3.7 RESISTIVITY OF VARIOUS MATERIALS The materials are classified as conductors, semiconductors and insulators 89depending on their resistivities, in an increasing order of their values. Reprint 2025-26 Physics Metals have low resistivities in the range of 10–8 Wm to 10–6 Wm. At the other end are insulators like ceramic, rubber and plastics having resistivities 1018 times greater than metals or more. In between the two are the semiconductors. These, however, have resistivities characteristically decreasing with a rise in temperature. The resistivities of semiconductors can be decreased by adding small amount of suitable impurities. This last feature is exploited in use of semiconductors for electronic devices. 3.8 TEMPERATURE DEPENDENCE OF RESISTIVITY The resistivity of a material is found to be dependent on the temperature. Different materials do not exhibit the same dependence on temperatures. Over a limited range of temperatures, that is not too large, the resistivity of a metallic conductor is approximately given by, rT = r0 [1 + a (T–T0)] (3.26) where rT is the resistivity at a temperature T and r0 is the same at a reference temperature T0. a is called the temperature co-efficient of resistivity, and from Eq. (3.26), the dimension of a is (Temperature)–1. For metals, a is positive. The relation of Eq. (3.26) implies that a graph of rT plotted against T would be a straight line. At temperatures much lower than 0°C, the graph, however, deviates considerably from a straight line (Fig. 3.8). Equation (3.26) thus, can be used approximately over a limited range of T around any reference temperature T0, where the graph can be approximated as a straight line.  FIGURE 3.8 FIGURE 3.9 Resistivity FIGURE 3.10 Resistivity rT of rT of nichrome as a Temperature dependence copper as a function function of absolute of resistivity for a typical of temperature T. temperature T. semiconductor. Some materials like Nichrome (which is an alloy of nickel, iron and chromium) exhibit a very weak dependence of resistivity with temperature (Fig. 3.9). Manganin and constantan have similar properties. These materials are thus widely used in wire bound standard resistors since 90 their resistance values would change very little with temperatures. Reprint 2025-26 Current Electricity Unlike metals, the resistivities of semiconductors decrease with increasing temperatures. A typical dependence is shown in Fig. 3.10. We can qualitatively understand the temperature dependence of resistivity, in the light of our derivation of Eq. (3.23). From this equation, resistivity of a material is given by 1 m ρ= = 2 (3.27) σ n e τ r thus depends inversely both on the number n of free electrons per unit volume and on the average time t between collisions. As we increase temperature, average speed of the electrons, which act as the carriers of current, increases resulting in more frequent collisions. The average time of collisions t, thus decreases with temperature. In a metal, n is not dependent on temperature to any appreciable extent and thus the decrease in the value of t with rise in temperature causes r to increase as we have observed. For insulators and semiconductors, however, n increases with temperature. This increase more than compensates any decrease in t in Eq.(3.23) so that for such materials, r decreases with temperature. Example 3.3 An electric toaster uses nichrome for its heating element. When a negligibly small current passes through it, its resistance at room temperature (27.0 °C) is found to be 75.3 W. When the toaster is connected to a 230 V supply, the current settles, after a few seconds, to a steady value of 2.68 A. What is the steady temperature of the nichrome element? The temperature coefficient of resistance of nichrome averaged over the temperature range involved, is 1.70 × 10–4 °C–1. Solution When the current through the element is very small, heating effects can be ignored and the temperature T1 of the element is the same as room temperature. When the toaster is connected to the supply, its initial current will be slightly higher than its steady value of 2.68 A. But due to heating effect of the current, the temperature will rise. This will cause an increase in resistance and a slight decrease in current. In a few seconds, a steady state will be reached when temperature will rise no further, and both the resistance of the element and the current drawn will achieve steady values. The resistance R2 at the steady temperature T2 is 230 V R2 = 2.68 A = 85.8 Ω Using the relation R2 = R1 [1 + a (T2 – T1)] with a = 1.70 × 10–4 °C–1, we get (85.8 – 75.3) T2 – T1 = –4 = 820 °C (75.3) × 1.70 × 10 that is, T2 = (820 + 27.0) °C = 847 °C Thus, the steady temperature of the heating element (when heating EXAMPLE effect due to the current equals heat loss to the surroundings) is 3.3 91 847 °C. Reprint 2025-26 Physics Example 3.4 The resistance of the platinum wire of a platinum resistance thermometer at the ice point is 5 W and at steam point is 5.23 W. When the thermometer is inserted in a hot bath, the resistance of the platinum wire is 5.795 W. Calculate the temperature of the bath. Solution R0 = 5 W, R100 = 5.23 W and Rt = 5.795 W Rt − R 0 Now, t = × 100, Rt = R 0 (1 + αt ) R100 − R 0 3.4 5.795 − 5 = × 100 5.23 − 5 0.795 EXAMPLE = × 100 = 345.65 °C 0.23 3.9 ELECTRICAL ENERGY, POWER Consider a conductor with end points A and B, in which a current I is flowing from A to B. The electric potential at A and B are denoted by V(A) and V(B) respectively. Since current is flowing from A to B, V(A) > V(B) and the potential difference across AB is V = V(A) – V(B) > 0. In a time interval Dt, an amount of charge DQ = I Dt travels from A to B. The potential energy of the charge at A, by definition, was Q V(A) and similarly at B, it is Q V(B). Thus, change in its potential energy DUpot is DUpot = Final potential energy – Initial potential energy = DQ[(V (B) – V (A)] = –DQ V = –I VDt < 0 (3.28) If charges moved without collisions through the conductor, their kinetic energy would also change so that the total energy is unchanged. Conservation of total energy would then imply that, DK = –DUpot (3.29) that is, DK = I VDt > 0 (3.30) Thus, in case charges were moving freely through the conductor under the action of electric field, their kinetic energy would increase as they move. We have, however, seen earlier that on the average, charge carriers do not move with acceleration but with a steady drift velocity. This is because of the collisions with ions and atoms during transit. During collisions, the energy gained by the charges thus is shared with the atoms. The atoms vibrate more vigorously, i.e., the conductor heats up. Thus, in an actual conductor, an amount of energy dissipated as heat in the conductor during the time interval Dt is, DW = I VDt (3.31) The energy dissipated per unit time is the power dissipated P = DW/Dt and we have, 92 P = I V (3.32) Reprint 2025-26 Current Electricity Using Ohm’s law V = IR, we get P = I 2 R = V 2/R (3.33) as the power loss (“ohmic loss”) in a conductor of resistance R carrying a current I. It is this power which heats up, for example, the coil of an electric bulb to incandescence, radiating out heat and light. Where does the power come from? As we have reasoned before, we need an external source to keep a steady current through the conductor. It is clearly this source which must supply this power. In the simple circuit shown with a cell (Fig.3.11), it is the chemical energy of the cell which supplies this power for as long as it can. The expressions for power, Eqs. (3.32) and (3.33), show the dependence of the power dissipated in a resistor R on the current through it and the voltage FIGURE 3.11 Heat is produced in the across it. resistor R which is connected across Equation (3.33) has an important application to the terminals of a cell. The energy power transmission. Electrical power is transmitted dissipated in the resistor R comes from from power stations to homes and factories, which the chemical energy of the electrolyte. may be hundreds of miles away, via transmission cables. One obviously wants to minimise the power loss in the transmission cables connecting the power stations to homes and factories. We shall see now how this can be achieved. Consider a device R, to which a power P is to be delivered via transmission cables having a resistance Rc to be dissipated by it finally. If V is the voltage across R and I the current through it, then P = V I (3.34) The connecting wires from the power station to the device has a finite resistance Rc. The power dissipated in the connecting wires, which is wasted is Pc with Pc = I 2 Rc P 2 R c = 2 (3.35) V from Eq. (3.32). Thus, to drive a device of power P, the power wasted in the connecting wires is inversely proportional to V 2. The transmission cables from power stations are hundreds of miles long and their resistance Rc is considerable. To reduce Pc, these wires carry current at enormous values of V and this is the reason for the high voltage danger signs on transmission lines — a common sight as we move away from populated areas. Using electricity at such voltages is not safe and hence at the other end, a device called a transformer lowers the voltage to a value suitable for use.

14.3 Carbon, silicon and germanium have four valence electrons each. These are characterised by valence and conduction bands separated 341 Reprint 2025-26 Physics 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 14.4 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. 14.5 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. 14.6 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. Reprint 2025-26 Notes Reprint 2025-26 Physics APPENDICES APPENDIX A 1 THE GREEK ALPHABET APPENDIX A 2 COMMON SI PREFIXES AND SYMBOLS FOR MULTIPLES AND SUB-MULTIPLES Reprint 2025-26 AppendicesAnswers APPENDIX A 3 SOME IMPORTANT CONSTANTS OTHER USEFUL CONSTANTS 345 Reprint 2025-26 Physics ANSWERS CHAPTER 9 9.1 v = –54 cm. The image is real, inverted and magnified. The size of the image is 5.0 cm. As u ® f, v ® ¥; for u < f, image is virtual. 9.2 v = 6.7 cm. Magnification = 5/9, i.e., the size of the image is 2.5 cm. As u ® ¥; v ® f (but never beyond) while m ® 0. 9.3 1.33; 1.7 cm 9.4 nga = 1.51; nwa = 1.32; ngw = 1.144; which gives sin r = 0.6181 i.e., r ~ 38°. 9.5 r = 0.8 × tan ic and sin ci = 1/1.33 ≅ 0.75 , where r is the radius (in m) of the largest circle from which light comes out and ic is the critical angle for water-air interface, Area = 2.6 m2 9.6 n ≅ 1.53 and Dm for prism in water ≅ 10° 9.7 R = 22 cm 9.8 Here the object is virtual and the image is real. u = +12 cm (object on right; virtual) (a) f = +20 cm. Image is real and at 7.5 cm from the lens on its right side. (b) f = –16 cm. Image is real and at 48 cm from the lens on its right side. 9.9 v = 8.4 cm, image is erect and virtual. It is diminished to a size 1.8 cm. As u ® ¥, v ® f (but never beyond f while m ® 0). Note that when the object is placed at the focus of the concave lens (21 cm), the image is located at 10.5 cm (not at infinity as one might wrongly think). 9.10 A diverging lens of focal length 60 cm 9.11 (a) ve = –25 cm and fe = 6.25 cm give ue = –5 cm; vO = (15 – 5) cm = 10 cm, fO = uO = – 2.5 cm; Magnifying power = 20 (b) uO = – 2.59 cm. Magnifying power = 13.5. 9.12 Angular magnification of the eye-piece for image at 25 cm 25 25   1  11; | u e |= cm = 2 .27cm ; vO = 7.2 cm 2.5 11 Separation = 9.47 cm; Magnifying power = 88 9.13 24; 150 cm 9.14 (a) Angular magnification = 1500 346 (b) Diameter of the image = 13.7 cm. Reprint 2025-26 Answers

14.4 EXTRINSIC SEMICONDUCTOR The conductivity of an intrinsic semiconductor depends on its temperature, but at room temperature its conductivity is very low. As such, no important electronic devices can be developed using these semiconductors. Hence there is a necessity of improving their conductivity. This can be done by making use of impurities. When a small amount, say, a few parts per million (ppm), of a suitable impurity is added to the pure semiconductor, the conductivity of the semiconductor is increased manifold. Such materials are known as extrinsic semiconductors or impurity semiconductors. The deliberate addition of a desirable impurity is called doping and the impurity atoms are called dopants. Such a material is also called a doped semiconductor. The dopant has to be such that it does not distort the original pure semiconductor lattice. It occupies only a very few of the original semiconductor atom sites in the crystal. A necessary condition to attain this is that the sizes of the dopant and the semiconductor atoms should be nearly the same. There are two types of dopants used in doping the tetravalent Si or Ge: (i) Pentavalent (valency 5); like Arsenic (As), Antimony (Sb), Phosphorous 329 (P), etc. Reprint 2025-26 Physics (ii) Trivalent (valency 3); like Indium (In), Boron (B), Aluminium (Al), etc. We shall now discuss how the doping changes the number of charge carriers (and hence the conductivity) of semiconductors. Si or Ge belongs to the fourth group in the Periodic table and, therefore, we choose the dopant element from nearby fifth or third group, expecting and taking care that the size of the dopant atom is nearly the same as that of Si or Ge. Interestingly, the pentavalent and trivalent dopants in Si or Ge give two entirely different types of semiconductors as discussed below. (i) n-type semiconductor Suppose we dope Si or Ge with a pentavalent element as shown in Fig. 14.7. When an atom of +5 valency element occupies the position of an atom in the crystal lattice of Si, four of its electrons bond with the four silicon neighbours while the fifth remains very weakly bound to its parent atom. This is because the four electrons participating in FIGURE 14.7 (a) Pentavalent donor atom (As, Sb, bonding are seen as part of the effective core P, etc.) doped for tetravalent Si or Ge giving n- of the atom by the fifth electron. As a result type semiconductor, and (b) Commonly used the ionisation energy required to set this schematic representation of n-type material electron free is very small and even at room which shows only the fixed cores of the temperature it will be free to move in the substituent donors with one additional effective lattice of the semiconductor. For example, thepositive charge and its associated extra electron. energy required is ~ 0.01 eV for germanium, and 0.05 eV for silicon, to separate this electron from its atom. This is in contrast to the energy required to jump the forbidden band (about 0.72 eV for germanium and about 1.1 eV for silicon) at room temperature in the intrinsic semiconductor. Thus, the pentavalent dopant is donating one extra electron for conduction and hence is known as donor impurity. The number of electrons made available for conduction by dopant atoms depends strongly upon the doping level and is independent of any increase in ambient temperature. On the other hand, the number of free electrons (with an equal number of holes) generated by Si atoms, increases weakly with temperature. In a doped semiconductor the total number of conduction electrons ne is due to the electrons contributed by donors and those generated intrinsically, while the total number of holes nh is only due to the holes from the intrinsic source. But the rate of recombination of holes would increase due to the increase in the number of electrons. As a result, the number of holes would get reduced further. Thus, with proper level of doping the number of conduction electrons 330 can be made much larger than the number of holes. Hence in an extrinsic Reprint 2025-26 Semiconductor Electronics: Materials, Devices and Simple Circuits semiconductor doped with pentavalent impurity, electrons become the majority carriers and holes the minority carriers. These semiconductors are, therefore, known as n-type semiconductors. For n-type semiconductors, we have, ne >> nh (14.3) (ii) p-type semiconductor This is obtained when Si or Ge is doped with a trivalent impurity like Al, B, In, etc. The dopant has one valence electron less than Si or Ge and, therefore, this atom can form covalent bonds with neighbouring three Si atoms but does not have any electron to offer to the fourth Si atom. So the bond between the fourth neighbour and the trivalent atom has a vacancy or hole as shown in Fig. 14.8. Since the neighbouring Si atom in the lattice wants an electron in place of a hole, an electron in the outer orbit of an atom in the neighbourhood may jump to fill this vacancy, leaving a vacancy or hole at its own site. Thus the hole is available for conduction. Note that the trivalent foreign atom becomes effectively negatively charged when it shares fourth electron with neighbouring Si atom. Therefore, the dopant atom of p-type material can be treated as core of one negative charge along with its associated hole as shown in Fig. 14.8(b). It is obvious that one acceptor atom gives one hole. These holes are in addition to the intrinsically generated holes while the source of conduction electrons is only intrinsic generation. Thus, for such a material, the holes are the majority carriers and electrons are minority carriers. Therefore, extrinsic semiconductors doped FIGURE 14.8 (a) Trivalent acceptor atom (In, Al, B etc.)with trivalent impurity are called p-type semiconductors. For doped in tetravalent Si or Gep-type semiconductors, the recombination process will reduce lattice giving p-type semicon- the number (ni)of intrinsically generated electrons to ne. ductor. (b) Commonly used We have, for p-type semiconductors schematic representation of nh >> ne (14.4) p-type material which shows Note that the crystal maintains an overall charge neutrality only the fixed core of the substituent acceptor withas the charge of additional charge carriers is just equal and one effective additional opposite to that of the ionised cores in the lattice. negative charge and its In extrinsic semiconductors, because of the abundance of associated hole. majority current carriers, the minority carriers produced thermally have more chance of meeting majority carriers and thus getting destroyed. Hence, the dopant, by adding a large number of current carriers of one type, which become the majority carriers, indirectly helps to reduce the intrinsic concentration of minority carriers. The semiconductor’s energy band structure is affected by doping. In the case of extrinsic semiconductors, additional energy states due to donor impurities (ED) and acceptor impurities (EA) also exist. In the energy band diagram of n-type Si semiconductor, the donor energy level ED is slightly below the bottom EC of the conduction band and electrons from this level move into the conduction band with very small supply of energy. At room temperature, most of the donor atoms get ionised but very few (~1012) atoms of Si get ionised. So the conduction band will have most electrons coming from the donor impurities, as shown in Fig. 14.9(a). Similarly, 331 Reprint 2025-26 Physics for p-type semiconductor, the acceptor energy level EA is slightly above the top EV of the valence band as shown in Fig. 14.9(b). With very small supply of energy an electron from the valence band can jump to the level EA and ionise the acceptor negatively. (Alternately, we can also say that with very small supply of energy the hole from level EA sinks down into the valence band. 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 nenh = ni 2 (14.5) Though the above description is grossly approximate and hypothetical, it helps in understanding the difference between metals, insulators and semiconductors (extrinsic and intrinsic) in a simple manner. The difference in the resistivity of C, Si and Ge depends upon the energy gap between their conduction and valence bands. For C (diamond), Si and Ge, the energy gaps are 5.4 eV, 1.1 eV and 0.7 eV, respectively. Sn also is a group IV element but it is a metal because the energy gap in its case is 0 eV. FIGURE 14.9 Energy bands of (a) n-type semiconductor at T > 0K, (b) p-type semiconductor at T > 0K. Example 14.2 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. 14.2 Therefore, ne » ND. Since nenh = ni 2, The number of holes nh = (2.25 × 1032)/(5 ×1022) EXAMPLE332 ~ 4.5 × 109 m–3 Reprint 2025-26 Semiconductor Electronics: Materials, Devices and Simple Circuits