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ESE319 Introduction to Microelectronics
Output Stages
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Power amplifier classification
Class A amplifier circuits
Class A Power conversion efficiency
Class B amplifier circuits
Class B Power conversion efficiency
Class AB amplifier circuits
Class AB Power conversion efficiency
2008 Kenneth R. Laker updated 29Nov10
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ESE319 Introduction to Microelectronics
Output Stage Functions




Provide amplifier with low output resistance
Handle large signals with low THD
Deliver power to the load efficiently
Output stages are classified according to
the iC waveform due to input vI waveform
2008 Kenneth R. Laker updated 29Nov10
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ESE319 Introduction to Microelectronics
Amplifier Classifications
VCC
iC
RC
RB
vI
VB
Class A amplifier – amplifier BJT
conducts for entire vI cycle. For all vI:
v I V B ≥ 0.7 V
where V B max v I 0.7 V
Ic = amplitude of current
due to vi.
RE
IC = dc current
Transistor cut off (iC = 0) if:
v I V B 0.7 V
2008 Kenneth R. Laker updated 29Nov10
NOTE: when vI = 0, iC = IC
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ESE319 Introduction to Microelectronics
Amplifier Classifications - cont.
Class B – Amplifier BJT conducts positive-half
of vI cycle.
Amp BJT conducts for all vI s.t.:
V B =0 V ⇒ v I ≥0.7 V
Transistor cut off (iC = 0) if:
v I V B 0.7 V
2008 Kenneth R. Laker updated 29Nov10
NOTE: 1. when vI < 0.7V, iC = 0
2. a 2nd class B BJT is needed to
conduct for the negative vI cycle.
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ESE319 Introduction to Microelectronics
Amplifier Classifications - cont.
Class AB – Amplifier BJT conducts for positive
vI swing + part of negative vI swing s.t.:
v I V B ≥ 0.7 V where 0V B max v I 0.7 V
Conducts for:
v I ≥0.7−V B
Cut-off for rest of
negative vI swing:
Transistor cut-off (iC = 0) if:
v I V B 0.7 V
2008 Kenneth R. Laker updated 29Nov10
NOTE: 1. when vI = 0, iC = IC
2. a 2nd class AB BJT is needed to
conduct for interval slightly larger
than the negative vI cycle.
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ESE319 Introduction to Microelectronics
Class A Power Amplifier Design
Used as op amp output stage and some audio output power amps.
Basic considerations for low (audio) frequency operation.
1. Power usually delivered to a low impedance load.
2. Signal usually has little, preferably no dc content.
3. May have low frequency content, as low as 20 Hz.
Emitter follower circuit has best power transfer efficiency, since its
output impedance is low. As a bonus, its input impedance is
relatively high.
Principal advantage – lower distortion than Class B & AB.
Principal disadvantage – lower power efficiency than Class B & AB.
2008 Kenneth R. Laker updated 29Nov10
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ESE319 Introduction to Microelectronics
Current Biased Class A Emitter Follower
+VCC
Q3
diode
connected
transistor
I
-VCC
-VCC
current mirror
2008 Kenneth R. Laker updated 29Nov10
A current mirror establishes the
bias current.
To operate reliably,
1. Q1 and Q2 must be forward active.
2. Current Mirror Q3 and Q2 need to be
matched as well as possible and be at
the same ambient temperature.
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ESE319 Introduction to Microelectronics
Class A Amplifier Analysis
Consider the case when vI ≥ vBE1 = 0.7 V (pos. swing of vI):
if vCE1 < VCE1-sat => Q1 sat.
+
vCE1
-
i L =i E1 −I
vO =v I −0.7V =i L R L
vCE1=V CC −v O
For Q1 ≠ sat: v CE1V CE1− sat
v CE1=V CC −v O V CE1− sat
−v O −V CC V CE1− sat
v O V CC −V CE1− sat ⇒ v O−max=V CC −V CE1− sat
vCE2 =v OV CC
-VCC
-VCC
v I ≥v BE1=0.7V
vO =v I −v BE1=v I −0.7 V
vCE1=V CC −v O
2008 Kenneth R. Laker updated 29Nov10
Max values: Q1 ≠ sat.
v O vO−max =V CC −V CE1−sat
v I v I −max=V CC −V CE1− sat 0.7V
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ESE319 Introduction to Microelectronics
Class A Amplifier Analysis - cont.
Consider the case where vI < 0.7 V (neg. swing of vI):
vCE1=V CC −v O
i L =i E1 −I
if iE1 = 0 => Q1 off
i E1=i L I 0
v CE2V CE2−sat
vO
v CE2=v O V CC V CE2−sat
i L = −I
RL
⇒ v O −V CC V CE2− sat
v O −I R L
v I −V CC V CE2− sat 0.7
v I −I R L 0.7 V
Min values: Q1 and Q2 forward-active
vO =v I −0.7V =i L R L
vCE2 =v OV CC
+
vCE2
-VCC
-VCC
if vCE2 < VCE2-sat => Q2 sat.
v I 0.7 V
v O v O−min =max {−IR L ,−V CC V CE2−sat }
vO =v I −0.7 & vO =i L R L
v I v I −min=max {−IR L0.7 V ,−V CC V CE2−sat 0.7 V }
vCE2 =v OV CC =v I −0.7V CC
2008 Kenneth R. Laker updated 29Nov10
NOTE: max means least negative
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ESE319 Introduction to Microelectronics
Class A Amplifier VTC – Plot
vO
Q1 saturated
VCC - VCE1-sat
max vO swing iff
RL ≥
or
Iff −IR L−V CC −V CE2−sat 
V CC −V CE2−sat
I
V CC −V CE2−sat
I≥
RL
Q1 cutoff
Q2 saturated
Bias current I & RL set limits
on negative vO = vO-min swing
vO =v I −V BE1
slope = 1
VBE1
-IRL
vI
V CC −V CE2− sat
R L
I
-VCC + VCE2-sat
v O v O−min=−IR L ⇒
V CC −V CE2−sat
R L
I
If −IR L−V CC −V CE2−sat 
v O v O−min=−V CC V CE2−sat ⇒
V CC −V CE2−sat
R L
I
v O =v I −0.7 where max {−IR L ,−V CC −V CE2−sat }v O V CC −V CE1−sat
2008 Kenneth R. Laker updated 29Nov10
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ESE319 Introduction to Microelectronics
Class A Stage VTC Simulation
I =120 mA
IR L =V CC =12 V ⇒ R L =100 
11.8 V
Q1 cutoff
ideal
current
source
i.e. no
Q2
Q1 saturated
- 12 V
VEE = - VCC
Q1 forward active => no clipping
v O v O−max =V CC −V CE1−sat =11.8V Q1 not Sat
v O v O−min =−IR L=−12 V
Q1 not Cut-off
2008 Kenneth R. Laker updated 29Nov10
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ESE319 Introduction to Microelectronics
Class A Stage VTC Simulation - cont.
I =120 mA
R L =75 
IR L =9V V CC −V CE1−sat
Q1 saturated
11.8 V
Q1 cutoff
-9V
2008 Kenneth R. Laker updated 29Nov10
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ESE319 Introduction to Microelectronics
Quick Review
Class A Amp VTC
vCE1=V CC −v O
max vo swing i.f.f.
or
I≥
V CC −V CE2−sat
RL
v O =v I −0.7 where max {−IR L ,−V CC −V CE2−sat }v O V CC −V CE1−sat
2008 Kenneth R. Laker updated 29Nov10
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ESE319 Introduction to Microelectronics
+VCC = +15 V
Example
I
R L =1 k 
-VCC = -15 V
2008 Kenneth R. Laker updated 29Nov10
Let VCE1-sat = VCE2-sat = 0.2 V, VBE1 = VBE2 = 0.7 V
and 1 =2=large
.
1. Determine the value for resistor R that will set the
bias current I sufficiently large to allow the largest
possible output voltage vO swing.
2. Determine the resulting output voltage swing and
the maximum and minimum Q1 emitter currents.
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ESE319 Introduction to Microelectronics
Example cont.
I
SOLUTION:
R L =1 k 
-VCC
1. For maximum output voltage swing:
IR L =V CC −V CE2−sat where RL = 1 kΩ
V CC −V CEsat 15 V −0.2V
I=
=
=14.8 mA
RL
1k
R=
2008 Kenneth R. Laker updated 29Nov10
V CC −V BE 15 V −0.7V
=
=0.97 k 
I
14.8 mA
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ESE319 Introduction to Microelectronics
Example - cont.
From Part 1:
I =14.8 mA
SOLUTION:
2. Output voltage swing:
V o− peak = I R L =14.8 V ⇒−14. 8V v O 14.8 V
-VCC = -15 V
−14.8 V v O 14. 8V ⇒− I i L  I
0−V BE2−V CC 
I=
R
Max and min Q1 emitter currents:
2008 Kenneth R. Laker updated 29Nov10
i E1= I i L ⇒ 0 mAi E1 2I=29.6 mA
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ESE319 Introduction to Microelectronics
Instantaneous and Average Power
The source of power to the amplifier load, RL, comes from the
supplies,VCC and – VCC. The supplies deliver power, and the
load and the transistors absorb it.
Instantaneous power absorbed by resistor RL:
2
2
pa t =v ab t ia b t =iab t  R L =va b t / R L
i ab
RL
Instantaneous power delivered by battery VCC:
pd t=v ab t i ba t=V CC I
Average power: i ab =I ab− peak sin  t for period T
T
2
2
I
V
1
ab− peak
ab− peak
P ab av = ∫ i 2ab t  R L dt =I 2ab−rms R L=
R L=
T 0
2
2 RL
P D av =V CC I
2008 Kenneth R. Laker updated 29Nov10
i ba = I
V CC
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ESE319 Introduction to Microelectronics
Emitter Follower Power Relationships
I. Average power delivered by the batteries:
For the current mirror transistor side:
P −VCC =V CC I
For the amplifier transistor side:
T
1
PVCC = ∫ V CC i C1 dt where iC1 =I  Ic sin t 
T 0
T
Ic
1
P VCC =V CC I ∫ 1 sin  t  dt=V CC I
T 0
I
Total delivered power:
iC
-VCC
-VCC
P D =P D av =P−VCC PVCC =2 V CC I
II. Average power to the load:
2
2
2
V
V
/
2
V

o−rms
o− peak
o− peak
v O =V o− peak sin  t 
P L av =
=
=
RL
RL
2 RL
2008 Kenneth R. Laker updated 29Nov10
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ESE319 Introduction to Microelectronics
Class A Power Conversion Efficiency
Using the power delivered to transistors from the batteries PDav
and the power delivered to the load PLav:
P D av =2V CC I
and
V 2o− peak
P L av =
2 RL
Note:
1. Average currents and PD av from the power supply do not change with
the signal level Vo-peak.
2. PL av increases with the square of the signal level Vo-peak.
power conversion efficiency
2
2
P L av V o−
/2
R
V
1 o− peak 1 V o− peak V o− peak
peak
L
=
=
=
=
P D av
2V CC I
4 V CC I R L 4 I R L
V CC
2008 Kenneth R. Laker updated 29Nov10
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ESE319 Introduction to Microelectronics
Power Conversion Efficiency
2
V
1 o− peak 1 V o− peak V o− peak
=
=
4 I R L V CC 4 I R L
V CC
Since V o− peak V CC and V o− peak  I R L :
Maximum power conversion efficiency is realized when
V o− peak =V CC =I R L
Hence:
ignoring the VCE1-sat and VCE2-sat
1 V CC V CC 1
max =
=
or 25 %
4 V CC V CC 4
2008 Kenneth R. Laker updated 29Nov10
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ESE319 Introduction to Microelectronics
Class A Power Efficiency Simulation
V o− peak =V i− peak −0.7
V o− peak =12−0.7=11.3V V CC −V CE−sat = I R L
98 Ω
P L av =692.36 mW
P D av=1.33W 1.44 W =2.77 W
P L av 0.69 W
=
=
=0.249≈0.25
P D av 2.77 W
2008 Kenneth R. Laker updated 29Nov10
R L =98
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ESE319 Introduction to Microelectronics
Class A Power Simulation - cont.
V o− peak =V i− peak −0.7
V o− peak =6−0.7=5.3 V V CC −V CE −sat = I R L
98 Ω
P L av =182.28 mW
P D av=1.33W 1.44 W =2.77 W
P L av 0.18 W
=
=
=0.0650.25
P D av 2.77 W
2008 Kenneth R. Laker updated 29Nov10
R L =98
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ESE319 Introduction to Microelectronics
Conclusions
1. The class A amplifier provides the most “nearly linear”
amplification of its input, but this comes at a price:
The best power conversion efficiency that can be obtained
is 25%.
2. That is 75% of the power supplied by the sources is dissipated
in the transistors. This is a waste of power, and it leads to
a potentially serious heating problems with the transistors.
All of this constant battery power is dissipated in the transistors
even when no signal is applied – zero percent efficiency!
Next we will consider a much more efficient amplifier configuration
– the class B amplifier.
2008 Kenneth R. Laker updated 29Nov10
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