Semiconductors and pn Junction Diode PDF

Unit-I Semiconductors and pn Junction Diode 1. Explain N-type and P-type semiconductors. Ans: n- type semiconductor  When a small amount of pentavalent impurity is added to a pure semiconductor, it is called n-type semiconductor.  The pentavalent impurity is also cal...

Unit-I Semiconductors and pn Junction Diode 1. Explain N-type and P-type semiconductors. Ans: n- type semiconductor  When a small amount of pentavalent impurity is added to a pure semiconductor, it is called n-type semiconductor.  The pentavalent impurity is also called donor impurity has five valence electrons.  Examples: Phosphorous, Arsenic, Antimony, Bismuth.  Consider the formation of n type material by adding Arsenic(As) into Silicon(Si).  The As atom has five valence electrons. An As atom fits in the silicon crystal in such a way that its four valence electrons from covalent bonds with four adjacent Si atoms.  The fifth electron has no chance of forming a covalent bond. It enters the conduction band as a free electron.  One donor impurity atom donates one free electron in n- type material. The free electrons are majority charge carriers. p- type semiconductor  When a small amount of trivalent impurity is added to a pure semiconductor, it is called p-type semiconductor.  The trivalent impurity is also called acceptor impurity has three valence electrons.  Examples: Boron, Aluminium, Gallium, Indium.  Consider the formation of p- type material by adding Gallium(Ga) into Silicon(Si).  The Ga atom has three valence electrons, the fourth covalent bond in the valence shell is incomplete. The resulting vacancy is called a hole.  This means that each Ga atom added into Si atom gives one hole.  One acceptor impurity creates one hole in a p- type material. The holes are majority charge carriers. 1  2. Write short notes on Diffusion and Drift current. Ans: Diffusion current: The flow of charge carriers from a high density region to low density region constitute diffusion current. This flow occurs until the distribution of charges becomes uniform, in both regions. The diffusion current density due to hole is given by, dp Jp = −qDp dx The diffusion current density due to electron is given by dn Jn = qDn dx Where q = charge of electron Dp = diffusion constant for hole m2/sec Dn = diffusion constant for electron m2/sec dp = concentration of gradient for holes dx dn = concentration of gradient for electrons. dx Drift current : Whenever voltage is applied to the pn junction, there exist a flow of current because of applied voltage and this current is known as drift current. The drift current density due to hole is given by, Jp = qpμpE The drift current density due to free electron is given by, Jn = qnμnE Total drift current density Jtotal = qpμpE + qnμnE Where p = no.of holes/cm3 n = no.of free electrons /cm3 μp = mobility of hole μn = mobility of electron E = electric field intensity 3.State law of mass action and the Einstein’s relationship for semiconductor. Ans: Law of mass action: It states that the product of concentrations of electrons and holes is always constant, at a fixed temperature. 𝑛𝑝 = 𝑛𝑖2 Where n = concentration of electrons p = concentration of holes ni = intrinsic concentration For n-type material , n = nn while p = pn hence law can be stated as 𝑛𝑛 𝑝𝑛 = 𝑛𝑖2 For p-type material , n = np while p = pp hence law can be stated as 𝑛𝑝 𝑝𝑝 = 𝑛𝑖2 Einstein’s relationship: It states that, at a fixed temperature, the ratio of diffusion constant to the mobility contant. Dp Dn = = KT = VT μp μn Where K= Boltzmann‟s constant = 8.62×10−5 ev/ok T = temperature in ok KT= VT = voltage equivalent of temperature. T At room temperature VT = 0.02586V or VT = 11600 N.Madhu GRIET-ECE 2 4. Explain the Fermi’s level in intrinsic semiconductor. (or) Show that the Fermi energy level lies in the center of forbidden energy band for an intrinsic semiconductor. Ans: In intrinsic semiconductor the probability of finding an electron in the conduction band is zero and probability of finding a hole in the valence band is zero at T=0 ok. At temperature increases, equal number of electrons and holes get generated. In such a case the Fermi level E F is given by EC + EV EF = 2 Proof: Taking logarithm on both sides If NC = NV, EC + EF −2EF = 0 N.Madhu GRIET-ECE 3  5. Explain about Fermi level in extrinsic semiconductor. Ans: Fermi level in n type semiconductor:  The donor impurity added is just below the conduction band. This donor level is indicated as ED.  As this distance is very small, even at room temperature almost all the extra electrons from the donor impurity atoms jump into the conduction band.  The probability of occupying the energy level by the electrons, towards the conduction band is more.  So in n- type material, the Fermi level EF gets shifted towards the conduction band. But it is below the donor energy level. Fermi level in P type semiconductor:  In p-type material, acceptor impurity is added just above the valence band.  Due to this, large number of holes gets generated in the valence band.  At room temperature, the electrons from valence band jump to acceptor energy level, leaving behind the holes in valence band.  This shifts the Fermi level EF towards the valence band. It lies above the acceptor energy level. N.Madhu GRIET-ECE 4  6.Explain the formation of depletion region in an open circuited p-n junction with neat sketches. (or) With the help of necessary sketches, explain the potential distribution in an open circuited p-n junction. Ans:  When two extrinsic semiconductors, one p-type and another n-type are joined, a p-n junction is formed.  The holes from p-side and electrons from n-side which form majority carriers in their respective sections are diffused in either direction.  At this stage p-side will loose its holes and hence a negative field exists towards left side of the junction.  Similarly, n-side will loose its electrons and hence a positive field exists towards right side of the junction.  The net result is that an electrical field exists across the junction. This field is known as potential barrier or contact potential.  Potential barrier will be limited because the negative field on p- side and positive of n- side prevents further displacement of electrons and holes on either side respectively.  The charge distribution is shown in figure. Fig: Potential Distribution in p-n junction Diode N.Madhu GRIET-ECE 5   The concentration of charge decrease as we move away from the junction. Because at the junction, holes and electrons are combined and become neutral.  The junction is depleted of mobile charges. The region is called space charge region and its thickness is about few microns.  The electric field intensity, which is the integral of density function ρ is shown in the figure.  Contact potential as shown in the figure will be of the order of few tenths of volts. Its value of germanium is 0.3V and silicon is 0.7V. 7. Draw the energy band diagram of p-n junction under open circuit conditions and explain. (or) Derive expression for pn junction diode barrier potential. (or) Explain how the built-in potential difference exists at pn junction without the application of an external voltage across it. Ans:  It is known that the Fermi level in n-type material lies just below the conduction band while in p-type material, it lies just above the valence band.  When p-n junction is formed, the diffusion starts. The changes get adjusted so as to equalize the Fermi level in the two parts of p-n junction.  This is similar to adjustment of water levels in two tanks of unequal level, when connected each other.  The changes flow from p to n and n to p side till, the Fermi level on two sides get lined up.  In n-type semi conductor , EF is close to conduction band Ecn and it is close to valence band edge EVP on p-side.  So the conduction band edge of n-type semiconductor can‟t be at the same level as that of p-type semiconductor.  Hence, as shown, the energy band diagram for p-n junction is where a shift in energy levels E0 is indicated. Fig:Energy Band Diagram for p-n Junction under open-circuit Conditions  NC we know that, for n-type, EF = ECn − kT ln ND NC ECn − EF = kT ln ⋯ ⋯ ⋯ ⋯ ⋯ ⋯ ⋯ (1) ND NV for p-type, EF = EVp + kT ln NA NV EF − EVp = kT ln ⋯ ⋯ ⋯ ⋯ ⋯ ⋯ ⋯ (2) NA Since, 𝑛 = 𝑝 = 𝑛𝑖 𝑎𝑛𝑑 𝑛𝑝 = 𝑛𝑖2 In N-type, 𝑛𝑛 ≅ 𝑁𝐷 𝑛𝑛 𝑝𝑛 ≅ 𝑛𝑖2 𝑛2 𝑝𝑛 = 𝑁 𝑖 ⋯ ⋯ ⋯ ⋯ ⋯ ⋯ ⋯ ⋯ ⋯ ⋯ ⋯ ⋯ ⋯ ⋯ ⋯ ⋯ (3) 𝐷 In P-type 𝑝𝑝 ≅ 𝑁𝐴 𝑛𝑝 𝑝𝑝 ≅ 𝑛𝑖2 𝑛2 𝑛𝑝 = 𝑁𝑖 ⋯ ⋯ ⋯ ⋯ ⋯ ⋯ ⋯ ⋯ ⋯ ⋯ ⋯ ⋯ ⋯ ⋯ ⋯ ⋯ (4) 𝐴 From the energy band diagram EG ECn − EF = − E2 ⋯ ⋯ ⋯ ⋯ ⋯ ⋯ ⋯ ⋯ ⋯ ⋯ ⋯ ⋯ 5 2 EG EF − EVp = − E1 ⋯ ⋯ ⋯ ⋯ ⋯ ⋯ ⋯ ⋯ ⋯ ⋯ ⋯ ⋯ 6 2 From equation (5)and equation (6), we get, E1 + E2 = EG − ECn − EVp ⋯ ⋯ ⋯ ⋯ ⋯ ⋯ ⋯ ⋯ (7) Also from band structure, E0 = E1 + E2 ⋯ ⋯ ⋯ ⋯ ⋯ ⋯ ⋯ ⋯ ⋯ ⋯ (8) From equation (7) and equation (8), we get, E0 = EG − ECn − EVp ⋯ ⋯ ⋯ ⋯ ⋯ 9 By adding equation (1) and (2), we get, NC NV ECn − EF + EF − EVp = kT ln + kT ln ND NA  NC NV ECn − EVp = kT ln + kT ln ⋯ ⋯ ⋯ ⋯ ⋯ ⋯ ⋯ ⋯ ⋯ ⋯ (10) ND NA Since, np = e− E C −E F /kT. NC e− E F −E V /kT. NV 𝑛𝑖2 = 𝑁𝐶 𝑁𝑉 𝑒 − E C −E V +E F −E F /𝑘𝑇 𝑛 𝑖2 = 𝑒 − E C −E V /kT NC NV EG 𝑛 𝑖2 ≫ = 𝑒− kT ∴ EC − EV = EG NC NV Taking ln on both sides we get, n 2i EG ln NC NV =− kT NC NV kT ln = EG ⋯ ⋯ ⋯ ⋯ ⋯ ⋯ ⋯ ⋯ ⋯ ⋯ ⋯ (11) n 2i substituting the values of equation (10) and equation (11) in equation (9), we get, NC NV NC NV E0 = kT ln − kT ln − kT ln n 2i ND NA NC NV ND NA E0 = kT ln × × n 2i NC NV NA ND E0 = kT ln ⋯ ⋯ ⋯ ⋯ ⋯ ⋯ ⋯ ⋯ ⋯ ⋯ ⋯ (12) n 2i 𝑛2 𝑛2 Further for p-type, 𝑝𝑝 0 = 𝑁𝐴 , 𝑛𝑝0 = 𝑁𝑖 , 𝑁𝐴 = 𝑛 𝑖 𝐴 𝑝0 𝑛2 𝑛2 Further for N-type, 𝑛𝑛 0 = 𝑁𝐷 , 𝑝𝑛0 = 𝑁 𝑖 , 𝑁𝐷 = 𝑝 𝑖 𝐷 𝑛0 nn 0 Pp 0 Therefore E0 = kT ln and E0 = kT ln np0 pn 0 Where the subscript „0‟ indicate that the above relations are obtained under thermal conditions of equilibrium. 8. What do you understand by depletion region at pn junction? What is the effect of forward and reverse biasing of pn junction on the depletion region? Explain with necessary diagrams. Ans: Depletion region: In a PN junction p-type consists of holes and n-type consists of electrons. Due to diffusion, the large number of holes from p-side diffuse to n-side and similarly, the large number of electrons from n-side diffuse to p-side.  Due to this displacement, p-side looses holes and forms a negative electric field to the left side of junction and n-side looses electrons and forms a positive electric field to the right side of the junction. Because of this large movement of holes and electrons, a barrier potential is developed across the junction. Finally, the holes will combine with free electrons and gets disappear leaving negative potential at p-side near the junction. Similarly the free electrons will combine with holes and gets disappear leaving positive potential at n-side near the junction. This region at the junction is known as depletion region. The thickness of depletion region is the order of few microns. Where, 1 micron=10-4cm. Forward bias: In forward bias, the thickness of depletion layer is very thin because p-type is connected to positive terminal and the n-type is connected to the negative terminal. This causes the holes and electrons to move freely across the junction, hence resulting in a large current. Reverse bias:  In reverse bias, as the p-type is connected to the negative terminal and n-type is connected to the positive terminal, the force of attraction takes place, so the holes from p-side and the electrons from n-side moves away from the junction, thus increasing the width of depletion region. This results in a very little current, almost equal to zero. Therefore, in reverse bias the thickness of the depletion region is large. 9. Explain about various current components in a forward biased pn junction diode. Ans:  When a p-n junction is forward biased a large forward current flows, which is mainly due to majority carriers. The depletion region near the junction is very very small, under forward biased condition.  In forward biased condition holes get diffused into n side from p side while electrons get diffused into p-side from n side.  So on p-side, the current carried by electrons which is diffusion current due to minority carriers, decreases exponentially with respect to distance measured from the junction. This current due to electrons, on p-side which are minority carriers is denoted as Inp.  Similarly holes from p side diffuse into n-side carry current which decreases exponentially with respect to distance measured from the junction. This current due to holes on n-side, which are minority carriers is denoted as Ipn. Fig: Current componets  If distance is denoted by then, Inp(x) = Current due to electrons in p side as a function of x Ipn(x) = Current due to holes in n side as a function of x  At the junction i.e. at x = 0, electrons crossing from n side to p side constitute a current, Inp(0) in the same direction as holes crossing the junction from p side to n side constitute a current , Ipn(0). Hence the current at the junction is the total conventional current I flowing through the circuit.  ∴ I = Ipn(0) + Inp(0)  But as the entire circuit is a series circuit, the total current must be maintained at I, independent of x. This indicates that on p side there exists one more current component which is due to holes on p side which are majority carriers. It is denoted by Ipp(x). Ipp(x) = current due to holes in p side  Similarly on n side, there exists, there exists one more current component which is due to electrons on n side, which are the majority carriers. It is denoted by Inn(x). Inn(x) = Current due to electrons in n side.  On p side the total current is I = Ipp(x) + Inp(x)  On n side the total current is I = Inn(x) + Ipn(x) 10. Define law of junction. Explain about the term cut-in voltage associated with pn junction diode. How do you obtain cut-in voltage from forward V-I characteristics? Ans: Law of Junction: This law states that for a forward biased junction diode the injected hole concentration 𝑝𝑛 (0) in the n-region increases over thermal equilibrium value 𝑃𝑛 0. It is given by, 𝑉 𝑝𝑛 0 = 𝑝𝑛 0 𝑒 𝑉𝑇 𝑉 Similarly for electron concentration, 𝑛𝑝 0 = 𝑛𝑝 0 𝑒 𝑉𝑇 Cut-in voltage 𝑉𝛾 is the minimum bias voltage that should be applied across a diode for conduction to take place. The amount of current below 𝑉𝛾 is very small and above 𝑉𝛾 the current raises rapidly. Typically values of 𝑉𝛾 for germanium and silicon diodes are 0.2 V and 0.6V respectively. The forward V-I characteristics of a junction diode are shown below. Cut-in voltage can be obtained from the V-I characteristics by selecting a point on voltage axis where the current raises rapidly. 11.What are the applications of PN diode and Explain the volt-ampere characteristics of PN diode. Ans: PN diode applications:  Rectifiers in DC power supplies  Switching in digital logic circuits  Clippers in wave shaping circuits  Clampers in TV receivers  Diode gates  Comparator Fig: V-I Characteristics of a Diode  In forward characteristics, it is seen that initially forward current is small as long as the bias voltage is less than the barrier potential.  At a certain voltage close to barrier potential, current increases rapidly.  The voltage at which diode current starts increasing rapidly is called as cut-in voltage. It is denoted by 𝑉𝛾.  Below this voltage, current is less than 1% of maximum rated value of diode current.  The cut-in voltage for germanium is about 0.2V while for silicon it is 0.6V.  In reverse characteristics , reverse current increases initially as reverse voltage is increased. But after a certain voltage, the current remains constant equal to reverse saturation current 𝐼0.  The voltage at which breakdown occurs is called reverse breakdown voltage denoted as 𝑉𝐵𝑅.  Reverse current before the breakdown is very small and can be neglected practically.  It is important to note that the breakdown voltage is much higher and practically diodes are not operated in the breakdown condition. 12. Explain the temperature dependence of V-I characteristics. Ans: Dependency on V-I characteristics on temperature The generation of electron hole pairs in semiconductors is increased due to the rise in temperature and which further leads to increase in their conductivity. Thus, the variation of current through the pn junction diode with temperature can be obtained using diode current equation i.e., 𝐼 = 𝐼0 𝑒 (𝑉/𝜂𝑉𝑇 ) − 1 Where, 𝐼 = Diode current 𝐼0 = Diode reverse saturation current 𝑉 = External applied voltage to the diode 𝜂 = Constant = 1 for germanium = 2 for silicon 𝑇 𝑉𝑇 = Thermal voltage = 11600 𝑇 = Temperature of diode junction(oK) For the diodes both germanium and silicon, the reverse saturation current 𝐼0 increases approximately 7 percent/oC. The reverse saturation current approximately doubles for every 10oC rise in temperature. Since (1.07)10≈2. Thus current I increases, if the temperature is increased at fixed voltage. By only reducing V, we can bring back current I to its original value. 𝑑𝑉 In order to maintain a constant current I value, the value of is found to be −2.5mV/oC at room 𝑑𝑇 temperature for either germanium or silicon. Basically, the value of cut-in voltage(or barrier voltage) is about 0.3V for germanium and 0.7 V for silicon. For both germanium and silicon diodes, the barrier voltage is decreased by 2mV/oC. This is due to dependency of barrier voltage on temperature. Mathematically, 𝐼02 = 𝐼01 × 2 𝑇2 −𝑇1 /10 Where, 𝐼01 − Saturation current of the diode at temperature, T1 𝐼02 − Saturation current of the diode at temperature, T2.  Fig: Effect of Temperature on the Diode charactristics The effect of increased temperature on the pn junction diode characteristic curve is shown in figure. The maximum limit of temperature upto which a germanium and silicon diodes can used are 75o and 175oC respectively. 13. Explain about static and dynamic resistance in pn diode. Ana: DC or Static Resistance  The application of a dc voltage to a circuit containing a semiconductor diode will result in an operating point on the characteristic curve that will not change with time. The resistance of the diode at the operating point can be found simply by finding the Corresponding levels of 𝑉𝐷 and 𝐼𝐷 as shown in Fig.(1) and applying the following Equation: 𝑉𝐷 𝑅𝐷 = 𝐼𝐷  The dc resistance levels at the knee and below will be greater than the resistance levels obtained for the vertical rise section of the characteristics. The resistance levels in the reverse-bias region will naturally be quite high. Since ohmmeters typically employ a relatively constant-current source, the resistance determined will be at a preset current level (typically, a few mill amperes). Figure(1): Determining the dc resistance of a diode at a particular operating point. N.Madhu GRIET-ECE 14  AC or Dynamic Resistance  It is obvious from above equation that the dc resistance of a diode is independent of the shape of the characteristic in the region surrounding the point of interest. If a sinusoidal rather than dc input is applied, the situation will change completely. The varying input will move the instantaneous operating point up and down a region of the characteristics and thus defines a specific change in current and voltage. With no applied varying signal, the point of operation would be the Q-point determined by the applied dc levels as shown in fig(2). Figure(2): determining the ac resistance at a Q-point  A straight line drawn tangent to the curve through the Q-point as shown in Fig. 2 will define a particular change in voltage and current that can be used to determine the ac or dynamic resistance for this region of the diode characteristics. In equation form, ∆𝑉 𝑟𝐷 = ∆𝐼 𝐷 𝐷 14. Write short notes on transition capacitance and Diffusion capacitance. Ans: Transition capacitance:  Consider a reverse biased p-n junction diode as shown in figure. Fig: Transition capacitance in reverse biased condition   When a diode is reversed biased, reverse current flows due to minority carriers.  The width of the depletion region increases as reverse bias voltage increases.  As the charged particles move away from the junction there exists a change in charge with respect to the applied reverse voltage.  So change in charge dQ with respect to the change in voltage dV is nothing but a capacitive effect.  Such a capacitance which into the picture under reverse biased condition is called Transition capacitance or space charge capacitance denoted as CT. 𝑑𝑄 𝜀𝐴 𝐶𝑇 = = 𝑑𝑉 𝑊 Where, dQ is the increases in charge caused by a change dV in voltage. Diffusion Capacitance  When a p-n junction is forward biased, the depletion layer almost completely disappears.  The electrons move and stored in p-region and holes stored in n-region.  As applied voltage increases, the amount of charge stored on both sides of junction also increases.  It is observed that the charge stored varies directly as the applied forward bias voltage.  This effect is similar to a capacitor in which amount of charge stored varies with applied voltage. 𝑑𝑄 𝑅𝑎𝑡𝑒 𝑜𝑓 𝑐𝑕𝑎𝑛𝑔𝑒 𝑜𝑓 𝑐𝑕𝑎𝑟𝑔𝑒 𝑎𝑡 𝑗𝑢𝑛𝑐𝑡𝑖𝑜𝑛 𝐶𝑇 = 𝑑𝑉 = 𝑅𝑎𝑡𝑒 𝑜𝑓 𝑐𝑕𝑎𝑛𝑔𝑒 𝑜𝑓 𝑎𝑝𝑝𝑙𝑖𝑒𝑑 𝑓𝑜𝑟𝑤𝑎𝑟𝑑 𝑣𝑜𝑙𝑡𝑎𝑔𝑒  The flow of charge Q results in a diode current I is given by 𝑄 𝐼= 𝜏 𝑄 =𝐼×𝜏 ⋯ ⋯ (1) Where 𝜏 is average life time of charge carrier The diode current equation 𝐼 = 𝐼0 𝑒 𝑉/𝜂𝑉𝑇 − 1 The charge 𝑄 = 𝐼0 𝑒 𝑉/𝜂 𝑉𝑇 − 1 𝜏 Differentiating above equation with respect to V 𝑑𝑄 𝜏𝐼0 𝑉/𝜂 𝑉 = 𝑒 𝑇 ⋯ ⋯ ⋯ ⋯ ⋯ ⋯ (2) 𝑑𝑉 𝜂𝑉𝑇 From diode current equation 𝐼 = 𝐼0 𝑒 𝑉/𝜂 𝑉𝑇 − 𝐼0 𝐼 + 𝐼0 = 𝐼0 𝑒 𝑉/𝜂 𝑉𝑇 ⋯ ⋯ ⋯ (3) Substitute equation (3) in equation (2) 𝑑𝑄 𝜏 = 𝜂 𝑉 (𝐼 + 𝐼0 ) 𝑑𝑉 𝑇 𝑑𝑄 𝜏𝐼 ≈ 𝜂𝑉 𝑠𝑖𝑛𝑐𝑒 𝐼 ≫ 𝐼0 𝑑𝑉 𝑇 𝜏𝐼 Diffusion capacitance, 𝐶𝐷 = 𝜂 𝑉 𝑇 15. Derive an expression for transition capacitance.  Consider a p-n junction diode, the two sides of which are not equally doped.  Assume that p-side is lightly doped and n side is heavily doped.  If NAWn. Fig: Unequally doped p-n junction diode  Hence the width of depletion region on n-side is negligible small compared to width of depletion region on p-side.  Hence the entire depletion region can be assumed to be on the p-side only  The relationship between potential and charge density is given by poisson‟s equation, 𝑑2𝑉 𝑞𝑁𝐴 = ⋯ ⋯ ⋯ ⋯ ⋯ (1) 𝑑𝑥 2 𝜀 Integrating equation (1) twice 𝑞𝑁𝐴 𝑤 2 ∴ 𝑉= ⋯ ⋯ ⋯ ⋯ ⋯ (2) 𝜀 2 At 𝑥 = 𝑤, 𝑉 = 𝑉𝐵 which is barrier potential 𝑞𝑁𝐴 𝑤 2 𝑉𝐵 = ⋯ ⋯ ⋯ (3) 𝜀 2 From the above expression it can be observed that 𝑤 ∝ 𝑉𝐵 The width of barrier i.e depletion layer increases with applied reverse bias. Differentiating equation (3) w.r.toV, 1 𝑞𝑁𝐴 𝑑𝑤 1=2 2𝑤 𝜀 𝑑𝑉 𝑑𝑤 𝜀 = 𝑞𝑁 ⋯ ⋯ (4) 𝑑𝑉 𝐴𝑤 If A is the area of cross section of the junction, then net charge Q in the distance w is 𝑄 = 𝑛𝑜. 𝑜𝑓 𝑐𝑕𝑎𝑟𝑔𝑒𝑑 𝑝𝑎𝑟𝑡𝑖𝑐𝑙𝑒𝑠 × 𝑐𝑕𝑎𝑟𝑔𝑒 𝑜𝑛 𝑒𝑎𝑐𝑕 𝑝𝑎𝑟𝑡𝑖𝑐𝑙𝑒 = 𝑁𝐴 × 𝑉𝑜𝑙𝑢𝑚𝑒 𝑞  = 𝑁𝐴 𝐴𝑊𝑞 ⋯ ⋯ (5) Differentiating equation (5) w.r.to 𝑉, 𝑑𝑄 𝑑𝑊 = 𝑁𝐴 𝐴𝑞 𝑑𝑉𝑅 𝑑𝑉 𝜀 = 𝑁𝐴 𝐴𝑞 𝑞𝑁 𝐴𝑊 𝜀𝐴 𝐶𝑇 = 𝑊 16. Explain in brief about diode equivalent circuit. Ans: An equivalent circuit is combination of elements, inserted in the place of a device without changing behavior of system. Different equivalent circuit models of diode are given below.  Piece wise linear model  Simplified model  Ideal Diode model Piece wise linear model: Figure shows the V-I characteristics of a diode in piece wise linear model. The slope of straight line represents the reciprocal of a diode resistance, when the diode is ON. Whenever the voltage across the diode reaches 𝑉𝛾 then the diode will be ON. Fig: V-I Characteristics and its Equivalent Circuit of Piece Wise Linear Model Simplified model: If we consider above piece wise linear model, the resistance 𝑟𝑑 is practically smaller. So by removing the 𝑟𝑑 , we get the diode equivalent in simplified model. Fig: V-I Characteristics and its Equivalent Circuit of Simplified Model Ideal Diode model: An ideal diode is the first and simplest approximation of a real diode. It has no forward voltage drop, no reverse current, and no breakdown. In fact, an ideal diode is only a theoretical approximation of a real diode. However, in a well defined circuit, a real diode behaves almost like an ideal diode because forward voltage across the diode is small as compared to the input and output voltages. Fig: V-I Characteristics and its Equivalent Circuit of ideal Diode 17. Explain about breakdown mechanisms in semiconductor diodes. A: Avalanche breakdown  Though reverse current is not dependent on reverse voltage, if reverse voltage is increased, at a particular value, velocity of minority carriers increases.  Due to kinetic energy associated with the minority carriers, more minority carriers are generated when there is collision of minority carriers with the atoms.  The collision makes the electrons to break the covalent bonds.  These electrons are available as minority carriers and get accelerated due to high reverse voltage.  They again collide with another atoms to generate more minority carriers. This is called carrier multiplication.  Finally large number of minority carriers move across the junction, breaking the p-n junction.  These large number of minority carriers give rise to a very high reverse current.  This effect is called avalanche effect and the mechanism of destroying the junction is called reverse breakdown of a p-n junction.  The voltage at which breakdown of a p-n junction occurs is called reverse breakdown voltage. Zener breakdown  When a p-n junction is heavily doped the depletion region is very narrow. So under reverse bias conditions, the electric field across the depletion layer is very intense.  Such an intense field is enough to pull the electrons out of the valance bands of the stable atoms.  So this is not due to the collision of carriers with atoms.  Such a creation of free electrons is called zener effect.  These minority carriers constitute very large current and mechanism is called zener breakdown. 18 Explain about Zener diode and its characteristics. A Zener diode  The diodes designed to work in breakdown region are called zener diode.  It is heavily doped than ordinary diode.  Operates in reverse breakdown region.  It can be used as voltage regulator.  When biased in the forward direction it behaves just like a normal signal diode passing the rated current, but when a reverse voltage is applied to it the reverse saturation current remains fairly constant over a wide range of voltages. Fig: V-I Characteristics of a Zener diode  Under forward bias condition, zener diode is same as that of ordinary pn diode.  In reverse bias, as the reverse voltage VR is increased the reverse saturation current remains extremely small up to knee of the curve.  If the reverse voltage increased further, breakdown occurs and the current increase rapidly.  The reverse voltage at which the breakdown occurs is known as reverse breakdown voltage.  The breakdown voltage has a sharp knee, followed by an almost vertical increase in current, during which period the voltage across the device remains almost constant.  Thus zener diode is most suited for voltage regulators. Problems: Problem 1 Problem 2 Problem 3 Problem 4 Problem5. In the case of an open circuited p-n junction, the acceptor atom concentration is 2.5×1016/m3, donor atom concentration is 2.5×1022/m3 and intrinsic concentration is 2.5×1019/m3. Determine the value of the contact difference of potential. Sol: 𝑁𝐴 = 2.5 × 1016 /m3 𝑁𝐷 = 2.5 × 1022 /m3 𝑛𝑖 = 2.5 × 1019 /m3 𝑁𝐴 𝑁𝐷 Contact potential 𝑉𝐽 = 𝑉𝑇 ln 𝑛 𝑖2 Assume 𝑉𝑇 = 26𝑚𝑉 2.5×1016 ×2.5×1022 𝑉𝐽 = 26 × 10−3 ln 2.5×1019 2 = 0𝑉 Problem6. A diode operating at 300K at a forward voltage of 0.4V carriers a current of 10mA. When voltage is changed to 0.42V the current becomes thrice. Calculate the value of reverse leakage current and Ƞ for the diode(assume VT = 26 mV). Sol: Given data: T = 3000K V1= 0.4V I1 = 10mA V2= 0.42V I2= 30 mA ( given the current becomes thrice) VT = 26mV To find: I0 = ? 𝝶= ? The expression for diode current is given by 𝐼 = 𝐼0 𝑒 𝑉/𝜂 𝑉𝑇 − 1 For V1= 0.4V, I1 = 10mA and VT = 26mV, we get, −3 10 × 10−3 = 𝐼0 𝑒 0.4/𝜂×26×10 −1  400 400 = 𝐼0 𝑒 26 𝜂 − 1 𝑠𝑖𝑛𝑐𝑒 𝑒 26 𝜂 >> 1 400 10 × 10−3 = 𝐼0 𝑒 26 𝜂 ⋯⋯⋯ (1) For V2= 0.42V, I2 = 30mA and VT = 26mV, we get, −3 30 × 10−3 = 𝐼0 𝑒 0.42/𝜂×26×10 −1 420 400 = 𝐼0 𝑒 26 𝜂 − 1 𝑠𝑖𝑛𝑐𝑒 𝑒 26 𝜂 >> 1 420 30 × 10−3 = 𝐼0 𝑒 26 𝜂 ⋯ ⋯ ⋯ (2) Dividing equation (2) with equation (1), we get, 420 30×10 −3 𝐼0 𝑒 26 𝜂 = 400 10×10 −3 𝐼0 𝑒 26 𝜂 20 3 = 𝑒 26 𝜂 10 3 = 𝑒 13 𝜂 10 = 𝑙𝑛3 = 1.099 13𝜂 ∴ 𝜂 = 0.7 Then , substituting the above 𝜂 value in equation (1), we get, 400 10 × 10−3 = 𝐼0 𝑒 26 ×0.7 ∴ 𝐼0 = 2.85 𝑝𝐴𝑚𝑠 Problem7. A pn junction diode has a reverse saturation current of 30 μA at a temperature of 1250 C. At the same temperature, find the dynamic resistance for 0.2 V bias in forward and reverse direction. Sol: Given data: 𝐼0 = 30𝜇𝐴 𝑇 = 1250 𝐶 𝑉𝑓 = 0.2 𝑓𝑜𝑟𝑤𝑎𝑟𝑑 𝑏𝑖𝑎𝑠  −𝑉𝑓 = −0.2 𝑟𝑒𝑣𝑒𝑟𝑠𝑒 𝑏𝑖𝑎𝑠 To find: 𝑟𝑓 =? 𝑟𝑟 =? Consider 𝝶 =2 for silicon 𝑇 398 We know that, 𝑉𝑇 = 11600 = 11600 = 34.3 × 10−3 𝑉 𝑠𝑖𝑛𝑐𝑒 𝑇 = 125 + 273 = 3980 𝐾 𝜂 𝑉𝑇 2×34.3×10 −3 Forward dynamic resistance 𝑟𝑓 = 𝐼 = 0.2 = 123.9Ω 0 𝑒 𝑉 /𝜂 𝑉 𝑇 30×10 −6 ×𝑒 2×34.3×10 −3 𝜂 𝑉𝑇 2×34.3×10 −3 Reverse dynamic resistance 𝑟𝑟 = = −0.2 = 42.18𝐾Ω 𝐼0 𝑒 −𝑉/𝜂 𝑉 𝑇 30×10 −6 ×𝑒 2×34.3×10 −3