Fundamentals of Microelectronics
CH1
CH2
CH3
CH4
CH5
CH6
CH7
CH8
Chapter 5
Why Microelectronics?
Basic Physics of Semiconductors
Diode Circuits
Physics of Bipolar Transistors
Bipolar Amplifiers
Physics of MOS Transistors
CMOS Amplifiers
Operational Amplifier As A Black Box
Bipolar Amplifiers
5.1 General Considerations
5.2 Operating Point Analysis and Design
5.3 Bipolar Amplifier Topologies
5.4 Summary and Additional Examples
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2
Bipolar Amplifiers
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.
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1
Input/Output Impedances
Rx
Input Impedance Example I
vx
= rπ
ix
V
= x
ix
The figure above shows the techniques of measuring input
and output impedances.
CH5 Bipolar Amplifiers
When calculating input/output impedance, small-signal
analysis is assumed.
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Impedance at a Node
6
Impedance at Collector
Rout = ro
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
With Early effect, the impedance seen at the collector is
equal to the intrinsic output impedance of the transistor (if
emitter is grounded).
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2
Impedance at Emitter
Three Master Rules of Transistor Impedances
vx
1
=
ix g + 1
m
rπ
Rout ≈
1
gm
(VA = ∞)
Rule # 1: looking into the base, the impedance is rπ if
emitter 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.
The impedance seen at the emitter of a transistor is
approximately equal to one over its transconductance (if
the base is grounded).
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Biasing of BJT
10
DC Analysis vs. Small-Signal Analysis
Transistors and circuits must be biased because (1)
transistors must operate in the active region, (2) their smallsignal parameters depend on the bias conditions.
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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.
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3
Notation Simplification
Example of Bad Biasing
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.
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.
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Another Example of Bad Biasing
14
Biasing with Base Resistor
IB =
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.
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VCC −VBE
V −V
, IC = β CC BE
RB
RB
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 β variations.
15
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4
Improved Biasing: Resistive Divider
VX =
Accounting for Base Current
R2
V CC
R1 + R 2
I C = I S exp(
R 2 V CC
)
R1 + R 2 V T
Using resistor divider to set VBE, it is possible to produce
an IC that is relatively independent of β if base current is
small.
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 V − I B RThev 
I C = I S exp  Thev

VT


With proper ratio of R1 and R2, IC can be insensitive to β ;
however, its exponential dependence on resistor deviations
makes it less useful.
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Emitter Degeneration Biasing
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Design Procedure
Choose an IC to provide the necessary small signal
parameters, gm, rπ, etc.
Considering the variations of R1, R2, and VBE, choose a
value for VRE.
With VRE chosen, and VBE calculated, Vx can be
determined.
The presence of RE helps to absorb the error in VX so VBE
stays relatively constant.
This bias technique is less sensitive to β (I1 >> IB) and VBE
variations.
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Select R1 and R2 to provide Vx.
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20
5
Self-Biasing Technique
Self-Biasing Design Guidelines
RB
(1) RC >> β
(2) ∆VBE << VCC − VBE
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.
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(1) provides insensitivity to β .
(2) provides insensitivity to variation in VBE .
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Summary of Biasing Techniques
22
PNP Biasing Techniques
Same principles that apply to NPN biasing also apply to
PNP biasing with only polarity modifications.
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6
Possible Bipolar Amplifier Topologies
Study of Common-Emitter Topology
Analysis of CE Core
Inclusion of Early Effect
Emitter Degeneration
Inclusion of Early Effect
CE Stage with Biasing
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.
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Common-Emitter Topology
Small Signal of CE Amplifier
Av =
−
vout
vin
vout
= g m vπ = g m vin
RC
Av = − g m RC
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7
Limitation on CE Voltage Gain
Av =
ICRC
VT
Av =
VRC
VT
Tradeoff between Voltage Gain and Headroom
Av <
VCC −VBE
VT
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.
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I/O Impedances of CE Stage
Rout =
R in =
vX
= rπ
iX
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CE Stage Trade-offs
vX
= RC
iX
When measuring output impedance, the input port has to
be grounded so that Vin = 0.
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8
Inclusion of Early Effect
Intrinsic Gain
Av = − g m rO
Av =
VA
VT
Av = −gm(RC || rO)
Rout = RC || rO
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.
Early effect will lower the gain of the CE amplifier, as it
appears in parallel with RC.
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Current Gain
AI =
iout
iin
AI
=β
CE
Emitter Degeneration
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 β .
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35
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.
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9
Small-Signal Model
Emitter Degeneration Example I
Av = −
Av = −
g m RC
1 + g m RE
RC
1
+ RE
gm
Av = −
RC
1
+ RE || rπ 2
gm1
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.
The input impedance of Q2 can be combined in parallel with
RE to yield an equivalent impedance that degenerates Q1.
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Emitter Degeneration Example II
Av = −
38
Input Impedance of Degenerated CE Stage
VA = ∞
RC || rπ 2
1
+ RE
g m1
vX = rπ iX + RE (1+ β )iX
Rin =
vX
= rπ + (β + 1)RE
iX
In this example, the input impedance of Q2 can be
combined in parallel with RC to yield an equivalent collector
impedance to ground.
With emitter degeneration, the input impedance is
increased from rπ to rπ + (β
β +1)RE; a desirable effect.
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40
10
Output Impedance of Degenerated CE Stage
Capacitor at Emitter
VA = ∞
v

vin = 0 = vπ +  π + gmvπ RE ⇒ vπ = 0

 rπ
v
Rout = X = RC
iX
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.
Emitter degeneration does not alter the output impedance
in this case. (More on this later.)
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Example: Design CE Stage with Degeneration as a Black Box
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Degenerated CE Stage with Base Resistance
VA = ∞
VA = ∞
iout = g m
Gm =
v out v A v out
= .
v in
vin v A
vin
1 + ( rπ−1 + g m ) RE
v out
− β RC
=
v in
rπ + ( β + 1) R E + R B
iout
gm
≈
vin 1 + g m RE
Av ≈
If gmRE is much greater than unity, Gm is more linear.
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43
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− RC
1
R
+ RE + B
gm
β +1
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11
Input/Output Impedances
Emitter Degeneration Example III
VA = ∞
Rin1 = rπ + ( β + 1) RE
Rin 2 = RB + rπ 2 + ( β + 1) RE
Rout = RC
Av =
Rin =r π +(β +1)R2
Rin1 is more important in practice as RB is often the output
impedance of the previous stage.
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− (RC || R1 )
1
R
+ R2 + B
β +1
gm
Rout = RC || R1
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Output Impedance of Degenerated Stage with VA< ∞
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Two Special Cases
1)
Rout = [1 + g m ( RE || rπ ) ]rO + RE || rπ
2)
Rout = rO + ( g m rO + 1)( RE || rπ )
Rout ≈ rO [1 + g m ( RE || rπ ) ]
Emitter degeneration boosts the output impedance by a
factor of 1+gm(RE||rπ).
This improves the gain of the amplifier and makes
the
circuit a better current source.
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CH5 Bipolar Amplifiers
R
E
R
out
R
E
R
out
>>
≈
<<
≈
rπ
r
O
(1 +
g
rπ ) ≈
m
β rO
rπ
(1 +
g
m
R
E
) r
O
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12
Analysis by Inspection
Rout 1 = [1 + g m ( R2 || rπ )]rO
Rout = R1 || Rout 1
Example: Degeneration by Another Transistor
Rout = [1 + g m ( R2 || rπ ) ]rO || R1
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.
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Rout = [1 + gm1 (rO 2 || rπ 1 )]rO1
Called a “cascode”, the circuit offers many advantages that
are described later in the book.
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CH5 Bipolar Amplifiers
Study of Common-Emitter Topology
50
Bad Input Connection
Analysis of CE Core
Inclusion of Early Effect
Emitter Degeneration
Inclusion of Early Effect
CE Stage with Biasing
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.
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13
Use of Coupling Capacitor
DC and AC Analysis
Av = − g m ( RC || rO )
Rin = rπ || RB
Capacitor isolates the bias network from the microphone at
DC but shorts the microphone to the amplifier at higher
frequencies.
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Rout = RC || rO
Coupling capacitor is open for DC calculations and shorted
for AC calculations.
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Bad Output Connection
Still No Gain!!!
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.
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55
In this example, the AC coupling indeed allows correct
biasing. However, due to the speaker’s small input
impedance, the overall gain drops considerably.
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14
CE Stage with Biasing
CE Stage with Robust Biasing
VA = ∞
Av =
A
R
R
v
in
out
=
− g
=
=
m
( R
C
Rin = [rπ + (β +1)RE ] || R1 || R2
|| r O )
r π || R 1 || R
R C || r O
− RC
1
+ RE
gm
2
Rout = RC
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CH5 Bipolar Amplifiers
Removal of Degeneration for Signals at AC
58
Complete CE Stage
Av = − g m R C
R in = rπ || R1 || R 2
R out = R C
Av =
Capacitor shorts out RE at higher frequencies and
removes degeneration.
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CH5 Bipolar Amplifiers
− RC || RL
1
R || R || R
+ RE + s 1 2
gm
β +1
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15
Summary of CE Concepts
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.
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CB Core
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Tradeoff between Gain and Headroom
Av =
=
Av = g
m
VCC − VBE
VT
RC
To maintain the transistor out of saturation, the maximum
voltage drop across RC cannot exceed VCC-VBE.
The voltage gain of CB stage is gmRC, which is identical to
that of CE stage in magnitude and opposite in phase.
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IC
.RC
VT
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16
Simple CB Example
Input Impedance of CB
Rin =
Av = g m RC = 17.2
1
gm
R1 = 22.3KΩ
The input impedance of CB stage is much smaller than that
of the CE stage.
R2 = 67.7 KΩ
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Practical Application of CB Stage
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Output Impedance of CB Stage
Rout = rO || RC
To avoid “reflections”, need impedance matching.
CB stage’s low input impedance can be used to create a
match with 50 Ω .
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The output impedance of CB stage is similar to that of CE
stage.
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17
CB Stage with Source Resistance
Av =
Practical Example of CB Stage
RC
1
+ RS
gm
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.
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An antenna usually has low output impedance; therefore, a
correspondingly low input impedance is required for the
following stage.
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Realistic Output Impedance of CB Stage
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Output Impedance of CE and CB Stages
Rout 1 = [1 + g m ( R E || rπ ) ]rO + (R E || rπ )
Rout = RC || Rout 1
The output impedance of CB stage is equal to RC in parallel
with the impedance looking down into the collector.
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.
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72
18
Fallacy of the “Old Wisdom”
CB with Base Resistance
vout
RC
≈
vin R + RB + 1
E
β + 1 gm
The statement “CB output impedance is higher than CE
output impedance” is flawed.
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With an addition of base resistance, the voltage gain
degrades.
73
Comparison of CE and CB Stages with Base
Resistance
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Input Impedance of CB Stage with Base Resistance
vX rπ + RB 1 RB
=
≈ +
iX
β +1 gm β +1
The input impedance of CB with base resistance is equal to
1/gm plus RB divided by (β +1). This is in contrast to
degenerated CE stage, in which the resistance in series
with the emitter is multiplied by (β +1) when seen from the
base.
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.
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19
Input Impedance Seen at Emitter and Base
Input Impedance Example
RX =
1
1  1 RB 
 +
+

gm2 β +1 gm1 β +1
To find the RX, we have to first find Req, treat it as the base
resistance of Q2 and divide it by (β +1).
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CH5 Bipolar Amplifiers
Bad Bias Technique for CB Stage
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.
Unfortunately, no emitter current can flow.
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79
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20
Proper Biasing for CB Stage
Rin =
CH5 Bipolar Amplifiers
Reduction of Input Impedance Due to RE
1
|| RE
gm
vout
1
=
g m RC
vin 1 + (1 + g m RE )RS
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) .
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Creation of Vb
82
Example of CB Stage with Bias
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.
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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.
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21
Emitter Follower (Common Collector Amplifier)
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.
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CH5 Bipolar Amplifiers
Small-Signal Model of Emitter Follower
Unity-Gain Emitter Follower
VA = ∞
VA = ∞
vout
1
R
=
≈ E
vin 1+ rπ ⋅ 1 R + 1
β +1 RE E gm
Av = 1
The voltage gain is unity because a constant collector
current (= I1) results in a constant VBE, and hence Vout
follows Vin exactly.
As shown above, the voltage gain is less than unity and
positive.
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87
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88
22
Analysis of Emitter Follower as a Voltage Divider
Emitter Follower with Source Resistance
VA = ∞
vout
RE
=
vin R + RS + 1
E
β +1 gm
VA = ∞
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Input Impedance of Emitter Follower
90
Emitter Follower as Buffer
VA = ∞
vX
= rπ + (1+ β )RE
iX
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.
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.
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23
Output Impedance of Emitter Follower
Emitter Follower with Early Effect
Av =
RE || rO
RS
1
+
β +1 gm
RE || rO +
Rin = rπ + (β + 1)( RE || rO )
 R
1
Rout =  s +  || RE
β
+
1
g

m
 R
1 
Rout =  s +  || RE || rO
 β +1 gm 
Emitter follower lowers the source impedance by a factor of
β+1 improved driving capability.
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Since rO is in parallel with RE, its effect can be easily
incorporated into voltage gain and input and output
impedance equations.
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Current Gain
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.
There is a current gain of (β +1) from base to emitter.
Effectively speaking, the load resistance is multiplied by
(β +1) as seen from the base.
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95
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96
24
Supply-Independent Biasing
Summary of Amplifier Topologies
By putting a constant current source at the emitter, the bias
current, VBE, and IBRB are fixed regardless of the supply
value.
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97
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.
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Amplifier Example I
98
Amplifier Example II
v out
R 2 || RC
R1
=−
⋅
1
R1 || R S
v in
+
+ R E R1 + R S
β +1 gm
v out
RC
R1
=−
⋅
R S || R1
1
v in
+
+ R 2 R1 + R S
β +1
gm
The keys in solving this problem are recognizing the AC
ground between R1 and R2, and Thevenin transformation of
the input network.
Again, AC ground/short and Thevenin transformation are
needed to transform the complex circuit into a simple stage
with emitter degeneration.
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100
25
Amplifier Example III
Amplifier Example IV
Rin = rπ 1 + R1 + rπ 2
Av =
− RC
1
R
1
+ 1 +
g m1 β + 1 g m 2
Av =
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.
CH5 Bipolar Amplifiers
R C || R1
1
gm
RS +
101
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.
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Amplifier Example V
102
Amplifier Example VI
vout
RE || R2 || rO
R
=
⋅ 1
vin R || R || r + 1 + RS ||R1 R1 + RS
E
2
O
gm β +1
 R || R 1 
Rout =  S 1 +  || RE || R2 || rO
 β +1 gm 
 1
1  RB
1 
Rin =

+
 || R +
β + 1  β + 1 g m 2  E  g m1
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.
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103
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.
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26
Amplifier Example VII

R
1 

Rin = rπ 1 + ( β + 1) R E + B1 +
β + 1 g m 2 

R
1
Rout = RC + B 2 +
β + 1 g m3
RC +
Av = −
RB 1
RB 2
+
1
β + 1 g m3
+
1
β + 1 gm2
+
1
g m1
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.
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105
27