is to produce an output signal that is a variable
fraction of the ac input signal.
* Two large coupling capacitors: behave like short circuit for ac signal and
open circuit for dc, thus the Q point of the diode is unaffected by the ac
input and the load.
ZC 
1
j C
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10. Diodes – Linear Small-Signal Equivalent Circuits
Voltage-Controlled Attenuator
First apply dc analysis to find the diode Q point,
n VT
determine I DQ , then the rd of the diode : rd 
I DQ
Next,we perform small ac signal analysis :
(note : the dc voltage source has an ac component of current but no ac voltage,
the dc voltage source is equivalent to a short circuit for ac signal.)
Rp
vo
1
Rp 
, based on voltage divider : Av 

1
1 RC  1 RL  1 rd
vin R  R p
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10. Diodes – Linear Small-Signal Equivalent Circuits
Exercise 10.20 - Voltage-Controlled Attenuator
Given : R  100 Ω , RC  R L  2kΩ , diode n  1 at 300K
Find : the Q - point values assuming V f  0.6V
and Av for VC  1.6 and 10.6V
First apply dc analysis to find the diode Q point,
VC - 0.6
nVT
I DQ 
, rd 
withVT  0.026V
RC
I DQ
Next, we perform small ac signal analysis :
Rp
vo
1
Rp 
, Av 

1 RC  1 R L  1 rd
v in R  R p
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Chapter 11: Amplifiers:
Specifications and External Characteristics
11. Amplifiers – Basic Amplifier Concepts
11.1 Basic Amplifier Concepts
11.1.1 For Starting
* Ideally, an amplifier produces an output signal with identical waveshape as the input signal but with a larger amplitude.
v o t   Av v i t 
Av is the Voltage Gain
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11. Amplifiers – Basic Amplifier Concepts
11.1.1 For Starting
* inverting and
non-inverting amplifiers
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11. Amplifiers – Basic Amplifier Concepts
11.1.1 For Starting
* Often, one of the amplifier input
terminals and one of the output
terminals are connected to a
common ground.
* The common ground serve as the
return path for signal and the dc
power supply currents.
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11. Amplifiers – Basic Amplifier Concepts
11.1.1 For Starting
* Another example for common ground
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11. Amplifiers – Basic Amplifier Concepts
11.1.2 Voltage-Amplifier Model
* Amplification can be modeled by a controlled source.
Amplifier
Av > 0, non inverting amplifier
Av < 0, inverting amplifier
Av = 1, 1:1 buffer (but may Ai > 1)
Ri : the input resistance (or impedance), is the equivalent resistance seen
when looking into the input terminals.
Ro : in series with the output terminals,
is the output resistance (or impedance)
Avo  v oc /v i : the open - circuit voltage gain
(note : the real gain is smaller tham Avo )
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11. Amplifiers – Basic Amplifier Concepts
11.1.3 Current and Voltage Gains
i i is the current delivered
into the input terminals
of the amplifier;
i o is the current flowing
through the load.
io
The current gain Ai is the ratio between output and input currents : Ai 
ii
Furthermor e, Ai 
i o v o RL
R
v

 Av i , where Av  o is the voltage gain
ii
v i Ri
RL
vi
The voltage gain Av is usually smaller than the open - circuit voltage gain Avo
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11. Amplifiers – Basic Amplifier Concepts
11.1.3 Power Gain
The power gain is the ratio of the output power to the input power : G 
Po
Pi
Since the average power is the product of the rms values Vrms and I rms ,
we have : G 
Po Vrms o I rms o
2 R

 Av Ai   Av  i
Pi Vrms i I rms i
RL
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11. Amplifiers – Basic Amplifier Concepts
Example 11.1 Calculating Amplifier Performance
Find the voltage gains
Avs  Vo /V s and Av  Vo /V i
Also, find the current gain
and the power gain
Vo AvoVi  R L /(R o  R L )
RL
8
4
Av 

 Av o
 10
 8000
Vi
Vi
Ro  R L
28
Av s 
Vo
Vo
Ri

 Av
 5333
V s Vi (Ri  Rs )/R i 
Ri  R s
Note : due to the loading effect, A
vs
Ai  Av
Ri
 2  10 9 ,
RL
 Av  Av,o

G  Av Ai  16  10 12
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11. Amplifiers – Cascade Amplifiers
11.2 Cascade Amplifiers
v o 2 v o1 v o 2 v o1 v o 2
Av 




 Av  Av 1 Av 2
v i 1 v i 1 v 01 v i 1 v i 2
Av o 
vo c 2
vi 1

Av o 2 v i 2
vi 1
 Av o 2
In addition, Ai  Ai 1 Ai 2 ,
vo 1
vi 2
 Av o  Av 1 Av o 2
G  G 1G 2
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11. Amplifiers – Cascade Amplifiers
Example 11.2 – Calculating Performance of Cascade Amplifier
Ri 2  R L 1
Find the current gain, voltage gain, and power gain
Av 1  Avo 1
Ri 2
RL
 150 , Av 2  Avo 2
 50  Av  Av 1 Av 2  7500
Ri 2  Ro 1
R L  Ro 2
Ri 1
Ri 2
5
Ai 1  Av 1
 10 , Ai 2  Av 2
 750  Ai  Ai 1 Ai 2  75  10 6
Ri 2
RL
G 1  Av 1 Ai 1  1.5  10 7 , G 2  Av 2 Ai 2  3.75  10 4
 G  G 1 G 2  5.625  10 11
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11. Amplifiers – Cascade Amplifiers
Example 11.3– Simplified Model for Amplifier Cascade
Av 1  Av o 1
Ri 2
 150
R i 2  Ro 1
Av o 2  100
Av o  Av 1 Av o 2  15  10 3
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11. Amplifiers – Power Supplies and Efficiency
11. 3 Power Supplies and Efficiency
* The power gain of an amplifier is usually large, the additional
power is taken from the power supply.
The total average power
supplied to the amplifier is
Ps  V AA I A  V BB I B
Conservation of energy :
Pi  Ps  Po  Pd
Pi is the power of input signal
Po is the power of output signal
Pd is the power dissipated in
internal carcuits
The efficiency of an amplifier :
Po
η
 100%
Ps
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11. Amplifiers – Power Supplies and Efficiency
Example 11.4 Amplifier Efficiency
2
V
Pi  i  10  11 W  10 pW
Ri
RL
Vo  Avo V i
 8 V rms
R L  Ro
2
Vo
Po 
 8W
RL
Ps  V AA I A  V BB I B  22.5 W
Pd  Ps  Pi  Po  14.5 W
η
Po
 35.6%
Ps
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11. Amplifiers – Additional Amplifier Models
11.4 Additional Amplifier Models
Current-Amplifier Model
Thevenin vs. Norton approach
vi
Avo v i
ii 
and i osc 
Ri
Ro
the short circuit current gain :
i
R
Aisc  osc  Avo i
ii
Ro
Ri and Ro are the same
for the two models.
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11. Amplifiers – Additional Amplifier Models
Example 11.5 – Transform Voltage-Amplifier to
Current-Amplifier Model
vi
ii 
Ri
Aisc
and
i osc
Avo v i

Ro
i osc
Ri

 Avo
 10 3
ii
Ro
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11. Amplifiers – Additional Amplifier Models
11.4 Additional Amplifier Models
Transconductance-Amplifier Model
Gmsc
iosc

vi
Transresistance-Amplifier Model
Rmsc
vosc

ii
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11. Amplifiers – Ideal Amplifiers
11.6 Ideal Amplifiers
Voltage Amplifier :
Ri   , v i  v s max.
Ro  0, v o  Avo v i max
 Maximum Voltage Gain!
Current Amplifier :
Ri  0, i i  i i sc max.
Ro   , i o  Ai sc i i max.
 Maximum Current Gain!
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11. Amplifiers – Frequency Response
11.7 Frequency Response
* The gain of an amplifier is a function of frequency. Av  Vo
Vi
* Both amplitude and phase are affected.
Example 11.8
The input for a certain amplifier is : v i t   0.1 cos 2000 π t  30 
the outpot is : v 0 t   10 cos2000 π t  15 
Find the complex voltage gain and express the amplitude in dB
Vo
1015
 100 45

Vi  0.1  30 , Vo  10 15  Av 
0.1  30
Vi
Note : there is a 45 phase shift.
Av
dB
 20 log Av  20 log( 100 )  40 dB
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11. Amplifiers – Frequency Response
Gain as a Function of Frequency
66
11. Amplifiers – Frequency Response
AC Coupled Amplifiers
The gain drops to zero at dc (low frequency).
67
11. Amplifiers – Frequency Response
DC Coupled Amplifiers
Amplifiers that are realized as integrated circuits are often dc coupled,
because capacitors or transformers can not be fabricated in integrated
form.
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11. Amplifiers – Frequency Response
High-Frequency Drop Off
* The small amount of capacitance in parallel or inductances in
series with the signal path in the amplifier circuit will cause the
gain of the amplifier to drop at high frequencies.
j2 πfL, open as f 
1/j2 πfC, short as f 
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11. Amplifiers – Frequency Response
Half-Power Frequencies and
Bandwidth
* Bandwidth is the distance
between the half-power
frequencies.
* Half-power frequencies:
A  Amid / 2  3dB from Amid
20log(1/ 2 )  -3.01 dB
* Wideband (Baseband) Amp.
* Narrowband (Bandpass) Amp.
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11. Amplifiers – Linear Waveform Distortion
11.8 Linear Waveform Distortion
* Distortion may occur even though the amplifier is linear (i.e., obeys
superposition principle).
Amplitude Distortion
If a signal contains components of various frequencies, the output
waveform may be distorted due to the frequency response of the
amplifier gain.
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11. Amplifiers – Linear Waveform Distortion
Phase Distortion
* If the phase shift of an amplifier is not
proportional to the frequency, phase
distortion may occur.
Assume we have an input like :
v i t   3 cos2000 π t   cos6000 π t 
and three amplifiiers having the gains :
The output would look like :
v oA t   30 cos2000 π t   10 cos6000 π t 
v oB t   30 cos2000 π t  45   10 cos6000 π t  135 
v oC t   30 cos2000 π t  45   10 cos6000 π t  45 
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11. Amplifiers – Linear Waveform Distortion
11.8 Linear Waveform Distortion
* In order to avoid distortion:
(1) the magnitude of the gain must be constant against frequency.
(2) The phase response must be proportional to the frequency.
θ 1 : θ 2  ω1 : ω2  T2 : T1
θ1
θ
 T1 : 2  T2
2π
2π
θ
θ ( ω / ω2 )
 1  T1 : 1 1
 T1 ( ω2 / ω1 )
2π
2π
1:1
Δt 1 : Δt 2 
The same time delay!
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11. Amplifiers – Pulse Response
11.9 Pulse Response
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11. Amplifiers – Nonlinear Distortion
11.10 Nonlinear Distortion
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11. Amplifiers – Differential Amplifiers
11.11 Differential Amplifiers
* A Differential amplifier has two input sources, the output voltage is
proportional to the difference between the two input voltages.
v o t   Ad v i 1 t   v i 2 t 
 Ad v i 1 t   Ad v i 2 t 
Non-inverting input
Inverting input
Define the differential signal :
v id  v i 1  v i 2
and Ad the differentail gain
We can write the output of
a differentail amplifier as :
v o  Ad v id
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11. Amplifiers – Differential Amplifiers
Electrocardiogram
* The desired waveform is given by the difference between the
potentials measured by the two electrodes, i.e., the output of an
ideal differential amplifier.v o  Ad v id  Ad (v i1 - v i2 )
* While both electrodes (also act like antennas) pick a commonmode signal (noise) from the 60 Hz power line:
v i1  Vn cos(377t  φn ), v i2  Vn cos(377t  φn )
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11. Amplifiers – Differential Amplifiers
Electrocardiogram
v o  Ad v id  Ad (v i1 - v i2 )
v i1  Vn cos(377t  φn ),
v i2  Vn cos(377t  φn )
* Ideally, the common-mode signal was nullified by the differential amplifier.
* However, real differential amplifier responds to both differential-mode and
common-mode signals:
v o  Ad v id  Acm v icm where Acm is the common - mode gain,
v id is the differential - mode signal and v icm the common - mode signal
Common-Mode Rejection Ratio (CMRR): CMRR  20 log
Ad
Acm
* At 60 Hz, CMRR of 120 dB is considered good.
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11. Amplifiers – Differential Amplifiers
Example 11.12 Determination of Minimum CMRR Specification
* Find the minimum CMRR for an electrocardiogram amplifier if the
differential gain is 1000, the desired differential input signal has a
peak amplitude of 1 mV, the common-mode signal is 100 V-peak
60 Hz sine wave, and it is desired that output contain a peak
common-mode signal that is 1% or less of the peak output caused
by the differential signal.
peak v i d  1 mV, Ad  1000  v o d  1 V
peak v o cm  1 V  1%  0.01 V  Acm
CMRR  20 log
Ad
Acm
 20 log
0.01V

 10  4
100 V
1000
 140 dB
4
10
Ans : CMRR  140 dB
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