What is Transistor ?
When a third doped element is added to a crystal diode in such a way that two pn junctions are formed, the resulting device is known as a transistor. The transistor—an entirely new type of electronic device—is capable of achieving amplification of weak signals in a fashion comparable and often superior to that realised by vacuum tubes. Transistors are far smaller than vacuum tubes, have no filament and hence need no heating power and may be operated in any position. They are mechanically strong, have practically unlimited life and can do some jobs better than vacuum tubes.
Invented in 1948 by J. Bardeen and W.H. Brattain of Bell Telephone Laboratories, U.S.A.; the transistor has now become the heart of most electronic applications. Though transistor is only slightly more than 58 years old, yet it is fast replacing vacuum tubes in almost all applications. In this article, we shall focus our attention on the various aspects of transistors and their increasing applications in the fast-developing electronics industry.
A transistor consists of two pn junctions formed by *sandwiching either p-type or n-type semiconductor between a pair of opposite types. Accordingly ; there are two types of transistors, namely;
(i) n-p-n (ii) p-n-p
An n-p-n transistor is composed of two n-type semiconductors separated by a thin section of p-type as shown in Fig. 8.1 (i). However, a p-n-p transistor is formed by two p-sections separated by a thin section of n-type as shown in Fig. 8.1 (ii).
In each type of transistor, the following points may be noted :
(i) These are two pn junctions. Therefore, a transistor may be regarded as a combination of two
diodes connected back to back.
(ii) There are three terminals, one taken from each type of semiconductor.
(iii) The middle section is a very thin layer. This is the most important factor in the function of a
Origin of the name “Transistor”. When new devices are invented, scientists often try to devise a name that will appropriately describe the device. A transistor has two pn junctions. As discussed later, one junction is forward biased and the other is reverse biased. The forward biased junction has a low resistance path whereas a reverse biased junction has a high resistance path. The weak signal is introduced in the low resistance circuit and output is taken from the high resistance circuit. Therefore, a transistor transfers a signal from a low resistance to high resistance. The prefix ‘trans’ means the signal transfer property of the device while ‘istor’ classifies it as a solid element in the same general family with resistors.
A transistor (pnp or npn) has three sections of doped semiconductors. The section on one side is the emitter and the section on the opposite side is the collector. The middle section is called the base and forms two junctions between the emitter and collector.
(i) Emitter. The section on one side that supplies charge carriers (electrons or holes) is called the emitter. The emitter is always forward biased w.r.t. base so that it can supply a large number of *majority carriers. In Fig. 8.2 (i), the emitter (p-type) of pnp transistor is forward biased and supplies hole charges to its junction with the base. Similarly, in Fig. 8.2 (ii), the emitter (n-type) of npn transistor has a forward bias and supplies free electrons to its junction with the base.
(ii) Collector. The section on the other side that collects the charges is called the collector. The collector is always reverse biased. Its function is to remove charges from its junction with the base. In Fig. 8.2 (i), the collector (p-type) of pnp transistor has a reverse bias and receives hole charges that flow in the output circuit. Similarly, in Fig. 8.2 (ii), the collector (n-type) of npn transistor has reverse bias and receives electrons.
(iii) Base. The middle section which forms two pn-junctions between the emitter and collector is called the base. The base-emitter junction is forward biased, allowing low resistance for the emitter circuit. The base-collector junction is reverse biased and provides high resistance in the collector circuit.
Before discussing transistor action, it is important that the reader may keep in mind the following
facts about the transistor :
(i) The transistor has three regions, namely ; emitter, base and collector. The base is much thinner than the emitter while **collector is wider than both as shown in Fig. 8.3. However, for the sake of convenience, it is customary to show emitter and collector to be of equal size.
(ii) The emitter is heavily doped so that it can inject a large number of charge carriers (electrons or holes) into the base. The base is lightly doped and very thin ; it passes most of the emitter injected charge carriers to the collector. The collector is moderately doped.
(iii) The transistor has two pn junctions i.e. it is like two diodes. The junction between emitter and base may be called emitter-base diode or simply the emitter diode. The junction between the base and collector may be called collector-base diode or simply collector diode.
(iv) The emitter diode is always forward biased whereas collector diode is always reverse biased.
(v) The resistance of emitter diode (forward biased) is very small as compared to collector diode (reverse biased). Therefore, forward bias applied to the emitter diode is generally very small whereas reverse bias on the collector diode is much higher
The emitter-base junction of a transistor is forward biased whereas collector-base junction is reverse biased. If for a moment, we ignore the presence of emitter-base junction, then practically* no current would flow in the collector circuit because of the reverse bias. However, if the emitter-base junction is also present, then forward bias on it causes the emitter current to flow. It is seen that this emitter current almost entirely flows in the collector circuit. Therefore, the current in the collector circuit depends upon the emitter current. If the emitter current is zero, then collector current is nearly zero. However, if the emitter current is 1mA, then collector current is also about 1mA. This is precisely what happens in a transistor. We shall now discuss this transistor action for npn and pnp transistors.
Working of npn transistor
Fig. 8.4 shows the npn transistor with forward bias to emitter base junction and reverse bias to collector-base junction. The forward bias causes the electrons in the n-type emitter to flow towards the base. This constitutes the emitter current IE. As these electrons flow through the p-type base, they tend to combine with holes. As the base is lightly doped and very thin, therefore, only a few electrons (less than 5%) combine with holes to constitute base** current IB. The remainder (***more than 95%) cross over into the collector region to constitute collector current IC. In this way, almost the entire emitter current flows in the collector circuit. It is clear that emitter current is the sum of collector and base currents i.e
IE = IB + IC
Working of pnp transistor
Fig. 8.5 shows the basic connection of a pnp transistor. The forward bias causes the holes in the p-type emitter to flow towards the base. This constitutes the emitter current IE. As these holes cross into n-type base, they tend to combine with the electrons. As the base is lightly doped and very thin, therefore, only a few holes (less than 5%) combine with the electrons.
The remainder (more than 95%) cross into the collector region to constitute collector current IC. In this way, almost the entire emitter current flows in the collector circuit. It may be noted that current conduction within PNP transistor is by holes. However, in the external connecting wires, the current is still by electrons. Importance of transistor action. The input circuit (i.e. emitter-base junction) has low resistance because of forward bias whereas output circuit (i.e. collector-base junction) has high resistance due to reverse bias. As we have seen, the input emitter current almost entirely flows in the collector circuit. Therefore, a transistor transfers the input signal current from a low-resistance circuit to a high-resistance circuit. This is the key factor responsible for the amplifying capability of the transistor.
Note. There are two basic transistor types : the bipolar junction transistor (BJT) and field effect transistor (FET). As we shall see, these two transistor types differ in both their operating characteristics and their internal construction. Note that when we use the term transistor, it means bipolar junction transistor (BJT). The term comes from the fact that in a bipolar transistor, there are two types of charge carriers (viz. electrons and holes) that play part in conductions. Note that bi
means two and polar refers to polarities. The field-effect transistor is simply referred to as FET
In the earlier diagrams, the transistors have been shown in diagrammatic form. However, for the sake of convenience, the transistors are represented by schematic diagrams. The symbols used for npn and pnp transistors are shown in Fig. 8.6.
Note that emitter is shown by an arrow which indicates the direction of conventional current flow
with forward bias. For npn connection, it is clear that conventional current flows out of the emitter as indicated by the outgoing arrow in Fig. 8.6 (i). Similarly, for pnp connection, the conventional current flows into the emitter as indicated by inward arrow in Fig. 8.6 (ii).
Transistor Circuit as an Amplifier
A transistor raises the strength of a weak signal and thus acts as an amplifier. Fig. 8.7 shows the basic circuit of a transistor amplifier. The weak signal is applied between emitter-base junction and output is taken across the load RC
connected in the collector circuit. In order to achieve faithful amplification, the
input circuit should always remain forward biased. To do so, a d.c. voltage VEE is applied in the input circuit in addition to the signal as shown. This d.c. voltage is known as bias voltage and its magnitude is such that it always keeps the input circuit forward biased regardless of the polarity of the signal. As the input circuit has low resistance, therefore, a small change in signal voltage causes an appreciable change in emitter current. This causes almost the *same change in collector current due to transistor action. The collector current flowing through a high load resistance RC produces a large voltage across it. Thus, a weak signal applied in the input circuit appears in the amplified form in the collector circuit. It is in this way that a transistor acts as an amplifier.
Illustration. The action of a transistor as an amplifier can be made more illustrative if we consider typical circuit values. Suppose collector load resistance RC = 5 kΩ. Let us further assume that a change of 0.1V in signal voltage produces a change of 1 mA in emitter current. Obviously, the change in collector current would also be approximately 1 mA. This collector current flowing through collector load RC would produce a voltage = 5 kΩ × 1 mA = 5 V. Thus, a change of 0.1 V in the signal has caused a change of 5 V in the output circuit. In other words, the transistor has been able to raise the voltage level of the signal from 0.1 V to 5 V i.e. voltage amplification is 50.
Example 8.1. A common base transistor amplifier has an input resistance of 20 Ω and output resistance of 100 kΩ. The collector load is 1 kΩ. If a signal of 500 mV is applied between emitter and base, find the voltage amplification. Assume αac to be nearly one.
Solution. **Fig. 8.8 shows the conditions of the problem. Note that output resistance is very
high as compared to input resistance. This is not surprising because input junction (base to emitter)
of the transistor is forward biased while the output junction (base to collector) is reverse biased.
Comments. The reader may note that basic amplifying action is produced by transferring a current from a low-resistance to a high-resistance circuit. Consequently, the name transistor is given to the device by combining the two terms given in magenta letters below :
Transfer + Resistor ⎯→ Transistor
What are Transistor Connections
There are three leads in a transistor viz., emitter, base and collector terminals. However, when a
transistor is to be connected in a circuit, we require four terminals; two for the input and two for the output. This difficulty is overcome by making one terminal of the transistor common to both input and output terminals. The input is fed between this common terminal and one of the other two terminals. The output is obtained between the common terminal and the remaining terminal.
Accordingly; a transistor can be connected in a circuit in the following three ways :
(i) common base connection (ii) common emitter connection
(iii) common collector connection
Each circuit connection has specific advantages and disadvantages. It may be noted here that regardless of circuit connection, the emitter is always biased in the forward direction, while the collector always has a reverse bias.
Common Base Connection
In this circuit arrangement, input is applied between emitter and base and output is taken from collector and base. Here, base of the transistor is common to both input and output circuits and hence the name common base connection. In Fig. 8.9 (i), a common base npn transistor circuit is shown whereas Fig. 8.9 (ii) shows the common base pnp transistor circuit.
Current amplification factor (α).
It is the ratio of output current to input current. In a common base connection, the input current is the emitter current IE and output current is the collector current IC. The ratio of change in collector current to the change in emitter current at constant collector base voltage VCB is known as current amplification factor i.e.
It is clear that current amplification factor is less than **unity this value can be increased (but not more than unity) by decreasing the base current. This is achieved by making the base thin and doping it lightly. Practical values of α in commercial transistors range from 0.9 to 0.99.
Expression for collector current
The whole of emitter current does not reach the collector. It is because a small percentage of it, as a result of electron-hole combinations occurring in base area, gives rise to base current. Moreover, as the collector-base junction is reverse biased,
therefore, some leakage current flows due to minority carriers. It follows, therefore, that total collector current consists of :
(i) That part of emitter current which reaches the collector terminal i.e. ***α IE
(ii) The leakage current I leakage. This current is due to the movement of minority carriers across
base-collector junction on account of it being reverse biased. This is generally much smaller than
Relation (i) or (ii) can be used to find IC.
It is further clear from these relations that the collector current of a transistor can be controlled by either the emitter or base current. Fig. 8.11 shows the concept of ICBO. In CB configuration, a small collector current flows even when the emitter current is zero. This is the leakage collector current (i.e. the collector current when emitter is open) and is denoted by ICBO. When the emitter voltage VEE is also applied, the various currents are as shown in Fig. 8.11 (ii).
Note. Owing to improved construction techniques, the magnitude of ICBO for general-purpose and low-powered transistors (especially silicon transistors) is usually very small and may be neglected in calculations. However, for high power applications, it will appear in microampere range. Further, ICBO is very much temperature dependent; it increases rapidly with the increase in temperature. Therefore, at higher temperatures, ICBO plays an important role and must be taken care of in calculations.
Example 8.2. In a common base connection, IE = 1mA, IC = 0.95mA. Calculate the value of IB
Example 8.3. In a common base connection, current amplification factor is 0.9. If the emitter
current is 1mA, determine the value of base current.
Example 8.4. In a common base connection, IC = 0.95 mA and IB = 0.05 mA. Find the value of α.
Example 8.5. In a common base connection, the emitter current is 1mA. If the emitter circuit is
open, the collector current is 50 µA. Find the total collector current. Given that α = 0.92.
Example 8.6. In a common base connection, α = 0.95. The voltage drop across 2 kΩ resistance which is connected in the collector is 2V. Find the base current. Solution. Fig. 8.12 shows the required common base connection. The voltage drop across RC (=2 kΩ) is 2V.
Example 8.7. For the common base circuit shown in Fig. 8.13, determine IC and VCB. Assume the transistor to be of silicon.
Solution. Since the transistor is of silicon, VBE = 0.7V. Applying Kirchhoff’s voltage law to the emitter-side loop, we get,
Characteristics of Common Base Connection
The complete electrical behaviour of a transistor can be described by stating the interrelation of the
various currents and voltages. These relationships can be conveniently displayed graphically and the curves thus obtained are known as the characteristics of transistor. The most important characteristics of common base connection are input characteristics and output characteristics.
It is the curve between emitter current IE and emitter-base voltage VEB at constant collector-base voltage VCB. The
emitter current is generally taken along y-axis
and emitter-base voltage along x-axis. Fig. 8.14
shows the input characteristics of a typical transistor in CB arrangement . The following points
may be noted from these characteristics :
(i) The emitter current IE increases rapidly with small increase in emitter-base voltage VEB. It means that input resistance is very small.
(ii) The emitter current is almost independent of collector-base voltage VCB. This leads to the conclusion that emitter current (and hence collector current) is almost independent of collector voltage.
Input resistance of common base connection
It is the ratio of change in emitter-base voltage (ΔVEB) to the resulting
In fact, input resistance is the opposition offered to the signal current. As a very small VEB is sufficient to produce a large flow of emitter current IE , therefore, input resistance is quite small, of the order of a few ohms. 2. Output characteristic. It is the curve between collector current IC and collector-base voltage VCB at *constant emitter current IE . Generally, collector current is taken along y-axis and collector-base voltage along x-axis. Fig. 8.15 shows the output characteristics of a typical transistor in CB arrangement. The following points may be noted from the characteristics :
(i) The collector current IC varies with VCB only at very low voltages ( < 1V). The transistor is never operated in this region.
(ii) When the value of VCB is raised above 1 − 2 V, the collector current becomes constant as indicated by straight horizontal curves. It means that now IC is independent of VCB and depends upon IE only. This is consistent with the theory that the emitter current flows almost entirely to the collector terminal. The transistor is always operated in this region.
(iii) A very large change in collector-base voltage produces only a tiny change in collector current. This means that output resistance is very high.
It is the ratio of change in collector-base voltage (ΔVCB) to the resulting change in collector current (ΔIC ) at constant emitter current i.e.
The output resistance of CB circuit is very high, of the order of several tens of kilo-ohms. This is
not surprising because the collector current changes very slightly with the change in VCB.
What is Common Emitter?
In this circuit arrangement, input is applied between base and emitter and output is taken from the
collector and emitter. Here, emitter of the transistor is common to both input and output circuits and hence the name common emitter connection. Fig. 8.16 (i) shows common emitter npn transistor circuit whereas Fig. 8.16 (ii) shows common emitter pnp transistor circuit.
- Base current amplification factor ( β). In common emitter connection, input current is IB
and output current is IC
The ratio of change in collector current (ΔIC) to the change in base current (ΔIB) is known as base current amplification factor i.e.
In almost any transistor, less than 5% of emitter current flows as the base current. Therefore, the value of β is generally greater than 20. Usually, its value ranges from 20 to 500. This type of connection is frequently used as it gives appreciable current gain as well as voltage gain. Relation between β and α. A simple relation exists between β and α. This can be derived as follows :
It is clear that as α approaches unity, β approaches infinity. In other words, the current gain in
common emitter connection is very high. It is due to this reason that this circuit arrangement is used in about 90 to 95 percent of all transistor applications.
Expression for collector current
In common emitter circuit, IB is the input current and IC is the output current
Concept of ICEO. In CE configuration, a small collector current flows even when the base current is zero [See Fig. 8.17 (i)]. This is the collector cut off current (i.e. the collector current that flows when base is open) and is denoted by ICEO. The value of ICEO is much larger than ICBO.
Measurement of Leakage Current in Common Emitter
A very small leakage current flows in all transistor circuits. However, in most cases, it is quite small
and can be neglected.
(i) Circuit for ICEO test. Fig. 8.18 shows the circuit for measuring ICEO. Since base is open (IB = 0), the transistor is in cut off. Ideally, IC = 0 but actually there is a small current from collector to emitter due to minority carriers. It is called ICEO (collector-to-emitter current with base open). This current is usually in the nA range for silicon. A faulty transistor will often have excessive leakage current.
(ii) Circuit for ICBO test. Fig. 8.19 shows the circuit for measuring ICBO. Since the emitter is open (IE = 0), there is a small current from collector to base. This is called ICBO (collector-to-base current with emitter open). This current is due to the movement of minority carriers across base collector junction. The value of ICBO is also small. If in measurement, ICBO is excessive, then there is a possibility that collector-base is shorted.
Example 8.8. Find the value of β if (i) α = 0.9 (ii) α = 0.98 (iii) α = 0.99.
Example 8.9. Calculate IE in a transistor for which β = 50 and IB = 20 µA.
Example 8.10. Find the α rating of the transistor shown in Fig. 8.20. Hence determine the value of IC using both α and β rating of the transistor.
Solution. Fig. 8.20 shows the conditions of the problem
Example 8.11. For a transistor, β = 45 and voltage drop across 1kΩ which is connected in the collector circuit is 1 volt. Find the base current for common emitter connection.
Solution. Fig. 8.21 shows the required common emitter connection. The voltage drop across RC (= 1 kΩ) is 1volt.
Example 8.12. A transistor is connected in common emitter (CE) configuration in which collector supply is 8V and the voltage drop across resistance RC connected in the collector circuit is 0.5V. The value of RC = 800 Ω. If α = 0.96,
(i) collector-emitter voltage
(ii) base current
Solution. Fig. 8.22 shows the required common emitter connection with various values.
Example 8.13. An n-p-n transistor at room temperature has its emitter disconnected. A voltage of 5V is applied between collector and base. With collector positive, a current of 0.2 µA flows. When the base is disconnected and the same voltage is applied between collector and emitter, the current is found to be 20 µA. Find α, IE and IB when collector current is 1mA.
Solution. When the emitter circuit is open [See Fig. 8.23 (i)], the collector-base junction is
reverse biased. A small leakage current ICBO flows due to minority carriers.
Example 8.14. The collector leakage current in a transistor is 300 µA in CE arrangement. If now
the transistor is connected in CB arrangement, what will be the leakage current? Given that β = 120
Example 8.15. For a certain transistor, IB = 20 µA; IC = 2 mA and β = 80. Calculate ICBO..
Example 8.16. Using diagrams, explain the correctness of the relation ICEO = (β + 1) ICBO.
Solution. The leakage current ICBO is the current that flows through the base-collector junction
when emitter is open as shown is Fig. 8.24. When the transistor is in CE arrangement, the *base
current (i.e. ICBO) is multiplied by β in the collector as shown in Fig. 8.25.
Example 8.17 Determine VCB in the transistor * circuit shown in Fig. 8.26 (i). The transistor is
of silicon and has β = 150.
Example 8.18. In a transistor, IB = 68 µA, IE = 30 mA and β = 440. Determine the α rating of the transistor. Then determine the value of IC using both the α rating and β rating of the transistor.
Example 8.19. A transistor has the following ratings : IC (max) = 500 mA and βmax = 300.
Determine the maximum allowable value of IB for the device.
For this transistor, if the base current is allowed to exceed 1.67 mA, the collector current will
exceed its maximum rating of 500 mA and the transistor will probably be destroyed.
Example 8.20. Fig. 8.27 shows the open circuit failures in a transistor. What will be the circuit
behaviour in each case ?
Solution. *Fig 8.27 shows the open circuit failures in a transistor. We shall discuss the circuit behaviour in each case.
Fig. 8.27 (i) shows an open emitter failure in a transistor. Since the collector diode is not forward biased, it is OFF and there can be neither collector current nor base current. Therefore, there will be no voltage drops across either resistor and the voltage at the base and at the collector leads of the transistor will be 12V.
Fig. 8.27 (ii) shows an open base failure in a transistor. Since the base is open, there can be no base current so that the transistor is in cut-off. Therefore, all the transistor currents are 0A. In this case, the base and collector voltages will both be at 12V
Note. It may be noted that an open failure at either the base or emitter will produce similar
Fig. 8.27 (iii) shows an open collector failure in a transistor. In this case, the emitter diode is still ON, so we expect to see 0.7V at the base. However, we will see 12V at the collector because there is no collector current.
Example 8.21. Fig. 8.28 shows the short circuit failures in a transistor. What will be the circuit
behaviour in each case ?
Solution. Fig. 8.28 shows the short circuit failures in a transistor. We shall discuss the circuit
behaviour in each case.
Fig. 8.28 (i) shows a short between collector and emitter. The emitter diode is still forward biased, so we expect to see 0.7V at the base. Since the collector is shorted to the emitter, VC= VE = 0V.
Base -emitter short
Fig 8.28 (ii) shows a short between base and emitter. Since the base is now directly connected to ground, VB = 0. Therefore, the current through RB will be diverted to ground and there is no current to forward bias the emitter diode. As a result, the transistor will be cutoff and there is no collector current. So we will expect the collector voltage to be 12V.
Fig. 8.28 (iii) shows a short between the collector and the base. In this case, the emitter diode is still forward biased so VB = 0.7V. Now, however, because the collector is shorted to the base, VC = VB = 0.7V.
Note. The collector-emitter short is probably the most common type of fault in a transistor. It is
because the collector current (IC) and collector-emitter voltage (VCE) are responsible for the major
part of the power dissipation in the transistor. As we shall see (See Art. 8.23), the power dissipation in a transistor is mainly due to IC and VCE (i.e. PD = VCE IC). Therefore, the transistor chip between the collector and the emitter is most likely to melt first.
Characteristics of Common Emitter Connection
The important characteristics of this circuit arrangement are the input characteristics and output
It is the curve between base current IB and base-emitter voltage VBE at constant collector-emitter voltage VCE. The input characteristics of a CE connection can be determined by the circuit shown in Fig. 8.29. Keeping VCE constant (say at 10 V), note the base current IB for various values of VBE. Then plot the readings obtained on the graph, taking IB along yaxis and VBE along x-axis. This gives the input characteristic at VCE = 10V as shown in Fig. 8.30. Following a similar procedure, a family of input characteristics can be drawn.
The following points may be noted from the characteristics :
(i) The characteristic resembles that of a forward biased diode curve. This is expected since the base-emitter section of transistor is a diode and it is forward biased.
(ii) As compared to CB arrangement, IB increases less rapidly with VBE. Therefore, input resistance of a CE circuit is higher than that of CB circuit.
It is the ratio of change in base-emitter voltage (ΔVBE) to the change in base current (ΔIB) at constant VCE i.e.
The value of input resistance for a CE circuit is of the order of a few hundred ohms.
It is the curve between collector current IC and collector-emitter voltage VCE at constant base current IB. The output characteristics of a CE circuit can be drawn with the help of the circuit shown in Fig. 8.29. Keeping the base current IB fixed at some value say, 5 µA, note the collector current IC
for various values of VCE. Then plot the readings on a graph, taking IC along y-axis and VCE along x-axis. This gives the output characteristic at IB = 5 µA as shown in Fig. 8.31 (i). The test can be repeated for IB = 10 µA to obtain the new output characteristic as shown in Fig. 8.31 (ii). Following similar procedure, a family of output characteristics can be drawn as shown in Fig. 8.31 (iii).
The following points may be noted from the characteristics:
(i) The collector current IC varies with VCE for VCE between 0 and 1V only. After this, collector
current becomes almost constant and independent of VCE. This value of VCE upto which collector current IC changes with VCE is called the knee voltage (Vknee). The transistors are always operated in the region above knee voltage.
(ii) Above knee voltage, IC is almost constant. However, a small increase in IC with increasing VCE is caused by the collector depletion layer getting wider and capturing a few more majority carriers before electron-hole combinations occur in the base area.
(iii) For any value of VCE above knee voltage, collector current IC is approximately equal toβ × IB
.Output resistance. It is the ratio of change in collector-emitter voltage (ΔVCE) to the change in
collector current (ΔIC) at constant IB i.e.
It may be noted that whereas the output characteristics of CB circuit are horizontal, they have noticeable slope for the CE circuit. Therefore, the output resistance of a CE circuit is less than that of CB circuit. Its value is of the order of 50 kΩ.
Common Collector Connection
In this circuit arrangement, input is applied between base and collector while output is taken between the emitter and collector. Here, collector of the transistor is common to both input and output circuits and hence the name common collector connection. Fig. 8.32 (i) shows common collector npn transistor circuit whereas Fig. 8.32 (ii) shows common collector pnp circuit.
Current amplification factor γ
In common collector circuit, input current is the base current IB and output current is the emitter current IE. Therefore, current amplification in this circuit arrangement can be defined as under :
The ratio of change in emitter current (ΔIE) to the change in base current (ΔIB) is known as current amplification factor in common collector (CC) arrangement i.e.
This circuit provides about the same current gain as the common emitter circuit as ΔIE~ ΔIC
However, its voltage gain is always less than 1.
The common collector circuit has very high input resistance (about 750 kΩ) and very low output resistance (about 25 Ω). Due to this reason, the voltage gain provided by this circuit is always less than 1. Therefore, this circuit arrangement is seldom used for amplification. However, due to relatively high input resistance and low output resistance, this circuit is primarily used for impedance matching i.e. for driving a low impedance load from a high impedance source.
Comparison of Transistor Connections
The comparison of various characteristics of the three connections is given below in the tabular
The following points are worth noting about transistor arrangements :
Common Base Circuit
The input resistance (ri) of CB circuit is low because IE is high. The output resistance (ro ) is high because of reverse voltage at the collector. It has no current gain (α < 1) but voltage gain can be high. The CB circuit is seldom used. The only advantage of CB circuit is that it provides good stability against increase in temperature.
Common Emitter Circuit
The input resistance (ri ) of a CE circuit is high because of small IB . Therefore, ri for a CE circuit is much higher than that of CB circuit. The output resistance (ro ) of CE circuit is smaller than that of CB circuit. The current gain of CE circuit is large because IC is much larger than IB . The voltage gain of CE circuit is larger than that of CB circuit. The CE circuit is generally used because it has the best combination of voltage gain and current gain. The disadvantage of CE circuit is that the leakage current is amplified in the circuit, but bias stabilisation methods can be used.
Common Collector Circuit
The input resistance (ri ) and output resistance (ro ) of CC circuit are respectively high and low as compared to other circuits. There is no voltage gain (Av < 1) in a CC circuit. This circuit is often used for impedance matching.
Commonly Used Transistor Connection
Out of the three transistor connections, the common emitter circuit is the most efficient. It is used in about 90 to 95 percent of all transistor applications. The main reasons for the wide spread use of this circuit arrangement are :
(i) High current gain
In a common emitter connection, IC is the output current and IB is the input current. In this circuit arrangement, collector current is given by :
IC= β IB\ICEO
As the value of β is very large, therefore, the output current IC is much more than the input current IB . Hence, the current gain in CE arrangement is very high. It may range from 20 to 500.
(ii) High voltage and power gain.
Due to high current gain, the common emitter circuit has the highest voltage and power gain of three transistor connections. This is the major reason for using the transistor in this circuit arrangement.
(iii) Moderate output to input impedance ratio.
In a common emitter circuit, the ratio of output impedance to input impedance is small (about 50). This makes this circuit arrangement an ideal one for coupling between various transistor stages. However, in other connections, the ratio of output impedance to input impedance is very large and hence coupling becomes highly inefficient due to gross mismatching.
Transistor as an Amplifier in Common Emitter (CE) Arrangement
Fig. 8.33 shows the common emitter npn amplifier circuit. Note that a battery VBB is connected in the input circuit in addition to the signal voltage. This d.c. voltage is known as bias voltage and its
magnitude is such that it always keeps the emitter-base junction forward *biased regardless of the
polarity of the signal source.
During the positive half-cycle of the **signal, the forward bias across the emitter-base junction is increased. Therefore, more electrons flow from the emitter to the collector via the base. This causes an increase in collector current. The increased collector current produces a greater voltage drop across the collector load resistance RC. However, during the negative half-cycle of the signal, the forward bias across emitter-base junction is decreased. Therefore, collector current decreases. This results in the decreased output voltage (in the opposite direction). Hence, an amplified output is obtained across the load.
Analysis of collector currents
When no signal is applied, the input circuit is forward biased by the battery VBB. Therefore, a d.c. collector current IC flows in the collector circuit. This is called zero signal collector current. When the signal voltage is applied, the forward bias on the emitterbase junction increases or decreases depending upon whether the signal is positive or negative.
During the positive half-cycle of the signal, the forward bias on emitter-base junction is increased,
causing total collector current iC to increase. Reverse will happen for the negative half-cycle of the
signal. Fig. 8.34 shows the graph of total collector current iC versus time. From the graph, it is clear that total collector current consists of two components, namely ;
(i) The d.c. collector current IC (zero signal collector current) due to bias battery VBB. This is the current that flows in the collector in the absence of signal.
(ii) The a.c. collector current ic due to signal.
∴ Total collector current, iC= ic + IC
The useful output is the voltage drop across collector load RC due to the a.c. component Ic. The purpose of zero signal collector current is to ensure that the emitter-base junction is forward biased at all times. The table below gives the symbols usually employed for currents and voltages in transistor applications.
Transistor Load Line Analysis
In the transistor circuit analysis, it is generally required to determine the collector current for various collector-emitter voltages. One of the methods can be used to plot the output characteristics and determine the collector current at any desired collector-emitter voltage. However, a more convenient method, known as load line method can be used to solve such problems. As explained later in this section, this method is quite easy and is frequently used in the analysis of transistor applications. d.c. load line. Consider a common emitter NPN transistor circuit shown in Fig. 8.35 (i) where no signal is applied. Therefore, d.c. conditions prevail in the circuit. The output characteristics of this circuit are shown in Fig. 8.35 (ii). The value of collector-emitter voltage VCE at any time is given by ;
As VCC and RC are fixed values, therefore, it is a first degree equation and can be represented by
a straight line on the output characteristics. This is known as d.c. load line and determines the locus
of VCE − IC points for any given value of RC. To add load line, we need two end points of the straight line. These two points can be located as under :
(i) When the collector current IC = 0, then collector-emitter voltage is maximum and is equal to
This gives the second point A (OA= VCC /RC) on the collector current axis as shown in Fig. 8.35 (ii). By joining these two points, d.c. *load line AB is
Importance. The current (IC) and voltage (VCE) conditions in the transistor circuit are represented by some point on the output characteristics. The same information can be obtained from the load line. Thus when IC is maximum (= VCC /RC), then VCE = 0 as shown in Fig. 8.36. If IC = 0, then VCE is maximum and is equal to VCC. For any other value of collector current say OC, the collector-emitter voltage VCE = OD. It follows, therefore, that load line gives a far more convenient and direct solution to the problem.
Note. If we plot the load line on the output characteristic of the transistor, we can investigate the behaviour of the transistor amplifier. It is because we have the transistor output current and voltage specified in the form of load line equation and the transistor behaviour itself specified implicitly by the output characteristics.
Operating Point And Load Line
The zero signal values of IC and VCE are known as the operating point. It is called operating point because the variations of IC and VCE take place about this point when signal is applied. It is also called quiescent (silent) point or Q-point because it is the point on IC−VCE characteristic when the transistor is silent i.e.in the absence of the signal. Suppose in the absence of signal, the base current is 5µA. Then IC and VCE conditions in the circuit must be represented by some point on IB = 5 µA characteristic. But IC and VCE conditions in the circuit should also be represented by some point on the d.c. load line AB. The point Q where the
load line and the characteristic intersect is the only point which satisfies both these conditions. Therefore, the point Q describes the actual state of affairs in the circuit in the zero signal conditions and is called the operating point. Referring to Fig. 8.37, for IB = 5 µA, the zero signal values are :
VCE = OC volts
IC = OD mA
It follows, therefore, that the zero signal values of IC and VCE (i.e. operating point) are determined by the point where d.c. load line intersects the proper base current curve.
Example 8.22. For the circuit shown in Fig. 8.38 (i), draw the d.c. load line.
Solution. The collector-emitter voltage VCE is given by ;
This locates the point A of the load line on the collector current axis. By joining these two points,
we get the d.c. load line AB as shown in Fig. 8.38 (ii).
Example 8.23. In the circuit diagram shown in Fig. 8.39 (i), if VCC = 12V and RC = 6 kΩ, draw
the d.c. load line. What will be the Q point if zero signal base current is 20µA and β = 50 ?
Solution. The collector-emitter voltage VCE is given by :
Example 8.24. In a transistor circuit, collector load is 4 kΩ whereas quiescent current (zero
signal collector current) is 1mA.
(i) What is the operating point if VCC = 10 V ?
(ii) What will be the operating point if RC = 5 kΩ ?
Example 8.25. Determine the Q point of the transistor circuit shown in Fig. 8.40. Also draw the
d.c. load line. Given β = 200 and VBE = 0.7V
Solution. The presence of resistor RB in the base circuit should not disturb you because we can
apply Kirchhoff’s voltage law to find the value of IB and hence IC (= βIB). Referring to Fig. 8.40 and
applying Kirchhoff’s voltage law to base-emitter loop, we have,
Therefore, the Q-point is IC = 39.6 mA and VCE = 6.93V.
D.C. load line
In order to draw the d.c. load line, we need two end points.VCE = VCC – ICRC When IC = 0, VCE=VCC = 20V. This locates the point B of the load line on the collector-emitter voltage axis as shown in Fig. 8.41. When VCE = 0, IC= VCC/RC= 20V/330Ω = 60.6 mA. This locates the point A of the load line on the collector current axis. By joining these two points, d.c. load line AB
is constructed as shown in Fig. 8.41.
Example 8.26. Determine the Q point of the transistor circuit shown in *Fig. 8.42. Also draw
the d.c. load line. Given β = 100 and VBE = 0.7V
Solution. The transistor circuit shown in Fig. 8.42 may look complex but we can easily apply Kirchhoff’s voltage law to find the various voltages and currents in the * circuit. Applying Kirchhoff’s voltage law to the base-emitter loop, we have,
This locates the second point A (OA = 3.51 mA) of the load line on the collector current axis. By
joining points A and B, d.c. load line AB is constructed as shown in Fig. 8.43.
Example 8.27. In the above example, find (i) emitter voltage w.r.t. ground (ii) base voltage w.r.t. ground (iii) collector voltage w.r.t. ground.
Practical Way of Drawing Common Emitter (CE) Circuit
The common emitter circuits drawn so far can be shown in another convenient way. Fig. 8.45 shows the practical way of drawing CE circuit. In Fig. 8.45 (i), the practical way of drawing common emitter npn circuit is shown. Similarly, Fig. 8.45 (ii) shows the practical way of drawing common emitter pnp circuit. In our further discussion, we shall often use this scheme of presentation.
Output from Transistor Amplifier
A transistor raises the strength of a weak signal and thus acts as an amplifier. Fig. 8.46 shows the common emitter amplifier. There are two ways of taking output from this transistor connection. The output can be taken either across RC or across terminals 1 and 2. In either case, the magnitude of output is the same. This is clear from the following discussion :
(i) First method. We can take the output directly by putting a load resistance RC in the collector circuit i.e. Output = voltage across RC = icRC …(i)
This method of taking output from collector load is used only in single stage of amplification.
(ii) Second method. The output can also be taken across terminals 1 and 2 i.e. from collector and emitter end of supply.
Output = Voltage across terminals 1 and 2
= VCC − ic RC
As VCC is a direct voltage and cannot pass through capacitor CC, therefore, only varying voltage
ic RC will appear across terminals 1 and 2.
∴ Output = − ic RC
From exps. (i) and (ii), it is clear that magnitude of output is the same whether we take output across collector load or terminals 1 and 2. The minus sign in exp. (ii) simply indicates the phase reversal. The second method of taking output is used in multistages of amplification.
Performance of Transistor Amplifier
The performance of a transistor amplifier depends upon input resistance, output resistance, effective collector load, current gain, voltage gain and power gain. As common emitter connection is universally adopted, therefore, we shall explain these terms with reference to this mode of connection.
(i) Input resistance. It is the ratio of small change in base-emitter voltage (ΔVBE) to the resulting change in base current (ΔIB) at constant collector-emitter voltage i.e.
The value of input resistance is quite small because the input circuit is always forward biased. It
ranges from 500 Ω for small low powered transistors to as low as 5 Ω for high powered transistors. In fact, input resistance is the opposition offered by the base-emitter junction to the signal flow. Fig. 8.47 shows the general form of an amplifier. The input voltage VBE causes an input current IB
Thus if the input resistance of an amplifier is 500 Ω and the signal voltage at any instant is 1 V, then, Base current, ib =1v/500Ω= 2 mA
(ii) Output resistance. It is the ratio of change in collectoremitter voltage (ΔVCE) to the resulting change in collector current (ΔIC) at constant base current i.e.
The output characteristics reveal that collector current changes very slightly with the change in collector-emitter voltage. Therefore, output resistance of a transistor amplifier is very high– of the order of several hundred kilo-ohms. The physical explanation of high output resistance is that collector-base junction is reverse biased.
(iii) Effective collector load. It is the total load as seen by the a.c. collector current. In case of single stage amplifiers, the effective collector load is a parallel combination of RC and RO as shown in Fig. 8.48 (i).
It follows, therefore, that for a single stage amplifier, effective load is equal to collector load RC. However, in a multistage amplifier (i.e. having more than one amplification stage), the input resistance Ri of the next stage also comes into picture as shown in Fig. 8.48 (ii). Therefore, effective
collector load becomes parallel combination of RC, RO and Ri i.e.
As input resistance Ri is quite small (25 Ω to 500 Ω), therefore, effective load is reduced.
(iv) Current gain. It is the ratio of change in collector current (ΔIC) to the change in base current (ΔIB) i.e.
The value of β ranges from 20 to 500. The current gain indicates that input current becomes β
times in the collector circuit.
Voltage gain. It is the ratio of change in output voltage (ΔVCE) to the change in input
voltage (ΔVBE) i.e.
Example 8.28. A change of 200 mV in base-emitter voltage causes a change of 100 µA in the
base current. Find the input resistance of the transistor.
Solution. Change in base-emitter voltage is
Example 8.29. If the collector current changes from 2 mA to 3mA in a transistor when collector-emitter voltage is increased from 2V to 10V, what is the output resistance?
Example 8.30. For a single stage transistor amplifier, the collector load is RC = 2kΩ and the
input resistance Ri = 1kΩ. If the current gain is 50, calculate the voltage gain of the amplifier.
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