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Transcript
ELE 2110A Electronic Circuits
Week 12:
Output Stages, Frequency Response
(2 hours only)
ELE2110A © 2008
Lecture 12 - 1
Topics to cover …
z
Output Stages
z
Amplifier Frequency Response
Reading Assignment:
Chap 15.3, 16.1 of Jaeger and Blalock or
Chap 14.1 – 14.4 of Sedra & Smith
ELE2110A © 2008
Lecture 12 - 2
Multistage Amplifiers
Practical amplifiers usually consist of a number of stages connected in
cascade.
z
The first (input) stage is usually required to provide
– a high input resistance
– a high common-mode rejection for a differential amplifier
z
Middle stages are to provide
– majority of voltage gain
– conversion of the signal from differential mode to single-end
mode
– shifting of the dc level of the signal
z
The last (output) stage is to provide
– a low output resistance in order to
z
z
avoid loss of gain and
provide the current required by the load (power amplifiers)
ELE2110A © 2008
Lecture 12 - 3
Example
z
The input stage (Q1, Q2) is
differential-in and differentialout
– biased by current source Q3
z
(Q4, Q5) is a differential-in
and single-ended-out stage
– biased by current source Q6
z
Q7 provides
– additional gain
– shifting the dc level of the
signal
z
ELE2110A © 2008
The output stage Q8 is an
emitter follower
Lecture 12 - 4
Output Stages
z
Function of an output stage is
– To provide a low output resistance so that it can deliver
the output signal to the load without loss of gain
z
Requirements of an output stage:
– Large input signal range
z
z
b/c it is the final stage of the amplifier, and usually deals with
relatively large signals.
Small-signal approximations and models either are not
applicable or must be used with care.
– Low distortion
– High power efficiency
ELE2110A © 2008
Lecture 12 - 5
Classification of Output Stages
z
Class A
Class B
z
z
Class AB
Class C
Class A: the transistor
conducts for the entire cycle
of the input signal
Class B: the transistor
conducts for only half the
cycle
Class AB: conduction cycle is
greater than 180o and less
than 360o
– Used for opamp output stage
and audio power amplifiers
z
Collector or Drain current waveforms
of different output stages
ELE2110A © 2008
Class C: conduction cycle is
less than 180o
– Used for radio-frequency
(RF) power amplifications
(mobile phones, radio and
TV)
Lecture 12 - 6
Class-A Amplifier: Source/Emitter Follower
For a source follower biased by an ideal
current source, vGS is fixed and
⎛
2 I SS
⎜
vO = vI − VGS1 = vI − ⎜VTN +
Kn
⎝
⎞
⎟
⎟
⎠
Input range:
− VSS + VGS ≤ vI ≤ VDD + VTN
Output range:
− VSS ≤ vO ≤ VDD
The largest output voltage is
vo ≅ V
sin ω t
DD
ELE2110A © 2008
(if VSS=VDD)
Lecture 12 - 7
Source Emitter with Load
To maintain class A operation,
is > 0 at all times:
v
∴ i =I
+ o ≥0
S
SS R
L
vo ≥ − I SS RL
For largest output amplitude:
We have:
vo ≅ V
sin ω t
DD
VDD sin ωt ≥ − I SS RL for all t
The lowest value for the LHS occurs when sin ωt = -1,
∴VDD ≤ I SS RL
∴ I SS
ELE2110A © 2008
VDD
≥
RL
Lecture 12 - 8
Power Efficiency
Average power supplied to the source follower:
⎡
⎛ VDD sin ωt ⎞ ⎤
(
)
I
V
V
+
+
∫0 ⎢⎢ SS DD SS ⎜⎜⎝ RL ⎟⎟⎠VDD ⎥⎥ dt
⎣
⎦
= I SS (VDD + VSS ) = 2 I SSVDD
(if VSS=VDD)
Pav =
1
T
T
Average power delivered to the load:
2
The largest output
voltage is
vo ≅ V
sin ω t
DD
ELE2110A © 2008
⎛ VDD ⎞
⎜⎜
⎟⎟
2
VDD
2⎠
⎝
=
Pac =
RL
2 RL
Efficiency of amplifier is:
V 2 /( 2 R )
P
L
ζ = ac = DD
P
2I V
av
SS DD
V 2 /( 2 R )
L
≤ ⎛ DD
= 25 %
⎞
⎟
⎜
2⎜⎜V
/ R ⎟⎟V
L ⎠ DD
⎝ DD
I SS ≥
VDD
RL
- Low efficiency
Lecture 12 - 9
Push-Pull Operation: Class B
When a push-pull amplifier is operated in Class B, all of the output current
comes either from the current-sourcing transistor or from the currentsinking device but never from both at the same time.
Source: B. Putzeys, “Digital Audio’s final frontier”, IEEE Spectrum, Mar 2003.
ELE2110A © 2008
Lecture 12 - 10
Class-B Amplifier
z
z
A complementary pair of source followers
biased at zero source current
When
V ≤ v ≤V
I
TN
TP
z
ELE2110A © 2008
neither transistor conducts
No quiescent (DC) current consumption!
Lecture 12 - 11
Class-B Amplifier
z
When VI > VTN,
– M1 turns on and acts as an source
follower, vo ≈ vI - VTN
– M2 off
z
When VI < VTP,
– M2 turns on and acts as an source
follower, vo ≈ vI – VTP
– M1 off
ELE2110A © 2008
z
Power efficiency is high, can be up to
about 80%
z
Disadv.: Output waveform suffers from a
dead-zone Æ Large distortion
Lecture 12 - 12
Class AB
Class AB exhibits less distortion by allowing the transistors to work
together when the output signal is near zero, in what is called the
crossover region.
Source: B. Putzeys, “Digital Audio’s final frontier”, IEEE Spectrum, Mar 2003.
ELE2110A © 2008
Lecture 12 - 13
Class-AB Amplifiers
z
Remove dead zone by biasing transistors into conduction but at a low
quiescent current level
– Distortion less than Class-B but worse than Class-A amplifier
z
z
For each transistor, 1800<conduction angle<3600 Æ Class AB amplifier
Power efficiency lower than Class-B but higher than Class-A amplifier
ELE2110A © 2008
Lecture 12 - 14
Class-AB Amplifiers
Biasing examples:
DC currents:
⎞
K n ⎛⎜V
⎟
⎜ GG −V
⎟
I =
⎜
⎟
D
TN
⎟
2 ⎜⎝ 2
⎠
ELE2110A © 2008
2
I R ⎞⎟
I = I exp B B ⎟⎟
C S
2V ⎟⎟
T ⎠
⎛
⎜
⎜
⎜
⎜
⎜
⎝
Lecture 12 - 15
Topics to cover …
z
Output Stages
z
Amplifier Frequency Response
ELE2110A © 2008
Lecture 12 - 16
Frequency Response of Amplifiers
A typical amplifier:
Block DC and low frequency signals
By-pass high frequency currents
Amplifier’s gain is frequency dependent!
ELE2110A © 2008
Lecture 12 - 17
Typical Amplifier Transfer Function
Mid-band gain
Lower cut-off frequency
z
z
Upper cut-off frequency
In low frequency side, drop in gain is caused by coupling and
bypass capacitors
In high frequency side, drop in gain is caused by transistor’s
parasitic capacitors
– More on this topic later
z
In the mid-band range, no capacitors are in effect:
– Coupling and bypass capacitors are short circuits
– Transistor parasitic capacitors are open circuits
ELE2110A © 2008
Lecture 12 - 18
Estimate fL: Short-Circuit Time Constant Method
z
Lower cutoff frequency for a network
with n coupling and bypass
capacitors can be estimated by:
n
1
L i =1 R C
iS i
f = ω / 2π
L L
ω ≅
∑
RiS = resistance at terminals of ith
capacitor Ci with all other capacitors
replaced by short circuits.
Product RiS Ci is “short-circuit time
constant” associated with Ci.
ELE2110A © 2008
Lecture 12 - 19
SCTC: Example
β= 100 and VA=75V
Q-point: (1.73mA, 2.32V)
AC equivalent with finite
coupling capacitances
BJT small signal parameters:
rπ =
VT
25mV
=
= 1.45kΩ
I B 1.73mA / 100
ELE2110A © 2008
VA + VCE 75V + 2.32V
ro =
=
= 44.7kΩ
IC
1.73mA
Lecture 12 - 20
Time Constant Associated with C1
To find the time constant associated
with C1:
(C2 and C3 are short-circuited and set vi = 0)
) = R + ( R rπ )
R = R + ( R RCE
I
B
I
B in
1S
R1s = 1000Ω + (7500Ω || 1450Ω) = 2220Ω
1
1
=
= 225 rad/s
R1s C1 2.22kΩ ⋅ 2 µF
ELE2110A © 2008
Lecture 12 - 21
Time Constant Associated with C2
To find the time constant associated
with C2:
(C1 and C3 are short-circuited and set vi = 0)
R = R + ( R RCE
) = R + (R ro ) ≅ R + R
C
2S
3
C out
3
3 C
R2 s = 100kΩ + (4.3kΩ || 44.7kΩ) = 104kΩ
1
1
=
= 96.1 rad/s
R2 s C2 104kΩ ⋅ 0.1µF
ELE2110A © 2008
Lecture 12 - 22
Time Constant Associated with C3
rπ + R
th
CC
R = R R out = R
E
E β +1
3S
rπ + (R R )
I B
=R
E
β +1
To find the time
constant associated
with C3:
1450V + 1000Ω || 7500Ω
R3s = 1300kΩ ||
101
= 22.7Ω
1
1
=
= 4410 rad/s
R3 s C2 22.7kΩ ⋅10 µF
(C1 and C2 are short-circuited and set vi = 0)
ELE2110A © 2008
Lecture 12 - 23
Lower Cutoff Frequency
The lower cutoff frequency is:
ω ≅
L
and
3
1
= 225 + 96.1+ 4410 = 4730 rad/s
∑
i =1 RiS Ci
ωL
fL =
= 753 Hz
2π
In this example the time constant associated with the bypass capacitor
C3 is dominant.
ELE2110A © 2008
Lecture 12 - 24
High Frequency Response
Mid-band gain
Lower cut-off frequency
z
Upper cut-off frequency
At high frequency side, drop in gain is caused by transistor’s
parasitic capacitors
ELE2110A © 2008
Lecture 12 - 25
High Frequency Small Signal Model for BJT
B’
Model for active mode BJT
Cπ: diffusion capacitance of the forward-biased base-emitter junction.
Cµ: depletion capacitance of the reverse-biased base-collector junction.
rx: the resistance of the silicon material of the base region between the
base terminal and the intrinsic base terminal B’ that is right under the
emitter region.
ELE2110A © 2008
Lecture 12 - 26
High-frequency Small Signal Model for
MOSFET
Lov
Model for saturation mode MOSFET
2
C gs = WLCox = the capacitance between the Gate and the conducting channel.
3
C gd = Cov = WLov Cox = the overlap capacitance (very small).
ELE2110A © 2008
Lecture 12 - 27
Open-Circuit Time Constant Method to
Determine fH
fH can be estimated by open-circuit time constant method:
ω ≅ m 1
H
∑ RioCi
i =1
,
f
H
= ω / 2π
H
where Rio is resistance at terminals of ith capacitor Ci with all other
capacitors open-circuited.
ELE2110A © 2008
Lecture 12 - 28
High Frequency Analysis of C-E
Amplifier
R = R || R = 30kΩ 10kΩ
B 1 2
R = R R = 100kΩ 4.3kΩ
L 3 C
β = 100 and VA=75V
Let Q-point be (1.6mA, 3V)
Cπ = 19.9 pF
C µ = 0.5 pF
rx = 250Ω
ELE2110A © 2008
R
B
v =v
th i R + R
I
B
R R
R = I B = 882Ω
th R + R
I
B
Lecture 12 - 29
High Frequency Small Signal Equivalent
Norton source transformation
v
th
is =
R + rx
th
rπo = rπ ( R + rx )
th
ELE2110A © 2008
Lecture 12 - 30
Determine Amid
v2 = − g m (is rπ 0 ) RL
v
th
is =
R + rx
th
rπo = rπ ( R + rx )
th
rπ 0
rπ
v2
= − g m RL
= − g m RL
vth
Rth + rx
Rth + rx + rπ
Amid = −
ELE2110A © 2008
β o RL
Rth + rx + rπ
=−
100(4120)
= −153
882 + 250 + 1560
Lecture 12 - 31
OCTC: Time Constant Associated with Cπ
To find the time constant
associated with Cπ:
(Cµ is open-circuited and set is = 0)
Rπo = rπo = rπ || ( R + rx ) = 1.56kΩ || (882Ω + 250Ω) = 656Ω
th
Cπ = 19.9 pF
Cπ Rπ 0 = 1.3 ×10 −8
ELE2110A © 2008
Lecture 12 - 32
OCTC: Time Constant Associated with Cµ
To find the time constant
associated with Cµ:
(Cπ is open-circuited and set is = 0)
v x = ix rπ 0 + iL RL = ix rπ 0 + (ix + g m v) RL
⎛
R ⎞⎟
⎜
= rπo ⎜1+ g m R + L ⎟
Rµo =
⎜
L r ⎟⎟
ix
⎜
πo ⎠
⎝
⎛
∴ R o = 656Ω⎜⎜1+ 0.064(4120) +
⎝
vx
µ
v = i x rπ 0
4120 ⎞⎟
= 178kΩ
⎟
656 ⎠
Cµ Rµ 0 = 89.0 ×10 −9
ELE2110A © 2008
Lecture 12 - 33
Upper Cutoff Frequency
Upper cutoff frequency:
ω
H
≅
fH =
ELE2110A © 2008
1
1
=
= 9.8×106 rad/s
RπoCπ + RµoCµ 1.3×10− 8 + 89×10− 9
ωH
= 1.56 MHz
2π
Lecture 12 - 34