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Fundamentals of Microelectronics
CH1 Why Microelectronics?
CH2 Basic Physics of Semiconductors
CH3 Diode Circuits
CH4 Physics of Bipolar Transistors
CH5 Bipolar Amplifiers
CH6 Physics of MOS Transistors
CH7 CMOS Amplifiers
CH8 Operational Amplifier As A Black Box
2
Chapter 5 Bipolar Amplifiers
5.1 General Considerations
5.2 Operating Point Analysis and Design
5.3 Bipolar Amplifier Topologies
5.4 Summary and Additional Examples
CH5 Bipolar Amplifiers 3
Bipolar Amplifiers
CH5 Bipolar Amplifiers 4
Voltage Amplifier
In an ideal voltage amplifier, the input impedance is infinite and the output impedance zero.
But in reality, input or output impedances depart from their ideal values.
2
CH5 Bipolar Amplifiers 5
Input/Output Impedances
The figure above shows the techniques of measuring input
and output impedances.
x
x
xi
VR =
CH5 Bipolar Amplifiers 6
Input Impedance Example I
When calculating input/output impedance, small-signal
analysis is assumed.
πri
v
x
x =
CH5 Bipolar Amplifiers 7
Impedance at a Node
When calculating I/O impedances at a port, we usually ground one terminal while applying the test source to the
other terminal of interest.
CH5 Bipolar Amplifiers 8
Impedance at Collector
With Early effect, the impedance seen at the collector is equal to the intrinsic output impedance of the transistor (if
emitter is grounded).
ooutrR =
3
CH5 Bipolar Amplifiers 9
Impedance at Emitter
The impedance seen at the emitter of a transistor is approximately equal to one over its transconductance (if the base is grounded).
)(
1
1
1
∞=
≈
+
=
A
m
out
mx
x
V
gR
rg
i
v
π
CH5 Bipolar Amplifiers 10
Three Master Rules of Transistor Impedances
Rule # 1: looking into the base, the impedance is rππππ ifemitter is (ac) grounded.
Rule # 2: looking into the collector, the impedance is ro if emitter is (ac) grounded.
Rule # 3: looking into the emitter, the impedance is 1/gm if base is (ac) grounded and Early effect is neglected.
CH5 Bipolar Amplifiers 11
Biasing of BJT
Transistors and circuits must be biased because (1)
transistors must operate in the active region, (2) their small-signal parameters depend on the bias conditions.
CH5 Bipolar Amplifiers 12
DC Analysis vs. Small-Signal Analysis
First, DC analysis is performed to determine operating point and obtain small-signal parameters.
Second, sources are set to zero and small-signal model is used.
4
CH5 Bipolar Amplifiers 13
Notation Simplification
Hereafter, the battery that supplies power to the circuit is replaced by a horizontal bar labeled Vcc, and input signal is simplified as one node called Vin.
CH5 Bipolar Amplifiers 14
Example of Bad Biasing
The microphone is connected to the amplifier in an attempt to amplify the small output signal of the microphone.
Unfortunately, there’s no DC bias current running thru the
transistor to set the transconductance.
CH5 Bipolar Amplifiers 15
Another Example of Bad Biasing
The base of the amplifier is connected to Vcc, trying to establish a DC bias.
Unfortunately, the output signal produced by the microphone is shorted to the power supply.
CH5 Bipolar Amplifiers 16
Biasing with Base Resistor
Assuming a constant value for VBE, one can solve for both IB and IC and determine the terminal voltages of the transistor.
However, bias point is sensitive to ββββ variations.
B
BECC
C
B
BECC
BR
VVI
R
VVI
−=
−= β,
5
CH5 Bipolar Amplifiers 17
Improved Biasing: Resistive Divider
Using resistor divider to set VBE, it is possible to produce an IC that is relatively independent of ββββ if base current is small.
)exp(21
2
21
2
T
CC
SC
CCX
V
V
RR
RII
VRR
RV
+=
+=
CH5 Bipolar Amplifiers 18
Accounting for Base Current
With proper ratio of R1 and R2, IC can be insensitive to ββββ; however, its exponential dependence on resistor deviations makes it less useful.
−=
T
ThevBThev
SCV
RIVII exp
CH5 Bipolar Amplifiers 19
Emitter Degeneration Biasing
The presence of RE helps to absorb the error in VX so VBE
stays relatively constant.
This bias technique is less sensitive to ββββ (I1 >> IB) and VBE
variations.
20
Design Procedure
Choose an IC to provide the necessary small signal
parameters, gm, rππππ, etc.
Considering the variations of R1, R2, and VBE, choose a
value for VRE.
With VRE chosen, and VBE calculated, Vx can be determined.
Select R1 and R2 to provide Vx.
6
CH5 Bipolar Amplifiers 21
Self-Biasing Technique
This bias technique utilizes the collector voltage to provide the necessary Vx and IB.
One important characteristic of this technique is that collector has a higher potential than the base, thus guaranteeing active operation of the transistor.
CH5 Bipolar Amplifiers 22
Self-Biasing Design Guidelines
(1) provides insensitivity to ββββ .
(2) provides insensitivity to variation in VBE .
BECCBE
B
C
VVV
RR
−<<∆
>>β
)2(
)1(
CH5 Bipolar Amplifiers 23
Summary of Biasing Techniques
CH5 Bipolar Amplifiers 24
PNP Biasing Techniques
Same principles that apply to NPN biasing also apply to
PNP biasing with only polarity modifications.
7
CH5 Bipolar Amplifiers 25
Possible Bipolar Amplifier Topologies
Three possible ways to apply an input to an amplifier and three possible ways to sense its output.
However, in reality only three of six input/output
combinations are useful.26
Study of Common-Emitter Topology
Analysis of CE Core
Inclusion of Early Effect
Emitter Degeneration
Inclusion of Early Effect
CE Stage with Biasing
CH5 Bipolar Amplifiers 27
Common-Emitter Topology
CH5 Bipolar Amplifiers 28
Small Signal of CE Amplifier
Cmv
inmm
C
out
in
out
v
RgA
vgvgR
v
v
vA
−=
==−
=
π
8
CH5 Bipolar Amplifiers 29
Limitation on CE Voltage Gain
Since gm can be written as IC/VT, the CE voltage gain can be written as the ratio of VRC and VT.
VRC is the potential difference between VCC and VCE, and VCE cannot go below VBE in order for the transistor to be in active region.
T
CC
vV
RIA =
T
RC
vV
VA =
T
BECC
vV
VVA
−<
CH5 Bipolar Amplifiers 30
Tradeoff between Voltage Gain and Headroom
CH5 Bipolar Amplifiers 31
I/O Impedances of CE Stage
When measuring output impedance, the input port has to
be grounded so that Vin = 0.
πri
vR
X
X
in==
C
X
X
outR
i
vR ==
CH5 Bipolar Amplifiers 32
CE Stage Trade-offs
9
CH5 Bipolar Amplifiers 33
Inclusion of Early Effect
Early effect will lower the gain of the CE amplifier, as it
appears in parallel with RC.
OCout
OCmv
rRR
rRgA
||
)||(
=
−=
CH5 Bipolar Amplifiers 34
Intrinsic Gain
As RC goes to infinity, the voltage gain reaches the product of gm and rO, which represents the maximum voltage gain
the amplifier can have.
The intrinsic gain is independent of the bias current.
T
A
v
Omv
V
VA
rgA
=
−=
CH5 Bipolar Amplifiers 35
Current Gain
Another parameter of the amplifier is the current gain, which is defined as the ratio of current delivered to the load to the current flowing into the input.
For a CE stage, it is equal to ββββ.
β=
=
CEI
in
out
I
A
i
iA
CH5 Bipolar Amplifiers 36
Emitter Degeneration
By inserting a resistor in series with the emitter, we “degenerate” the CE stage.
This topology will decrease the gain of the amplifier but improve other aspects, such as linearity, and input impedance.
10
CH5 Bipolar Amplifiers 37
Small-Signal Model
Interestingly, this gain is equal to the total load resistance
to ground divided by 1/gm plus the total resistance placed in series with the emitter.
E
m
C
v
Em
Cm
v
Rg
RA
Rg
RgA
+
−=
+−=
1
1
CH5 Bipolar Amplifiers 38
Emitter Degeneration Example I
The input impedance of Q2 can be combined in parallel with
RE to yield an equivalent impedance that degenerates Q1.
2
1
||1
πrRg
RA
E
m
C
v
+
−=
CH5 Bipolar Amplifiers 39
Emitter Degeneration Example II
In this example, the input impedance of Q2 can be combined in parallel with RC to yield an equivalent collector impedance to ground.
E
m
C
v
Rg
rRA
+
−=
1
2
1
|| π
CH5 Bipolar Amplifiers 40
Input Impedance of Degenerated CE Stage
With emitter degeneration, the input impedance is
increased from rππππ to rππππ + (ββββ+1)RE; a desirable effect.
E
X
X
in
XEXX
A
Rri
vR
iRirv
V
)1(
)1(
++==
++=
∞=
β
β
π
π
11
CH5 Bipolar Amplifiers 41
Output Impedance of Degenerated CE Stage
Emitter degeneration does not alter the output impedance
in this case. (More on this later.)
C
X
X
out
Emin
A
Ri
vR
vRvgr
vvv
V
==
=⇒
++==
∞=
00 ππ
π
π
π
CH5 Bipolar Amplifiers 42
Capacitor at Emitter
At DC the capacitor is open and the current source biases the amplifier.
For ac signals, the capacitor is short and the amplifier is
degenerated by RE.
CH5 Bipolar Amplifiers 43
Example: Design CE Stage with Degeneration as a Black Box
If gmRE is much greater than unity, Gm is more linear.
Em
m
in
out
m
Em
in
mout
A
Rg
g
v
iG
Rgr
vgi
V
+≈=
++=
∞=
−
1
)(1 1
π
CH5 Bipolar Amplifiers 44
Degenerated CE Stage with Base Resistance
1
1
)1(
.
+++
−≈
+++
−=
=
∞=
β
β
β
π
B
E
m
C
v
BE
C
in
out
A
out
in
A
in
out
A
RR
g
RA
RRr
R
v
v
v
v
v
v
v
v
V
12
CH5 Bipolar Amplifiers 45
Input/Output Impedances
Rin1 is more important in practice as RB is often the output
impedance of the previous stage.
Cout
EBin
Ein
A
RR
RrRR
RrR
V
=
+++=
++=
∞=
)1(
)1(
22
1
β
β
π
π
CH5 Bipolar Amplifiers 46
Emitter Degeneration Example III
1
2
2
1
||
)1(
1
1
)||(
RRR
RrR
RR
g
RRA
Cout
in
B
m
C
v
=
++=
+++
−=
β
β
π
CH5 Bipolar Amplifiers 47
Output Impedance of Degenerated Stage with VA<
Emitter degeneration boosts the output impedance by a factor of 1+gm(RE||rππππ).
This improves the gain of the amplifier and makes the circuit a better current source.
∞
[ ]
[ ])||(1
)||)(1(
||)||(1
π
π
ππ
rRgrR
rRrgrR
rRrrRgR
EmOout
EOmOout
EOEmout
+≈
++=
++=
CH5 Bipolar Amplifiers 48
Two Special Cases
OEmout
E
OmOout
E
rRgR
rR
rrgrR
rR
)1(
)1(
+≈
<<
≈+≈
>>
π
π
π
β
)2
)1
13
CH5 Bipolar Amplifiers 49
Analysis by Inspection
This seemingly complicated circuit can be greatly simplified by first recognizing that the capacitor creates an AC short to ground, and gradually transforming the circuit to a known topology.
[ ]12
||)||(1 RrrRgROmout π+=[ ]
OmoutrrRgR )||(1
21 π+=11
||outout
RRR =
CH5 Bipolar Amplifiers 50
Example: Degeneration by Another Transistor
Called a “cascode”, the circuit offers many advantages that
are described later in the book.
[ ]1121
)||(1OOmout
rrrgR π+=
51
Study of Common-Emitter Topology
Analysis of CE Core
Inclusion of Early Effect
Emitter Degeneration
Inclusion of Early Effect
CE Stage with Biasing
CH5 Bipolar Amplifiers 52
Bad Input Connection
Since the microphone has a very low resistance that connects from the base of Q1 to ground, it attenuates the
base voltage and renders Q1 without a bias current.
14
CH5 Bipolar Amplifiers 53
Use of Coupling Capacitor
Capacitor isolates the bias network from the microphone at DC but shorts the microphone to the amplifier at higher
frequencies.
CH5 Bipolar Amplifiers 54
DC and AC Analysis
Coupling capacitor is open for DC calculations and shorted
for AC calculations.
OCout
Bin
OCmv
rRR
RrR
rRgA
||
||
)||(
=
=
−=
π
CH5 Bipolar Amplifiers 55
Bad Output Connection
Since the speaker has an inductor, connecting it directly to the amplifier would short the collector at DC and therefore push the transistor into deep saturation.
CH5 Bipolar Amplifiers 56
Still No Gain!!!
In this example, the AC coupling indeed allows correct biasing. However, due to the speaker’s small input impedance, the overall gain drops considerably.
15
CH5 Bipolar Amplifiers 57
CE Stage with Biasing
OCout
in
OCmv
rRR
RRrR
rRgA
||
||||
)||(
21
=
=
−=
π
CH5 Bipolar Amplifiers 58
CE Stage with Robust Biasing
[ ]
Cout
Ein
E
m
C
v
RR
RRRrR
Rg
RA
=
++=
+
−=
21 ||||)1(
1
βπ
∞=A
V
CH5 Bipolar Amplifiers 59
Removal of Degeneration for Signals at AC
Capacitor shorts out RE at higher frequencies and
removes degeneration.
Cout
in
Cmv
RR
RRrR
RgA
=
=
−=
21 ||||π
CH5 Bipolar Amplifiers 60
Complete CE Stage
1
||||1
||
21
+++
−=
β
RRRR
g
RRA
s
E
m
LC
v
16
CH5 Bipolar Amplifiers 61
Summary of CE Concepts
CH5 Bipolar Amplifiers 62
Common Base (CB) Amplifier
In common base topology, where the base terminal is biased with a fixed voltage, emitter is fed with a signal, and collector is the output.
CH5 Bipolar Amplifiers 63
CB Core
The voltage gain of CB stage is gmRC, which is identical to
that of CE stage in magnitude and opposite in phase.
CmvRgA =
CH5 Bipolar Amplifiers 64
Tradeoff between Gain and Headroom
To maintain the transistor out of saturation, the maximum
voltage drop across RC cannot exceed VCC-VBE.
T
BECC
C
T
C
v
V
VV
RV
IA
−=
= .
17
CH5 Bipolar Amplifiers 65
Simple CB Example
Ω=
Ω=
==
KR
KR
RgACmv
7.67
3.22
2.17
2
1
CH5 Bipolar Amplifiers 66
Input Impedance of CB
The input impedance of CB stage is much smaller than that of the CE stage.
m
ing
R1
=
CH5 Bipolar Amplifiers 67
Practical Application of CB Stage
To avoid “reflections”, need impedance matching.
CB stage’s low input impedance can be used to create a match with 50 ΩΩΩΩ.
CH5 Bipolar Amplifiers 68
Output Impedance of CB Stage
The output impedance of CB stage is similar to that of CE stage.
COoutRrR ||=
18
CH5 Bipolar Amplifiers 69
CB Stage with Source Resistance
With an inclusion of a source resistor, the input signal is attenuated before it reaches the emitter of the amplifier; therefore, we see a lower voltage gain.
This is similar to CE stage emitter degeneration; only the phase is reversed.
S
m
C
v
Rg
RA
+
=1
CH5 Bipolar Amplifiers 70
Practical Example of CB Stage
An antenna usually has low output impedance; therefore, a correspondingly low input impedance is required for the following stage.
CH5 Bipolar Amplifiers 71
Realistic Output Impedance of CB Stage
The output impedance of CB stage is equal to RC in parallel
with the impedance looking down into the collector.
[ ] ( )
1
1
||
||)||(1
outCout
EOEmout
RRR
rRrrRgR
=
++= ππ
CH5 Bipolar Amplifiers 72
Output Impedance of CE and CB Stages
The output impedances of CE, CB stages are the same if both circuits are under the same condition. This is because when calculating output impedance, the input port is grounded, which renders the same circuit for both CE and CB stages.
19
CH5 Bipolar Amplifiers 73
Fallacy of the “Old Wisdom”
The statement “CB output impedance is higher than CE output impedance” is flawed.
CH5 Bipolar Amplifiers 74
CB with Base Resistance
With an addition of base resistance, the voltage gain degrades.
m
B
E
C
in
out
g
RR
R
v
v
1
1+
++
≈
β
CH5 Bipolar Amplifiers 75
Comparison of CE and CB Stages with Base Resistance
The voltage gain of CB amplifier with base resistance is exactly the same as that of CE stage with base resistance and emitter degeneration, except for a negative sign.
CH5 Bipolar Amplifiers 76
Input Impedance of CB Stage with Base Resistance
The input impedance of CB with base resistance is equal to 1/gm plus RB divided by (ββββ+1). This is in contrast to degenerated CE stage, in which the resistance in series with the emitter is multiplied by (ββββ+1) when seen from the base.
1
1
1 ++≈
+
+=
ββπ B
m
B
X
XR
g
Rr
i
v
20
CH5 Bipolar Amplifiers 77
Input Impedance Seen at Emitter and Base
CH5 Bipolar Amplifiers 78
Input Impedance Example
To find the RX, we have to first find Req, treat it as the base
resistance of Q2 and divide it by (ββββ+1).
++
++=
1
1
1
11
12ββ
B
mm
X
R
ggR
CH5 Bipolar Amplifiers 79
Bad Bias Technique for CB Stage
Unfortunately, no emitter current can flow.
CH5 Bipolar Amplifiers 80
Still No Good
In haste, the student connects the emitter to ground, thinking it will provide a DC current path to bias the amplifier. Little did he/she know that the input signal has been shorted to ground as well. The circuit still does not amplify.
21
CH5 Bipolar Amplifiers 81
Proper Biasing for CB Stage
( ) Cm
SEmin
out
E
m
in
RgRRgv
v
Rg
R
++=
=
11
1
||1
CH5 Bipolar Amplifiers 82
Reduction of Input Impedance Due to RE
The reduction of input impedance due to RE is bad because it shunts part of the input current to ground instead of to Q1
(and Rc) .
CH5 Bipolar Amplifiers 83
Creation of Vb
Resistive divider lowers the gain.
To remedy this problem, a capacitor is inserted from base to
ground to short out the resistor divider at the frequency of interest.
CH5 Bipolar Amplifiers 84
Example of CB Stage with Bias
For the circuit shown above, RE >> 1/gm.
R1 and R2 are chosen so that Vb is at the appropriate value and the current that flows thru the divider is much larger than the base current.
Capacitors are chosen to be small compared to 1/gm at the required frequency.
22
CH5 Bipolar Amplifiers 85
Emitter Follower (Common Collector Amplifier)
CH5 Bipolar Amplifiers 86
Emitter Follower Core
When the input is increased by ∆∆∆∆V, output is also increased by an amount that is less than ∆∆∆∆V due to the increase in collector current and hence the increase in potential drop across RE.
However the absolute values of input and output differ by a VBE.
CH5 Bipolar Amplifiers 87
Small-Signal Model of Emitter Follower
As shown above, the voltage gain is less than unity and
positive.
m
E
E
E
in
out
gR
R
R
rv
v
11
11
1
+
≈
⋅+
+
=
βπ
∞=A
V
CH5 Bipolar Amplifiers 88
Unity-Gain Emitter Follower
The voltage gain is unity because a constant collector current (= I1) results in a constant VBE, and hence Vout
follows Vin exactly.
1=v
A
∞=A
V
23
CH5 Bipolar Amplifiers 89
Analysis of Emitter Follower as a Voltage Divider
∞=A
VCH5 Bipolar Amplifiers 90
Emitter Follower with Source Resistance
m
S
E
E
in
out
g
RR
R
v
v
1
1+
++
=
β
∞=A
V
CH5 Bipolar Amplifiers 91
Input Impedance of Emitter Follower
The input impedance of emitter follower is exactly the
same as that of CE stage with emitter degeneration. This is not surprising because the input impedance of CE with
emitter degeneration does not depend on the collector resistance.
E
X
X Rri
v)1( βπ ++=
∞=A
V
CH5 Bipolar Amplifiers 92
Emitter Follower as Buffer
Since the emitter follower increases the load resistance to a
much higher value, it is suited as a buffer between a CE
stage and a heavy load resistance to alleviate the problem of gain degradation.
24
CH5 Bipolar Amplifiers 93
Output Impedance of Emitter Follower
Emitter follower lowers the source impedance by a factor of ββββ+1 improved driving capability.
E
m
s
out Rg
RR ||
1
1
+
+=
β
CH5 Bipolar Amplifiers 94
Emitter Follower with Early Effect
Since rO is in parallel with RE, its effect can be easily
incorporated into voltage gain and input and output impedance equations.
( )( )
OE
m
s
out
OEin
m
S
OE
OE
v
rRg
RR
rRrR
g
RrR
rRA
||||1
1
||1
1
1||
||
+
+=
++=
++
+
=
β
β
β
π
CH5 Bipolar Amplifiers 95
Current Gain
There is a current gain of (ββββ+1) from base to emitter.
Effectively speaking, the load resistance is multiplied by
(ββββ+1) as seen from the base.
CH5 Bipolar Amplifiers 96
Emitter Follower with Biasing
A biasing technique similar to that of CE stage can be used for the emitter follower.
Also, Vb can be close to Vcc because the collector is also at Vcc.
25
CH5 Bipolar Amplifiers 97
Supply-Independent Biasing
By putting a constant current source at the emitter, the bias
current, VBE, and IBRB are fixed regardless of the supply value.
CH5 Bipolar Amplifiers 98
Summary of Amplifier Topologies
The three amplifier topologies studied so far have different properties and are used on different occasions.
CE and CB have voltage gain with magnitude greater than one, while follower’s voltage gain is at most one.
CH5 Bipolar Amplifiers 99
Amplifier Example I
The keys in solving this problem are recognizing the AC ground between R1 and R2, and Thevenin transformation of
the input network.
SE
m
S
C
in
out
RR
R
Rg
RR
RR
v
v
+⋅
+++
−=1
1
1
2
1
1
||
||
β
CH5 Bipolar Amplifiers 100
Amplifier Example II
Again, AC ground/short and Thevenin transformation are
needed to transform the complex circuit into a simple stage
with emitter degeneration.
S
m
S
C
in
out
RR
R
Rg
RR
R
v
v
+⋅
+++
−=1
1
2
1 1
1
||
β
26
CH5 Bipolar Amplifiers 101
Amplifier Example III
The key for solving this problem is first identifying Req, which is the impedance seen at the emitter of Q2 in parallel with the infinite output impedance of an ideal current source. Second, use the equations for degenerated CE stage with RE replaced by Req.
2
1
1
211
1
1
1
mm
C
v
in
g
R
g
RA
rRrR
++
+
−=
++=
β
ππ
CH5 Bipolar Amplifiers 102
Amplifier Example IV
The key for solving this problem is recognizing that CB at frequency of interest shorts out R2 and provide a ground for R1.
R1 appears in parallel with RC and the circuit simplifies to a simple CB stage.
m
S
C
v
gR
RRA
1
||1
+
=
CH5 Bipolar Amplifiers 103
Amplifier Example V
The key for solving this problem is recognizing the equivalent base resistance of Q1 is the parallel connection of RE and the impedance seen at the emitter of Q2.
12
1||
1
11
1
m
E
m
B
ing
Rg
RR +
+
++=
ββ
CH5 Bipolar Amplifiers 104
Amplifier Example VI
The key in solving this problem is recognizing a DC supply is actually an AC ground and using Thevenin transformation to simplify the circuit into an emitter follower.
SS
m
OE
OE
in
out
RR
R
RR
grRR
rRR
v
v
+⋅
+++
=1
1
1
2
2
1
||1||||
||||
β
OE
m
S
outrRR
g
RRR ||||||
1
1
||2
1
+
+=
β
27
CH5 Bipolar Amplifiers 105
Amplifier Example VII
Impedances seen at the emitter of Q1 and Q2 can be lumped with RC and RE, respectively, to form the equivalent emitter and collector impedances.
( )
12
1
3
2
3
2
2
1
1
11
1
1
1
1
1
1
11
mm
B
m
B
C
v
m
B
Cout
m
B
Ein
gg
R
g
RR
A
g
RRR
g
RRrR
+++
++
+
−=
++
+=
+
++++=
β
β
β
ββπ
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