Semi Conductor

1 LESSON SEMICONDUCTOR ELECTRONICS Introduction The devices used before the discovery of Transistor, like vacuum tubes and valves, were bulky, consumed high power, had high operating voltages, limited life and low reliability. However, with the realization that some solid-state conductors and their junctions offer the possibility of controlling the number and the direction of flow of charge carriers through them led to the development of modern solid-state semiconductor electronics. Classification of Solids (based on the relative values of electrical conductivity (σ) or resistivity (ρ)) (a) Metals (Conductors): They possess very low resistivity (or high conductivity). ρ ∼ 10−2 -10−8 Ωm σ ∼ 102 -108 m −1 (b) Insulators: They have high resistivity (or low conductivity). ρ ∼ 1011 -1019 Ωm σ ∼ 10−11 -10−19 m −1 (c) Semiconductors: They have resistivity or conductivity intermediate to metals and insulators. ρ ∼ 10−5 -106 Ωm σ ∼ 105 -10−6 m −1 Classification of Semiconductors There are two schemes of classifying semiconductors. They are as follows: (I) Based on chemical composition (a) Elemental semiconductors: Si and Ge (b) Compound semiconductors: Examples are (i) Inorganic: CdS, GaAs, CdSe, InP etc. (ii) Organic: Anthracene, doped pthalocyanines etc. (iii) Organic polymers: polypyrrole, polyaniline, polythiophene etc. Note: Most of the currently available semicondutor devices are based on elemental semiconductors and inorganic semiconductors. (II) Based on source and nature (a) Intrinsic semiconductor: These are pure semiconducting materials (impurity less than 1 part in 1010). The electrical conduction is by means of mobile electrons and holes (positive charge carriers). The presence of these mobile charge carriers is the intrinsic property of the material. (b) Extrinsic semiconductor: These are obtained by adding (doping) the semiconductor material with small amounts of certain specific impurities with valency different from that of the atoms of the parent material. The electrical conductivity is essentially due to the foreign atoms and hence they are called extrinsic. The number of mobile charge carrier changes. Energy-Band Model Due to strong overlapping of the atomic orbitals, it is difficult to ascribe any one electron (or hole) belonging to any particular atom. As a result, electron energies in solids will be more like an energy

2 Semiconductor Electronics band instead of discrete energy levels of single isolated atoms. This alternative approach is termed as Energy-band description of solids. Consider that the Si or Ge crystal contains N atoms. Electrons of each atom will have discrete energies in different orbits. The electron energy will be same if all the atoms are isolated, i.e., separated from each other by a large distance. But, in a crystal the atoms are close to each other (2 to 3 A ) and therefore the electrons interact with each other and also with the neighbouring atomic cores. This overlap will be more felt by the electrons in the outermost orbit while the inner orbit or core electron energies may remain unaffected. Therefore, for understanding electron energies in Si or Ge crystal, we need to consider the changes in the energies of the electrons in the outermost orbit only. For Si, the outermost orbit is the third orbit (n = 3), while for Ge it is the fourth orbit (n = 4). The number of electrons in the outermost orbit is 4 (2s and 2p electrons). Hence, the total number of outer electrons in the crystal is 4N. The maximum possible number of outer electrons in the orbit is 8 (2s + 6p electrons). So, out of the 4N electrons, 2N electrons are in the 2N s-states (orbital quantum number l = 0) and 2N electrons are in the available 6N p-states. Obviously, some p-electrons states are empty. This is the case of well separated or isolated atoms. (i) There are exactly as many states in the lower band (4N) as there are available valence electrons from the atoms (4N). (ii) Therefore, this band (valence band) is completely filled while the upper band (conduction band) is completely empty. It is important to realize that at equilibrium spacing, the lowest conduction band energy is EC and highest valence band energy is EV. Above EC or below EV there are a large number of closely spaced energy levels as shown in figure. Maximum number of two electrons can be in each energy level due to Pauli's exclusion principle. The gap between the top of the valence band and bottom of the conduction band is called the energy band gap. It may be large, small, or zero, depending upon the material.

These different situations are depicted in figure and discussed below: Case (I): This refers to a situation (figure (a)) where the conduction and valence bands are overlapping.

3 This is the case of a conductors where Eg = 0. This situation makes a large number of electrons available for electrical conduction and, therefore, the resistance of such materials is low or the conductivity is high. Case (II): In this case (figure (b)), a large band gap Eg exists (Eg>3 eV). There are no electrons in the conduction band, and therefore no electrical conduction is possible. Note that the energy gap is so large that electrons cannot be easily excited from the valence band cannot be easily excited to the conduction band by any external stimuli (electrical, thermal or optical). This is the case of insulators. Case (III): Here a finite but small band gap (Eg>nh or n>>p. (b) p-type semiconductor This is obtained when Si or Ge is doped with trivalent impurities (Al, B, In etc.) as shown in figure. The dopant has one outer electron less than Si or Ge and, therefore, this atom can form bonds from three sides with Si (or Ge) and fails to form bond on one side. To hold the dopant atom tightly within the Si (or Ge) lattice, some of the outer bound electrons in the neighbourhood slide into this vacant bond leaving a vacancy or hole at its own site. This hole is available for conduction. The trivalent foreign atom becomes effectively negatively charged when all its valence bonds are filled and hence, the p-type material is designated as fixed core of one negative charge alongwith its associated hole (shown in figure). One ionized acceptor atom (NA) gives one hole. These holes are in addition to the thermally generated holes while the source of conduction electrons is only thermal generation. Thus, for such a material, the

holes are the majority carriers and electrons are minority carriers. ∴nh>>ne or p>>n In a nutshell • Adding donor (pentavalent impurity (ND) in Si or Ge, additional conduction electrons are generated, while the dopant atoms themselves become ionized +ve cores. • Similarly, additional holes are created by adding acceptors or trivalent impurity (NA) and they themselves become ionized −ve cores. • The crystal maintains an overall charge neutrality. • Apart from the process of generation of conduction electrons ne and holes nh, a simultaneous process of destruction occurs in which the electrons recombine with the holes. At equilibrium, the rate of generation of charge carriers is equal to the rate of destruction of charge carriers. Recombination process of electrons and holes The recombination process of electron and hole occurs due to an electron colliding with a hole. Larger the value of ne or nh, higher is the probability of their recombination with each other.

6 Semiconductor Electronics Electrons rise up and holes fall down when they gain external energy. At room temperature, most of the acceptor atoms get ionised leaving holes in the valence band. Thus at room temperature the density of holes in the valence band is predominantly due to impurity in the extrinsic semiconductor. The electron and hole concentration in a semiconductor in thermal equilibrium is given by n e n h = n i2 This equation is very important in semiconductor physics because of the following reasons: (i) For n-type semiconductor: ne is necessarily greater than ni and yet its product with nh continues to be n i2 . This is possible only if nh becomes less than ni. This means that the number of holes gets suppressed. (ii) For p-type semiconductor: Here ne would be less than ni since nh is more than ni. Thus, in ptype material the number of electrons is suppressed to a value lower than even ni. Energy band description of semiconductors Energy band description of an Intrinsic semiconductor The conduction band is completely empty in the absence of thermal energy (i.e., T = 0K). Hence, an intrinsic semiconductor will behave like an insulator at T = 0K. It is the thermal energy at high temperature (T>0K) which excites some electrons from the valence band (thus creating an equal number of holes in the valence band) to the conduction band. These thermally excited electrons at T>0K, partially occupy some states in the conduction band. Therefore, the energy-band diagram of an intrinsic semiconductor will be as shown in figure (a). Here, some electrons are shown in the conduction band, which have come from the valence band leaving equal number of holes there. Energy band description of an Extrinsic semiconductor In the case of extrinsic semiconductors, additional energy states, apart from EC and EV due to donor impurities (ED) and acceptor impurities (EA) also exist. Very small energy (~ 0.1 eV) is required for the electrons to be released from the donor impurity in the n-type semiconductor. Hence the donor energy level ED lies very near the bottom of the conduction band which can pump electrons into the conduction band. The conduction band will now have more electrons as they have come from thermal excitation as well as from the donor impurities as shown in figure (b). Similarly, for p-type semiconductor, an acceptor energy level EA is very near to the top of the valence band because an electron added to the acceptor impurity to complete its bonding within the semiconductor structure comes easily from the valence band electrons of some other semiconductor atoms in the lattice.

p-n junction A p-n junction is the basic building block of many semiconductor devices like diodes, transistor, etc. A clear understanding of the junction behaviour is important to analyse the working of other

7 semiconductor devices. We will now try to understand how a junction is formed and how the junction behaves under the influence of external applied voltage (also called bias). Formation of p-n junction Consider a thin p-type silicon (p-Si) semiconductor wafer. By adding precisely a small quantity of pentavelent impurity, part of the p-Si wafer can be converted into n-Si. There are several processes by which a semiconductor can be formed. The wafer now contains p-region and n-region and a metallurgical junction between p-, and n- region. Two important processes occur during the formation of a p-n junction: diffusion and drift. Diffusion and drift We know that in an n-type semiconductor, the concentration of electrons (number of electrons per unit volume) is more compared to the concentration of holes. Similarly, in a p-type semiconductor, the concentration of holes is more than the concentration of electrons. Diffusion current- During the formation of p-n junction, and due to the concentration gradient across p-, and n- sides, holes diffuse from p-side to n-side (p → n) and electrons diffuse from n-side to p-side (n → p). This motion of charge carries gives rise to diffusion current across the junction. Depletion region- When an electron diffuses from n → p, it leaves behind an ionized donor on nside. This ionised donor (positive charge) is immobile as it is bonded to the surrounding atoms. As the electrons continue to diffuse from n → p, a layer of positive charge (or positive space-charge region) on n-side of the junction is developed. Similarly, when a hole diffuses from p → n due to the concentration gradient, it leaves behind an ionised acceptor (negative charge) which is immobile. As the holes continue to diffuse, a layer of negative charge (or negative space-charge region) on the pside of the junction is developed. This space-charge region on either side of the junction together is known as depletion region as the electrons and holes taking part in the initial movement across the junction depleted the region of its free charges (see figure). The thickness of depletion region is of the order of one-tenth of a micrometre. Drift current- Due to the positive space-charge region on n-side of the junction and negative space charge region on p-side of the junction, an electric field directed from positive charge towards negative charge develops. Due to this field, an electron on p-side of the junction moves to n-side and a hole on n-side of the junction moves to p-side. The motion of charge carriers due to the electric field is called drift. Thus a drift current, which is opposite in direction to the diffusion current (see figure) starts. Barrier Potential Initially, diffusion current is large and drift current is small. As the diffusion process continues, the space-charge regions on either side of the junction extend, thus increasing the electric field strength and hence drift current. This process continues until the diffusion current equals the drift current. Thus a p-n junction is formed. In a p-n junction under equilibrium there is no net current. The loss of electrons from the n-region and the gain of electron by the p-region causes a difference of potential across the junction of the two regions. The polarity of this potential is such as to oppose further flow of carriers so that a condition of equilibrium exists.

8 Semiconductor Electronics The figure shows the p-n junction at equilibrium and the potential across the junction. The n-material has lost electrons, and p material has acquired electrons. The n material is thus positive relative to the p material. Since this potential tends to prevent the movement of electron from the n region into the p region, it is often called a barrier potential. Semiconductor diode A semiconductor diode [Fig. (a)] is basically a p-n junction with metallic contacts provided at the ends for the application of an external voltage. It is a two terminal device. A p-n junction diode is symbolically represented as shown in Fig. (b). The direction of arrow indicates the conventional direction of current (when the diode is under forward bias). The equilibrium barrier potential can be altered by applying an external voltage V across the diode. The situation of p-n junction diode under equilibrium (without bias) is shown in Fig. (a) and (b).

p-n junction diode under forward bias When an external voltage V is applied across a semiconductor diode, such that p-side is connected to the positive terminal of the battery and n-side to the negative terminal [Fig. (a)], it is said to be forward biased. The applied voltage mostly drops across the depletion region and the voltage drop across the p-side and n-side of the junction is negligible. (This is because the resistance of the depletion region – a region where there are no charges – is very high compared to the resistance of n-side and p-side.) The direction of the applied voltage (V) is opposite to the built-in potential V0. As a result, the depletion layer width decreases and the barrier height is reduced [Fig. (b)]. The effective barrier height under forward bias is (V0 – V ). Application of small voltage If the applied voltage is small, the barrier potential will be reduced only slightly below the equilibrium value, and only a small number of carriers in the material—those that happen to be in the uppermost energy levels—will possess enough energy to cross the junction. So the current will be small. Application of higher voltage If we increase the applied voltage significantly, the barrier height will be reduced and more number of carriers will have the required energy. Thus the current increases. Due to the applied voltage, electrons from n-side cross the depletion region and reach p-side (where they are minority carries). Similarly, holes from p-side cross the junction and reach the n-side (where they are minority carries). This process under forward bias is known as minority carrier

9 injection. At the junction boundary, on each side, the minority carrier concentration increases significantly compared to the locations far from the junction. Due to this concentration gradient, the injected electrons on p-side diffuse from the junction edge of p-side to the other end of p-side. Likewise, the injected holes on n-side diffuse from the junction edge of n-side to the other end of nside (see figure). This motion of charged carriers on either side gives rise to current. The total diode forward current is sum of hole diffusion current and conventional current due to electron diffusion. The magnitude of this current is usually in mA. p-n junction diode under reverse bias When an external voltage (V) is applied across the diode such that n-side is positive and p-side is negative, it is said to be reverse biased [Fig. (a)]. The applied voltage mostly drops across the depletion region. The direction of applied voltage is same as the direction of barrier potential. As a result, the barrier height increases and the depletion region widens due to the change in the electric field. The effective barrier height under reverse bias is (V0 + V), [Fig. (b)]. This suppresses the flow of electrons from n → p and holes from p → n. Thus, diffusion current, decreases enormously compared to the diode under forward bias. The electric field direction of the junction is such that if electrons on p-side or holes on n-side in their random motion come close to the junction, they will be swept to its majority zone. This drift of carriers gives rise to current. The drift current is of the order of a few μA. This is quite low because it is due to the motion of carriers from their minority side to their majority side across the junction. The drift current is also there under forward bias but it is negligible (μA) when compared with current due to injected carriers which is usually in mA. The diode reverse current is not very much dependent on the applied voltage. Even a small voltage is sufficient to sweep the minority carriers from one side of the junction to the other side of the junction. The current is not limited by the magnitude of the applied voltage but is limited due to the concentration of the minority carrier on either side of the junction. The current under reverse bias is essentially voltage independent upto a critical reverse bias voltage, known as breakdown voltage (Vbr ). When V = Vbr, the diode reverse current increases sharply. Even a slight increase in the bias voltage causes large change in the current. If the reverse current is not limited by an external circuit below the rated value (specified by the manufacturer) the p-n junction will get destroyed. Once it exceeds the rated value, the diode gets destroyed due to overheating. This can happen even for the diode under forward bias, if the forward current exceeds the rated value. V-I chracterstic of diode The circuit arrangement for studying the V-I characteristics of a diode, (i.e., the variation of current as a function of applied voltage) are shown in Fig. (a) and (b). The battery is connected to the diode through a potentiometer (or reheostat) so that the applied voltage to the diode can be changed. For different values of voltages, the value of the current is noted. A graph between V and I is obtained as in Fig. (c). Note that in forward bias measurement, we use a milliammeter since the expected current is large (as explained in the earlier section) while a micrometer is used in reverse bias to measure the current. You can see in Fig. (c) that in forward bias, the current first increases very slowly, almost negligibly, till the voltage across the diode crosses a certain value. After a certain voltage, the diode

10 Semiconductor Electronics current increases significantly (exponentially), even for a very small increase in the diode bias voltage. This voltage is called the threshold voltage or cut-in voltage (~0.2V for germanium diode and ~0.7 V for silicon diode). For the diode in reverse bias, the current is very small (~μA) and almost remains constant with change in bias. It is called reverse saturation current. However, for special cases, at very high reverse

bias (break down voltage), the current suddenly increases. This special action of the diode is discussed later in the lesson. The general purpose diode are not used beyond the reverse saturation current region. Diode as device The above discussion shows that the p-n junction diode primarily allows the flow of current only in one direction (forward bias). The forward bias resistance is low as compared to the reverse bias resistance. This property is used for rectification of ac voltages as discussed in the next section. For diodes, we define a quantity called dynamic resistance as the ratio of small change in voltage ΔV to ΔV a small change in current ΔI: rd = ΔI Application of p-n Diode as a Rectifier (I) Half wave Rectifier The half wave rectifier circuit is made of only one diode (figure (a)). Consider an ideal diode in which the reverse bias resistance is infinite. The secondary of the transformer supplies the desired ac voltage across A and B. When the voltage at A is positive, the diode is forward biased and it conducts. When A is negative, the diode is reverse biased and it does not conduct. Therefore, in the positive half-cycle of ac there is a current through the load resistor RL and output voltage is obtained across it (figure (b)). But there is negligibly small current in the negative half-cycle. The output voltage, though still varying, is restricted to only one direction and is said to be rectified. Since output voltage is obtained for only one-half cycle, such a circuit is known as half wave rectifier.

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(II) Full wave Rectifier The circuit using two diodes as shown in the Fig. (c), gives output rectified voltage corresponding to both the positive as well as negative half of the ac cycle. Hence, it is known as full-wave rectifier. Here the p-side of the two diodes are connected to the ends of the secondary of the transformer. The n-side of the diodes are connected together and the output is taken between this common point of diodes and the midpoint of the secondary of the transformer. So for a full-wave rectifier the secondary of the transformer is provided with a centre tapping and so it is called centre-tap transformer. As can be seen from Fig.(e) the voltage rectified by each diode is only half the total secondary voltage. Each diode rectifies only for half the cycle, but the two do so for alternate cycles. Thus, the output between their common terminals and the centretap of the transformer becomes a full-wave rectifier output. (Note that there is another circuit of full wave rectifier which does not need a centretap transformer but needs four diodes.) Suppose the input voltage to A with respect to the centre tap at any instant is positive. It is clear that, at that instant, voltage at B being out of phase will be negative as shown in Fig. (d). So, diode D1 gets forward biased and conducts (while D2 being reverse biased is not conducting). Hence, during this positive half cycle we get

12 Semiconductor Electronics an output current (and a output voltage across the load resistor RL ) as shown in Fig. (e). In the course of the ac cycle when the voltage at A becomes negative with respect to centre tap, the voltage at B would be positive. In this part of the cycle diode D1 would not conduct but diode D2 would, giving an output current and output voltage (across RL) during the negative half cycle of the input ac. Thus, we get output voltage during both the positive as well as the negative half of the cycle. Obviously, this is a more efficient circuit for getting rectified voltage or current than the halfwave rectifier. Full wave rectifier with capacitor filter Unlike the varying ac input voltage, the rectified voltage is still varying voltage but restricted to only one direction. It is known that all the varying signals can be considered as the sum of a dc signal superimposed with many ac signals of different harmonic frequencies. So, this would also be the case for the rectified voltages. From this rectified voltage, you can obtain dc voltage (or dc power supply) by filtering out the ac components. Figure (f) shows a rectifier circuit with a single capacitor C across it which filters out the ac component. In general, a high value of C provides a low impedance path to ac but high, almost infinite, impedance to dc. A

predominantly dc like voltage appears at the load and ac is bypassed through C or filtered(figure (g)). Special purpose p-n Junction Diodes (I) Zener Diodes It is a special purpose semiconductor diode, named after its inventor C. Zener. It is designed to operate under reverse bias in the breakdown region and used as a voltage regulator. The symbol for Zener diode is shown in Fig. (a). Zener diode is fabricated by heavily doping both p-, and n- sides of the junction. Due to this, depletion region formed is very thin ( Vz, then the Zener diode is in the breakdown condition and so for a wide range of values of load (RL) the current in the circuit or through the Zener diode may change but the voltage across it remains unaffected by load. Thus, the output voltage across the Zener diode is a regulated voltage. Note: Zener diodes with different breakdown voltages (for regulations of different voltages) can be obtained by changing the doping concentration on its p- and n- sides since they would change junction width (and junction field).

(II) Photonic p-n Junction Devices In the devices discussed so far the phenomena of the electron excitation as well as the resulting diode current were electrical in nature. But, there are some semiconductor electronic devices, in which the light photons also play a role in the overall performance of the device. Such devices are called photonic or opto-electronic devices which can be classified as: • Photo-detectors for detecting optical signals (e.g., photodiodes and photoconducting cell). • Photovoltaic devices for converting optical radiation into electricity (e.g., solar cells). • Devices for converting electrical energy into light (e.g., light emitting diodes and diode lasers) (a) Photodiode Photodetectors: The general principle of all photodetectors is the electron-excitation from the valence band to the conduction band by photons. Suppose an optical photon of frequency ν is incident on a semiconductor, such that its energy is greater than the band gap of the semiconductor (i.e., hν>Eg). This photon will excite an electron from the valence band to the conduction band leaving a vacancy (hole) in the valence band. Thus, an electron-hole pair is generated. These are additional photogenerated charge carriers which increase the conductivity of the semiconductor. Larger the number of incident photons (incident intensity of light), larger would be the change in the conductivity. By measuring the change in the conductance of the semiconductor one can measure the intensity of the optical signal. Such photodetectors are known as photoconductive cells. However, more commonly used photo detecting devices are photodiodes. A Photodiode is a special type of photo-detector where a reverse biased p-n junction diode is used as shown in figure.

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As already discussed, at p-n junction there exists a junction field which, at equilibrium, does not permit the flow of charge carriers across the junction. The current can only flow with applied bias. A reverse biased diode is used as photodiode. Suppose light of intensities I1, I2, I3 etc. is used to illuminate a diode. The electron and hole pairs generated in the depletion layer will be separated by the junction field and made to flow across the junction. The variation of reverse saturation current with intensities is shown in figure. Measurement of the change in the reverse saturation current on illumination can give the values of the light intensity. Note: Other devices used as photodetectors are p-i-n diodes (i stands for the insulating layer); metalsemiconductor diode; p-n avalanche photodiode etc. (b) Solar Cell Photovoltaic devices: They are based on similar principle as junction photodiode except that: (i) they are operated in the photovoltaic mode (generation of voltage due to the bombardment of optical photons), (ii) no external bias is applied, and (iii) the active junction area is kept large for obtaining more power. Solar cell is a device for converting solar energy into electricity. Different types of solar cells can be obtained by using different types of junctions like p-n or MetalSemiconductor (M-S) junction or Metal-Oxide Semiconductor (MOS) or Metal-insulatorSemiconductor (M-I-S) junction. A simple p-n junction solar cell is made by an n-type semiconductor substrate (backed with a current collecting metal electrode) over which a thin p-layer is grown (by diffusion of a suitable acceptor impurity or by vapour deposition). On top of the p-layer, metal finger electrodes are prepared so that there is enough space between the fingers for the light to reach p-layer (and the underlying p-n junction) to be able to produce photogenerated holes and electrons.

15 Consider the p-n junction as shown in figure. When light (with hν>Eg) falls at the junction, electron-hole pairs are generated which move in opposite directions due to the junction field. Photo-generated electrons and holes respectively move towards n-side and p-side. If no external load is connected, they would be collected at the two sides of the junction giving rise to a photo-voltage. When external load is connected a photo-current IL flows. The materials most commonly used for solar cells are silicon (Si) and gallium arsenide (GaAs). Other important materials are cadmium sulfide (CdS), cadmium telluride (CdTe), cadmium selenide (CdSe), copper indium selenide (CuInSe2) etc. Efficiency and cost are two important criteria for the selection of materials. (c) Light Emitting Diode (LED) These are forward-biased p-n junctions which emit spontaneous radiation. Radiation is emitted whenever an exited electron falls from higher excited energy state to a lower energy state. For excitation, thermal energy (as in incandescent lamps) or light energy (as in photo-luminescent panels in road signs) or electron bombardment (as in cathodeluminescent screens of TV or cathode ray oscillograph) or electric field (electroluminescence) can be used. The possible mechanisms of spontaneous emission are shown in figure (a). There are two distinct energy bands in a semiconductor, the conduction (higher energy) and the valence (lower energy) bands. There may also be energy bands due to donor impurities (ED) near the conduction band or acceptor impurities (EA) near the valence band. When electron falls from the higher to lower energy level containing holes, the energy in the form of light radiation is released. Generally, radiative transitions occur (sometimes non-radiative transition may also occur). The energy of radiation emitted by LED is equal to or less than the band gap of the semiconductor used. Schematically, the construction of a LED is shown in figure. Note: • The semiconductor used in LED is chosen according to the required wavelength of emitted radiation. Visible LED's (visible wavelength is from 0.45 μ m (2.8 eV) to 0.7 μ m (1.8 eV)) are available for red, green and orange.. Thus, the least band gap of the semiconductor for use in the visible region is 1.8 eV. • • Phosphorus doped GaAs (GaAs1−xPx) and GaP are the preferred materials. • Si (Eg ~ 1.1 eV) or Ge (Eg ~ 0.7 eV) are not suitable, since Eg is less than the minimum required Eg of ~ 1.8 eV. • GaAs (with Eg ~ 1.5 eV) alone or in conjunction with aluminium doped GaAs (AlxGa1-xAs) are commonly used for infrared LED's. Advantages of LED over conventional incandescent lamps: (i) Low operational voltage and less power. (ii) Fast action and no warm up time required. (iii) The bandwidth of emitted light is 100 A to 500 A or in other words, it is nearly (but not exactly) monochromatic. (iv) Long life and ruggedness.

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Transistors Construction and Action It consists of two p-n junctions back-to-back and is obtained by sandwiching either p-type or n-type semiconductor between a pair of opposite type of semiconductors as shown in figure. There are two types of transistors: • p-n-p transistor: Here two blocks of p-type semiconductor termed as emitter and collector are separated by a thin block of n-type semiconductor (termed as base). • n-p-n transistor: Here two blocks of n-type semicondcutor (emitter and collector) are separated by a thin block of p-type semiconductor (base). Note: All the three blocks of a transistor are not equal.

Further, for getting transistor action, the doping levels in the different blocks are kept different as under: Emitter: This is the left hand block of the transistor. It is of moderate size and heavily doped. This supplies a large number of majority carriers for the current flow through the transistor. Base: This is the central block. It is very thin and lightly doped. Collector: This collects a major portion of the majority carriers supplied by the emitter. The collector side is moderately doped and larger in size as compared to the emitter. Generally, the emitter-base junction of a transistor is forward biased while collector base junction is reverse biased. The depletion layer formed at the emitter-base junction is narrow as this junction is forward biased as well as due to the heavy doping of the emitter. Whereas collector-base junction being reverse biased the depletion layer at this junction is relatively wider. The forward bias voltage VEB is small (0.5 to 1 V) while the reverse bias voltage VCB is considerably high (5 to 15 V). p-n-p Transistor Consider the case of a biased p-n-p transistor as shown in figure (a). As the emitter base junction is forward biased, a large number of holes (majority carriers) from p-type emitter block flow towards the base. These constitute the current through the emitter, IE. These holes have a tendency to combine with the electrons in the n-region of the base. Only a few holes (less than 5%) are able to combine with the electrons in the base-region (giving only a very small base current, IB) because the base is lightly doped and very thin (this constructional feature is the key of transistor action). Most of the holes

17 coming from the emitter are able to diffuse through the base region to the collector region. In the collector region, these holes see the favourable negative potential at the collector and hence they easily reach the collector terminal to constitute the collector current, IC. Thus, the emitter current is the sum of collector current and base current, i.e. IE = IC + IB (IC >> IB). n-p-n Transistor A biased n-p-n transistor is shown in figure (b). Here, the electrons (instead of holes as in p-n-p transistor) are the majority carriers supplied by the n-type emitter region which cross the thin p-base region and are able to reach the collector to give the collector current, IC. Basic Transistor circuit configurations and Transistor characteristics In any electronic circuit there has to be two terminals for input

and two terminals for output. There are three terminals in a transistor, viz., Emitter (E), Base (B) and Collector (C). Therefore, in a transistor circuit the input/output connections have to be such that one of these (E, B

or C) is common to both the input and the output. Based on this there are three configurations: (a) Common Emitter (CE) (b) Common Base (CB) (c) Common Collector (CC) These configurations are shown in figure. Any variation in the voltages on the input and output sides results in a change in the input and output currents. The variation of current on the input side with input voltage (IE versus VBE) is known as input characteristics while the variation in the output current with output voltage (IC versus VCE) is known as output characteristics. Input/Output characteristics of a CE configuration

18 Semiconductor Electronics A simple circuit for drawing the input and output characteristics of an n-p-n transistor is shown in figure. To draw the input characteristics a curve is plotted between the base current (IB) and the base-emitter voltage VBE. First the collector voltage VCE is adjusted with the help of a rheostat, R1 so that measurable values of IB and IC are obtained. Then VCE is kept fixed and rheostat, R2 is varied to obtain different values of VBE. The values of IB are noted for different values of VBE (keeping VCE fixed). A graph IB versus VBE is plotted to get input characteristics. To draw the output characteristics, the value of IB is kept fixed (with the help of VBE). For fixed IB, the value of VCE is changed and the values of IC are noted. The plot IC versus VCE for different fixed values of IB gives the output characteristics as shown in figure. Note: A microammeter is needed to read IB since base currents are small while a milliammeter is used for reading the collector current. These characteristics are used to calculate the important transistor parameters as follows: (i) Input resistance (ri): This is defined as the ratio of change in base-emitter voltage ( Δ VBE) to the resulting change in base current ( Δ IB) at constant collector-emitter voltage (VCE). ⎛ ΔV ⎞ ri = ⎜ BE ⎟ (1) ⎝ ΔI B ⎠ V CE

The value of ri is of the order of a few hundred ohms. (ii) Output resistance (ro): This is defined as the ratio of change in collector-emitter voltage ( Δ VCE) to the change in collector current ( Δ IC) at constant base current IB. ⎛ ΔV ⎞ ro = ⎜ CE ⎟ (2) ⎝ ΔIC ⎠ I B

Since IC changes very little with VCE (after initial rise in IC), the values of ro are very high (of the order of 50 to 100 k Ω ). (iii) Current amplification factor ( β ) : This is defined as the ratio of the change in collector current (output current) to the change in base current. ⎛ ΔI ⎞ β=⎜ C ⎟ (3) ⎝ ΔI B ⎠V CE

This is also known as current gain. We normally work in the region in which IC is almost independent of VCE (or varies very slowly with VCE). This is called the active region. Transistor as a device The transistor can be used as a device application depending on the configuration used (namely CB, CC and CE), the biasing of the E-B and B-C junction and the operation region namely cutoff, active region and saturation. As mentioned earlier we have confined only to the CE configuration and will be concentrating on the biasing and the operation region to understand the working of a device. When the transistor is used in the cutoff or saturation state it acts as a switch. On the other hand for using the transistor as an amplifier, it has to operate in the active region. (i). Transistor as a switch

19

We shall try to understand the operation of the transistor as a switch by analysing the behaviour of the base-biased transistor in CE configuration as shown in Figure (a). Applying Kirchhoff’s voltage rule to the input and output sides of this circuit, we get VBB = IBRB + VBE (4) and VCE = VCC – ICRC (5) We shall treat VBB as the dc input voltage Vi and VCE as the dc output voltage VO. So, we have Vi = IBRB + VBE and Vo = VCC – ICRC Let us see how Vo changes as Vi increases from zero onwards. In the case of Si transistor, as long as input Vi is less than 0.6 V, the transistor will be in cut off state and current IC will be zero. Hence Vo = VCC When Vi becomes greater than 0.6 V the transistor is in active state with some current IC in the output path and the output Vo decrease as the term ICRC increases. With increase of Vi , IC increases almost linearly and so Vo decreases linearly till its value becomes less than about 1.0 V. Beyond this, the change becomes non linear and transistor goes into saturation state. With further increase in Vi the output voltage is found to decrease further towards zero though it may never become zero. If we plot the Vo vs Vi curve, [also called the transfer characteristics of the base-biased transistor [Fig. (b)], we see that between cut off state and active state and also between active state and saturation state there are regions of non-linearity showing that the transition from cutoff state to active state and from active state to saturation state are not sharply defined. Let us see now how the transistor is operated as a switch. As long as Vi is low and unable to forward-bias the transistor, Vo is high (at VCC). If Vi is high enough to drive the transistor into saturation, then Vo is low, very near to zero. When the transistor is not conducting it is said to be switched off and when it is driven into saturation it is said to be switched on. This shows that if we define low and high states as below and above certain voltage levels corresponding to cutoff and saturation of the transistor, then we can say that a low input switches the transistor off and a high input switches it on. Alternatively, we can say that a low input to the transistor gives a high output and a high input gives a low output. The switching circuits are designed in such a way that the transistor does not remain in active state. (ii). Transistor as an amplifier For using the transistor as an amplifier we will use the active region of the Vo versus Vi curve. The slope of the linear part of the curve represents the rate of change of the output with the input. It is negative because the output is VCC – ICRC and not ICRC. That is why as input voltage of the CE amplifier increases its output voltage decreases and the output is said to be out of

20 Semiconductor Electronics phase with the input. If we consider ΔVo and ΔVi as small changes in the output and input voltages then ΔVo/ΔVi is called the small signal voltage gain AV of the amplifier. If the VBB voltage has a fixed value corresponding to the mid point of the active region, the circuit will behave as a CE amplifier with voltage gain ΔVo/ ΔVi. We can express the voltage gain AV in terms of the resistors in the circuit and the current gain of the transistor as follows. We have, Vo = VCC – ICRC Therefore, ΔVo = 0 – RC ΔIC Similarly, from Vi = IBRB + VBE ΔVi = RB ΔIB + ΔVBE But ΔVBE is negligibly small in comparison to ΔIBRB in this circuit. So, the voltage gain of this CE amplifier (Figure) is given by AV = – RC Δ IC / RB ΔIB = –βac(RC /RB ) (6) where βac is equal to ΔIC/ΔIB from Eq. (3). Thus the linear portion of the active region of the transistor can be exploited for the use in amplifiers. Transistor as an amplifier (CE configuration) is discussed in detail in the next section. Transistor as an Amplifier (CE-Configuration) To operate the transistor as an amplifier it is necessary to fix its operating point somewhere in the middle of its active region. If we fix the value of VBB corresponding to a point in the middle of the linear part of the transfer curve then the dc base current IB would be constant and corresponding collector current IC will also be constant. The dc voltage VCE = VCC ICRC would also remain constant. The operating values of VCE and IB determine the operating point, of the amplifier. If a small sinusoidal voltage with amplitude vs is superposed on the dc base bias by connecting the source of that signal in series with the VBB supply, then the base current will have sinusoidal variations superimposed on the value of IB. As a consequence the collector current also will have sinusoidal variations superimposed on the value of IC, producing in turn corresponding change in the value of VO. We can measure the ac variations across the input and output terminals by blocking the dc voltages by large capacitors. In the discription of the amplifier given above we have not considered any ac signal.

In general, amplifiers are used to amplify alternating signals. Now let us superimpose an ac input signal vi (to be amplified) on the bias VBB (dc). The output is taken between the collector and the ground. The working of an amplifier can be easily understood, if we first assume that vi = 0. Then applying Kirchhoff’s law to the output loop, we get Vcc = VCE + IcRL (7) Likewise, the input loop gives VBB = VBE + IBRB (8) When vi is not zero, we get VBE + vi = VBE + IBRB + ΔIB(RB + ri)

21 The change in VBE can be related to the input resistance ri [see Eq. (1)] and the change in IB. Hence vi = ΔIB (RB + ri) = rΔIB The change in IB causes a change in Ic. We define a parameter βac, which is ΔI i βac = c = c (9) ΔI B i b which is also known as the ac current gain Ai. Usually βac is close to βdc in the linear region of the output characteristics. The change in Ic due to a change in IB causes a change in VCE and the voltage drop across the resistor RL because VCC is fixed. These changes can be given by Eq. (7) as ΔVCC = ΔVCE + RLΔIC = 0 or ΔVCE = –RLΔIC The change in VCE is the output voltage v0. From Eq. (3), we get v0 = ΔVCE = –βacRLΔIB The voltage gain of the amplifier is ΔVCE v Av = o = vi rΔI B β R = − ac L (10) r The negative sign represents that output voltage is opposite with phase with the input voltage. From the discussion of the transistor characteristics you have seen that there is a current gain βac in the CE configuration. Here we have also seen the voltage gain Av. Therefore the power gain Ap can be expressed as the product of the current gain and voltage gain. (11) Mathematically Ap = βac × Av Since βac and Av are greater than 1, we get ac power gain. However it should be realised that transistor is not a power generating device. The energy for the higher ac power at the output is supplied by the battery. Transistor as an Oscillator In an amplifier, when a sinusoidal signal is applied at the input, an amplified signal appears in the output. This implies that an external input is necessary to sustain ac signal in the output for an amplifier. However, in an oscillator, we get ac output without any external input signal i.e. the output is self-sustained. This is achieved by returning back (feedback) a portion of the output power to the input of an amplifier. The output power is in phase with the input power and so this process is termed as positive feedback (figure (a)). The feedback can be achieved by inductive coupling (through mutual inductance) or LC or RC networks. Construction and Working Consider the circuit shown in figure (b). Here the feedback is accomplished by inductive coupling from one coil winding (T1) to another coil winding (T2). Since the coils T2 and T1 are wound on the same core, they are inductively coupled through mutual inductance. As in an amplifier, the baseemitter junction is forward biased while the base-collector junction is reverse biased. Building up of oscillations • Suppose switch S1 is put on to apply proper bias for the first time, a surge of collector current flows in the transistor. This current flows through the coil T2 where terminals are numbered 3 and 4. This current does • not reach full amplitude instantaneously but increases from X to Y (figure (c)).

22 Semiconductor Electronics • The inductive coupling between coil T2 and coil T1 now causes a current to flow in the emitter circuit (this is the feedback from input to output). • •

• • •

• •

• •

•

•

•

As a result of this positive feedback, this current (in T1; emitter current) also increases from X' to Y'. The current in T2 (collector current) connected in the collector circuit acquires the value Y when the transistor becomes saturated. This means that maximum collector current is flowing and can increase no further. Since there is no further change in collector current, the magnetic field around T2 ceases to grow. As soon as the field becomes static, there will be no further feedback from T2 to T1. Without continued feedback, the emitter current begins to fall, consequently, collector current decreases from Y towards Z. However, a decrease of collector current causes the magnetic field to decay around the coil T2. Thus, T1 is now seeing a decaying field in T2 (opposite from what it saw when the field was growing at the initial start operation). This causes a further decrease in the emitter current till it reaches Z when the transistor is cut-off. This means that both IE and IC cease to flow and the transistor has reverted back to its original state (when the power was first switched on). The whole process now repeats itself, i.e. the transistor is driven to saturation, then to cut-off, and then back to saturation. The time for change from saturation to cut-off and back is determined by the constants of the tank circuit or tuned circuit (inductance L of coil T2 and C connected in parallel to it). The resonance frequency (f) of this tuned circuit determines the frequency at which the oscillator will oscillate. ⎛ 1 ⎞ f =⎜ ⎟ ⎝ 2π LC ⎠

Note: (i) Different types of oscillators essentially use different methods of coupling the output to the input (feedback network) apart from the resonant circuit for obtaining oscillation at a particular frequency. (ii) In the circuit of figure, the tank or tuned circuit is connected in the collector side. Hence, it is known as tuned collector oscillator. (iii) If the tuned circuit is on the base side, it will be known as tuned base oscillator. (iv) There are many other types of tank circuits (say RC) or feedback circuits giving different types

23 of oscillators like Colpitt's oscillator, Hartley oscillator, RC-oscillator etc. Digital Electronics and Logic Gates Consider the voltage waveforms shown in figure Waveforms of figure (a) are analog signals where continuous range of values of voltages are possible. The electronic circuits like amplifier, oscillator are grouped together as analog circuits. Figure (b) shows a pulse waveform in which only discrete values of voltages are possible. The high level is termed as '1' while the low level is called '0'. This is closely related to the binary system of digits (0 and 1). In Digital Electronics, two levels of a signal are used to represent the binary digits 0 and 1 (called bits). The basic building blocks of digital electronics are called Logic Gates, which process the 0 and 1 level signals in a specific manner. (i). NOT gate This is the most basic gate, with one input and one output. It produces a ‘1’ output if the input is ‘0’ and vice-versa. That is, it produces an inverted version of the input at its output. This is why it is also known as an inverter. The commonly used symbol together with the truth table for this gate is given in Fig. Truth Table of NOT gate A Y

0 1

1 0

(ii). OR Gate An OR gate has two or more inputs with one output. The logic symbol and truth table are shown in the figure. The output Y is 1 when either input A or input B or both are 1s, that is, if any of the input is high, the output is high. A 0 0 1 1

Truth Table of two input OR gate B Y 0 0 1 1 0 1 1 1

Apart from carrying out the above mathematical logic operation, this gate can be used for modifying the pulse waveform as explained in the following example. (iii). AND Gate An AND gate has two or more inputs and one output. The output Y of AND gate is 1 only when input A and input B are both 1. The

24 Semiconductor Electronics logic symbol and truth table for this gate are given in Fig. Truth Table of two input AND gate A B Y 0 0 0 0 1 0 1 0 0 1 1 1

(iv). NAND Gate This is an AND gate followed by a NOT gate. If inputs A and B are both ‘1’, the output Y is not ‘1’. The gate gets its name from this NOT AND behaviour. Figure shows the symbol and truth table of NAND gate. A 0 0 1 1

Truth Table for NAND gate B Y 0 1 1 1 0 1 1 0

NAND gates are also called Universal Gates since by using these gates you can realise other basic gates like OR, AND and NOT. (v). NOR Gate It has two or more inputs and one output. A NOT- operation applied after OR gate gives a NOT-OR gate (or simply NOR gate). Its output Y is ‘1’ only when both inputs A and B are ‘0’, i.e., neither one input nor the other is ‘1’. The symbol and truth table for NOR gate is given in Fig. NOR gates are considered as universal gates because you can obtain all the gates like AND, OR, NOT by using only NOR gates.

A 0 0 1 1

Truth Table for NOR gate B 0 1 0 1

Y 1 0 0 0

NAND gates are also called Universal Gates since by using these gates you can realise other basic gates like OR, AND and NOT. Integrated Circuits The conventional method of making circuits is to choose components like diodes, transistor, R, L, C etc., and connect them by soldering wires in the desired manner. Inspite of the miniaturisation introduced by the discovery of transistors, such circuits were still bulky. Apart from this, such circuits were less reliable and less shock proof. The concept of fabricating an entire circuit (consisting of many passive components like R and C and active devices like diode and transistor) on a small single block (or chip) of a semiconductor has revolutionised the electronics

25 technology. Such a circuit is known as Integrated Circuit (IC). The most widely used technology is the Monolithic Integrated Circuit. The word monolithic is a combination of two greek words, monos (means single) and lithos (means stone). This, in effect, means that the entire circuit is formed on a single silicon crystal (or chip). The chip dimensions are as small as 1mm× 1mm or it could even be smaller. Depending upon the level of integration (i.e., the number of circuit components or logic gates), the ICs are termed as Small Scale Integration, SSI (logic gates ≤ 10); Medium Scale Integration, MSI (logic gates ≤ 100); Large Scale Integration, LSI (logic gates ≤ 1000); and Very Large Scale Integration, VLSI (logic gates > 1000). The technology of fabrication is very involved but large scale industrial production has made them very inexpensive.

Semiconductor Electronics

26 SOLVED EXAMPLES

NCERT Solved Examples NCERT 1: C, Si and Ge have same lattice structure. Why is C insulator while Si and Ge intrinsic semiconductors? Solution: The 4 bonding electrons of C, Si or Ge lie, respectively, in the second, third and fourth orbit. Hence, energy required to take out an electron from these atoms (i.e. ionisation energy Eg) will be least for Ge, followed by Si and highest for C. Hence, number of free electrons for conduction in Ge and Si are significant but negligibly small for C. NCERT 2: Suppose a pure Si crystal has 5 × 1028 atoms m–3. It is doped by 1 ppm concentration of pentavalent As. Calculate the number of electrons and holes. Given that ni =1.5 × 1016 m–3. Solution: Note that thermally generated electrons (ni ~1016 m–3) are negligibly small as compared to those produced by doping. Therefore, ne ≈ ND. Since nenh = ni2, The number of holes nh = (2.25 × 1032)/(5 ×1022) ~ 4.5 × 109 m–3 NCERT 3: Can we take one slab of p-type semiconductor and physically join it to another n-type semiconductor to get p-n junction? Solution: No! Any slab, howsoever flat, will have roughness much larger than the inter-atomic crystal spacing (~2 to 3 Å) and hence continuous contact at the atomic level will not be possible. The junction will behave as a discontinuity for the flowing charge carriers. NCERT 4: The V-I characteristic of a silicon diode is shown in the Figure. Calculate the resistance of the diode at (a) ID = 15 mA and (b) VD = –10 V.

Solution: Considering the diode characteristics as a straight line between I = 10 mA to I = 20 mA passing through the origin, we can calculate the resistance using Ohm’s law. (a) From the curve, at I = 20 mA, V = 0.8 V, I = 10 mA, V = 0.7 V rfb = ΔV/ΔI = 0.1V/10 mA = 10 Ω (b) From the curve at V = –10 V, I = –1 μA, Therefore, rrb = 10 V/1μA = 1.0 × 107 Ω NCERT 5: In a Zener regulated power supply a Zener diode with VZ = 6.0 V is used for regulation. The load current is to be 4.0 mA and the unregulated input is 10.0 V. What should be the value of

27 series resistor RS? Solution: The value of RS should be such that the current through the Zener diode is much larger than the load current. This is to have good load regulation. Choose Zener current as five times the load current, i.e. IZ = 20 mA. The total current through RS is, therefore, 24 mA. The voltage drop across RS is 10.0 – 6.0 = 4.0 V. This gives RS = 4.0V/(24 × 10–3) A = 167 Ω. The nearest value of carbon resistor is 150 Ω. So, a series resistor of 150 Ω is appropriate. Note that slight variation in the value of the resistor does not matter, what is important is that the current IZ should be sufficiently larger than IL. NCERT 6: The current in the forward bias is known to be more (~mA) than the current in the reverse bias (~μA). What is the reason then to operate the photodiodes in reverse bias? Solution: Consider the case of an n-type semiconductor. Obviously, the majority carrier density (n) is considerably larger than the minority hole density p (i.e., n >> p). On illumination, let the excess electrons and holes generated be Δn and Δp, respectively: n′ = n + Δn p′ = p + Δp Here n′ and p′ are the electron and hole concentrations at any particular illumination and n and p are carriers concentration when there is no illumination. Remember Δn = Δp and n >> p. Hence, the fractional change in the majority carriers (i.e. Δn/n) would be much less than that in the minority carriers (i.e. Δp/p). In general, we can state that the fractional change due to the photo-effects on the minority carrier dominated reverse bias current is more easily measurable than the fractional change in the forward bias current. Hence, photodiodes are preferably used in the reverse bias condition for measuring light intensity. NCERT 7: Why are Si and GaAs are preferred materials for solar cells? Solution: The solar radiation spectrum received by us is shown in Figure.

The maxima is near 1.5 eV. For photo-excitation, hν > Eg. Hence, semiconductor with a band gap ~1.5 eV or lower is likely to give better solar conversion efficiency. Silicon has Eg ~ 1.1 eV while for GaAs it is ~1.53 eV. In fact, GaAs is better (in spite of its higher band gap) than Si because of its relatively higher absorption coefficient. If we choose materials like CdS or CdSe (Eg ~ 2.4 eV), we can use only the high energy component of the solar energy for photo-conversion and a significant part of energy will be of no use. The question arises: why we do not use material like PbS (Eg ~ 0.4 eV) which satisfy the condition hν > Eg for ν maxima corresponding to the solar radiation spectra? If we do so, most of the solar

28 Semiconductor Electronics radiation will be absorbed on the top-layer of solar cell and will not reach in or near the depletion region. For effective electron-hole separation, due to the junction field, we want the photogeneration to occur in the junction region only. NCERT 8: From the output characteristics shown in Figure, calculate the values of βac and βdc of the transistor when VCE is 10 V and IC = 4.0 mA.

⎛ ΔI ⎞ I βac = ⎜ C ⎟ , βdc = C IB ⎝ ΔI B ⎠ VCE For determining βac and βdc at the stated values of VCE and IC one can proceed as follows. Consider any two characteristics for two values of IB which lie above and below the given value of IC. Here IC = 4.0 mA. (Choose characteristics for IB= 30 and 20 μA.) At VCE = 10 V we read the two values of IC from the graph. Then ΔIB = (30 – 20) μA = 10 μA, ΔIC = (4.5 – 3.0) mA = 1.5 mA Therefore, βac = 1.5 mA/ 10 μA = 150 For determining βdc, either estimate the value of IB corresponding to IC = 4.0 mA at VCE = 10 V or calculate the two values of βdc for the two characteristics chosen and find their mean. Therefore, for IC = 4.5 mA and IB = 30 μA, βdc = 4.5 mA/ 30 μA = 150 and for IC = 3.0 mA and IB = 20 μA βdc =3.0 mA / 20 μA = 150 Hence, βdc =(150 + 150) /2 = 150 Solution:

NCERT 9: In Figure, the VBB supply can be varied from 0V to 5.0 V. The Si transistor has βdc = 250 and RB = 100 kΩ, RC = 1 KΩ, VCC = 5.0V. Assume that when the transistor is saturated, VCE = 0V and VBE = 0.8V. Calculate (a) the minimum base current, for which the transistor will reach saturation. Hence, (b) determine V1 when the transistor is ‘switched on’. (c) find the ranges of V1 for which the transistor is ‘switched off’ and ‘switched on’.

29 Solution: Given at saturation VCE = 0V, VBE = 0.8V VCE = VCC – ICRC IC = VCC/RC = 5.0V/1.0kΩ = 5.0 mA Therefore IB = IC/β = 5.0 mA/250 = 20μA The input voltage at which the transistor will go into saturation is given by VIH = VBB = IBRB + VBE = 20μA × 100 kΩ + 0.8V = 2.8V The value of input voltage below which the transistor remains cutoff is given by VIL = 0.6V, VIH = 2.8V Between 0.0V and 0.6V, the transistor will be in the ‘switched off’ state. Between 2.8V and 5.0V, it will be in ‘switched on’ state. Note that the transistor is in active state when IB varies from 0.0mA to 20mA. In this range, IC = βIB is valid. In the saturation range, IC ≤ βIB. NCERT 10: For a CE transistor amplifier, the audio signal voltage across the collector resistance of 2.0 kΩ is 2.0 V. Suppose the current amplification factor of the transistor is 100, What should be the value of RB in series with VBB supply of 2.0 V if the DC base current has to be 10 times the signal current. Also calculate the DC drop across the collector resistance.

Solution: The output AC voltage is 2.0 V. So, the ac collector current iC = 2.0/2000 = 1.0 mA. The signal current through the base is, therefore given by iB = iC /β = 1.0 mA/100 = 0.010 mA. The DC base current has to be 10 × 0.010 = 0.10 mA. RB = (VBB - VBE) /IB. Assuming VBE = 0.6 V, RB = (2.0 – 0.6)/0.10 = 14 kΩ. The DC collector current IC = 100 × 0.10 = 10 mA. NCERT 11: Justify the output waveform (Y) of the OR gate for the following inputs A and B given in Figure. Solution: Note the following: At t < t1; A = 0, B = 0; Hence Y = 0

30

Semiconductor Electronics For t1 to t2; A = 1,

For t2 to t3; For t3 to t4; For t4 to t5; For t5 to t6; For t > t6;

A = 1, A = 0, A = 0, A = 1, A = 0,

B = 0;

B = 1; B = 1; B = 0; B = 0; B = 1;

Hence Y = 1

Hence Y = 1 Hence Y = 1 Hence Y = 0 Hence Y = 1 Hence Y = 1

Therefore the waveform Y will be as shown in the Figure. NCERT 12: Take A and B input waveforms similar to that in Problem 11. Sketch the output waveform obtained from AND gate. Solution: A = 0, B = 0; Hence Y = 0 For t ≤ t1; For t1 to t2; A = 1, B = 0; Hence Y = 0 For t2 to t3; A = 1, B = 1; Hence Y = 1 For t3 to t4; A = 0, B = 1; Hence Y = 0 For t4 to t5; A = 0, B = 0; Hence Y = 0 For t5 to t6; A = 1, B = 0; Hence Y = 0 For t > t6; A = 0, B = 1; Hence Y = 0 Based on the above, the output waveform for AND gate can be drawn as given below.

NCERT 13: Sketch the output Y from a NAND gate having inputs A and B given below:

31

Solution: For t < t1; For t1 to t2; For t2 to t3; For t3 to t4; For t4 to t5; For t5 to t6; For t > t6;

A = 1, B = 1; A = 0, B = 0; A = 0, B = 1; A = 1, B = 0; A = 1, B = 1; A = 0, B = 0; A = 0, B = 1;

Hence Y = 0 Hence Y = 1 Hence Y = 1 Hence Y = 1 Hence Y = 0 Hence Y = 1 Hence Y = 1

Additional Solved Examples Example 1: Suppose a pure Si crystal has 5 × 1028 atoms m−3. It is doped by 1 ppm concentration of pentavalent Arsenic (As). Calculate the number of electrons and holes. given that n1 = 1.5 × 1016 m−3. Solution: Note that thermally generated electrons ( n1 ∼ 1016 m −3 ) are negligibly small as compared to

those produced by doping. Therefore, n e ≈ N D . Since nenh = n i2 , The number of holes 2.25 × 1032 5 × 1022 9 −3 ∼ 4.5 × 10 m

nh =

Example 2: A p-type semiconductor has acceptor levels 57 meV above the valence band. Find the maximum wavelength of light which can create a hole. Solution: To create a hole, an electron from the valence band should be given sufficient energy to go into one of the acceptor levels. Since the acceptor levels are 57 meV above the valence band, at least 57 meV is needed to create a hole. hc If λ be the wavelength of light, its photon will have an energy . To create a hole, λ hc ≥ 57 meV λ hc ∴λ ≥ 57 meV 1242 eV-nm = 57 ×10−3 eV = 2.18 × 10−5 m Example 3: The energy of a photon of sodium light ( λ = 589 nm) equals the band gap of a semiconducting material. Find the minimum energy E required to create a hole-electron pair. hc Solution: The energy of the photon is E = λ 1242eV-nm = 589nm = 2.1eV Thus the band gap is 2.1 eV. This is also the minimum energy E required to push an electron from the valence band into the conduction band. Hence, the minimum energy required to create a holeelectron pair is 2.1 eV.

Semiconductor Electronics

32

Example 4: The V-I characteristic of a silicon diode is given in figure. Calculate the diode resistance in: (a) forward bias at V = +2V and V = +1V, and (b) reverse bias V = −1V and –2V. Solution: The method of calculating resistance from Ohm's law by calculating V/I is not strictly valid for the forward bias case of a diode. The Ohm's law assumes the V-I plot to be a straight line passing through the origin. This is not true for forward bias as seen in figure. So, we calculate the forward bias diode resistance (rfb) in the following manner: ΔV ΔI where Δ V and Δ I are incremental changes in the voltage

(a) rfb =

and current near the values of interest. 0.4 rfb (at +2V) = 20 ×10−3 = 200 Ω 0.4 rfb (at +1V) = 10 ×10−3 = 400 Ω (b) In the reverse bias characteristics, the non-linearity in V-I curve is very small and, therefore, the slopes of V-I plot at –1V or –2V are almost similar. Hence, calculated rrb at –2V and –1V will be nearly (but not exactly) the same. The following gives the approximated value of rrb at –2V. 2 rrb = = 8 × 106 Ω . −6 0.25 ×10 Example 5: The V-I characteristic of a p-n junction diode is shown in figure. Find the approximate dynamic resistance of the p-n junction when (a) a forward bias of 1 volt is applied, (b) a forward bias of 2 volt is applied. Solution: (a) The current at 1 volt is 10 mA and at 1.2 volt it is 15 mA. The dynamic resistance in this region is ΔV 0.2volt R= = = 40Ω. Δi 5 mA (b) The current at 2 volt is 400 mA and at 2.1 volt it is 800 mA. The dynamic resistance in this region is ΔV 0.1volt R= = = 0.25 Ω. Δi 400 mA Example 6: A photodiode is prepared from a semiconductor having 2.8 eV band gap. Will it be able to respond to a 6620 nm wavelength radiation? (h = 6.62×10−34 Js) Solution: Eg = 2.8 eV = 2.8×1.6×10−19 = 4.48×10−19 J The wavelength of the radiation λ = 6620 nm = 6.620×10−6 m hc 6.62 × 10−34 × 3 × 108 = The energy of the radiation, E = hf = λ 6620 × 10−6 = 3 ×10−20 J Here, E (0.7 + 0.3 = 1V). Hence both the diodes will be forward biased. The equivalent circuit in this case can be drawn as per the figure (ii). Applying kirchoff’s law, E – VD1 – VD2 – IDR = 0 10 − 0.7 − 0.3 ∴ ID = 103 ∴ I D = 9 mA ∴ Output voltage Vo = IDR = 9 × 10−3 × 103 = 9 V Example 9: From the output characteristics shown in figure, calculate the value of current amplification factor of the transistor when VCE is 2 V.

Semiconductor Electronics

34

⎛ ΔIC ⎞ ⎟ ⎝ ΔI B ⎠ V

Solution: β = ⎜

CE

Consider any characteristics for any two values of IB (say, 10 and 60 μ A). Then for VCE = 2 V from the graph we have: ΔI B ≈ ( 60 − 10 ) μA = 50μA

ΔIC ≈ ( 9.5 − 2.5 ) mA = 7.0mA ⎛ 7 × 10−3 A ⎞

Therefore β = ⎜ ⎟ = 40. −6 ⎝ 50 × 10 A ⎠ Example 10: Calculate the input resistance of the transistor operating at VCE = 4 V in CE configuration having its input characteristics as shown in figure.

Solution: From the figure considering the dotted lines shown therein, we get ΔVBE ≈ (1.1 – 0.9) V ≈ 0.2 V ΔI B ≈ ( 68 − 34 ) μA = 34μA ⎛ ΔV ⎞ r1 = ⎜ BE ⎟ ⎝ ΔI B ⎠ VCE ≈ 6000 Ω

Example 11: In a n-p-n transistor about 1010 electrons enter the emitter when it is connected to a battery. About 2% electrons recombine with the holes in the base. Calculate the values of I E , I B , IC and βdc . (e = 1.6 × 10−19 C) Solution: As per the definition of the current, Q ne 1010 × 1.6 × 10−19 = = 100 μA Emitter current I E = = t t 10−6 2% of the total electrons entering the base from the emitter recombine with the holes which constitutes the base current IB. The rest of the 98 % electrons reach the collector and constitute the collector current. ∴ I B = 0.02 I E = 0.02 × 1600 = 32 μA ∴ IC = 0.98 I E = 0.98 × 1600 = 1568 μA

35

βdc =

IC 1568 × 10−6 = = 49 IB 32 × 10−6

Example 12: A transistor has a current amplification factor (current gain) of 50. In a CE-amplifier circuit, the collector resistance is chosen as 5 k Ω and the input resistance is 1 k Ω . Calculate the output voltage if input voltage is 0.01 V. Solution: ⎛v ⎞ R Av = ⎜ 0 ⎟ = β C RB ⎝ vi ⎠ R v ∴ v0 = β C i RB

β = 50, RC = 5 k Ω , RB = 1 k Ω , and vi = 0.01 V.

Therefore, 5 × 103 × 0.01 V 1 × 103 = 2.5V

v 0 = 50

Example 13: A transistor is used in common-emitter mode in an amplifier circuit. When a signal of 20 mV is added to the base-emitter voltage, the base current changes by 20 μ A and the collector current changes by 2 mA. The load resistance is 5 k Ω . Calculate (a) the factor β (b) the input resistance RBE, (c) the transconductance and (d) the voltage gain. ΔI 2mA = 100. Solution: (a) β = C = ΔI B 20μA ΔVBE (b) The input resistance R BE = ΔI B 20mV = 20μA = 1kΩ ΔIC (c) Trans conductance = ΔVBE 2mA = 20mV = 0.1mho (d) The change in output voltage is R L ΔIc = ( 5kΩ )( 2mA ) = 10V The applied signal voltage = 20 mV. 10V Thus, the voltage gain = 20mV = 500 Example 14: The collector supply voltage in a CE transistor amplifier is equal to 10V. The base current is equal to 10 μA in the absence of the signal voltage and the voltage between the collector and the emitter is equal to 4 V. The current gain (β) of the transistor is equal to 300. Calculate the value of the load resistance RL.

Semiconductor Electronics

36

Solution: Here, VCC = 10V, I B = 10μA = 10 × 10−6 A, VCE = 4 V, β = 300

Now, IC = β I B = (300)(10 × 10−6 ) = 3mA VCC = VCE + IC R L V − VCE ∴ R L = CC IC 10 − 4 = 3 × 10−3 = 2 kΩ Example 15: Two amplifier circuits are connected in series. The voltage gain of the first amplifier is equal to 15 and that of the second amplifier is equal to 10. If a 10 mV signal is applied at the input of the amplifier, then what will be the output signal? Solution: When two amplifiers are connected in series the output of the first is connected to the input of the second.

Vo1 = 15 Vi V The voltage gain of the second amplifier, A v2 = o = 10 Vo1 The combined voltage gain can be calculated as follows: V Av = o Vi V V = o × o1 Vo1 Vi = A V 2 × A V1

The voltage gain of the first amplifier, A vl =

∴ A v = 10 × 15 = 150 V Now, A v = o = 150 Vi ∴ Vo = 150 × Vi

= 150 × 10 × 10−3 ∴ Vo = 1.5 V Example 16: In a tuned collector oscillator circuit an output signal of 1 MHz frequency is obtained. The value of capacitance C = 100 pF. What should be the value of the capacitor if a signal of 2 MHz frequency is to be obtained. Solution: C1 = 100 pF = 100 × 10−12 F, f1 = 1 MHz = 106 Hz f2 = 2 MHz = 2 × 106 Hz

37

f1 = ∴

1 2π LC1

and f 2 =

1 2π LC2

LC2 f1 C2 = = f2 C1 LC1 2

2

⎛f ⎞ ⎛1⎞ ∴ C2 = ⎜ 1 ⎟ × C1 = ⎜ ⎟ × 100 × 10−12 ⎝2⎠ ⎝ f2 ⎠ C2 = 25 pF Example 17: Do 0 and 1 in digital electronics designate voltages equal to 0 V and 1 V, respectively? If not, then what are the commonly used voltage levels?

Solution: No! Remember that 0 and 1 levels are not the same as 0 and 1 V. In practice, the bit 0 or 1 is recognised by the presence or absence of the pulse (i.e., either at high or at low levels). Further, the digital high and low levels are specified in certain voltage levels as shown in figure. To avoid confusion, the voltage level of 1 and 0 are widely separated (high level is at 4 ± 1 V, low level is at 0.2 ± 0.2 V). For some devices in use, the high level may be higher than 4 V as well (like 12 V). Example 18: Figure shows the digital signals for the two input OR gate. Draw the shape of the output signal for the OR gate.

Solution: When t < t1; A = 0, B = 0; Y = 0 t1 < t < t 2 ; A = 0, B = 1; Y = 1 t 2 < t < t 3 ; A = 0, B = 1; Y = 1 t 3 < t < t 4 ; A = 0, B = 1; Y = 1 t 4 < t < t 5 ; A = 0, B = 1; Y = 0

38

Semiconductor Electronics t 5 < t < t 6 ; A = 1, B = 1; Y = 1 t 6 < t < t 7 ; A = 0, B = 0; Y = 0 t 7 < t < t 8 ; A = 0, B = 1; Y = 1

Example 19: A logic circuit is shown in the diagram. Draw the output signal at the point X and Y for the input signal shown in the figure at point A.

Solution: Here the output of the NOT gate will be equal to X = A. Hence the output obtained will be equal to opposite of A. This signal is given as input to another NOT gate. The signal again gets

inverted. As a result we again get back the signal A. ( Y = X = A = A )

Example 20: Write the truth table for the circuit given below.

Solution: Here Y’ is the output of the AND gate having A and B as the input. The input to the OR gate is A and Y’ (= A ⋅ B). Hence the output Y = A + Y’ = A + (A ⋅ B) The truth table of the circuit can be given as under:

A 0 0 1 1

B 0 1 0 1

Y' = A⋅B 0 0 0 1

Y = A + Y' 0 0 1 1

Example 21: Construct the truth table for the function X of A and B represented by figure.

Solution: Here an AND gate and an OR gate are used. Let the output of the OR gate be Y. Clearly,

39 Y = A + B. The AND gate receives A and A + B as input. The output of this gate is X. So X = A.(A + B). The following table evaluates X for all combinations of A and B. The last three columns give the truth table. A B Y=A+B X = A(A + B) A B X 0 0 0 0 0 0 0 0 1 1 0 0 1 0 1 0 1 1 1 0 1 1 1 1 1 1 1 1

Semiconductor Electronics

40 PROBLEMS

Exercise I Q 1. In an n-type silicon, which of the following statement is true: (a) Electrons are majority carriers and trivalent atoms are the dopants. (b) Electrons are minority carriers and pentavalent atoms are the dopants. (c) Holes are minority carriers and pentavalent atoms are the dopants. (d) Holes are majority carriers and trivalent atoms are the dopants. Q 2.

Which of the statements given in problem 1 is true for p-type semiconductos?

Q 3.

Carbon, silicon and germanium have four valence electrons each. These are characterised by valence and conduction bands separated by energy band gap respectively equal to (Eg)C, (Eg)Si and (Eg)Ge. Which of the following statements is true? (a) (Eg)Si < (Eg)Ge < (Eg)C (b) (Eg)C < (Eg)Ge > (Eg)Si (c) (Eg)C > (Eg)Si > (Eg)Ge (d) (Eg)C = (Eg)Si = (Eg)Ge

Q 4.

In an unbiased p-n junction, holes diffuse from the p-region to n-region because (a) free electrons in the n-region attract them. (b) they move across the junction by the potential difference. (c) hole concentration in p-region is more as compared to n-region. (d) All the above.

Q 5.

When a forward bias is applied to a p-n junction, it (a) raises the potential barrier. (b) reduces the majority carrier current to zero. (c) lowers the potential barrier. (d) None of the above.

Q 6.

For transistor action, which of the following statements are correct: (a) Base, emitter and collector regions should have similar size and doping concentrations. (b) The base region must be very thin and lightly doped. (c) The emitter junction is forward biased and collector junction is reverse biased. (d) Both the emitter junction as well as the collector junction are forward biased.

Q 7.

For a transistor amplifier, the voltage gain (a) remains constant for all frequencies. (b) is high at high and low frequencies and constant in the middle frequency range. (c) is low at high and low frequencies and constant at mid frequencies. (d) None of the above.

Q 8.

In half-wave rectification, what is the output frequency if the input frequency is 50 Hz. What is the output frequency of a full-wave rectifier for the same input frequency.

Q 9.

For a CE-transistor amplifier, the audio signal voltage across the collected resistance of 2 kΩ is 2 V. Suppose the current amplification factor of the transistor is 100, find the input signal voltage and base current, if the base resistance is 1 kΩ.

41 Q 10. Two amplifiers are connected one after the other in series (cascaded). The first amplifier has a voltage gain of 10 and the second has a voltage gain of 20. If the input signal is 0.01 volt, calculate the output ac signal. Q 11. A p-n photodiode is fabricated from a semiconductor with band gap of 2.8 eV. Can it detect a wavelength of 6000 nm? Q 12. The number of silicon atoms per m3 is 5 × 1028. This is doped simultaneously with 5 × 1022 atoms per m3 of Arsenic and 5 × 1020 per m3 atoms of Indium. Calculate the number of electrons and holes. Given that ni = 1.5 × 1016 m–3. Is the material n-type or p-type? Q 13. In an intrinsic semiconductor the energy gap Eg is 1.2eV. Its hole mobility is much smaller than electron mobility and independent of temperature. What is the ratio between conductivity at 600K and that at 300K? Assume that the temperature dependence of intrinsic carrier concentration ni is given by ⎛ Eg ⎞ n i = n 0 exp ⎜ − ⎟ ⎝ 2k B T ⎠ where n0 is a constant. Q 14. In a p-n junction diode, the current I can be expressed as where I0 is called the reverse saturation current, V is the voltage across the diode and is positive for forward bias and negative for reverse bias, and I is the current through the diode, kB is the Boltzmann constant (8.6 ×10–5 eV/K) and T is the absolute temperature. If for a given diode I0 = 5 × 10–12 A and T = 300 K, then (a) What will be the forward current at a forward voltage of 0.6 V? (b) What will be the increase in the current if the voltage across the diode is increased to 0.7 V? (c) What is the dynamic resistance? (d) What will be the current if reverse bias voltage changes from 1 V to 2 V? Q 15. You are given the two circuits as shown in Figure. Show that circuit (a) acts as OR gate while the circuit (b) acts as AND gate.

Q 16. Write the truth table for a NAND gate connected as given in Figure.

Hence identify the exact logic operation carried out by this circuit.

42 Semiconductor Electronics Q 17. You are given two circuits as shown in Figure which consist of NAND gates. Identify the logic operation carried out by the two circuits.

Q 18. Write the truth table for circuit given in Figure below consisting of NOR gates and identify the logic operation (OR, AND, NOT) which this circuit is performing.

(Hint: A = 0, B = 1 then A and B inputs of second NOR gate will be 0 and hence Y=1. Similarly work out the values of Y for other combinations of A and B. Compare with the truth table of OR, AND, NOT gates and find the correct one.) Q 19. Write the truth table for the circuits given in Figure consisting of NOR gates only. Identify the logic operations (OR, AND, NOT) performed by the two circuits.

Exercise II Q 1. Name the charge carriers in the following at room temperature : (i) Conductor (ii) Intrinsic semiconductor (iii) Insulator. Q 2.

What is doping?

Q 3.

Explain the effect of temperature variation on the resistivity of pure semiconductors.

Q 4.

Why is a semiconductor damaged by a strong current?

Q 5.

What type of impurity is added to obtain n-type semiconductor?

Q 6.

Distinguish between n-type and p-type semiconductors.

Q 7.

Explain, with neat diagram, formation of energy bands in solids.

Q 8.

Draw the energy band diagrams of the following substances: (a) conductor (b) n-type semiconductor (c) p-type semiconductor (d) insulator.

Q 9.

Show diagrammatically a forward biased and a reverse biased p-n junction.

Q 10. Distinguish between forward biasing and reverse biasing in p-n junction. Discuss its use.

43 Q 11. What is depletion region in a p-n-junction? Q 12. Which type of biasing results in very high resistance of a p-n-junction diode? Draw a diagram showing this bias. Q 13. Draw the circuit diagram of a half wave rectifier using a junction diode. Q 14. Draw a labeled circuit diagram showing use of p-n junctions as full wave rectifier. Q 15. Draw the V-I characteristics of a p-n junction diode. Show the Zener voltage on the characteristics curve. Q 16. State the function of a Zener diode in a circuit. Q 17. Represent symbolically Zener diode, photodiode, n-p-n and p-n-p transistors. Q 18. How is an n-p-n transistor represented symbolically? Q 19. What is a p-n-p transistor? How does it differ from a n-p-n transistor? Q 20. Why a transistor cannot be used as a rectifier? Q 21. Draw the circuit diagram of a transistor oscillator. Q 22. Explain the working and amplifying action of a transistor. Q 23. Define current amplification factor in a common emitter mode of transistor. Q 24. What kinds of biasing are required to the collector and base of a transistor in a common emitter amplifier? Q 25. What is a logic gate? Q 26. Draw the logic symbol of OR gate and write its truth table. Q 27. Draw the logic symbol of AND gate. Q 28. Draw the logic symbol of NOT gate. Q 29. Draw the logic symbol for a NAND gate. Q 30. Write down the truth table for a NOR gate. Q 31. The number of electron-hole pairs in an intrinsic semiconductor is 2×1019 m−3 at 27 °C and Eg is 1 eV. Calculate the number of electron-hole pairs at 227 °C. Given that Boltzmann constant is 8.65×10−5eV. (Refer Q 13. Exercise I)

44 Semiconductor Electronics Q 32. If the above semiconductor is doped by a donor impurity such that the number of conduction electrons become 2×1024 m−3, calculate the number of holes at 27 °C. Also approximately calculate the dopant concentration. Q 33. The number of silicon atoms per m3 is 5 x 1028. This is doped simultaneously with 5 x 1022 atoms per m3 of Arsenic and 5 x 1020 per m3 atoms of Indium. Calculate the number of electrons and holes. Given that n1 = 1.5 x 1016 m−3. Is the material n-type or p-type? o

Q 34. Electron hole pairs are formed when 6000 A wavelength light is incident on the semiconductor. What will be the band gap energy of the semiconductor? [h = 6.62 ×10−34 Js ] Q 35. If a LED has to emit 662 nm wavelength light then what should be the band-gap energy of its semiconductor? Q 36. For an n-p-n transistor about 7% of the electron entering the base from the emitter recombines with the hole. This results in the collector current being 18.6 mA. Calculate the emitter current and the current gain. Q 37. The base current changes by 20 mA when a 200 μA signal is applied at the input of a CE amplifier. If the output voltage is equal to 2 V, what is the voltage gain? Q 38. In the circuit shown in the figure, on applying +5V at the base resistor RB, both the VBE and VCE voltages become zero. Then calculate the values of IC, IB and β.

Q 39. Calculate emitter current for which β = 100 and base current Ib = 20 μ A. Q 40. For a common-emitter emplifier, current gain is 60. If the emitter current is 7.7 mA, calculate the base current and collector current. Flashback CBSE 2002 Q 1. How does the width of the depletion region of p-n junction vary, if the reverse bias applied to it decreases? (1 out of 70) Q 2. In the following diagrams, write which of the diodes are forward biased and which are reverse biased.

45

(2 out of 70) The output of a 2-input NAND gate is fed to a NOT gate. Write down the truth table for the output of the combination for all possible inputs of A and B. (2 out of 70) CBSE 2003 Q 1. Name the gate obtained from the combination of gates shown in the figure. Draw its logic symbol. Write the truth table of the combination. Q 3.

(2 out of 70) Q 2. How is a p-type semiconductor formed? Name the major charge carriers in it. Draw the energy band diagram of a p-type semiconductor. (2 out of 70) Q 3. Draw a labelled circuit diagram of a common base amplifier using a p-n-p transistor. Define the term 'voltage gain' and write an expression for it. (3 out of 70) CBSE 2004 Q 1. Draw the voltage-current characteristic of a Zener diode. (1 out of 70) Q 2. Give the logic symbol for an OR gate. Draw the output wave form for input wave forms A (2 out of 70) and B for this gate.

Q 3. With the help of a labeled circuit diagram, explain how an n-p-n transistor can be used as an amplifier in common emitter configuration. Explain how the input and output voltages are out of phase by 180º for a common-emitter transistor amplifier. Or For an n-p-n transistor in the common-emitter configuration, draw a labeled circuit diagram of an arrangement for measuring the collector current as a function of collector-emitter voltage for at least two different values of base current. Draw the shape of the curves obtained. Define the terms: (i) output resistance and (ii) current amplification factor. (5 out of 70)

46 Semiconductor Electronics CBSE 2005 Q 1. On the basis of the energy band diagrams distinguish between metals, insulators and semiconductors. (3 out of 70) Q 2. (a) With the help of a circuit diagram explain the working of transistor as oscillator. (b) Draw a circuit diagram for a two input OR gate and explain its working with the help of input, output waveforms. OR (a) Explain briefly with the help of a circuit diagram how V-I characteristics of a p-n junction diode are obtained in (i) forward bias, and (ii) reverse bias. (b) A photodiode is fabricated from a semiconductor with a band gap of 2.8 eV. Can it detect wavelength of 6000 nm? Justify. (5 out of 70) CBSE 2006 Q 1. Explain (i) forward biasing, (ii) reverse biasing of a P-N junction diode. With the help of a circuit diagram, explain the use of this device as a half-wave rectifier. (3 out of 70) Q 2. What are energy bands? How are these formed? Distinguish between a conductor, an insulator and a semiconductor on the basis of energy band diagram. OR Explain the function of base region of a transistor. Why is this region made thin and lightly doped? Draw a circuit diagram to study the input and output characteristics of n-p-n transistor in a common 'emitter (CE) configuration. Show these characteristics graphically. Explain how current amplification factor of the transistor is calculated using output characteristics. (5 out of 70) CBSE 2007 Q 1. Two semiconductor materials X and Y shown in the given figure, are made by doping germanium crystal with indium and arsenic respectively. The two are joined end to end and connected to a battery as shown.

(i) Will the junction be forward biased or reverse biased? (ii) Sketch a V – I graph for this arrangement. Q 2.

Q 3.

(2 out of 70) Draw the circuit diagram of a common emitter amplifier using n-p-n transistor. What is the phase difference between the input signal and output voltage? State two reasons why a common emitter amplifier is preferred to a common base amplifier. (3 out of 70)

Explain the formation of energy band in solids. Draw energy band diagram for (i) a conductor (ii) an intrinsic semiconductor. (3 out of 70)

47 CBSE 2008 Q 1. State the reason, why GaAs is most commonly used in making of a solar cell. Q 2.

(1 out of 70) The input A and B are inverted by using two NOT gates and their outputs are fed to the NOR gate as shown below.

Analyse the action of the gates (1) and (2) and identify the logic gate of the complete circuit so obtained. Give its symbol and the truth table. (3 out of 70) Q 3. Draw the labelled circuit diagram of a common-emitter transistor amplifier. Explain clearly how the input and output signals are in opposite phase. (3 out of 70) Or State briefly the underlying principle of a transistor oscillator. Draw a circuit diagram showing how the feedback is accomplished by inductive coupling. Explain the oscillator action. (3 out of 70) CBSE 2009 Q.1 Give the logic symbol of NOR gate (1 out of 70) Q.2 With the help of a suitable diagram, explain the formation of depletion region in a p-n junction. How does its width change when the junction is (i) forward biased, and (ii) reverse biased ? (3 out of 70) Q.3 Give a circuit diagram of a common emitter emplifier using an n-p-n transistor. Draw the input and output waveforms of the signal. Write the expression for its voltage gain. (3 out of 70) CBSE 2010 Q.1 (a) Draw the circuit diagrams of a p-n junction diode in (i) forward bias, (ii) reverse bias. How are these circuits used to study the V–I characteristics of a silicon diode ? Draw the typical V – I characteristics. (b) What is a light emitting diode (LED) ? Mention two important advantages of LEDs over conventional lamps. (5 out of 70) OR (a) Draw the circuit arrangement for studying the input and output characteristics of an n=p-n transistor in CE configuration. With the help of these characteristics define (i) input resistance, (ii) current amplification factor. (b) Describe briefly with the help of a circuit diagram how an n-p-n- transistor is used to produce self-sustained oscillations. (5 out of 70)

48 Semiconductor Electronics Q.2 (i) Identify the logic gates marked P and Q in the given logic circuit.

(ii) Write down the output at X for the inputs A = 0, B = 0 and A = 1, B = 1. (3 out of 70) CBSE 2011 Q.1 What happens to the width of depletion layer of a p-n junction when its is (i) forward biased, (ii) reverse biased? (1 out of 70) Q.2 You are given a circuit below. Write its truth table. Hence, identify the logic operation carried out by this circuit. Draw the logic symbol of the gate it corresponds to.

Q.3

(3 out of 70) Draw a labelled diagram of a full wave rectifier circuit. State its working principle. Show the input-output waveforms. (3 out of 70)

49 ANSWERS Exercise I Q 1. (c) Q 2. (d) Q 3. (c) Q 4. (c) Q 5. (c) Q 6. (b), (c) Q 7. (c) Q 8. 50 Hz for half-wave, 100 Hz for full-wave Q 9. ν1 = 0.01 V; IB = 10 μA Q 10. 2V Q 11. No (hv has to be greater than Eg Q 12. n e ≈ 4.95 × 1022 ; nh = 4.75 × 109; n-type since ne >> nh For charge neutrality ND – NA = ne – nh; ne . nh = ni2 1 Solving these equations, n e = ⎡(N D − N A ) + (N D − N A ) 2 + 4n i2 ⎤ ⎦ 2⎣ 5 Q 13. About 1 × 10 Q 14. (a) 0.0629 A, (b) 2.97 A, (c) 0.336 Ω, (d) For both the voltages, the current I will be almost equal to I0, showing almost infinite dynamic resistance in the reverse bias. Q 16. NOT; A Y 0 1 1 0 Q 17. (a) AND (b) OR Q 18. OR gate Q 19. (a) NOT, (b) AND Exercise II Q 1. (i) Electrons (ii) Electrons and Holes (iii) None Q 5. Pentavalent atoms. Q 24. A collector is reverse biased, while a base is forward biased. Q 31. 4.45 × 1022 m−3 Q 32. Nh = 2 × 104 m−3; ND ≈ ne 22 9 Q 33. ne ≈ 4.95 × 10 ; nh = 4.75 × 10 ; n-type since ne >> nh Q 34. 2.07 eV Q 35. 1.875 eV Q 36. 20 mA, 0.93 Q 37. 10 Q 38. IC = 5 mA, IB = 50 μA, β = 100 Q 39. 2.02mA Q 40. 0.126mA, 7.574mA Flashback CBSE 2002 Q 1. Decreases. Q 2. Diodes (a) and (c) are forward biased while (b) and (d) are reverse biased.

Q 3. A 0 0 1 1

B 0 1 0 1

Output NAND gate, Y 1 1 1 0

Final output, (Y)’ = Y’ 0 0 0 1

Semiconductor Electronics CBSE 2003 Q 1. NOR gate. Truth Table of NOR gate Output Inputs Y A B 0 0 1 0 1 0 1 0 0 1 1 0

50

CBSE 2004 Q 2.

CBSE 2005 Ans. 2(b) No, because hν < Eg CBSE 2008 Q 2.

Truth Table Output Inputs Y A B 0 0 0 0 1 0 1 0 0 1 1 1

CBSE 2011 Q 1. (i) Width of depletion region/layer decreases when forward biased. (ii) Width of depletion layer increases when reverse bias. Q 2.

This gate is AND gate