Solutions – Tutorial Set 1 EMT 111/4 – Electronic Devices 1. What is the definition of semiconductor devices? Electronic component made by semiconductor materials and can be used in a lot of applications because it’s compact, suitable in large electromagnet spectrum (from dc to UV frequency), its ability to conduct current and voltage in large value and can be integrated in IC. Important elements of electronic system. 2 Explain with your own words an atom, electron, valence electron, free electron, ionization and shell. What is the maximum number of electrons that can exist in the third shell? An atom - the smallest particle of an element that retains the characteristics of that element. Electrons - The basic particles of negative charge. Electrons move around the atomic nuclei. Valence electrons - Electron in the valence shell and contribute to chemical reactions and bonding within the structure of a material and determine its electrical properties. Free electrons - The escape valence electron from valence shell. This phenomena happens when we gives the external energy to the materials. Ionization - The process of losing a valence electron. This phenomena happens when valence electron have a sufficient amount of energy and then escape from the outer shell and the atom’s influence. Shell - Orbits that have been grouped into energy bands. N e = 2n 2 = 2(3) 2 = 18 The maximum number of electrons in 3rd shell = 18 electrons. 1 3. Discuss the different in terms of energy band between insulator, semiconductor and conductor. Draw the figures to explain your answer. Insulator – very wide energy gap. Valence electrons do not jump into the conduction band except under breakdown conditions, i.e extremely high voltage, thus makes insulator doesn’t conduct electrical current under normal operation. Semiconductor – a much narrow energy gap compare to insulator. This gap permits some valence electrons to jump into conduction band and become free-electrons. This gap also makes semiconductor suitable for devices applications for a various temperature. Conductor – the valence band and conduction band are overlap. It is because a large number of free electrons in conduction band without any external energy. This overlap makes conductor are not suitable for devices applications for high temperature conditions. 4. Why does semiconductor have fewer free electrons than a conductor? The valence electrons of semiconductors are more tightly bound to the atoms than those of conductors. 5. How the covalent bond formed? By sharing of valence electrons with neighboring atoms. 2 6. Describe the properties of n and p-type semiconductors in terms of bonding diagram and energy diagram. In terms of bonding diagram: p-type n-type * to increase the number of conduction band electrons in intrinsic s/c (Si). * pentavalent (5 e valence) atom are added (As, P, Bi Sb). * 4 atom’s valence electrons are used to form the covalent bonds and leaving one extra electron – becomes a conduction electrons (do not leaves a hole in valence band) and can be carefully controlled by number of impurity atoms. - called donor atom. - electrons are the majority carriers - holes are the minority carriers. *to increase the number of holes in intrinsic s/c (Si). * trivalent (3 e valence) atom are added (B, In, Ga). * 3 atom’s valence electrons are used to form the covalent bonds and create one holes when each trivalent atom is added. * number of holes can be carefully controlled by number of impurity atoms. - called acceptor atom. - holes are the majority carriers. - electrons are the minority carriers. Energy diagram: Doping Process - an additional energy level is created in the energy band structure. n-type: - an additional energy level (donor energy level) is created near the conduction band. p-type: - an additional energy level (acceptor energy level) is created near the valence band. 3 7. Discuss the bias of a diode. Forward Bias - conditions that allows current through the pn junction. - The negative terminal of VBIAS is connected to n-region, while positive terminal connected to p-region. - VBIAS must be greater than the barrier potential. - The negative side of the bias-voltage source ‘pushes’ free electrons toward the pn junction. - Provides electron current and hole current. - As more electrons flow into the depletion region, the number of positive ions is reduced, and as more holes effectively flow into depletion region on the other side of the pn junction, the number of negative ions reduced. - Cause depletion region narrow. - In terms of barrier potential, free electrons provided enough energy to climb the energy hill and cross the depletion region. - Efree e = VBARRIER. - Additional small voltage drop occurs across p and n region due to internal resistance – dynamic resistance (doped semiconductive materials). Reverse Bias - conditions that essentially prevents current through the diode. - Positive terminal of VBIAS is connected to n region while the negative terminal is connected to p region of diode. - Because unlike charges attract, the positive side of the bias-voltage source ‘pulls’ the free-electrons away the pn junction – additional positive ions are created. - In the p-region, electrons from negative side of voltage source enter as valence electrons and move from hole to hole toward the depletion region and create additional negative ions. - Cause the depletion region widens. - Reverse current exist after the transitions current dies out- cause by the minority carriers in the n and p regions – produced thermally by electronhole pairs. - VREVERSE-BIAS = VBREAKDOWN , reverse current drastically increase and will damage the diode. 4 8. Analyze the voltage-current (V-I) characteristic curve of a diode. And what happens to the barrier potential when the temperature increases? FORWARD-BIASED - Forward current IF increases very little (mA) upward along the vertical axis until the forward voltage VF across pn junction reaches approximately 0.7V at the knee of the curve. - After this point (0.7V), IF increases rapidly but VF remains at approximately 0.7V. - Normal operation for forward-biased diode is above knee of the curve. - The resistance of the forward-biased is not constant over the entire curve – called dynamic or ac resistance, r’d. REVERSE-BIASED - Reverse voltage VR increases to the left along the horizontal axis, and the reverse current IR increases downward along the vertical axis. - Very little reverse current (µA or nA) until VR reaches approximately VBR at the knee of the curve. - After this point (VBR), IR increases rapidly resulting overheating and possible diode damage but VR remains at approximately VBR. The barrier potential decreases as temperature increases. 5 9. Discuss the operation of diodes and explain the three diode models. Discuss basic operation of diode - Diode is a single pn junction device with conductive contacts and wire leads connected to each region. Part of diode is n-type and other part is ptype semiconductor, respectively. - Diode is in forward-biased when voltage source is connected with the positive terminal at anode and negative terminal at cathode, respectively. - IF is from anode to cathode. - Diode in reverse-biased when the voltage source positive terminal at cathode and negative terminal at anode, respectively. Ideal Model - is a simple switch, with forward-biased for closed (on) switch and reversebiased in open (off) switch as shown in Figure 1 below. Fig. 1 (Ideal Diode Model) - Barrier potential, forward dynamic resistance and reverse current are neglected. VF = 0 because barrier potential and dynamic resistance are neglected. The forward current determined by Ohm’s law, IF = VBIAS / RLIMIT. IR = 0, since reverse current is neglected. The reverse voltage equals the bias voltage, VR = VBIAS Use the ideal model to troubleshoot or trying to figure out the operation of a circuit and are not concerned with more exact values of voltage or current. 6 Practical Model - Adds the barrier potential in series to the ideal switch model. When the diode is forward-biased, it is equivalent to closed (on) switch. - When diode in reverse, the switch is open (off) and just as in the ideal model. Fig. 2 below shows the practical model of diode. Fig. 2 (Practical Diode Model) - - Forward-dynamic resistance and reverse current are neglected. The diode assumed to have a voltage across it (right-shift from Fig. 2(c)) when forward-biased (VF = 0.7V (silicon) and VF = 0.3V (Ge)). The barrier potential does not affect reverse-bias, so it is not a factor. The forward current determine by Kirchhoff’s voltage law and − VF V I F = BIAS RLIMIT The diode assumed to have zero reverse current, IR = 0 and VR = VBIAS. Complete Model - complete model of a diode consists of the barrier potential, the small r’d, and large internal reverse resistance r’R. - Reverse current taken into account because it provides a path for the reverse current which is included in this diode model. - Fig. 3 shows the complete diode model. Fig. 3 (Complete Diode Model) 7 - - When diode is forward-biased, it acts as a closed (on) switch in series with the barier potential voltage and the small r’d. When diode in reverse, it acts as open switch (off) in parallel with the large r’R. Barrier potential does not affect reverse bias, so it is not a factor. The diode assumed to have a voltage across it when forward-biased (right-shift), since barrier potential and r’d are included. This voltage, VF, consists of the barrier potential voltage plus the small voltage drop across the dynamic resistance. Curve slopes because the voltage drop due to dynamic resistance increases as the current increases. The reverse current is indicated by the portion of the curve to the left of the origin. For the complete model of diode: V F = 0.7V + I F r ' d and I F = (VBIAS − 0.7V ) /( RLIMIT + r ' d ) 10. Assume that the diode in Fig. 1-36(a) (Flyod, pg. 32: Example 1-1) fails open. What is the voltage across the diode and the voltage across the limiting resistor? Diode is defective, then VD = 10V, thus no current, VLIMIT = 0V 11. Define the following term: (i) Laser = Light Amplification by Stimulated Emission of Radiation (ii) LED = Light Emitting Diode (iii) Varaktor = A variable Capacitance Diode 12. Sketch the symbols of the following diodes: (i) Varaktor : (ii) PIN 8 Part B : Calculation 01.(a) Determine the forward voltage and forward current for the germanium. Diode, if given Rlimit = 0.5 kΩ and Vbias = 20 V for each of the diode models. Also find the voltage across the limiting resistor in each case. Assume rd’ = 100 Ω at the determined value of forward current. Answer : a. Ideal model : VF = 0 V IF = Vbias 20V = = 40mA Rlim it 0.5 × 10 3 Ω Practical model : V F = 0.3 V IF = Vbias − VF (20 − 0.3)V = = 39.4 mA Rlim it 0.5 × 10 3 Ω V RLIMIT = I F RLIMIT = (39.4 × 10 −3 A) × (0.5 × 10 3 Ω) = 19.7 V Complete model : V − VF (20 − 0.3)V I F = bias = = 32.83 mA ' Rlim it + rd (0.5 × 10 3 Ω) + 100Ω V F = 0.3 + I F rd' = 0.3 + 32.83 10 −3 A × 100Ω = 3.583 V V RLIMIT = I F RLIMIT = 32.83 × 10 −3 A × 0.5 × 10 3 Ω = 16.415 V 9 (b) Determine the reverse voltage and reverse current for the silicon. Diode, if given Rlimit = 2 kΩ and Vbias = 8 V for each of the diode models. Also find the voltage across the limiting resistor in each case. Assume IR = 2 µA at the determined value of forward current. b. Ideal model : IR = 0 A V R = Vbias = 8 V V RLIMIT = 0 V Practical model : IR = 0 A V R = Vbias = 8 V V RLIMIT = 0 V Complete model : I R = 2 µA V R = I R RRLIMIT = 2 × 10 −6 A × 2 × 10 3 Ω = 4 mV V R = Vbias − VRLIMIT = 8V − 0.004V = 7.996 V 02. Determine the peak value of the output voltage for half–wave rectifier (use silicon diode), if given the input voltage is 400 V and the turn ratio is 0.8. Answer : V p ( pri ) = V p (in) = 400 V V p (sec ) = n V p (in) = 0.8 × 400 V = 320 V V p (out ) = V p (sec) − 0.7V = 320V − 0.7V = 319.3 V 10 03. Derive the average value of the half-wave rectifier output voltage (Vavg) as given in equation : Vp Vavg = π where : Vp is the peak value of the voltage Answer : The average value of a half-wave rectifier sine wave is the area under the curve divided by the period (2π). The equation for a sine wave is: V = V p sin θ V AVG = π Vp area 1 = V p sin θ dθ = − cos θ ∫ 2π 2π 0 2π ( [(− cos 2π V AVG = Vp V AVG = Vp π ) − (− cos 0)] = Vp 2π )] π 0 [− (−1) − (−1)] = Vp 2π [2] = π 11 04. Based on the filtered bridge rectifier in Figure 1, answer the following questions (All diodes are 1N4001). (i). Determine the peak primary voltage, the peak secondary voltage, and the unfiltered full-wave rectified peak voltage. Answer : Vp (pri) = √2 Vrms = √2 x 100V = 141.42 V Vp (sec) = n Vp (pri) n = Nsec/Npri = 1/10 = 0.1 Vp (sec) = n Vp (pri) = 0.1 x 141.42V = 14.142 V Vp (out) = V (rect) = Vp (sec) – 1.4V = 14.142 V– 1.4 V = 12.742 V (ii). Determine the peak to peak voltage ripple Vr(pp) at the output and the value of the dc output voltage,.VDC Answer : The frequency of a full-wave rectified voltage become is 120 Hz (from 60 Hz (frequency input) x 2 = 120 Hz). 1 1 Vr ( pp) = V p (rect ) = 12.742V = 2.94 V f RL C 120 Hz × 1.2 × 10 3 Ω × 30 × 10 −6 F ( ) ( ) ⎛ ⎞ 1 1 ⎞ ⎟⎟V p (rect ) = ⎛⎜1 − V DC = ⎜⎜1 − ⎟ 12.742V = 11.27 V 3 −6 2 f R C 2 120 Hz 1 . 2 10 30 10 F × × × Ω × × ⎝ ⎠ L ⎝ ⎠ (iii). Calculate the ripple percentage? Answer : V ( pp ) 2.94V r= r = = 0.26 VDC 11.27V The percentage ripple is 26 % (iv). How to reduce the ripple percentage in Figure 1? (i) increase the value of the filter capacitor (ii) increase the load resistance Figure 1 12 05. Figure 2 A loaded zener regulator is shown in Figure 2. Given VZ = 8 V at IZT = 0.05 A, IZK = 0.003 A, ZZ = 20 Ω and IZM = 0.01 A. Determine the maximum permissible load currents. Answer : VZ (min) = VZ − ∆I Z Z Z = 8V − (0.05 A − 0.003 A)(20Ω) = 7.06 V V R = Vin − VZ (min) = 25V − 7.06V = 17.94 V VR 17.94V = = 179.4 mA R 100 I L (max) = I T − I ZK = 0.1794 − 0.01 = 0.1694 A = 169.4 mA IT = 13