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From the study of subatomic particles to the laws of motion, Physics Topics offer insights into the workings of the world around us.
How Many Kinds of Transistors are There? What is the Main Function of Transistor?
In 1947 AD, John Bardeen, William Shockley and Walter Brat-tain invented transistor.
Out of different forms of transistors, the most widely used form is the bipolar Junction transistor (BJT). It ¡s a semiconductor device in which the current flow between the two end terminals (called the collector and the emitter), is controlled by an amount of current flowing through an intermediate third terminal (called the base). Transistors are used in almost all modern technologies. Thus, importance of semiconductors is immense in modern age. Hence, modern age is also known as ‘silicon age’.
Like a diode, a transistor is also made from a crystal. It is very small in size and is kept sealed inside a metallic or plastic covering in such way that it cannot come in contact with air or moisture. Transistors are of two types:
- p-n-p transistor and
- n-p-n transistor.
Structure of a p-n-p transistor:
i) A thin n -type layer Is Introduced by doping in between two p -type regions at the two ends of a semiconducting crystal [Fig.(a)]. The n -type layer at the middle is very small in thickness in comparison with the two p -type regions at the two ends. This layer at the middle is known as base (B) of the transistor.
ii) The two p -type regions at the two ends of the transistor are called emluer (E) and collector (C). Although these two regions are apparently identical, the rates of doping in them are different. The emitter region is heavily doped compared to the collector region. hence, in a circuit if the connection of emitter and collector be interchanged, the working of a transistor gets disturbed. The rate of doping of the base of a transistor is much less than that of its emitter and collector.
iii) The majority charge carriers of this kind of transistor are holes. Usually, holes are emitted from the emitter and after crossing the thin layer of the base, they are collected by the collector. The thin layer of the base controls this flow of holes.
Structure of an n-p-n transistor: Structure of an n-p-n transistor is almost identical to that of a p-n-p transistor as discussed above [Fig.(b)]. For n-p-n transistor, the
nature of doping of the different parts of the semiconducting crystal is just the opposite to the p-n-p transistor. In this case,
- the thin base layer at the middle is of p -type.
- the emitter and collector regions at the two ends are of n -type.
- For this kind of doping, the majority charge carriers are electrons.
Speed of electrons is more than that of holes as charge carriers. Hence in high frequency circuits and computer circuits, n-p-n transistors are used. Actually, during the transmission of signals through these circuits, greater the speed of the effective charge carriers, greater will be the rate of work done.
Circuit symbol: The circuit symbols of p-n-p and n-p-n transistors are shown in Fig. The arrow sign indicates the direction of conventional current flow between the emitter and the base. So, electrons flow in the direction opposite to the arrow sign.
Transistor in on open circuit: Let us assume that a transistor is a combination of two diodes. So, diffusion of electrons and holes takes place through the junctions just like that of a p-n junction diode.
As a result, each p-n junction becomes reverse biased without the presence of any external source [Fig.]. Just as in a p-n junction diode, depletion regions are formed around the junctions in a p-n-p transistor.
Common-Emitter or CE Configuration of a Transistor
Three kinds of circuit can be constructed using transistors:
- common-base (CB),
- common-emitter (CE) and
- common-collector (CC). Among these, common-emitter or CE circuit is widely used as amplifier circuit.
Flow of charge carriers in a CE-circuit: We take an n-p-n transistor and consider the flow of conduction electrons [Fig.].
It is to be noted that, by convention, the direction of currents is opposite to that of the moving electrons.
In Fig. a CE circuit is shown using an n-p-n transistor. In this case,
1. the circuit connecting the base and the emitter (left side circuit in the figure) Is used as input circuit and
2. the circuit connecting the collector and the emitter (right side circuit in the figure) is used as output circuit, Hence, in both the circuits, the emitter is common. In an alternating current (ac) circuit this emitter is grounded. So it is also called grounded emitter circuit.
Basing of a CE circuit:
- Keeping the emitter grounded, the base is kept at forward bias in the input circuit, i.e., the p -type base of n-p-n transistor is connected with the positive pole of the source battery VBB [Fig.].
- Keeping the emitter grounded, the collector is kept at reverse bias in the output circuit, i.e., the n -type collector of n-p-n transistor is connected with positive pole of the source battery VCC [Fig.].
Current in a CE circuit: In the input circuit, the emitter is at the negative potential with respect to the base. Hence, a large number of majority carriers, i.e., electrons are emitted from the emitter which are then attracted by the positive base. Since, the base layer is very thin, most of these moving electrons enter to the collector after crossing the thin base layer. Then they are attracted by the positive potential of the collector. The small number of electrons which fail to cross the base are attracted by the positive potential of the base. In this way, the flow of electrons from the emitter produces two currents :
- base current of the input circuit and
- collector current of the output circuit.
The conventional direction of electric current is opposite to the direction of electron flow. According to that, in Fig., the emitter current IE base current IB and collector current IC are shown.
Clearly, IE = IB + IC
The value of IB is much less than IE or IC.
For example IB = 10µA, IC = 2mA = 2000µA,
Then, IE = 2000 + 10 = 2010 µA
Discussions:
i) Keeping IE at a fixed value, if IB is increased, then from the relation IC = IIE – IIB, we see that the value of decreases. Hence in CE circuit, the phase difference between output signal and input signal is 180°.
ii) Usually power expended in the output circuit of a transistor is much greater than that of the input circuit. Hence for commercial purposes the area of the base-collector junction of a transistor is made much greater than the emitter base junction.
iii) As the emitter-base is forward biased, the input resistance, i.e., the resistance of the emitter-base junction becomes very small. Again, as the collector-base junction is reverse biased, the output resistance, i.e., the resistance of the emitter-collector junction becomes very high. The circuit with low Input resistance and high output resistance acts as the best current amplifier.
CE characteristics: In Fig.,
source voltage of the input circuit = VBB;
source voltage of the output circuit = VCC;
base current, IB = input current;
base-emitter voltage, VBE = input voltage;
collector current, IC = output current;
collector-emitter voltage, VCE = output voltage.
Among the input and output currents and voltages, only the input current IB and output voltage VCE can be changed easily according to need, i.e., in CE circuit IB and VCE should be taken as independent variables and VBE as well as IC as the two dependent functions of them. Out of these, VBE has less
importance in the analysis of the circuit.
So, IC = f(IB, VCE) …… (1)
Using the mathematical relation (1), two characteristic curves of CE circuit can be drawn:
Transfer characteristics: Keeping VCE at a fixed value, the graph of IC drawn with respect to IB is known as transfer characteristics [Fig.]. In this case, IB and IC are the input and output quantities respectively.
Generally, IC changes linearly with IB.
The ratio \(\frac{\Delta I_C}{\Delta I_B}\) is called current transfer ratio or current amplification factor or current gain and it is expressed by the symbol β. Usually, the range of β is 100 to 500, approximately. Fig.
Output characteristics: Keeping IB at different fixed values, the graphs of IC drawn with respect to VCE are called out put characteristics [Fig.]. In this case, both VCE and IC are output quantities. For different values of IB, a series of different output characteristics is obtained. This series is divided into three clear regions [Fig.]:
i) Active region: In this region, the base-emitter junction is forward biased and the collector-emitter junction is reverse biased. As a result, IB > 0 and VCE > 0 ; but in actual practice, the value of VCE should be more than 0.2V (approximately) to keep the collector junction in the actual reverse bias. If a transistor is to be used as a good amplifier without much distortion, it has to be operated in the active region.
ii) Cut-off region: In this region, both the base-emitter and the collector-emitter junctions are reverse biased.
iii) Saturation region: In this region, both the base-emitter and the base-collector junctions are the forward biased. Remember that, if the value of VCE be less than 0.2 V (approximately), the collector is forward biased effectively.
Use of transistor as a switch: An ideal switch, when it is made ‘on; makes a circuit closed. On the other hand, when it is made ‘off’ the circuit becomes an open one. Moreover all these are done by an ideal switch momentarily.
No transistor can satisfy these conditions of an ideal switch properly. In spite of that, on the whole, use of transistor as a switch in different electronic circuits Is very wide.
If a transistor is employed in common-emitter mode, in the cut-off region [Fig.] the base-emitter junction is reverse-biased. Under this condition, base current IB is negative and the magnitude of the collector current IC is very small. This is called ‘off’ condition of the transistor. On the other hand, if the base-current attains a high positive value, the collector-emitter junction is forward-biased and the transistor is placed in the saturation region [Fig.].
In this case, collector-emitter voltage VCE is nearly zero [Fig.]. So, almost the whole external bias VCC acts as the terminal potential difference of the load resistance RL. So, the collector current reaches a sufficiently high value.
This condition is treated as the ‘on’ condition of the transistor. In a switch system, arrangement is to be made to turn the base current of the transistor from positive to a negative value or from negative to a positive value very rapidly. As a result, the transistor can turn from ‘on’ to ‘off’ or from ‘off’ to ‘on’ respectively. But when it is ‘on: the collector current IC takes sometime to reach a high value and when it is ‘off the charge collected at base takes some time to decay. Hence, transistor as a switch can never act with the rapidity of an ideal switch. Only its efficiency can be increased by using some specially designed transistors.
Accordingly, since a transistor can be in either ‘on’ mode or ‘off’ mode, it has considerable use in digital circuits. The application of transistor to make NOT logic gate has been discussed in the chapter ‘Digital Circuits’.
Use of transistor as a current amplifier: In the circuit shown in Fig., dc biases, VBB and VCC have been applied
at the base-emitter junction and collector-emitter junction of an n-p-n transistor respectively. RL is the load resistance. In the collector-emitter circuit.
VCC = VCE + ICRL
[VCE is dc voltage and IC is dc current]
In the cut-off region, IC ≈ 0 ; So, VCE ≈ VCC; Fig., the point A indicates this condition.
Again in the saturation region. VCE ≈ 0; So, VCC ≈ ICRL or, IC = \(\frac{V_{C C}}{R_L}\). The point B in the Fig., indicates this condition.
The line AB is called load line of the referred circuit.
If a transistor is used as a current amplifier in common-emitter mode, dc bias voltage VBB and VCC and load resistance RL are so selected that, action of the transistor is confined in the active
region. Under this condition, if the output characteristics for the constant base current IB intersects the load line at the point Q [Fig.], this point is called dc operating point or, Q -point of the circuit. In case of amplifier circuit, generally, weak ac signal is supplied across the base-emitter circuit as input. For example, if a sound is made in front of a microphone, a weak ac signal is obtained. To apply this ac signal to a dc circuit, a condenser C1 is used.
Direct current (dc) cannot pass through the condenser C1 . So, the input signal is free from the influence of the battery VBB. The input ac signal is added to the constant dc base-current IB. So, the base-current oscillates between IB1 and IB2 [Fig.]. The point Q oscilates between P and R. It is under-stood easily from the output characteristics that, the collector current oscillates between IC1 and IC2.
So, the ac signal obtained is the output signal. In Fig., the values of IB are genearlly expressed in microampere (µA) scale and the values of IC are expressed in milliampere (mA) scale. So, the amplitude of the output signal is greater than that of the input signal by 100 to 500 times. This is the current amplification by transistor. Only the ac part of the amplified output signal is taken out from the two ends of the load resistance RL with the help of the condensar C2. The two parts of the circuit containing C1 and C2 are called filter circuits. From the mixture of ac and dc, the condensers filter out in ac stopping the dc.
The output ac signal can be applied again as an input signai in the amplifier circuit of another transistor. Hence, the signal is again amplified. Thus, using a successive amplifier circuits, input ac signals of small amplitude can be amplified many times. But this type of magnification has a limit. If magnification be very large, the waveform of the output ac is distorted. Output waveform does not resemble the input waveform. In that case, the output signal becomes useless. For example, if a man speaks out in a low voice in front of a microphone, the outcoming speech from the loudspeaker becomes distorted and hard to understand.
Current amplification factor or current gain: In CE circuit, the current amplification factor or current gain is defined as the ratio of ac component of output collector current to ac component of input base current. It is denoted by the symbol β. In different types of transistors, the value of β is in the range 20 to 200 approximately.
If ib = input ac base current and ic = output ac collector current, then
β = \(\frac{i_c}{i_b}\)
Initially, a stable dc biasing is applied in each transistor circuit. Now an input ac signal is applied at the base of the transistor. If IB and IC are dc base current and dc collector current respectively, then at any instant, total base current = IB + ib and total collector current = IC + ic.
Actually, both ac currents ib and ic are considered as instantaneous changes in constant dc currents IB and IC respectively. Hence we can write, ib = ∆IB and ic = ∆IC. Therefore current amplification can also be written as
β = \(\left(\frac{\Delta I_C}{\Delta I_B}\right)_{V_{C E}}=\text { constant }\) ….. (2)
On the other hand, the ratio of change in collector current IC to the change in emitter current IE is known as current transfer ratio of the transistor and it is denoted by the symbol α.
α = \(\left(\frac{\Delta I_C}{\Delta I_E}\right)_{V_{C B}=\text { constant }}\) …… (3)
Here, IC < IE so, Δ IC < ΔIE i.e., α < 1. Since in general, the value of IB is very small, so the value of α is less than unity but still very nearly equal to 1, α ≈ 1 (In most of the transistors, α is in the range 0.95 to 0.995 approximately). In CB circuit this α parameter is of great importance, but almost irrelevant to a CE circuit.
Relation between α and β: Now, from equation (2) we have,
Voltage gain and Power gain: For CE circuit, if ΔVi is the change in input voltage and ΔV0 is the corresponding change in output voltage, then
Voltage gain = \(\frac{\Delta V_o}{\Delta V_i}\) = \(\frac{\Delta V_{C E}}{\Delta V_{B E}}\) = \(\frac{\Delta I_C R_L}{\Delta I_B R_B}\) = β\(\frac{R_L}{R_B}\)
Here, RL = load resistance
and RB = base resistance or input resistance
Power gain = \(\frac{\Delta V_o}{\Delta V_i}\) = \(\frac{\Delta V_{C E}}{\Delta V_{B E}}\) = β\(\frac{R_L}{R_B}\) ᐧ B = β2\(\frac{R_L}{R_B}\)
Power gain = current gain × voltage gain
Another parameter, called transfer conductance or transconductance gm is defined as gm = \(\frac{\Delta I_C}{\Delta V_{B E}}\)
Numerical Examples
Example 1.
In a common-emitter circuit, collector-emitter voltage is fixed at 5V. For base currents 30 µA and 40 µA, the collector currents are 8.2 mA and 9.4 mA respectively. Calculate current gain of the circuit.
Solution:
The change in base current,
ΔIB = (40 – 30)µA = 10 µA
The change in collector current,
ΔIC = (9.4 – 8.2) = 1.2 mA = 1.2 × 103 µA = 1200 µA
∴ β = \(\frac{\Delta I_C}{\Delta I_B}\) = \(\frac{1200 \mu \mathrm{A}}{10 \mu \mathrm{A}}\) = 120
Example 2.
The collector current in an n-p-n transistor is 10 mA. If 99.5% of the emitted electrons reach the collector, determine the emitter current, base current and amplification factor of the transistor.
Solution:
In an n-p-n transistor, we know collector current = IC = 1o mA, emitter current = IE, base current = IB and amplification factor, β = \(\frac{I_C}{I_B}\).
According to the problem, IC = 99.5% of IE
∴ IC = \(\frac{995}{1000}\)IE or, IE = \(\frac{1000}{995}\) × 10 = 10.05 mA
∴ IB = IE – IC = (10.05 – 10) mA = 0.05 mA
∴ β = \(\frac{I_C}{I_B}\) = \(\frac{10}{0.05}\) = 200
Example 3.
An n-p-n transistor is kept ¡n common-emitter configuration. Amplification factor of the transistor is 100. If the collector current is changed by 1 mA, what will be the corresponding change in the emitter current? (AIIMS ‘05)
Solution:
Amplification factor, β = \(\frac{\Delta I_C}{\Delta I_B}\)
∴ ΔIB = \(\frac{\Delta I_C}{\beta}\) = \(\frac{1}{100}\) = 0.01mA
So, change in emitter current,
ΔIE = ΔIC + ΔIB = 1 + 0.01 = 1.01 mA
Example 4.
The input resistance of a silicon transistor is 100 Ω. Base current is changed by 40 µA which results in a change in collector current by 2 mA. This transistor is used as a common emitter amplifier with a load resistance of 4 kΩ. What is the voltage gain of the amplifier?
Solution:
Voltage gain:
Load resistance, RL = 4 kΩ = 4000Ω
Input resistance, RB = 100Ω
∴ Current gain, = \(\frac{\Delta I_C}{\Delta I_B}\) = \(\frac{2 \mathrm{~mA}}{40 \mu \mathrm{A}}\) = \(\frac{2 \times 10^3}{40}\) = 5
∴ Voltage gain = 50 × \(\frac{4000}{100}\) = 2000