(Latest) topic 4 bipolar_junction_transistors

1
Topic 4
Bipolar Junction Transistors

2
Introduction to BJT
a three-terminal component – Emitter, Collector and Base.
(a)npn transistor which has an n – type emitter and collector
and a p – type base, and
(b) pnp transistor which consist of p – type emitter and
collector and an n – type base.

3
The arrow on the schematic symbol
represents:
(1) the emitter terminal.
The terminal opposite the emitter is the
collector and the center terminal is the
base.
(2) the direction of the arrow is always point
towards the n – type of the material.
(3) the direction of the emitter current
(conventional).

4
Power Transistor Water Analogy
For transistor working at the linear
region, The current gain must be
obey.
For transistor working as
switches, this formula must be
abandon because Ic(sat) is almost
a constant value. The formula now
is,
B
C
I
I
=β
B
satc
force
I
I )(
=β

5
Transistor Currents
• IE, IC and IB – emitter, collector and base
currents.
IE > IC >> IB
(mA) (µA)
IE ≈ IC
• Current directions of the npn transistor are
opposite those for the pnp transistor.
• The direction of IE is always opposite that of IB
and IC.

6
BJT - Current-controlled device
• The values of the collector and emitter currents are
determined primarily by the value of the base current.
• The small increase in base current (from 10 µA to 20 µA)
produces a larger increase in IC and in IE (from 2 mA to 4 mA).
β = 200

7
BJT - Current-controlled device
• The value of IC is normally some multiple of the value of
IB.
β - forward current gain, of the device.
BC II β=

8
Example:
Assume that a transistor has values of IB = 50 µA and β
= 120. The collector current for the transistor is
Solution:
IC = 6 mA
)50)(120(C
BC
AI
II
µ
β
=
=

9
Example
Determine the values of collector current for the values of the
base current shown below.

10
Solution
The base current has an initial value of 20 µA. The beta rating
of the component is 300. The initial value of the collector
current is found as
mA
When IB increases to 50 µA, the collector current increases to
mA
Thus, a 30 µA change in base current causes a 9 mA change in
collector current.
( )( ) 6=== AII BC µβ 20300
( )( ) 15=== AII BC µβ 50300

11
Transistor Voltages
o VCC and VBB are dc voltage sources that are used
to bias the transistor.
o Some circuits also contain a dc biasing source in the
emitter circuit labeled VEE.

12
Transistor Voltages
o VC, VB and VE are transistor terminal voltages.
o Each voltage is measured from the identified
transistor terminal to ground.

13
Transistor Voltages
o VCE, VCB, and VBE are measured across the
identified transistor terminals.

14
Transistor Construction
• made up of three separate semiconductor materials
joined together so that they form two pn junctions.
the base – emitter junction.
the collector – base junction.

15
Transistor Operation
Base-emitter
junction
Collector-base
junction
Operating
region
BJT
Characteristics
Reverse biased Reverse biased Cutoff IB = IC = IE = 0,
VCE = VCC
Forward
biased
Reverse biased Active IC = βDCIB
Forward
biased
Forward
biased
Saturation
IC = IC(sat) =
C
CC
R
V
*Breakdown

16
Zero Bias
• no biasing potential applied.
• depletion layers due to recombination of
free carriers produced by thermal energy.
• both junctions are in reverse biased at
room temperature.
Note: depletion layers extend farther into
the base region due to its lower doping
level.

17
Cutoff
• both transistor junctions are
reverse biased.
• the depletion layers extend well
into the emitter, base and collector
regions.
• only an extremely small amount of
reverse current passes from the
emitter to the collector.
• the transistor is said to be in cutoff.

18
Example
A 2N3904 transistor with a collector –
emitter voltage (VCE) of 40 V and a reverse
base-emitter (VBE) as low as 3 V will allow
only 50 nA of collector current IC.
This is extremely small compared to the
200 mA value of IC that the component is
capable of handling when the base –
emitter junction is forward biased.

19
Saturation
• the opposite of cutoff.
• further increases in IB do not result
in further increases in IC – it has
reached its maximum possible
value.
• In an ideal situation (VCE = 0 V), IC
will depend completely on the
values of VCC, RC, and RE

20
Saturation
• As IB increases from 0 A, IC increases according to
IC = β IB
until it reaches its maximum value.
• At this point, IC cannot increase any further – additional
increases in IB do not increase the value of IC. The relationship
IC = βIB
no longer holds true.
• Both of the transistor junctions become forward biased.

21
Saturation
VCE for the transistor is shown to be
approximately 0.3 V (typical for a
saturated transistor).
With the 0.7 V value of VBE, the
collector-base junction is biased to
the difference between the two,
which is 0.4 V.
Note that this voltage indicates that
the collector-base junction of the
transistor is forward biased (even
though it is not fully).

22
Active Operation
• Current is generated in the emitter
and base region when VBE > φB.
• Base region is very lightly doped, its
resistance is greater than the
resistance of the reverse-biased
collector-base junction.
• A vast majority of the emitter current
continues through the reverse-biased
collector-base junction to the
collector circuit. Recall Zener diode.
• The collector-base junction is
designed to allow a reverse current
without damaging the junction.

23
Transistor Currents and Voltages
• The transistor is a current-controlled device.
• In many applications, the base current is varied to produce
variations in IC and IE
i.e. a small change in IB results in a large change in the other
terminal currents.

24
Example
Determine the values of collector
current for the values of the base
current shown below.
Solution
The initial value of the
collector current
When IB increases to 50
µA, the collector current
is
∆IB = 30 µA,
∆IC = 9 mA,
( )( )
mA
AII BC
=
== µβ 20300
( )( )
mA
AII BC
=
== µβ 50300

25
The Relationship Among IE, IC and IB
• Kirchhoff’s current law: the current leaving a component
must be equal to the current entering the component.
IE = IB + IC
• IB is normally much less that IC,
IC ≅ IE
• The current relationships shown
above hold true for both the npn
and pnp transistors.

26
• IB = 20 µA, IC = 6 mA
∴ IE = 6mA + 20 µA
= 6.02 mA
≈ IC
• IB = 50 µA, IC = 15 mA
IE = 15.05 mA

27
Transistor Currents & Voltages
• The BE junction is forward biased,
VBE ≈ 0.7 V.
• Emitter is at ground. The voltage across
RB is
VR(B) = VBB − VBE --(1)
• Also, from Ohm’s Law,
VR(B) = IBRB --(2)
• Let (1) = (2)
IBRB = VBB − VBE
• Solving for IB:
B
BEBB
B
R
VV
I
−
=

28
Transistor Currents & Voltages
• The voltage at the collector with respect
to grounded emitter is
VCE = VCC – VR(C)
• As the potential drop across RC is
VR(C) = ICRC
The voltage at the collector
VCE = VCC − ICRC
where IC = βDCIB.
• The voltage across the reverse biased
collector-base junction is
VCB = VCE − VBE

29
Example
Determine IB, IC, IE, VBE, VCE, and VCB in the following
circuit. The transistor has βDC = 150.

30
Solution
• Known VBE = 0.7 V (forward biased),
IB = (VBB - VBE) / RB = 430 µA
• The collector current,
IC = βDCIB = 64.5 mA
• Kirchhoff's current law:
IE = IC + IB = 64.9 mA
• Solving for VCE and VCB:
VCE = VCC – ICRC = 3.55 V
VCB = VCE – VBE = 2.85 V
• The collector is at higher potential than the base (VCB > 0 ⇒ VC >
VB), the collector-base junction is reverse-biased.

31
Current gain, DC Beta
• The dc beta (βDC) rating – ratio of dc collector current to dc
base current.
• It is a ratio of current values, thus it has no unit of measure.
• Typical beta ratings can be as high as 300 i.e. the IC can be
up to 300 times the value of IB.
• It can be used to define other terminal currents:
IC = βIB
IE = IC + IB = βIB + IB
IE = IB(β + 1)
B
C
I
I
=DCβ

32
DC Beta
• βDC is also designated by an equivalent hybrid (h)
parameter:
hFE = βDC
• Transistor data sheets do not provide βDC but hFE.
• βDC is not truly constant. It varies slightly with IC and
temperature.
• Transistors have both dc beta ratings and ac beta ratings
(which will be discussed later)

34
Example
Determine the values of IC and IE for the circuit shown.
Solution
IB = 125 µA
IC = βIB
IC = (200)(125 µA)
IC = 25 mA
IE is found as
IE = IC + IB
IE = 25.125 mA

35
Example
Determine the values of IC and IB for the circuit shown.
Solution
IB can be determine from
Now, IC can be found as
IC = IE − IB= 14.9 mA
or
IC = βIB = 14.9 mA
Aµ
β
6.74
1
=
+
= E
B
I
I

36
Example
Determine the values of IB and IE for the circuit shown.
Solution
…
The emitter current
IE = IC + IB = 50.125 mA

37
DC Alpha
• dc alpha (α) rating – the ratio of collector current to emitter
current.
• Typical values of αDC range from 0.95 to 0.99 or greater.
• Like beta, it has no units.
• It can be used to define other terminal currents, as follows:
IB = IE − IC = IE (1 − α)
( )1<=
E
C
I
I
α
α
α
C
E
EC
I
I
II
=
=

38
The Relationship between Alpha and Beta
• The spec. sheet for a given transistor lists the value
of beta for the device, but not the value of alpha as
beta is used far more commonly than alpha.
• Alpha can be determined using the value of beta
with the following equation:
1+
=
β
β
α

39
Example
Determine the value of alpha for the transistor shown.
Then, determine the value of IC using both the alpha and the
beta ratings of the transistor.
Solution:
IC = αIE = 29.9 mA
IC = βIB = 30.0 mA
9967.0
1
=
+
=
β
β
α

40
Maximum Current Ratings
• Most transistor spec sheets list maximum collector current
ratings for both saturation and cutoff.
• When the transistor is saturated, the collector current can go
as high as several hundred mA. High-power transistors typically
have current ratings as high as several amperes.
• The maximum allowable base current for a given transistor can
be found by dividing its maximum IC value by its maximum dc β
rating.
max
(max)
(max)
β
C
B
I
I =

41
Example
The transistor shown has the following ratings:
IC(max) = 500 mA and βmax = 300.
Determine the maximum allowable value of IB for the device.
Solution:
If IB > IB(max) then IC > IC(max)
The transistor will probably be
destroyed.

42
Maximum Cutoff Current Ratings
• These ratings are usually in the low nanoampere (nA) range and
are specified for exact values of VCE and reverse VBE.
• The 2N3904 has a maximum cutoff current rating of 50 nA
when the reverse value of VBE is 3 V and the value of VCE is 40 V.

44
Transistor Voltage Ratings
• It indicates the maximum amount of reverse bias that can be
applied to the collector-base junction (reverse biased for active
region operation) without damaging the transistor.
• The value of VCB is equal to the difference between the other two
voltages: 39.25 V. If this voltage exceeds the VCB rating of the
transistor, the component will probable be destroyed.

45
Transistor Voltage Ratings
• Every transistor has three breakdown voltage ratings.
• These ratings indicate the maximum reverse voltages that
the transistor can withstand.
• For the 2N3904, these voltage ratings are as follows:
Rating Value (Vdc
)
VCBO
60
VCEO
40
VEBO
6

46
Transistor Characteristic Curves
• The three curves are
(i) The collector curves,
(ii) The base curves and
(iii)The beta curves.
• The emitter curve is not part of the discussion as its
current characteristics are the same as those of the
collector.

47
Collector Curves
• The collector characteristics
curve illustrates the relationship
among IC, IB and VCE.
• Each collector curve is derived
for a specified value of IB.
• Note that the IB = 0 µA line
represents the operation of the
transistor when it is in cutoff.
• The collector curve is divided
into three parts – saturation,
active region, and breakdown.

48
Saturation Region
Let VBB > 0, then IB ≠ 0.
Set VCC = 0V  VCE = 0
• For this condition, both the BE junction and
CB junction are forward biased because the
base is approximately 0.7V while the emitter
and the collector are at 0V.
VCE = VCC – ICRC = 0 – 0(RC) = 0
• This represents the origin of the
characteristic curve and is independent of IB.
O V
O.7 V
O V

49
Saturation Region
VCE < VK, VCE ≠ 0
• As VCC increases, VCE ↑ gradually, thus IC
↑.
• However, VCE remains less than 0.7V as
VCE = VCC – ICRC
• This represents the portion of the graph
where VCE < VK.

50
Example
Determine whether or not the transistor in the circuit
below is in saturation. Assume VCE(sat) = 0.2 V.

51
Solution
First determine IC(sat):
IC(sat) = (VCC – VCE(sat)) / RC
IC(sat) = (10 V – 0.2V) / 1kΩ
= 9.8 mA
Now let’s determine whether IB is large enough to produce IC(sat):
IB = (VBB − VBE) / RB
= (3 V – 0.7 V) / 10kΩ = 0.23 mA
IC = βDCIB = (50) (0.23 mA) = 11.5 mA
• This shows that with the specified βDC, this base current is capable
of producing an IC greater than IC(sat).
• Thus, the transistor is saturated, and the collector current value
of 11.5 mA is never reached. The collector current remains at its
saturation value.

52
Active Region
• There is little change in the value IC when VCE increases from VK to VBR.
• IC is not controlled by the value of VCE when a transistor is operated in
its active region.
Let VCC = constant,
• IC increases with IB
• VCE decreases Until it reaches VCE(sat) 
somewhere below VK and it is usually
only a few tenths of a volt for a silicon
diode.
• IC cannot increase further
(IC = βDCIB is no longer valid).

53
Active Region
• Consider β value held constant at
100, and IB is increased to 150 µA:
IC increases proportionately with
increase in IB,
and it is still relatively independent
of changes in VCE.
• Alternatively, when VCE > 0.7V, the CB junction becomes reverse
biased (transistor in active region)
• IC remains essentially constant for a given value of IB as VCE continues
to increase.
• A slight increase in IC due to the widening of the CB depletion region.
In the active region, the relation IC = βDCIB holds.

54
Breakdown Region
• It happens when the value of VCE
exceeds the breakdown voltage
rating of the transistor.
• IC increases dramatically
• Note how VCE affects the CB
junction.
(1) If VCE < 0.7V, the CB junction is
forward-biased.
(2) If VCE > 0.7V, the CB junction
becomes reverse-biased.

55
Cutoff
• There is no current flowing in
the BE junction as VB = VE = 0.
Hence, IB = IE = 0.
• The CB junction is reverse
biased and hence, IC = 0.
• The two junctions are reverse-
biased and ideally, no current
should flow through the BE and
CB junctions.

56
Cutoff
• Similar cutoff state can also be obtained by introducing a
negative bias to the base.
• The negative bias is provided by VBB.
• The resistor RB is added to prevent the transistor from damage.
VC > 0
VE = 0RB
RC
VBB
VCC
VB < 0

57
Base Curves
• The base curve of a transistor plots IB as a function of VBE.
• This curve closely resembles the forward operating curve of a
typical pn junction diode.

59
Notations
• Amplifier circuits have both dc and ac quantities for current,
voltage and resistance.
• Italic capital letters are used for both dc and ac currents (I)
and voltages (V). Lowercase i and v for ac current and voltage
are reserved for instantaneous values.
• DC quantities always carry an UPPERCASE Roman, nonitalic
subscript. Example: IC, IB
• AC and all time-varying quantities always carry a lowercase
italic subscript. Example: Ic, Ib
• Internal transistor resistances are designated by lowercase r’
with an appropriate subscript e.g. r’e refers to the internal ac
emitter resistance.

60
Transistor as an Amplifier
• The figure above shows the basic transistor amplifier circuit.
• Vin is superimposed on VBB. They are in series with the base resistor
RB.
• VCC is connected to the collector through the collector resistor, RC.
Vc
Vb

61
VC
Vin
VBB VCC
• The ac input voltage produces an ac base current.
• The collector current is related to the base current by
IC = βDCIB
• This results in a much larger ac collector current.
• The ac collector current produces an ac voltage across RC,
producing an amplified, but inverted, reproduction of the ac
input voltage in the active region of operation.

62
• The forward-biased base-emitter junction presents a
very low resistance to the ac signal. This internal ac
emitter resistance is designated r’e. The ac emitter
current is
Ie ≅ Ic =
• The ac collector voltage, Vc equals the ac voltage drop
across RC.
Vc = IcRC ≅ IeRC
• The ac voltage gain, Av, of the transistor circuit
'
e
b
r
V
''
e
C
ee
Ce
b
c
v
r
R
rI
RI
V
V
A ≅≅=
Vb = Vin −IbRB

63
Example
• Determine the voltage gain and the ac output voltage in the circuit
on the right if r’e = 50 Ω.
Solution
The voltage gain is
Av =
Av =
Therefore the ac output
voltage is Vout = AvVb
RC
1.0 k
RB
VCC
VBB
Vin
Vout
100 mV
'
C
er
R
20=
Ω
Ω
50
0.1 k

64
Transistor as a Switch
• In part (a) the transistor is in the cutoff region because the base-
emitter junction is not forward-biased. There is ideally an open
between the collector and emitter.
• Neglecting leakage current, all of the currents are zero, and VCE is
equal to VCC.
VCE(cutoff) = VCC

65
Transistor as a Switch
• In part (b), the transistor is in the saturation region (base emitter and
base collector junctions are forward-biased) and the base current is
made large enough to cause the collector current to reach its
saturation value.
• There is a short between collector and emitter. Actually, a voltage
drop of up to a few tenths of a volt normally occurs, which is
saturation voltage, VCE(sat).
• The maximum collector current,
• The minimum value of base current
needed to produce saturation is
•
C
)(CECC
)(C
R
VV
I
sat
sat
−
=
DC
)(C
(min)B
β
satI
I =

66
A Simple Application of a Transistor
Switch
• The transistor is used as a switch to
turn the LED on and off.
• A square wave with a period of 2 s is
applied to the input.
• When the square wave is at 0 V, the
transistor is in cutoff and, there is no
collector current, the LED does NOT
emit light.
• Square wave goes to its high level, the
transistor saturates. It forward-biases
the LED. The resulting collector current
through the LED causes it to emit light.
RB
RC
VCC
ON ON
OFF1 s
Result: A blinking LED that
is on for 1 s and off for 1 s.

67
Example
The LED shown requires 30 mA to emit sufficient level of light.
Therefore, the collector current should be approximately 30 mA.
For the following circuit values: VCC = 9 V, VCE(sat) = 0.3 V,
RC = 270 Ω, RB = 3.3 kΩ, and βDC = 50.
IC(sat) =
IB(min) =
mA
VV
R
VV
2.32
270
3.09
C
CE(sat)CC
=
Ω
−
=
−
A
mAI
µ
β
644
50
2.32
DC
C(sat)
==

68
To ensure saturation, use twice the value of IB(min)
which gives 1.29 mA.
From the circuit,
IB =
Rearrange the expression above to obtain the
voltage amplitude of the square wave input Vin.
Vin − 0.7 V = 2IB(min)RB = (1.29 mA)(3.3 kΩ)
Vin = 4.96 V
Ω
−
=
−
=
k
VV
R
VV
R
V ininRB
3.3
7.0
B
BE
B

69
Example
For the circuit shown,
What is VCE when VIN = 0 V?
Sol: When VIN = 0 V, the transistor
is in cutoff (it acts like an open
switch) and
VCE = VCC = 10 V

70
Example
What minimum value of IB is required to saturate
this transistor if βDC is 200? Neglect VCE(sat).
Sol: Since VCE(sat) is neglected (assumed to be zero),
This is the value of IB necessary to drive the transistor to the
point of saturation. Any further increase in IB will drive the
transistor deeper into saturation but will not increase IC.
mA
k
V
R
V
I sat 10
0.1
10
C
CC
)(C =
Ω
==
A50µ
β
===
200
10
DC
)(C
(min)B
mAI
I
sat

71
Example
(c) Calculate the maximum value of RB
when VIN = 5 V.
Sol:When the transistor is on, VBE = 0.7 V. The voltage across RB is
VR(B) = VIN – VBE
= 5 V − 0.7 V
= 4.3 V
The maximum value of RB needed to allow a minimum IB of 50 µA
by Ohm’s law as follows:
Ω=== 86k
A
V
I
V
R
B
R
B
B
µ50
3.4
(min)
(max)

72
Chapter 2
BJT – DC Biasing Circuits
 DC Load Line
 Q-point
 Base Bias, Emitter Bias, Voltage Divider
Bias, and Collector Feedback Bias

73
Introduction
• The purpose of dc biasing is to set the initial values of IB, IC and
VCE. The ac operation of a transistor amplifier depends on
its initial dc values of IB, IC, and VCE.
• As IB varied from an initial value, IC and VCE are varied from

74
DC Operating Point
• DC operating point/Q-point (quiescent point)
• It is a point on the collector characteristic curve: (IC, VCE) with IB
= constant. It is set by properly biasing the transistor amplifier.
• Amplifiers are the most common linear devices. A linear
amplifier preserves its input/output waveform.

75
DC Operating Point
• If an amplifier is not properly biased, it will go either into cutoff or
saturation and the amplified signal will be distorted – nonlinear
amplifier.
• A nonlinear amplifier does not preserve its input/output waveform.
• Example of a nonlinear amplifier – a device that turns your voice
into a chipmunk (cartoon) voice. Novel purpose but in daily phone
conversation?

76
The DC Load Line
• Notice that when IB
increases, IC increases and
VCE decreases.
• When IB decreases, IC
decreases and VCE increases.
• As VBB is adjusted up or
down, the dc operating point
of the transistor moves along
a sloping straight line, called
the dc load line.

77
The DC Load Line
• The dc load line is a graph that represents all the possible
combinations of IC and VCE for a given amplifier.
• The dc load line represents all the IC-VCE combinations for
the circuit.

78
The DC Load Line
• The IC(sat) point represents the ideal value of saturation current
for the circuit.
• Consider a saturated transistor as being short-circuited from
emitter to collector, then VCE is zero, thus
VCC = ICRC (saturation)
• Rearranging the expression for IC, we get
• Recall:
C
CC
satC
R
V
I =)(

79
The DC Load Line
• When the transistor is in cutoff, it looks like an open circuit
from emitter to collector. Collector current, IC = 0
VCE(off) = VCC
• These equations are used to plot the end points of the dc load
line.
Recall:

80
Example
Plot the dc load line for the circuit shown.
Solution
From the circuit values shown,
VCE(off) = VCC = 12 V
and
The load line therefore has end points of
VCE(off) = 12 V and IC(sat) = 6 mA
6mA=
Ω
==
kR
V
I
C
CC
satC
2
12
)(

81
Example
(a) Plot the dc load line for the circuit shown below.
(b) Then, verify the load line VCE values for IC = 1 mA,
IC = 2 mA and IC = 5 mA.

82
Solution
From the circuit values
shown,
and
Using IC(sat) and VCE(off) as
the end points, the dc load
line for the circuit is
plotted as shown.
IC
(mA) 1 2 5
VCE
(V) 9 8 5
==
C
CC
satC
R
V
I )(
As shown in the figure, we would
expect to obtain:

83
Solution
These values are verified by the following equation
VCE = VCC − ICRC
For IC = 1 mA,
VCE = VCC − ICRC =1V − (1 mA)(1 kΩ) = 9V
For IC = 2 mA, VCE = = 8V
For IC = 5 mA,
VCE = = 5V
These calculations verify the values obtained from the dc load
line.

84
The Q-Point
• When a transistor does not have an input signal, its output rests
at specific dc values of IC and VCE. i.e. the Q-point on the dc
load line.
• The Q-point can be easily determined if the dc load line is
superimposed onto the collector curve for the transistor.

85
The Q-Point
(Linear Amp)
• For linear operation of an
amplifier, it is desirable to have
the Q-point centered on the load
line.
• VCE is half the value of VCC, and
IC is half the value of IC(sat).
• When a circuit is designed to
have a centered Q-point, the
amplifier is said to be midpoint
biased.

86
The Q-Point
(Midpoint biased)
• It allows optimum ac
operation of the amplifier.
• When an ac signal is applied to
the base of the transistor, IC
and VCE both vary around
their Q-point values.
• When the Q-point is centered,
IC and VCE can both make the
maximum possible transitions above
and below their initial dc values.

87
Example
The circuit on the right shows
the linear operation of a
transistor.
Assume a sinusoidal voltage
Vin is superimposed on VBB,
causing the base current to
vary sinusoidally 100 µA
above and below its Q-point
value of 300 µA.
Consider its operation as a
linear amplifier.

88
• The collector current varies 10 mA above and below its Q-
point value of 30 mA.
• The collector-to-emitter voltage varies 2.2 V above and below
its Q-point value of 3.4 V.
mAAII bDCC 20)200)(100( === µβ
mAAII bDCC 40)400)(100( === µβ
VmAVRIVV
mAAII
A
k
VV
R
VV
I
CCQCCCEQ
BQDCCQ
B
BB
BQ
4.3)220)(30(10
30)300)(100(
300
10
7.07.37.0
=Ω−=−=
===
=
Ω
−
=
−
=
µβ
µ
Q-point values
VmAVRIVV CCCCCE 2.1)220)(40(10 =Ω−=−=
VmAVRIVV CCCCCE 6.5)220)(20(10 =Ω−=−=

89
Point A on the load line corresponds to the positive peak of the
sinusoidal input voltage while point B corresponds to the
negative peak as shown.

90
• When the Q-point is above center of the dc load line, the input
may cause the transistor to saturate. When this happens, part of
the output sine wave is clipped off.
• If the Q-point is below midpoint on the load line, the input may
cause the transistor to go into cutoff. This can also cause a
portion of the output sine wave to be clipped.

91
Transistor is driven into saturation and cutoff because the input
signal is too large.
IBQ
ICQ
VCEQ

92
Example
Determine the Q-point for the circuit below and
find the maximum peak value of base current for
linear operation. Assume βDC= 200.

93
Solution
The Q-point is defined
by ICQ and VCEQ.
IBQ =
= 198 μA
ICQ = βDCIB = (200)(198 μA) = 39.6 mA
VCEQ = VCC – ICRC = 20 V – 13.07V = 6.93 V
Thus the Q-point is at I = 39.6 mA and V = 6.93 V.
B
BB
R
VV 7.0−

94
Solution
Since IC(cutoff) = 0, calculate IC(sat) to determine how much
variation in collector current can occur and still maintain linear
operation.
IC(sat) = 60.6 mA
Thus, IC can increase, ideally, by
IC(sat) – ICQ = 21 mA
And it can decrease by 39.6 mA before cutoff (IC = 0) is
reached.
The limiting value is then, 21 mA.
The maximum peak variation of the base current is:
Ib(peak) = Ic(peak) / βDC
= 105 μA

95
Bias Methods
(A) Base Bias / fixed bias
• Simplest form of biasing

96
Base Bias – circuit analysis
• The primary goal of biasing circuit analysis is to determine the Q-
point values (IC, VCE)todetermine whether the circuit is midpoint
biased.
Step 1:
Determining the base current, IB (Ohm’s law)
The voltage across RB equals the difference between
VCC and VBE. By formula,
VBB = VCC − VBE
Upon substitution, the base current
B
BECC
B
R
VV
I
−
=
B
BB
B
R
V
I =

97
Base Bias – circuit analysis
Step 2: The collector current is then calculated from
Step 3: VCE can be found as
VCE = VCC − ICRC
The last two equations give the Q-point values of IC and VCE.
BC IhI FE=

98
Example
(a) Determine the Q-point
values of IC and VCE for the
circuit shown below.
(b) Construct the dc load line
and plot the Q-point from
the values in part (a).
Determine whether the
circuit is midpoint biased.
(c) Determine whether the
circuit is midpoint biased
without drawing a dc load
line for the circuit.

99
Solution
First, IB is found as
Next, IC is found as
Finally, VCE is found as
VCE = VCC − ICRC = 3.94 V
A2820 µ.
360
7.08
=
Ω
−
=
−
=
k
VV
R
VV
I
B
BECC
B
028mA2.== BFEC IhI

100
Solution
(b) The end points of the dc load line are found as
IC(sat) =
and
4mA=
Ω
=
k
V
R
V
C
CC
2
8
The dc load line and the Q –
point is as plotted. The
amplifier is very nearly
midpoint biased.

101
Solution
By definition, a circuit is midpoint biased when the
Q-point value of VCE is ½ VCC.
From part (a), VCE = 3.94 V which is approximately
one-half of VCC.
The circuit is midpoint biased.

102
Base Bias – Q-point Shift
• Base bias circuits are easy to build and analyze but they are
extremely susceptible to a problem called Q-point shift (change
in the Q-point values).
• From circuit analysis performed earlier :
VCE = VCC − ICRC
(1) VCE will change if IC changes.
(2) IC is dependent on either IB or hFE changes.
• This circuit is referred to as a beta – dependent circuit.
• It is primarily used in switching applications where it mainly
operates in saturation and cutoff regions.
BC IhI FE=

103
Base Bias – Q-point Shift
• The dc gain varies with temperature.
• If temperature increases, hFE also increases, thus, IC increases.
This, in turn, causes a decrease in VCE. When this occurs, the
circuit will no longer be midpoint biased.

104
Example
The transistor below has values of hFE = 100 when T = 25 o
C
and hFE = 150 when T = 100 o
C. Determine the Q-point values
of IC and VCE at both of these temperatures.

105
Solution
• This is the same circuit as in the previous example.
• At T = 25 o
C, hFE = 100 (IB = 20.28 µA):
IC = 2.028 mA and VCE = 3.94 V
• When T = 100 o
C, hFE = 150 (assume IB does not change) :
IC = hFEIB = (150)(20.28 µA) = 3.04 mA
and VCE = VCC − ICRC = 1.92 V
• The percentage change in IC:
• The percentage change in VCE:
%50%100
028.2
028.204.3
% ≈




 −
=∆
mA
mAmA
IC
%51%100
94.3
94.392.1
% −≈




 −
=∆
V
VV
VCE

106
Voltage-Divider Bias
• It is the most commonly used biasing scheme, similar in form
to base bias, with the following exceptions:
(1) A resistor (R2) has been added between the base terminal
of the transistor and ground.
(2) A resistor (RE) has been added to the emitter circuit.
• These modifications results
in a biasing circuit with values
of ICQ and VCEQ that are
relatively stable against
variations in hFE.

107
Circuit Operation
• Resistors R1 and R2 form a simple voltage divider that sets
the value of base voltage, VB as follows:
• The value of VE can be found as
VE = VB − 0.7 V
• Then the value of IE is determined from Ohm’s law:
• Assume that ICQ ≅ IE, thus, VCEQ can be found as
VCEQ = VCC − ICQRC − ICQRE
VCEQ = VCC − ICQ(RC + RE)
CC
21
2
B V
RR
R
V
+
=
E
E
E
R
V
I =

108
Circuit Operation
• The value of IB can be found from
• Recall: hFE can be found from the spec sheet of the
transistor.
• A transistor spec sheet will list any combination of the
following values:
(a) a maximum value of hFE
(b) a minimum value of hFE
(c) a typical value of hFE
1
E
B
+
=
FEh
I
I

109
Circuit Operation
• If only one value of hFE is listed, it can be used in any
circuit analysis.
• When two or more values are listed, identify the typical
value and use the typical value in circuit analysis.
• If only a minimum value and a maximum value are
shown, use the geometrical average of these two values.
The geometrical average of hFE is found as
(max)(min))( FEFEaveFE hhh ×=

110
Example
Determine the value of ICQ and VCEQ for the circuit
shown below.
Solution
The base voltage is determined from
the voltage divider
VE is found as
VE = VB − 0.7 V = = 1.37 V
Because ICQ ≅ IE,
Finally,
VCEQ = VCC − ICQ(RC + RE)
= 4.87 V
VV
RR
R
V CCB 07.2
21
2
=
+
=
25mA1.
1.1
37.1
=
Ω
==
k
V
R
V
I
E
E
CQ

111
Base input resistance, RIN(base)
• From the previous example,
and I2 =
• As IB << I2, it can be ignored.
• However, if IB is NOT small enough to
be neglected compared to I2, then the
dc input resistance RIN(base) must be
A
kR
VB
µ440
7.4
07.2
2
=
Ω
=
A
mA
h
I
I
FE
µ5.24
51
25.1
1
E
B ==
+
=

112
Derivation of RIN(base)
• The input resistance of the transistor is
RIN(base) =
• Applying KVL and assuming VBE << IERE
VIN = VBE + IERE ≅ IERE
• Since IE ≈ IC = βDCIB, VIN becomes
VIN ≈ βDCIBRE
• The input current is the base current, IIN = IB.
IN
IN
I
V
Note: Arrows indicated
the direction of
conventional current
flow.
o Substituting VIN and IIN into the RIN(base):
The input resistance of the transistor is the gain βDC times the emitter
resistance RE.
EDCbaseIN
B
EBDC
baseIN
RR
I
RI
R
β
β
=
=
)(
)(

113
Example
Determine the dc input resistance looking in at the base
of the transistor in the circuit shown in previous slide
if RE = 1.0 kΩ and βDC = 125.
Solution
RIN(base) = βDC RE
RIN(base) = (125)(1.0 kΩ) = 125 kΩ

114
Analysis of a Voltage-Divider Bias
Circuit
When the base current, IB is NOT negligible (??), then the base
input resistance, RIN(base) must be considered in the determination
of the value of base voltage.
Note that the RIN(base) is parallel to the resistor R2 of the voltage-
divider.

115
Analysis of a Voltage-Divider Bias
Circuit
• The total resistance from base to ground is
R2 || RIN(base) = R2 || βDCRE
and it is in series with resistor R1. Applying the voltage
divider formula yields:
• From this point onwards, the analysis follows that the
procedure as in “circuit operation” section. Note that if
βDCRE > 10R2 then the expression above will revert to
CC
E21
E2
B
)||(
||
V
RRR
RR
V
DC
DC
β
β
+
=
CC
21
2
B V
RR
R
V
+
=

116
Example
Determine VCE and IC in the voltage-divider biased
transistor circuit on the right if βDC = 100.
Solution
Determine the dc input resistance at the base:
RIN(base) = βDCRE =(100)(560) = 56 kΩ
Since RIN(base) = 10R2, we may neglect RIN(base).
Thus, = 3.59 V
So,
VE = VB – VBE = 2.89 V
and from Ohm’s law,
IE = 5.16 mA
IC ≈ 5.16 mA
and VCE ≈ VCC – IC(RC + RE) = 1.95 V
Since VCE > 0, the transistor is NOT in saturation.
CC
21
2
B V
RR
R
V
+
=

117
Bias Stability
The Thevenin voltage and resistance are:
RTH = R1R2 / (R1 + R2)
Apply KVL around the equivalent base-emitter loop:
VTH – VR(th) – VBE – VR(E) = 0
VTH = IBRTH + VBE + IERE
But IB = IE / βDC,
VTH = IE (RE + RTH/βDC) + VBE
Solve for IE:
IE = (VTH – VBE) / (RE + RTH/βDC)
As usual, assume RE >> RTH / βDC, then
IE = (VTH – VBE)/RE
CCTh V
RR
R
V
21
2
+
=

118
Bias Stability
Note that the expression for IE is independent of βDC.
Varying βDC will not affect IE and IC is also unaffected by
βDC.
Remember that for this type of biasing (voltage-divider)
we must make sure RE is at least ten times RTH/βDC.
This is not an unreasonable assumption, since βDC is
generally large.

119
Emitter Bias
An emitter-bias circuit consists of
several resistors and a dual-polarity
power supply, a positive and a
negative supply voltage
Current action:
(i) the emitter current
originates at the emitter
power supply (− VEE).
(ii) A small portion of the
emitter current leaves
the transistor through the
base terminal. Base
current passes through
RB to ground.
(iii) the majority of the
emitter current continues
to the collector. Collector
current passes through
RC to VCC.

120
Circuit Currents and Voltage
• Emitter-bias circuits are always
designed so that VB ≅ 0 V.
• Assume IB = 100 µA, then the base
voltage
VB = IBRB = (100 µA)(100 Ω)
= 10 mV
• The circuit can be represented by
figure (a).
• Figure (b) shows an equivalent
circuit using a diode to represent
the base-emitter junction of the
transistor. As this circuit
illustrates,
VE = 0 V − VBE = − VBE
VE

121
• According to KVL, VR(E) must equal the difference between VE and
VEE and since, VE = − VBE,
VR(E) = VEE + VBE
• The value of the emitter current,
• Assuming that ICQ ≅ IE and VBE = 0.7 V,
• Note:
1) The above absolute value is used to obtain a positive value of ICQ.
2) hFE is not involved in the equation - the emitter bias provides
output values that are highly stable against variations in beta.
E
BEEE
E
R
VV
I
+
=
E
EE
CQ
R
VV
I
7.0+
=

122
• The approximate value of VCEQ can be determine using
VCEQ = VCC − ICRC
• This equation is based on the fact that VE = − 0.7 V. If we
assume this potential to be correct, then KVL tells us that
− 0.7 V + VCEQ + ICQRC = VCC
or
VCEQ = VCC − ICQRC + 0.7 V
• When ICQRC >> 0.7V, which is normally the case, we can ignore −
VE, leaving us with the
VCEQ = VCC − ICQRC

123
Example
Determine the values of ICQ and VCEQ for the amplifier shown
in figure below.
Solution
First, the value of ICQ is
approximated as
The value of VCEQ is found as
VCEQ = VCC − ICQRC
VCEQ = 12V−(7.53mA)(750Ω)
= 6.35 V
mA
k
VV
R
VV
I
E
EE
CQ 53.7
5.1
7.0127.0
=
Ω
+−
=
+
=

124
Collector – Feedback Bias
• The term feedback is used to
describe a circuit connection that
“feeds” a portion of the output
voltage or current back to the input
to control the circuit’s operating
characteristics.
• In this case, the circuit is constructed
so that the collector voltage, VC has
a direct effect on the base voltage,
VB.
• These feedback connections reduce
the effects that variations in hFE have
on Q-point values of each circuit.

125
Collector – Feedback Bias
• If IC tries to increases, it drops more
voltage across RC, thereby causing VC
to decrease.
• When VC decreases, there is a
decrease in voltage across RB, which
then decreases IB.
• The decrease in IB produces less IC
which, in turn, drops less voltage
across RC and thus offsets the
decrease in VC.

126
Circuit Analysis
• By Ohm’s law, the base current can be expressed as
• Substitute VC = VCC − ICRC and
into the first equation:
• Rearranging the above equation and solving for IC:
• Since the emitter is grounded, VCE = VC:VCE = VCC − ICRC
B
BEC
B
R
VV
I
−
=
DC
C
B
I
I
β
=
B
BECCCC
DC
C
R
VRIVI −−
=
β
( )DCBC
BECC
C
RR
VV
I
β+
−
=

127
Circuit Analysis
• The last 2nd
equation shows a dependency on βDC and
VBE. This can be minimized by making
RC >> and VCC >> VBE.
• This circuit essentially eliminates βDC and VBE
dependency even if the stated conditions are met.
• This circuit is able to maintain a relatively stable Q-
point values as temperature changes, either increases
or decreases.

128
Example
Calculate the Q-point values for the circuit below.
Solution
VCE = VCC − ICRC
VCE = 10 V − (2.82mA)(1.5 kΩ) = 5.77 V
( )DCBC
BECC
C
RR
VV
I
β+
−
=
( )
82mA2.
1001805.1
7.010
=
Ω+Ω
−
=
kk
VV
IC

129
Summary
1. The purpose of biasing a circuit is to establish a proper,
stable dc-operating point (Q-point) i.e. ICQ and VCEQ.
1. Graphically, the Q-point is the intersection of the dc load
line and the transistor’s collector curves
1. A dc load line intersects the vertical axis at approximately
IC(sat) and the horizontal axis at VCE(cutoff).
1. The linear (active) region of a transistor lies along the load
line, below saturation, and above cutoff.
1. The dc input resistance at the base of a BJT is
approximately βDCRE.

130
Summary

Voltage-divider provides good Q-point stability with single
polarity supply voltage. It is the most popular bias circuit.

Base bias circuit arrangement has poor stability because its Q-
point varies widely with βDC.

Emitter bias generally provides good Q-point stability, but
requires both positive and negative supply voltages.

Collector-feedback bias provides good stability using negative
feedback from collector to base.

131
Amplifier Operations
Introduction
Amplifiers are some of the most widely used circuits
encounter e.g. in audio, video, telecommunications systems,
digital systems, biomedical systems etc. etc.
For example: turning up the volume of a stereo.
What actually happens is that you are taking a relatively
weak signal and making it stronger i.e. increasing its power
level. This is a form of amplification.
Amplifiers are circuits that provide amplification.

132
BJT Amplifiers
• They are known as small-signal amplifiers - it uses signals that
take up a relatively small percentage of an amplifier’s
operational range.
• The biasing of a transistor is purely a dc operation (C2). It
establishes a Q-point about which variations in current and
voltage can occur in response to an ac signal.
• Variations about the Q-point are relative small.
• There are three types of BJT amplifiers:
- common-emitter (most commonly used),
- common-collector and
- common-base.

133
Linear Amplifier Operation
In an amplifier, there are two sources of currents and voltages:
dc and ac. They are mixed together in the circuit. When we
analyze the operation of an amplifier, we need to separate the dc
component from the ac component.

134
• The coupling capacitors block dc and prevent the internal source
resistance RS and the load resistance RL from changing the dc bias
voltages at the base and collector.
• The sinusoidal source voltage causes Vb to vary sinusoidally above and
below its VB. The resulting variation in Ib produces a larger variation in
the Ic because of the current gain of the transistor.

135
• Note that the voltage created at the
collector (VCE) is out-
of-phase with the input voltage
(common-emitter amplifier).
(CB and CC amplifiers do not exhibit this behaviour.)
• As IC increases, the collector voltage decreases and vice versa.
• This is caused by the relationship
VCE = VCC – IC(RC + RE).
• Note the negative sign on the right-hand-side of the equation.
This sign cause VCE and IC to be out-of-phase.

136
Transistor AC Equivalent
Circuits
• To visualize the operation of a transistor in an amplifier
circuit, it is often useful to represent the device by an ac
equivalent circuit.
• The elements forming the equivalent circuits relate the
changes in voltages and currents about the operating point.
• The hybrid-π equivalent circuit is as shown:

137
h-Parameters
(hybrid parameters)
• Data sheets only provide h parameters. In parenthesis are the
conditions required to get these parameters.
1. hi : Input impedance (Output shorted)
2. hr : Voltage feedback ratio (Input open)
3. hf : Forward current gain (Output shorted)
4. ho : Output admittance (Input open)
• A second subscript is added to indicate the configuration
(common-emitter, common-base, and common-collector) for
each of these parameters.
• E.g.: the h-parameters for the common-emitter (usually the
one given in data sheets) configuration are: hie, hre, hfe, and hoe.

138
r - parameters
The r-parameters are:
1. αac : ac alpha ( )
1. βac : ac beta ( )
1. : ac emitter resistance
1. : ac base resistance
1. : ac collector resistance
e
c
I
I
=
b
c
I
I
=
'
er
'
br
'
cr
Assume that:
is small enough to be neglected ⇒ replace r’b with a short.
is very large (hundreds of kΩ)⇒ replace r’c with an open.
'
br
'
cr

139
The conversion equations between r and h
parameters:
αac =
βac =
hfb
feh
='
er
oe
re
h
h
='
cr
oe
re
h
h 1+
='
br ( )fe
oe
re
ie h
h
h
h +− 1

140
Interpretation of simplified equivalent circuit:
• is the resistance seen looking into the
emitter of a forward-biased transistor. It
appears between the base and the emitter
terminals.
• The collector current Ic is equal to αac Ie.
Also, Ic = βacIb.
The most important parameter in the
simplified equivalent circuit is
'
er
'
er
EI
mV25
=

141
Common-Emitter (CE)
Amplifier
Common-emitter BJT amplifier with voltage – divider biasing

142
Common-Emitter (CE) Amplifier
• C1 and C3 are the bias and coupling capacitors on the input and
output, while C2 acts as bypass capacitor between the emitter
to ground.
• The circuit has a combination of dc and ac operation, both of
which must be considered.
• The input ac signal Vin is capacitively coupled into the base
and the output signal Vout is capacitively coupled from the
collector.
• It exhibit high voltage and current gains.
• The input is applied to base of transistor and the output is
taken from collector.

143
DC Analysis
A dc equivalent circuit is
developed by replacing the
coupling and bypass
capacitors with opens.
Recall: a capacitor is open to
dc.

144
• As seen previously, the input resistance is
RIN(base) = βDCRE = (150)(560) = 84 kΩ
which is 10 times more than R2 for this particular
case. Thus, we can ignore it.
• This yields the following base voltage:
• And the emitter voltage is
VE = VB – VBE = (2.83) – (0.7) = 2.13 V
• Therefore, the emitter current is
IE = VE / RE = (2.13) / (569) = 3.80 mA
• Since IC ≈ IE, we get the collector current
VC = VCC – ICRC = (12) – (2.80m)(1.0kΩ) = 8.20 V
• Finally, the common-emitter voltage is
VCE = VC – VE = (8.20) – (2.13)=6.07 V
VV
k
k
V
RR
R
V 83.212
8.28
8.6
CC
21
2
B =
Ω
Ω
=
+
=

145
AC equivalent circuit
• For ac operation of the amplifier, develop an ac equivalent
circuit by using the following rules:
1. Short all capacitors because we assume that their reactance
XC ≈ 0 at the signal frequency.
2. Replace all dc sources with a ground symbol.
• These rules apply to all amplifier circuits, not just common-
emitter amplifiers.

146
 Capacitors C1, C2 and C3 are replaced with shorts.
 The dc sources are replaced with ground (assuming the internal
resistance is very small). VCC terminal is at ac ground.
AC and dc grounds are both assumed to be at the same potential (0 V).

147
• The emitter bypass capacitor, C2 provides an effective short
to the ac signal around the emitter resistor, thus keeping the
emitter at ac ground.
• This allows the amplifier to have maximum gain (RC/ ). It
must be large enough so that its reactance over the
frequency range is very small compared to RE.
• A good rule-of-thumb is
10XC < RE
'
er

148
• If the internal resistance, Rs of the ac source is 0 Ω, then
all of the source voltage appears at the base terminal.
• If the ac source internal resistance Rs is non-zero, three
factors must be taken into account in determining the
actual signal voltage at the base.
They are:
1. the source resistance (Rs),
2. the bias resistance (R1||R2), and
3. the input resistance (Rin(base)).

149
Input Resistance, Rin(base)
• Rin(base) is an ac quantity (in contrast to RIN(base) which is a dc
quantity) and it is also known as the input impedance.
• The source voltage Vs is divided down by Rs and Rin(tot) so that
the signal voltage at the base of the transistor Vb is
• If Rs << Rin(tot) , then Vb ≈ Vs ，
s
)(
)(
V
RR
R
V
totins
totin
b
+
=

150
Derivation for the input resistance
Rin(base)
• The transistor is connected with the
external collector resistor RC. The input
resistance looking in at the base is
Rin(base) = Vin / Iin = Vb / Ib
• The base voltage is
Vb = Ie
• and the base current
Ib ≈ Ic / βac.
• Substituting for Vb and Ib in the first
equation above and since Ie ≈ Ic, we get
Rin(base) = βac
'
er
'
er

151
Example
Determine the signal voltage at the base of the
transistor in the circuit shown below. This circuit is
the ac equivalent of the amplifier. Assume a 10 mV,
300 Ω signal source. IE was found to be 3.80 mA.
r 'e=
25m
3.80m
=6.58Ω
Solution
Determine the ac emitter resistance,
Then, Rin(base) = βac
= 1.05 kW
Next determine the total input
resistance viewed from the source:
Rin(tot) = R1 || R2 || Rin(base) = 873 Ω

152
Solution
The source voltage is divided down by Rs and Rin(tot), so the
signal voltage at the base is the voltage across Rin(tot).
There is attenuation of the source voltage due to the source
resistance and amplifier’s input resistance acting as a voltage
divider.
Instead of getting the full 10 mV at the base, we get only 7.44
mV.
Vb=
Rin(tot)
Rs +Rin(tot )
V s= =7.44mV

153
Voltage Gain of the CE Amplifier
• Voltage gain (Av) of the CE amplifier is
the ratio between the ac output
voltage (Vc) and input voltage (Vb).
• From the figure,
Vb = Ie
Vc = αacIeRC ≈ IeRC ???
since αac ≈ 1.
• The voltage gain, from the base to
collector , is then
'
er
''
e
C
ee
Ce
b
c
v
r
R
rI
RI
V
V
A ===

154
Overall Gain
To get the overall gain of the amplifier from the source
voltage to collector voltage, the attenuation (due to the
internal source resistance Rs) of the input circuit must be
included.
• The attenuation from
source to base
multiplied by the gain
from base to collector is
called the overall
amplifier gain.
• The overall gain,
= (attenuation) Av
'
vA

155
Example
Calculate the base-to-collector voltage gain of
the amplifier shown next, with and without an
emitter bypass capacitor. There is no load
resistor.Solution
From a previous example we know that
= 6.58 Ω.
Without C2, the gain is
Av = RC / ( + RE) = 1.76
With C2 included, the gain is
Av = RC / = 152
The bypass capacitor makes a
big difference!
'
er
'
er
'
er

156
Further Notes
1. If a load of resistance RL is connected across the output of
the amplifier, the total resistance is then
Rtot = RCRL / (RC + RL).
The voltage gain is written as
Av = Rtot /
• If the load resistance RL >> RC, then Rtot ≈ RC and there is no
change in the gain.
• If RL << RC, then Rtot = RL. The voltage gain is reduced.
• Ideally, we require RL to be as large as possible.
'
er

157
Bypassing RE produces the maximum voltage
gain and stability problem.
The ac voltage gain is dependent on re’, IE and on
temperature, it will be unstable over temperature changes.
If we take the bypass off IE, the gain is decreased but RE
overpowers in the gain calculation. This actually makes the
circuit much less dependent on it.
We make a compromise by using a method called swamping.
In swamping, we only partially bypass RE as shown next.

158
Swamping
Note that the total external emitter resistance RE is
formed by two separate emitter resistors RE1 and RE2.
• One of the resistors RE2 is bypassed
and the other is not. Both resistors
affect the dc bias but only RE1 affects
the ac voltage gain:
Av = RC / (re’ + RE1)
• If we make RE1 at least ten times
larger than re’ (RE1 >> 10re’), then the
effect of re’ is minimized and the
approximate voltage gain for the
swamped amplifier is
Av ≈ RC / RE1

159
Example
For the amplifier shown next, determine the total
collector voltage and the total output voltage, both dc
and ac. Draw the waveforms.

160
Solution
Step (1): DC Analysis
RIN(base) = βDC(RE1 + RE2)
= 150 (940 Ω) = 141 kΩ
Since RIN(base) > 10R2, it can be neglected in the
dc base voltage calculation.
VB ≈ [R2 / (R1 + R2)]VCC
= [(10kΩ) / (47kΩ + 10kΩ)] 10V = 1.75 V
VE = VB – 0.7 V = 1.75 V – 0.7 = 1.05 V
IE = VE / (RE1 + RE2)
= 1.05 V / 940 Ω = 1.12 mA
VC = VCC − ICRC
= 10 V – (1.12 mA) (4.7 kΩ) = 4.74 V

161
Step(2) AC
Analysis
• Calculate re’:
re’ = 25 mV / IE = 25 mV / 1.12 mA = 22 Ω
• Determine the attenuation in the base circuit. Looking from
the 600 Ω source resistance, the total Rin is
Rin(tot) = R1||R2||Rin(base)
Rin(base) = βac(re’ + RE1) = 175 (492 Ω) = 86.1 kΩ
Rin(tot) = 47 kΩ||10 kΩ||86.1 kΩ = 7.53 kΩ

162
• The attenuation from source to base is
Attenuation = Vb / Vs = Rin(tot) /(Rs + Rin(tot))
= 7.53 kΩ / (600 Ω + 7.53 kΩ) = 0.93
Before Av can be determined, we need to know the ac collector
resistance:
Rc = RCRL/(RL + RC) = (4.7 kΩ) (47 kΩ) / (4.7 kΩ + 47 kΩ)
= 4.27 kΩ
• Now we are ready to calculate the gain from base to collector:
Av ≈ Rc /RE1 = 4.27 kΩ / 470 Ω = 9.09
• And the overall voltage gain is the attenuation times the
amplifier voltage:
A’v = (Vb / Vs) Av = (0.93) (9.09) = 8.45
• Since the source produces 10 mVrms-, the rms voltage at the
collector will be
Vc = A’v Vin = (8.45) (10 mV) = 84.5 mV

163
Step (3) Plot waveforms
The total collector voltage is the signal voltage of 84.5
mVrms riding on a dc level of 4.74 V. The peaks are
Max Vc(p) = 4.74 + (84.5 mV) (1.414) = 4.86 V
Min Vc(p) = 4.74 - (84.5 mV) (1.414) = 4.62 V

164
• The coupling capacitor C3 keeps the dc level from getting
to the output, so Vout is equal to the ac portion of the
collector voltage (Vout(p) = 119 mV).
• Source voltage is shown to emphasize phase inversion.

165
Common-Collector (CC) Amplifier
Emitter-follower amplifier (EF).
• The input is applied to base through a coupling capacitor and
the output is at the emitter. The voltage gain of a CC amplifier
is approximately 1.
• Its main advantage is its high input resistance and current
gain.

166
Voltage Gain
• From the ac equivalent circuit
Vout = IeRe
and
Vin = Ie(re’ + Re)
• The voltage gain is
• Note that Re = RE||RL and
Re = RE
when there is no load.
( ) ee
e
eee
ee
in
out
v
Rr
R
RrI
RI
V
V
A
+
=
+
== ''
In practice, Re >> re’ , hence
Av ≈ 1
that is, the voltage gain of a CC
amplifier is approximately unity.

167
b
eee
I
RrI )( '
+
b
eebac
I
RrI )( '
+β '
er
Input Resistance Rin(base)
• CC amplifier has high input resistance - it can be
used to minimize loading effects when a circuit is
driving a low resistance load. Note: the emitter resistance is
never bypassed.
• The derivation of the CC amplifier input resistance
is similar to that of the CE amplifier:
Rin(base) = Vin / Iin = Vb / Ib
=
• Since Ie ≈ Ic = βacIb ,
Rin(base) = = βac ( + Re)

168
• If Re >> , the input resistance becomes
Rin(base) = βacRe
• From the CC amplifier circuit, we can see that the bias
resistors appear in parallel with Rin(base), looking from the
input source. Thus, the total input resistance becomes
Rin(tot) = R1 || R2 || Rin(base)
'
er

169
Output Resistance
• The output resistance is very low. It is given by the
following expression:
Rout
where Rs is the resistance of the input source.
(FYI: Floyd/ pg. 937)
E
ac
s
R
R
||





≅
β

170
Current Gain
The current gain is
• The input current, Iin = Vin / Rin(tot). If , then
most of input current goes into the base. Thus, current gain
of amplifier is almost equal to current gain of the transistor,
βac, that is
Ai ≅ βac =
• This is because very little signal is diverted to the bias
resistors. Otherwise,
• βac is the maximum achievable current gain in both CC and
CE amplifiers. Recall Ie ≅ βacIb and Iin = Ib.
in
e
in
out
i
I
I
I
I
A ==
)(21 || baseinRRR >>
b
c
I
I
in
e
i
I
I
A =

171
Power Gain
• The CC power gain is
• Since Av ≈ 1, the overall power gain is
Ap ≈ Ai.
vi
inin
outout
in
out
p AA
VI
VI
P
P
A ===

172
Example
Determine the total input resistance of the emitter follower shown below.
Also find the voltage gain, current gain, and power gain in terms of
power delivered to the load, RL. Assume βac = 175. And that the
capacitive reactances are negligible at the frequency of operation.

173
Solution
• The ac emitter resistance external to the transistor, Re, is:
Re = RE || RL = 1 kΩ || 1kΩ = 500 Ω
• The approximate resistance, looking in at the base, is:
Rin(base) ≈ βac Re = (175) (500 Ω) = 87.5 kΩ
• The total input resistance is
Rin(tot) = R1||R2||Rin(base) = 18kΩ || 18kΩ || 87.56kΩ = 8.16 kΩ

174
Solution
• The voltage gain is almost unity (CC amplifier). By using
we can determine a more precise value for Av:
VE = [R2/(R1 + R2)]VCC – VBE
= (0.5)(10 V) – 0.7 V = 4.3 V
IE = VE / RE = 4.3 V / 1.0 kΩ = 4.3 mA
= 25 mV/IE = 25 mV / 4.3 mA = 5.8 Ω
So Av = Re / ( + Re) = 500 Ω / 505.8 Ω = 0.989
• The small difference in Av as a result of considering re’ is
insignificant in most cases.
'
er
'
er
'
er

175
Solution
• The current gain is Ai = Ie / Iin:
Ie = Ve / Re = AvVb / Re ≈ 1 V / 500 Ω = 2 mA
Iin = Vin / Rin(tot) = 1 V / 8.16 kΩ = 123 μA
Ai =Ie / Iin = 2 mA / 123 μA = 16.3
• The overall power gain is
Ap ≈ Ai = 16.3
Since RL = RE, one half of the total power is dissipated in
RL. So, in terms of power to the load, the power gain is
one half of the overall power gain.
Ap(load) = Ap / 2 = 16.3 / 2 = 8.15

176
The Darlington Pair
• The maximum achievable input
resistance you can get from a
given CC circuit is limited by
βac.One way to boost the input
resistance is to use a darlington
pair.
• The collectors of the two
transistors are connected and
the emitter of the first drives the
base of the second.

177
• This configuration achieves βac multiplication, i.e.,
Ie2 = βac2 βac1 Ib1.
• Thus, the effective current gain in the darlington pair is
βac = βac2 βac1
• Neglecting r’e by assuming that it is much smaller than RE,
the input resistance is
Rin = βac2βac1RE

178
• CC is often used as an interface between a circuit with a high output resistance
and a low resistance load. In such an application, it is called a buffer.
• E.g.: suppose a common-emitter amplifier with a 1.0 kΩ collector resistance
(output resistance) must drive a low-resistance load e.g. an 8 Ω low-power
speaker that is capacitively coupled to the output of amplifier. The 8 Ω load
(appears to the ac signal) in parallel with the 1 kΩ collector resistor. This
results in an ac collector resistance of
Rc = RC || RL = 1kΩ || 8Ω = 7.94 Ω.
• Obviously this is not acceptable, since most of the voltage gain is lost
Av = Rc / .
For example, if = 5 Ω, the voltage gain is reduced from
Av = RC / = 1 kΩ / 5 Ω = 200
to
Av = Rc / = 7.94 Ω / 5 Ω = 1.59
• We can add a darlington pair to interface the amplifier and the speaker.
'
er
'
er
'
er
'
er

179
For the CE amplifier : VCC = 12 V, RC = 1.0 kΩ, and re’ = 5 Ω.
For the Darlington EF : R1 = 10 kΩ, R2 = 22 kΩ, RE = 22 Ω, RL = 8
Ω. βDC = βac = 100 for both transistors.
(a) Determine the voltage gain of the CE amplifier.
(b) Determine the voltage gain of the Darlington EF amplifier.
(c) Determine the overall voltage gain and compare to the gain of the CE
amplifier driving the speaker directly without the Darlington EF.

180
Solution
The total input resistance of the Darlington emitter-
follower amplifier will act as a load to CE amplifier. In
order to determine the voltage gain for the CE amplifier
for this circuit, it is necessary to calculate Rin(tot) EF.VB = (R2 / R2 + R1)VCC = (22/32)12V = 8.25 V
VE = 8.25 V – 1.4 V = 6.85 V
IE = 6.85 V/22 Ω = 0.311 A
re’ = (25 mV/IE) = 80.3 mΩ
Re = RE || RL = 22 Ω || 8 Ω =5.87 Ω
The input resistance of the Darlington EF,
Rin(base) EF = βac1 βac2Re = (100)2
(80.3 mΩ + 5.87 Ω) = 59.5 kΩ
The total input resistance,
Rin(tot) EF = R1|| R2||Rin(base) EF = 6162.9 Ω
Thus, the ac output resistance of CE,

181
Solution cont.
Voltage gain for CE amplifier with Darlington EF,
Av = Rc /re’ = 860.4/5 = 172
Voltage gain for Darlington amplifier, Av EF = 0.987
Overall voltage gain, Av’ = 170
Voltage gain for CE amplifier w/out Darlington EF,
Av = Rc /re’ = 7.94/5 = 1.59
Using the same circuit determine Av’ is a single transistor is used
in the EF amplifier instead of Darlington pair.

182
Common-Base (CB) Amplifier
•The CB amplifier provides high voltage
gain with a maximum current gain of 1
and it has low input resistance.
•CB amplifiers are most appropriate for
certain applications where sources tend
to have very low-resistance outputs.
•The base is the common terminal and is
at ac ground because of the capacitor C2.
•The input signal is capacitively coupled
to the emitter.
•The output is capacitively coupled from
the collector to a load resistor.

183
Voltage Gain
'
er
Since the input voltage is Vin = Ve and
the output voltage is Vout = Vc, the
gain becomes
Av =
e
c
in
out
V
V
V
V
=
( )Eee
cc
RrI
RI
||'
=
( )Eee
ce
RrI
RI
||'
≈
Assuming that RE >> , we get
Av ≈ ≈ (RC||RL) /
Note: The gain expression is the same as for the common emitter
amplifier. However, there is no phase inversion from emitter to
collector.
'
e
c
r
R '
er

184
Input Resistance
• The resistance looking into the emitter is
Rin(emitter) =
• Assuming RE >> , then
Rin(emitter) ≈
• Typically, RE is much greater than , so the assumption
RE >> used is usually valid.
'
er
e
Eee
e
e
in
in
I
RrI
I
V
I
V )||( '
==
'
er
'
er
'
er

185
Output Resistance, Current Gain,
Power Gain
Output Resistance
• Looking into the collector, the ac collector resistance, rc’,
appears in parallel with RC. As rc’ is much larger than RC,
Rout ≈ RC
Current Gain
• It is defined as Ai = Iout / Iin. Thus we have
Ai = Ic / Ie ≈ 1
Power Gain
• Since the current gain is approximately unity, by Ap = Ai Av, we
have
Ap ≈ Av

186
Example
Find the input resistance, voltage gain, current
gain, and power gain for the amplifier shown
below. Firstly, find IE so that re’ can be determined.
Then Rin ≈ re’. Since βDCRE >> R2, then
VB = R2 / (R1 + R2) VCC = 1.76 V
VE = VB – 0.7 V = 1.06 V
IE = VE / RE = 1.06 mA
Therefore,
Rin ≈ re’ = 25 mV / IE = 23.6 Ω
The voltage gain is found as follows.
Rc = RC||RL = 1.8 kΩ
Av = Rc / re’ = 76.3
We also know Ai = 1 and
Ap ≈ Av = 76.3.

187
Amplifier Comparisons
CE CC (or EF) CB
Input Base Base Emitter
Output Collector Emitter Collector
Inversion Yes No No
Voltage gain High (RC
/r’e
) Low (≈ 1) High (RC
/r’e
)
Current gain High (βac
) High (Βac
) Low (≈ 1)
Power gain Very High
(Ai
Av
)
High (≈ Ai
) High (≈ Av
)
Input Resistance Low (βac
r’e
) High (βac
RE
) Very low (r’e
)
Output Resistance High
RC
Very low
(Rs
/βac
)||RE
High
RC
Frequency Range Medium Medium High

188
Summary
• A small signal amplifier uses only a small portion of its dc
load line.
• r-parameters are easily identifiable and applicable with a
transistor’s circuit operation.
• A common emitter amplifier has good voltage, current, and
power gains, but a relatively low input resistance.
• A common collector amplifier has high input resistance and
good current gain, but its voltage gain is approximately 1.
• The common base amplifier has a good voltage gain, but it
has a very low input resistance and its current gain is
approximately 1.
• A darlington pair provides β multiplication for increased
input resistance.

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