ac
ΔI E
This is also known as the short-circuit gain of a transistor, and written as −hfb
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2. Likewise β in a CE configurations is given by Eqs. (2.4).
ac
ΔI
β = h =
C
(2.4)
ac
fe
ΔI B
Example: 2.1
In a CB configuration, I and in a transistor are 1.5 mA and 30
B
I E
μ A. Calculate
the values of α and I .
C
Solution
= I − I = 1.5 × 10−3 − 30 × 10−6 = 1.47
I C
E
B
mA
I
1.47
α = C =
= 0.98
I
1.5
E
Example 2.2 Analysis of Common Collector configuration
I
I
E
E
PNP
NPN
IB
I
IC
C
IB
(a)
(b)
Figure 2.6 Analysis of Common Collector configuration
Note that the input is applied between base and collector, while ouput is taken out
from emitter-collector Fig. 2.4. I is the input current. Thus current gain is given
B
by Eq. 2.5
I
I
I
E
β
β
= E ⋅ C =
=
I
I
I
α
(
)
B
C
B
β / 1+ β
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I
∴ E = 1
( + β)
(2.5)
I B
Thus, Output current, I = 1
( + β)×input current
E
In both Fig. 2.6 (a) and(b) ,
.
I = I + I
E
B
C
For demonstration of common emitter amplifier (demonstration of its working)
see:
http://www.educypedia.be/electronics/javacollectors.htm. 10th August 2007.
http://www.educypedia.be/electronics/composemiconductors.htm. 10th August 2007.
Summary of Learning activity
Remember that relations between transistor currents are:
(
I
I
i)
β
α
C ;
C ;
α =
β =
α =
I
I
1
(
);and β = 1( )
(2.6)
E
B
+ β
− α
From Eq. 2.6
( )You should be able to show that:
β
(a)
I =
I
C
E
1+ β
(b)
I = 1
( −α)I
B
E
I
(c)
I =
B
E
1
(
) and that
− α
The three transistor dc currents are in the following ratios
(d)
I : I : I = 1: 1
( −α):α .
E
B
C
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Activity 2.1.7 Leakage Currents in Transistor
P
N
P
N
P
N
I
I
= I
I
I
= I
E
C
E
E
C
E
VEE
V
V
(1 ) IE
V
I
EE
CC
(1
) I
CO
I
CC
E
CO
I
I
=
E
C
I
I
I
B
(1
) E
CO
I = I I
B
(1
) E
CO
(a)
(b)
Figure 2.7 Leakage Currents in Transistor
In Fig. 2.7 (a) and (b),
is the supply voltage, and
is the emitter voltage. In
VCC
VEE
both circuits, we see that I splits into two parts, namely:
E
(i) 1
( −α)I which becomes base current, , in the external circuit and
E
I B
(ii) αI which becomes collector current, , in the external circuit.
E
I C
Though C/B is reversed biased for majority carriers, in Fig. 2.7 (a), it is forward
biased for thermally-generated electrons, which are minority carriers. This attributes
to leakage current, I , which flows in the same direction as the majority collector
CBO
current, I , even if
is disconnected. The subscripts CBO stand for ‘Collector
C
VEE
to Base with emitter open.
Note that I , is temperature-dependent because it is made of thermally-generated
CBO
minority carriers. If current due to minority carriers are taken into account, then
I = α I + I
(2.7)
C
E
CBO
= majority + minority
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Activity 2.1.8 Transistor Static Chracteristics
In this section, the three important characteristics of BJT are described. Study the
note provided along with other references in order to understand the concepts. You
will learn that a transistor has three important chracteristics: Input chracteristics;
output characteristics; and Constant-current transfer characteristics.
Let us use Fig. 2.8 to learn about these characteristics.
Common Base Static Characteristics
IC
IE E
C
R 2
R
V
1
VCB
V
EE
V
CC
BE
B
Figure 2.8 Common Base Static Characteristics
(1) Input Characteristics
This gives variation of I with when is constant.
E
VBE
VCB
(i) Use the references at your disposal and describe how sets of values I and
E
V are obtained when
is constant.
BE
VCB
(ii) Sketch graphs showing variations of of I with
for different values of
E
VBE
VCB
(iii) On a given graph obtain instantaneous input resistance, R , is obtained from
in
the reciprocal of the slope. i.e.
1
ΔV
R =
=
BE
(2.8)
in
/
ΔI
ΔV
ΔI
E
BE
E
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Note that, variation in R with usually gives rise to distortion of signals.
in
VBE
2
( ) Output Characteristics: (Best done through experiment)
This is a relation showing variation of I with when is constant.
C
VCB
I E
(i) The whole of activity 2.1.8 may be carried out experimentally. Get the com-
ponents shown in Fig. 2.8 and carry out this activity practically.
(ii) In order to obtain the output characteristics, record corresponding values of
I and
for different values of
.
C
VCB
I E
(iii) You should be able to note that the small amount of I flows even when
C
I E
=0.
(iv) Use the characteristics obtained to find α of the transistor.
ac
Learning points
(i) Beyond a certain value of V , rapidly increases to a near saturation level
CE I C
due to avalanche breakdown. This may damage the transistor.
(ii) The small amount of I which flows even when =0 is the collector leakage
C
I E
current I .
CBO
(iii) The reciprocal odf the near horizontal part of the characteristics gives the
output resistance, R of the transistor which it woulf offer to input signal.
out
3
( ) Current Transfer Characteristics
This is the relationship showing variation of I with when is constant.
C
I E
VCB
(i) Decribe how you may obtain corresponding values of I and when
C
I E
VCB
is constant.
(ii) Typical transfer characteristics is calculated using the diagram given in Fig.
2.9.
mA
I C
ΔI C
ΔI E
I
mA
E
ΔI C
The slope =α =
ac
ΔI E
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Figure 2.9 Current Transfer Characteristics
(iii) If you have carried out this activity practically, determine the values of
ΔI
α =
C .
ac
ΔI E
Task 2.3 Further readings and Note making
Repeat activity 2.1.8 for
(a)Common emitter static characterics.
(b)Common Colector Static Characteristics.
Activity 2.1.9
Different ways of Drawing Transistor Circuits
The essential concepts to learn here is how different transistor circuits can be drawn.
The important learning point to remember is that in an NPN transistor, both collector
and base have to be positive with respect to the emitter.
Figures 2.10-2.12 show how power supply voltage can be represented with only one
terminal of the battery, and the other terminal is understood to be grounded so as to
provide a complete path for the current.
V = + V
EE
10
V = V
CC
25
(i) Common Base configuration
I
I
E
C
IE
R
PNP
E
RL
RE 20 K
10 K
RL
20 K
10 K
V
PNP
EB
VCE
VCC
B
C
+
10 V
25 V
VEB
B VBC
+
(a)
(b)
Figure 2.10 Common Base configuration
Fig. 2.10 (a) can be redrawn as shown in Fig. 2.10 (b) in which the negative termi-
nal of V and positive terminal of are supposed to be grounded.
CC
VEE
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(ii) Common Emitter Configuration
V
+ BB
V
+
o
CC
o
R
I
L
C
I
I
B
C
I
R
NPN
10K
B
B
1M
RB
10K
R
L
1M
V
V
NPN
CE
V
V
BE
CC
BB
10V
I E
20V
VBE
I V
E
CE
(a)
( )
b
Figure 2.11 Common Emitter Configuration
A more popular way of indicating power supply voltages in Fig. 2.11(a) is given
in Fig. 2.11 (b) . Since both collector and base are positive with respect to emitter,
a single battery can be used.
• Sketch a new circuit for 2.11 (b) in which there is only one battery.
(iii) Common Collector Configuration
R
IB
I
E
I
E
E
R
NPN
I
B
R
RE
B
B
NPN
+
-
VEE
V
V
CE
+
CB
V
V
BB
I
CE
C
VCB
IC
+ + + +
++
++
Figure 2.12
(a)
(b)
The power supply voltages in Fig. 2.12 (a) for CC configuration can be redrawn as
shown in 2.12 (b)
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Numerical Examples
• Calculations of voltages and currents in the circuits
Consider Fig. 2.10 (b) . Starting from the ground and applying Kirchoff’s law for the
left part of the circuit, we have
V − V
−V − I R + V
= 0 ⇔ I = EE
BE
(a)
BE
E
E
EE
E
RE
10 − 0.7
For Si, V = 0.7 V ∴ I =
= 0.465 mA
BE
E
20
In most cases V ?V .
EE
E
Thus
V
10
I ≅ EE =
= 0.5 mA
E
R
20
E
= α I ≅ I = 0.5mA
(b)
I C
E
E
• State the reasons for this approximation.
(c) Similarly the circuit on the right, and starting from the ground, we have
V = V − I R ≅ V − I R = 25 − 0.5 × 10 = 20 V
CB
CC
C
L
CC
E
L
Activity 2.2
Field Effect Transistors (FET )
You will learn that:
(i) Field effect transistors are also three-terminal devices, which is widely used
in linear and digital integrated circuits.
