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Name of the file gives good info. This file intended for any ECE students wants to brush-up/refresh/read MOST basics of a BJT.
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BG/BJT2/C/ May 31, 2002 Chapter – 3/BJT 63
3.10 COMMON-COLLECTOR AMPLIFIER A typical common-collector amplifier using an npn transistor is shown in Figure 46. This
circuit is also known as emitter-follower. Its small-signal analysis circuit is given in Figure 47. The
corresponding dc and ac load lines are depicted in Figure 48.
VCC
R 1
RS
+
_v s
R 2 R E RL
+
_
v o
+
_v CE
i
i
i
i
1
i 2
in
i E
C
o
Figure 46: A typical common-collector configuration
R
+ _
R B
S
v s
r
RE RL
+
v o
_
π
i
i
i
i
i
i
in
a
b
e
o
c+
_
v in
Figure 47: Small-signal analysis circuit
BG/BJT2/C/ May 31, 2002 Chapter – 3/BJT 64
iC (mA) iACM AC Load Line IDCM Q-Point ICQ IBQ DC Load Line 0 vCE (V) 0 VCEQ vACM VDCM Figure 48: DC and ac load lines
The dc resistance in the collector-emitter loop is RE/α. For small-signal analysis, the
effective resistance in the emitter branch is the parallel combination of RE and RL. Let us denote
the parallel combination as REL. Then, REL = RE || RL.
The forward diode resistance in the base branch is rπ . The effective resistance as seen
from the base terminal is RibL such that
RibL = rπ + (β + 1) REL = (β + 1)(re + REL)
where re = rπ / (β + 1)
is the equivalent forward diode resistance in the emitter branch. Note that we have included the
subscript L to highlight the fact that the effective resistance as seen from the base terminal
depends upon the load resistance RL. As long as REL >> re (which is usually the case), the
resistance as seen from the base terminal is
RibL ≈ (β + 1) REL
This is a relatively high resistance. For instance, if REL is in kilo-ohms and β is high (say
100), then RibL is in hundreds of kilo-ohms. Thus, an emitter follower (common collector amplifier)
usually offers high input resistance as seen from the base terminal.
BG/BJT2/C/ May 31, 2002 Chapter – 3/BJT 65
It is evident from Figure 47 that the input resistance depends upon the load resistance. As
the load resistance changes so does the input resistance. Let us denote the input resistance
including the effect of the load resistance as RinL. From the small-signal analysis circuit, it is clear
that RinL is simply a parallel combination of RB and RibL. That is
RinL = RB || RibL
This equation states that the input resistance can be made very large by choosing a large
value of RB. We can, therefore, say that a common-collector configuration can be used as an
amplifier in such circuits where large input resistance is needed. In fact, we can use this
configuration as a pre-amplifier when the input signal has unusually large resistance to minimize
the loading effect.
We can now compute all the small-signal currents as follows:
ibL
inb R
vi =
ELe
in
ibL
inbe
ibL
inbc
Rrv
Rv)1(
i)1(i
Rv
ii
+=
+β=+β=
β=β=
Thus, the output voltage is
inELe
ELELeo v
RrR
Riv+
==
From this equation, it should be clear that the output voltage is nearly equal to the input voltage as
long as REL >> re (This is usually true). Thus, the voltage gain is
ELe
EL
in
oV Rr
Rvv
A+
==
When we neglect re in comparison with REL, we find that the voltage gain is unity.
BG/BJT2/C/ May 31, 2002 Chapter – 3/BJT 66
We can also express the output voltage and voltage gain in terms of vs as
sinLS
inL
ELe
ELo v
RRR
RrR
v
+
+=
and
+
+==
inLS
inL
ELe
EL
s
oVS RR
RRr
Rvv
A
From the voltage gain equation, it becomes evident that as long as re << REL and RS <<
RinL, the voltage gain approaches unity. Under these assumptions, we can label a common-emitter
configuration as unity gain amplifier. Since the voltage at the emitter is nearly equal to the
applied signal, it is for this reason the common-collector configuration is called the emitter
follower. The emitter voltage follows the base voltage.
From Figure 47, we see that the output resistance is a parallel combination of RE and the
total resistance in the base branch as seen from the emitter terminal. Note that the signal voltage
vs must be replaced with a short circuit. The total resistance in the base circuit is rπ + (RS || RB).
Thus, the output resistance is
RO = RE || [rπ + (RS || RB) / (β + 1)]
Let us assume that the input signal is ideal in the sense that its resistance is very small
and can be neglected, then RS → 0. The output resistance is now simply a parallel combination of
RE and re. Since re is very small in comparison with RE, the output resistance is nearly re. Thus, a
common-collector configuration has negligibly small output resistance. This fact enables us
to use the common-collector configuration as the output stage of an amplifier where the load
resistance is small (This is usually the case when the load is either a 4-Ω, an 8-Ω, or a 16-Ω
speaker).
We can summarize our findings for the common-collector or emitter follower circuit as follows:
(a) Its voltage gain is nearly unity.
(b) Its input resistance is usually quite high.
(c) Its output resistance is usually very low.
We have already discussed another circuit that has all of the above traits. Can you name
it? Well, it is an operational amplifier circuit called the buffer amplifier. Now you know that a single
transistor common-collector configuration behaves more like a buffer amplifier. However, it is not
as good as the buffer amplifier. But, what more can you expect from a single transistor?
BG/BJT2/C/ May 31, 2002 Chapter – 3/BJT 67
Design Example: Common-Collector Configuration __________________________________
Select the needed circuit components for the common-emitter amplifier of Figure 46.The following
information is known for the amplifier: RE = 5 kΩ, RL = 5 kΩ, β = 49, VBE = 0.7 V, VCE(sat) = 0.2
V, RS = 1 kΩ, VCC = 10 V, and vS = 2 sin(10,000t) V. The other components must be selected in
such a way that the input resistance is within ± 5% of 10 kΩ.
Design philosophy:
Let us first examine if we can design this amplifier for the maximum peak-to-peak voltage
swing exactly like the common-emitter design. To determine the collector current at the dc
operating point, we need to know RDC and RAC. For the dc analysis, the only resistance in the
collector-emitter loop is RE. Thus, the equivalent dc resistance as seen by the collector current is
RDC = RE/α = 5 /0.98 = 5.102 kΩ
where α = β/(β + 1) = 49/50 = 0.98.
The ac resistance in the emitter branch is simply the parallel combination of RE and RL. Since RE
= RL = 5 Ω, the equivalent ac load resistance is REL = 2.5 kΩ. Thus, the ac resistance in the
collector-emitter loop is
RAC = REL/α = 2.5/0.98 = 2.551 kΩ
We can now compute the collector current at the Q-point using (34) as
307.1102.5551.2
10RR
VI
DCAC
CCCQ =
+=
+= mA
The emitter current and the collector-to-emitter voltage at the Q-point are
IEQ = ICQ/α = 1.334 mA
VCEQ = VCC – IEQ RE = 10 – (1.334 mA)(5 kΩ) = 3.33 V
The power dissipated by the transistor at the operating point, neglecting the power loss in the base
circuit, is
PT = VCEQ ICQ = (3.33V)(1.307 mA) = 4.35 mW
The power supplied by the dc source at the operating point is
PDC = VCC ICQ = (10 V)(1.307 mA) = 13.07 mW
BG/BJT2/C/ May 31, 2002 Chapter – 3/BJT 68
The maximum peak-to-peak swing for the collector-to-emitter voltage is 2 VCEQ because
the collector-to-emitter voltage can theoretically be as low as 0 V and as high as 2 VCEQ. Stated
differently, the maximum swing in either direction for the collector-to-emitter voltage is VCEQ.
Therefore, the maximum value of the swing in either direction is 3.33 V. A quick glance at Figure
47 reveals that it is also the swing for the load voltage because vCE + vo = 0. It implies that the
maximum output voltage can be as high as 3.33 V. We already know that the voltage gain of the
common-collector circuit is nearly equal to unity and the maximum value of the input signal is 2 V.
Thus, the expected maximum value of the output signal is nearly equal to 2 V. We can reduce the
maximum undistorted swing to a lower value by moving the operating point to the right on the dc
load line. As we move down the dc load line, the collector current decreases. The decrease in the
collector current causes a decrease in the power supplied by the dc source. If the dc source is a
battery, as the power supplied by the battery decreases, the battery life increases. Therefore,
moving the operating point down the dc load line is good from the power supplied point of view.
As the operating point moves to the right, collector-to-emitter voltage increases while the
collector current decreases. The product of collector current and the collector-to-emitter voltage is
the power dissipated by the transistor. Note that the power dissipated by the transistor is zero
when the voltage drop across the transistor is zero. This is the point on the y-axis which we labeled
as (0, IDCM). On the other hand, the collector current is zero when the voltage drop across the
transistor is VDCM. This is the point on the x-axis and we labeled it as (VDCM, 0). At any other
point of operation on the dc load line, the transistor dissipates power. As we move from one
extreme point, say from (0, IDCM) on the dc load line toward the other point (VDCM,0), the power
dissipation by the transistor increases. At some value of the collector current IC(max) the power
dissipated by the transistor becomes maximum, and then begins to decrease until we reach the
other point where it is zero again.
Let us now determine the point at which the power dissipated by the transistor is at its
maximum. We already know that the voltage drop from collector to emitter is
VCE = VCC – IC RDC
Thus, the power dissipated by the transistor is
DC2CCCCT RIIVP −=
BG/BJT2/C/ May 31, 2002 Chapter – 3/BJT 69
The power dissipated by the transistor is at its maximum when the rate of change of PT
with respect to IC is zero. Thus, differentiating the above equation with respect to IC and setting it
to zero, we obtain the collector current at which the transistor dissipates maximum power as
DC
CC(max)C R2
VI =
In our case, RDC is 5.102 kΩ and VCC = 10 V. Thus, IC(max) = 0.98 mA.
The corresponding collector-to-emitter voltage is
VCE(max) = 10 – (0.98 mA)(5.102 kΩ) = 5 V
The maximum power dissipation occurs at a point where VCE is 5 V and IC is 0.98 mA. We
can call it as the maximum power dissipation point and represent it as (5 V, 0.98 mA). This
point is on the right side of the Q-point for the maximum undistorted peak-to-peak voltage swing
(3.334 V, 1.3067 mA) computed earlier.
As we said earlier, we need to decrease the collector current at the operating point in order
to reduce the power supplied by the dc voltage source. As we move the operating point to the right,
the power dissipation by the transistor is expected to increase. The question: Where should the
operating point be so that power supplied by the dc source is lower than 13.07 mW? Since
we expect the maximum value of the output voltage swing to be 2 V or so, we can select the
operating point such that vACM - VCEQ is about 2.5 V (using a safety factor of 1.25 to avoid going
into the cut-off region). This will give us a theoretical peak-to-peak output voltage swing of 5 V
whereas the expected peak-to-peak output voltage is only 4 V. This will surely keep us well above
the cut-off region. From Figure 48, it is clear that vACM – VCEQ must be equal to ICQ RAC. Since
RAC is 2.551 kΩ, the new Q-point collector current must be
ICQ = 2.5/2.551 = 0.98 mA.
It is merely a coincident that the new Q-point happens to be where the transistor dissipates
maximum power. Don’t always expect this to be the case. In fact, it would be nice if this really were
not the case so that the transistor does not have to operate at a point where it dissipates maximum
power.
The emitter and base currents at the new Q-point are
IEQ = ICQ / α = 0.98 / 0.98 = 1.0 mA
IBQ = ICQ / β = 0.98 / 49 = 0.02 mA
BG/BJT2/C/ May 31, 2002 Chapter – 3/BJT 70
Thus, the collector-to-emitter voltage at the new Q-point is
VCEQ = 10 – (1 mA)(5 kΩ) = 5 V.
The dc operating point selection appears to be satisfactory at this stage. At the Q-point,
ICQ = 0.98 mA, IBQ = 0.02 mA, IEQ = 1 mA, and VCEQ = 5 V
At the dc operating point, the base terminal voltage is
VBQ = VBE + IEQ RE = 0.7 + (1 mA)(5 kΩ) = 5.7 V
Selection of R1 and R2:
Since we need an input resistance of 10 kΩ, we select R1 and R2 on this basis. In the presence of
the load resistance RL, the input resistance is
RinL = RB || (β + 1)(re + REL) ≈ RB || (β + 1) REL
where we have neglected the contribution due to re at this time.
From the above equation, we get
5.250
1101
R)1(1
R1
R1
EinLB ×−=
+β−≈
Thus, RB = 10.87 kΩ.
Note that RB is the parallel combination of R1 and R2. We can compute base voltage as
VBB = VBQ + IBQ RB = 5.7 + (0.02 mA)(10.87 kΩ) = 5.92 V
You may have realized by now that VBB is the Thevenin equivalent voltage and RB is the Thevenin
equivalent resistance of the biasing circuit. Thus,
21
2CCBB RR
RVV
+= (46)
21
21B RR
RRR
+= (47)
From (46) and (47), we obtain
CC
BB1B V
VRR =
Substituting RB = 10.87 kΩ, VCC = 10 V and VBB = 5.92 V in the above equation, we get R1 =
18.36 kΩ. Let us select a standard resistor of 18 kΩΩΩΩ for R1.
BG/BJT2/C/ May 31, 2002 Chapter – 3/BJT 71
To compute R2, we can use the following equation:
21
111RRRB
+=
Substituting RB = 10.87 kΩ and R1 = 18 kΩ in the above equation, we obtain R2 = 27.44 kΩ.
Let us select a standard resistance of 27 kΩΩΩΩ for R2.
We should now verify the design by carrying out exact dc analysis and the small-signal analysis.
DC Analysis Using Selected Components
All the resistors in the amplifier circuit are:
R1 = 18 kΩ, R2 = 27 kΩ, RE = 5 kΩ, and RL = 5 kΩ.
The dc and ac resistances in the collector-emitter loop are
RDC = 5.102 kΩ
and RAC = 2.551 kΩ.
The Thevenin equivalent circuit of Figure 46 is given in Figure 49 where
645
2710VBB =×= V
8.1045
2718R B =×= kΩ
kΩ
kΩ 10.8
5
CI
EI
BI
+ _6 V
10 V
Figure 49: Thevenin equivalent circuit for dc analysis
BG/BJT2/C/ May 31, 2002 Chapter – 3/BJT 72
The currents in the circuit are
0203.05508.10
7.06IB =
×+−= mA
IC = β IB = 49 × 0.0203 = 0.995 mA
IE = (β + 1) IB = 50 × 0.0203 = 1.015 mA
The collector-to-emitter voltage: VCE = 10 – (1.015 mA)(5 kΩ) = 4.925 V
Since VCE is greater that VCE(sat), the transistor operates in the active region.
Thus, the currents and voltages at the dc operating point are as follows:
ICQ ≈≈≈≈ 1 mA, IEQ = 1.02 mA, IBQ = 0.02 mA, and VCEQ = 4.93 V
Small Signal Analysis The ac load line passes through the Q-point (4.93 V, 1 mA) and intersects the x-axis at
vACM where
vACM = VCEQ + ICQ RAC = 4.93 + (1 mA)(2.551 kΩ) = 7.48 V
Thus, the maximum positive swing in the collector-to-emitter voltage is 2.55 V. The maximum
negative swing, of course, is VCEQ = 4.93 V. Thus, the symmetrical undistorted peak-to-peak swing
is 5.1 V (2 × 2.55). The required peak-to-peak swing is about 4 V. Hence, the transistor is expected
to operate well above the cut-off region.
The forward resistance of the diode in the base branch, when n = 1, is
1250mA02.0
mV25IV
rBQ
T ===π Ω
Its counterpart in the emitter circuit is
2550
12501
rre ==
+β= π Ω
The input resistance, including the load resistance, is
RinL = RB || (β + 1)(re + REL) = 10.8 || 50 (2.5 + 0.025) = 9.95 kΩ.
The input resistance is well within the ±±±± 5% variations of the desired value of 10 kΩΩΩΩ. Thus,
our design meets the input resistance requirements.
BG/BJT2/C/ May 31, 2002 Chapter – 3/BJT 73
We can now compute the small signal base current and emitter current as
25.126
v)025.05.2(50
v)Rr)(1(
vi inin
ELe
inb =
+=
++β= mA
525.2v
i50i)1(i inbbe ==+β= mA
Thus, the output voltage is
inin
ELeo v99.05.2525.2v
Riv =
== V
Thus, the voltage gain is AV = 0.99. The output voltage is in phase with the input voltage.
We can also express vin in terms of vs as
ssinLS
inLin v
95.1095.9
vRR
Rv =
+=
Thus, the output voltage in terms of vs is
sso v9.0v95.1095.9
99.0v =
=
Thus, the overall voltage gain is AVS = 0.9.
As vs varies as 2 sin(10,000t) V, the output voltage varies as 1.8 sin(10,000t) V. The peak-
to-peak maximum swing of the output voltage is 3.6 V. It is well within the maximum undistorted
swing of 5.1 V.
We can, in fact, modify the design by redesigning the amplifier for a maximum peak-to-
peak swing of say 4.5 V using a safety factor of 1.25 (i.e. 1.25 × 3.6 V). This will certainly lower the
collector current at the new dc operating point and thereby further reduce the drain on the battery.
The new design is left as an exercise for you. Carry it out in order to build the necessary
confidence in designing the amplifiers.
The input and the output currents are
95.9
vRv
i in
inL
inin == mA
inin
L
oo v198.0
5v99.0
Rv
i === mA
BG/BJT2/C/ May 31, 2002 Chapter – 3/BJT 74
Thus, the current gain: 97.1ii
Ain
oI ==
Therefore, the power gain: 95.197.199.0AAA IVP =×==
The output resistance is
+β+
= π
1R//Rr
//RR BSEO
Substituting the values, we obtain
Ω=
+= k043.050
8.10//125.1//5R O or 43 Ω.
As stated earlier, the output resistance of an emitter follower is low. Thus, an emitter follower can
be used to match a high source resistance to a low load resistance.
Other quantities of interest such as the total collector current, total collector-to-emitter voltage drop,
power dissipated by all resistors, etc. can now be computed.
3.11 COMMON-BASE CONFIGURATION
A general single transistor common-base amplifier is given in Figure 50. In this case, the input is at
the emitter and the output is at the collector. The small-signal model of the common-base amplifier
is shown in Figure 51. The corresponding dc and ac load lines are depicted in Figure 52.
VCC
R C
R E
R1
R2
RS
+ _ vs
RL
+
_
v o
Figure 50: A common-base amplifier
BG/BJT2/C/ May 31, 2002 Chapter – 3/BJT 75
RB
r
RE
RS
+ _ v s
RC R L
ibi c
i ei in vin
+i o
_
+
_
v o
π
i RE
Figure 51: Small signal model of a common base amplifier iC IACM AC Load line IDCM Q-Point ICQ IBQ DC Load Line 0 vCE 0 VCEQ vACM VDCM
Figure 52: The dc and ac load lines
In order to draw the dc and ac load lines, we need the dc and ac resistances in the
collector-emitter branch. The examination of Figure 50 reveals that the dc resistance is
RDC = RC + RE/α (48)
where α = β/(β + 1)
BG/BJT2/C/ May 31, 2002 Chapter – 3/BJT 76
From the small signal model given in Figure 51, we find that the ac load resistance is
RCL = RC || RL (49)
In terms of RCL, the ac resistance in the collector-emitter branch is
RAC = RCL + RE/α (50)
When the transistor has a high β (≥100), we can make the first-order approximation that α is
nearly unity. Then, the approximate values of the dc and ac resistances are
RDC ≈ RC + RE
and RAC ≈ RCL + RE
The approximate expressions are good during the design process. Once we have selected all the
components, then we can use the exact expressions.
Let us now analyze the small-signal model given in Figure 51. The base current can be expressed
in terms of vin as
B
inb Rr
vi
+−=
π
(51)
Note that ib is inversely proportional to RB. The base current can be increase by setting
RB equal to zero for the small-signal analysis. This can, of course, be done by connecting a large
capacitor between the base terminal and the common terminal. The capacitor will then provide a
path for the ac variations in the base current.
The collector current is
B
inbc Rr
vii
+β
−=β=π
(52)
The emitter current is
B
inbe Rr
v)1(i)1(i
++β
−=+β=π
(53)
The current through RE is
E
inRE R
vi = (54)
BG/BJT2/C/ May 31, 2002 Chapter – 3/BJT 77
Hence, the input current is
++++β
=
+
++β=−=
π
π
π
)Rr(RRrR)1(
v
R1
Rr1
viii
BE
BEin
EBineREin
(55)
The output voltage: inB
CLCLco v
RrR
Riv+
β=−=
π
(56)
Therefore, the voltage gain: B
CLV Rr
RA
+β
=π
(57)
Once again, it is clear from this equation that the voltage gain can be increased by bypassing RB is
with a capacitor. This circuit, therefore, can provide a high voltage gain.
The output current: inLC
C
BL
oo v
RRR
RrRv
i
++β==
π
(58)
The current gain:
+β++
+β
==π EB
E
LC
C
in
oI R)1(Rr
RRR
Rii
A (59a)
When RB is bypassed by a capacitor and rπ is small in comparison with (β + 1) RE, then the
current gain becomes
LC
CI RR
RA
+α
≈ (59b)
Note that the output resistance for the common-base amplifier from Figure 51 is RC. If RC is
selected on the basis of maximum power transfer to the load, then RC = RL. In this case, the
current gain is less than 0.5 because α is usually less than 1.
The input resistance:
+β+
= π
1Rr
||RR BEin (60)
Once again, if RB is bypassed by a capacitor and RE is usually quite large in comparison with re,
then the input resistance is
ein r1
rR =
+β≈ π (61)
BG/BJT2/C/ May 31, 2002 Chapter – 3/BJT 78
From this equation, it is clear that the input resistance of a common-base configuration is
usually very low (few ohms). For this reason, this configuration is often used in high frequency
applications (above 10 MHz) where low impedances are very common. It is also widely used as a
differential amplifier in the design of integrated circuits.
In a nutshell, a common-base amplifier has the following characteristics:
* Low input resistance ( ≅≅≅≅ re)
* High output resistance (RC)
* Low current gain (usually less than 1)
* High voltage gain
Example: Analysis of a Common Base Amplifier ____________________________________
A common-base amplifier circuit is shown in Figure 53. The circuit elements and the parameters of
the transistor are RE = 10 kΩ, RC = 2.2 kΩ, RL = 2.2 kΩ, RS = 10 Ω, VCC = 15 V, – VEE = – 15
V, VBE = 0.7 V, VCE(sat) = 0.2 V, and β = 49. Determine the voltage gain, the current gain, the
input and output resistances.
VCC
VEE
RC
RE
RS
+ _ vs RL
vo
-
Figure 53: A common-base amplifier
BG/BJT2/C/ May 31, 2002 Chapter – 3/BJT 79
+ _
V CC
- V EE
R L
R C
R S
v s
+
_ vo
R E
Figure 54: Circuit of Figure 53 redrawn
The common-base amplifier is usually drawn as shown in Figure 53 in order to shown the input
signal on the left side and the output signal on the right side of the circuit. We can redraw this
circuit so that it appears familiar to us. Such a circuit is redrawn in Figure 54.
V
- V
R
R
I
I
I
CC
EE
E
C C
E
B
+
_V CE
Figure 55: Circuit for dc analysis
BG/BJT2/C/ May 31, 2002 Chapter – 3/BJT 80
DC Analysis
The amplifier circuit for the dc analysis is shown in Figure 55. The base voltage is zero because it
is directly connected to the common terminal (ground). The dc voltage at the emitter is – 0.7 V
because of VBE = 0.7 V. The voltage drop across RE is – 0.7 – (– VEE). Since, – VEE is – 15 V,
the voltage drop across RE is 14.3 V. Since RE is 10 kΩ, the emitter current is
43.1k10
V3.14IE =
Ω= mA
The corresponding base and collector currents, when β = 49, are
0286.050
mA43.11
II E
B ==+β
= mA
4014.1II BC =β= mA
The collector-to-emitter voltage is
VCE = VCC – VEE – IC RC – IE RE = 30 – (1.4014 mA)(2.2 kΩ) – (1.43 mA)(10 kΩ)
= 12.617 V
Since VCE is greater that VCE(sat), the transistor operates in the active region.
Therefore, at the dc operating point, we have ICQ = 1.4014 mA, IBQ = 0.0286 mA, IEQ = 1.43 mA, and VCEQ = 12.617 V
In order to sketch the dc load line, let us first determine the dc resistance in the collector-emitter
loop. From Figure 55, it is clear that the dc resistance is
RDC = RC + RE/α
where α = β / (β + 1) = 49 / 50 = 0.98
Thus: RDC = 2.2 + 10/0.98 = 12.404 kΩ
The dc load line equation can now be written as
VCE = VCC – VEE – IC RDC = 30 – 12.404 IC
The two points on this line are (30 V, 0) and (0, 2.42 mA). The line joining the two points yields the
dc load line as depicted in Figure 56. The intersection of the dc load line with the transistor
characteristic for IB = 0.0286 mA yields the Q-point.
BG/BJT2/C/ May 31, 2002 Chapter – 3/BJT 81
iC (mA)
IDCM = 2.42
ICQ = 1.40
0 12.62
30 vCE (V)
VCEQ VDCM
Q - Point
dc Load line
I BQ
= 0.0286 mA
Figure 56: The dc load line
Small signal analysis
Let us now compute the voltage gain, the current gain, the input resistance and the output
resistance. To do so, we have to sketch the small signal equivalent circuit as given in Figure 57.
+ _ vs
RS
RE
RC RL
i b
i c
i e
i RE
i o
v o
+
_
rπ
i in +
_
v in
Figure 57: Small signal circuit for the common-base amplifier
BG/BJT2/C/ May 31, 2002 Chapter – 3/BJT 82
From the above figure it is clear that the small signal collector current flows through the emitter
resistance RE and the parallel combination of RL and RC. Note that RC in parallel with RL is the ac
load resistance. That is,
RCL = RC || RL
For our circuit, RC = 2.2 kΩ, and RL = 2.2 kΩ. Substituting these values in the above equation, we
obtain RCL = 1.1 kΩ.
The forward resistance of the diode in the base branch, when n = 1, is
Ω=Ω===π k874.0874mA0286.0
mV25IV
rBQ
T
The equivalent resistance in the emitter branch is
Ω==+β
= π 48.1750
8741
rre
We can now calculate the base, collector, and emitter currents in Figure 57 as
ininin
b v144.1k874.0
vrv
i −=Ω
−=−=π
mA
ininbc v06.56)v144.1(49ii −=−=β= mA
inbe v20.57i)1(i −=+β= mA
The current through RE is
inin
E
inRE v1.0
k10v
Rv
i =Ω
== mA
The output and the input currents are
( ) ininLC
Cco v03.28v06.56
2.22.22.2
RRR
ii =−+
−=
+−= mA
ininineREin v3.57v2.57v1.0iii =+=−= mA (62)
The output voltage can now be computed as
ininLoo v67.61)2.2)(v03.28(Riv === V (63)
BG/BJT2/C/ May 31, 2002 Chapter – 3/BJT 83
Hence, the voltage and current gains are
67.61vv
Ain
oV ==
489.0v3.57v03.28
ii
Ain
in
in
oI ===
As expected, the voltage gain is high and the current gain is less than 0.5.
The power gain is
16.30489.067.61AAA IVP =×==
Let us now represent the output voltage in terms of the voltage of the applied signal. The input
voltage vin can be expressed in terms of vs as
ssin
inin v
RRR
v+
=
Since Rin is simply the ratio of vin to iin, it can be computed from (62) as 17.45 Ω.
We could have also computed it as the parallel combination of RE and re. Since RE >> re, the input
resistance is almost equal to re (17.5 Ω). The signal resistance RS is given as 10 Ω.
Substituting these values in the above equation, we obtain
sin v636.0v =
We can now express the output voltage from (63) in terms of vs as
vo = 61.67 (0.636 vs) = 39.22 vs V
The overall voltage gain can now be computed as
22.39vv
As
oVS == V
The output resistance is simply CR , i.e. 2.2 kΩ.