CENTURION UNIVERSITY OF TECHNOLOGY AND MANAGEMENT SCHOOL OF ENGINEERING & TECHNOLOGYDEPARTMENT OF ELECTRONICS & COMMUNICATION ENGINEERING ELECTRONIC DEVICES Section: ECE SEM: II PART-A 1. a) In a N-type silicon sample, the electron concentration 2.15 10 10 /cm 3. If the intrinsic carrier concentration is 1.5 10 8 /cm 3.Then calculate the hole concentration. Given that electron concentration n 2.15 10 10 /cm 3 intrinsic carrier concentration n i 1.5 10 8 /cm 3 according to the law of mass action n p n i 2 hole concentration p n i 2 n (1.5 108 cm 3)2 2.15 10 10 /cm 3 p 1.046 10 6 /cm 3 b) The intrinsic carrier density at 300K is 2.5 10 15 /cm 3, in silicon. For p-type silicon dopped to 1.25 10 12 atoms/cm 3 then find out the equilibrium electron and hole density. The question mentioned is wrong because the given doping concentration is less than intrinsic concentration which is practically impossible. Mathematically,if we calculate then the semiconductor will become N-type c) Determine the value of load resistor. R L 12v 0.3v 0.7v 2 ma R L 12v 1v 2 ma 11v 2 ma 5.5kΩ.
d) What is the ripple, 3.5 V RMS on average of 120 V? Given that R.M.S value of ac component V RMS 3.5V Average value of output voltage V dc or V avg 120V Ripple factor ( ᵧ) V RMS V dc 3.5 V 120 V 0.029 e) A dc voltage supply provides 36v when the output is unloaded. When connected to a load the output drops to 30v. Calculate the value of voltage regulation. No load voltage V NL 36v Full load voltage V FL 30v Voltage regulation V NL V FL V NL 36v 30v 36v 6 V 36 V 0.1667 f) If ἀ is 0.98, I B is 100 μa,i CO is 6 μa.then find out I C. β ἀ ( 1 ἀ ) 0.98 ( 1 0.98 ) 98 2 49. β I C I B I C β I B (49)(100 μa) 4.9mA. I E I C + I B (4900+100) μa 5000 μa 5mA. For common base transistor configuration: I C net current I Cmajority + I Cminority Where asi C majority ἀi E and I C minority I CBO I C ἀi E + I CBO (0.98 5mA)+(6 μa) 4.906mA.
g) Differentiate between fixed bias and self bias. Fixed Bias 1. Sβ + 1. It has poor stability due to large stability factor. 2.Absence of R E 3.Due to large value of s as β is greater,circuit is unstable. Self Bias 1. S<(β + 1). Stability is improved i.e. better than fixed biasing. 2. presence of R E 3.Due to less value of s which increases the stability. h) FET is known as voltage controlled device justify the statement. It is called as voltage controlled device because the output drain current (I D ) is controlled by input gate to source voltage (V GS ). i) What do you mean by intrinsic standoff ratio of UJT? The Greek letter η (eta) is called the intrinsic stand-off ratio of the device and is defined by R B1 η I R B1 + R E 0 R B1 B2 R BB j) What the different modes of operation of BJT? a. saturation mode b.active mode c. cut off mode
PART-B 2. a) Where is E F located in the energy band of silicon, at 300K with p10 14 cm 3 and N V 1.40 10 19. p N V exp [ (E F E V ) KT ] pn a 10 14 cm 3 and N V 1.40 10 19 exp[ (E F E V )] N V KT p [ (E F E V )] ln( N V ) KT p (E F E V ) KT ln( N V p ) (E F E V ) (26meV)ln ( 1.40 1019 (E F E V ) 0.31eV. Hence E F is above E V by 0.31eV 10 14 cm 3 ) b) Define mobility, conductivity and diffusion? Mobility :It is defined as the ratio between the particle drift velocity (V d ) per unit electric field (E). Mobility V d E Conductivity :reciprocal of resistivity. It is the measure of semiconductor s ability to conduct current. Depends on mobility of charge carriers and dopant concentration. Diffusion :It is a process of doping a semiconductor with impurities.it is the net movement of a substance (ae., an atom, ion or molecule) from a region of high concentration to a region of low concentration.
3. a) A sample of N-type semiconductor has a resistivity, of 0.1 Ω-cm and hall coefficient of 100 cm 3 /coulomb.assuming only electrons as carriers determine the electron density and mobility. Given that resistivity 0.1 Ω-cm hall coefficient R H 100 cm 3 /coulomb electronic charge (e) 1.6 10 19 coulomb. Conductivity Electron density 1 1 1 resistivity 0.1 Ω cm 1 R H 10 S/cm. 100 cm 3 /coulomb 0.01coulomb/cm3. Mobility (Conductivity) (R H ) (10)(100) (1000 S cm 2 )/ coulomb. b) Neatly draw and explain the V-I characteristics of a tunnel diode? Here forward current (I F ) increases sharply as applied voltage (V F ) increases. I F increases upto the point A on the curve i.e, the peak voltage. As the F/B is increases beyond this point,i F decreases and continuous to drop until the point B is reached. This is known as Valley Voltage V v. At B, the current starts to increase once again and does so rapidly as bias is increases further. Beyond this point, the Tunnel Diode will behaves as a P-N junction Diode.
4. a) A 230v 50Hz ac voltage is applied to the primary of 5:1 step down transformer, which is used in center tapped rectifier, having load resistance 800ohm assuming the diodes to be an ideal then determine, ac power input, dc power output,rectification efficiency and ripple factor. V 1 V 2 N 1 N 2 >V 2 ( N 2 N 1 )V 1 ( 1 5 )(230)46v V s Hence the peak value of the secondary voltage (V s ) rms (V s ) max 2 (V s ) rms 32.5269v >(V s ) rms (V s) max 2 46 2 A.C input power I 2 rms (R L + R F ) I max 2 (R L + R F ) 2 P in(a.c) I max 2 R L ( (V s ) max 2 R L ) 2 R L 2 P in(a.c) ( 46v 800Ω )2 800Ω 2 >P in(a.c) 1.3225watt Rectification efficiency (η) P dc(o/p) >η >η I 2 dc R L P ac(i/p) I 2 rms (R L + R F ) 8 R L (pi pi) (R L + R F ) ( 2 3.14 I max )2 R L Imax 2 (R L + R F ) 8 (pi pi) (1 + R F /R L ) 2 Assuming diode is ideal R L R F
>η 0.812 (1 + R F /R L ) 100% 0.812 100% 81.2% We know Rectification efficiency (η) P dc(o/p) P ac(i/p) 0.812 Dc power P dc(o/p) P ac(i/p) 0.812 0.812 1. 3225watt P dc(o/p) 1.0738watt Form factor (K F ) I rms I avg ( I max 2 ) ( 2I max pi ) pi 2 2 1.11 Ripple factor (ᵧ ) ((K F ) 2 ) 1 (1.11 1.11) 1 0.482 b) With a neat sketch, explain the zener regulator circuit. Under reverse bias condition,the voltage across the diode remains almost constant although the current through the diode increases. Hence, the voltage across the zener diode can be used as a reference voltage which can be used as voltage regulator or zener regulator. It is required to provide a constant voltage across load resistancs (R L ) where the input voltage may be varying over a range. Zener diode is reverse bias and as long as input voltage V in not less than V z. Voltage across the diode will be constant. Load voltage will also constant.
5. a) A full wave rectifier has a peak output voltage of 50 volts at 50 Hz and uses a shunt capacitor filter with c 45 μf.the connected load is of 8kΩ determine, 1) ripple voltage and 2) form factor. Given c 45 μf, R L 8kΩ, f 50Hz As per formula Ripple factor (ᵧ ) V ac(rms) V dc V r 2 3I dc R L Ripple factor (ᵧ ) 1 4 3f CR L 0.00801875 Maximum voltage across the load (V L ) max (V s ) max 50v We know that V dc V L max - ( I dc 4 f C ) I dc (8000 + I dc R L + ( 1 4 50 45 10 I dc 4 f C ) V L max 6 ) 50 I dc 0.00616438 Amp 6.164 ma V dc I dc * R L 49.31v (V ac ) rms Ripple voltage V r V r 2 3 0.3953 I dc 2 f C 1.3697v Ripple factor (ᵧ ) V ac(rms) V dc 0.3953v 49.312v 0.0080163 Ripple factor (ᵧ ) ((form factor) 2 ) 1 ᵧ ((K F ) 2 ) + 1 K F ((ᵧ) 2 ) + 1 K F ((0.0080163) 2 ) + 1 K F 1.00006426 1.
b) Differentiate between Half wave and full wave rectifier. Half waverectifier Full wave rectifier S.No Particulars Half waverectifier Center tapped full wave rectifier Bridge type full wave rectifier 1. Number of 1 2 4 diodes 2. Transformer Not essential Essential Not essential required 3. I dc I MAX π 2I MAX π 2I MAX π 4. I rms I MAX 2 I MAX 2 I MAX 2 5. Peak Inverse V MAX 2V MAX V MAX Voltage 6. η 40.6% 81.2% 81.2% 6. a) In a voltage divider biasing circuit of BJT, R1 47kΩ,R282k Ω,Rc2.2kΩ,Re1.2kΩ,and β 352,Vcc12V.Draw the load line for the above biasing circuit and locate the Q-point that is Ic and V CE.
Given that R1 47kΩ, R282k Ω, Rc2.2kΩ, Re1.2kΩ,and β 352, Vcc12V. Exact Analysis: on applying thevenins theorem, we have R TH (R 1 R 2 ) (47kΩ 82k Ω) 29.875kΩ V TH V cc R 2 (R 1 +R 2) 7.627V On applying the K.V.L to the input loop equation we have V TH R TH I B V BE R E I E 0 V TH R TH I B V BE R E I B (1 + β) 0 I B V TH V BE R TH +(1+β)R E 0.0153mA 15.3159μA. I C I B β 5.391mA On applying the K.V.L to the output loop equation we have V CC R C I C V CE R E I E 0 V CE V CC I C (R C + R E ) V CE ( 6.33)V As ve signifies that collector to emitter is reverse bias. Again from the output loop equation we know that V CC R C I C V CE R E I E 0 V CE V CC I C (R C + R E ) I C(sat) V CC (R C +R E) V CE 0
V CE V CC I C 0 Point A (0, V CC ) (R C +R E) Therefore Point A (0, 3.529mA) Point B (V CC, 0) Therefore Point B (12v,0) Therefore Point A and Point B provides the load line.
b) Draw the V-I characteristic of a UJT and explain how it could be use as an oscillator. Upto the peak point P, the diode is reverse bias and hence, the region to the left of the peak point is called cutoff region. There is a negative resistance region from peak point to valley point. After the valley point, the device is driven into saturation and behaves like a conventional forward bias pn junction diode. The region to the right of the valley point is called saturation region. In the valley point, the resistance changes from negative to positive. The resistance remains positive in the saturation region. Due to negative resistance property a UJT can be used to produce sawtooth waveform generator which is properly known as UJT Relaxation Oscillator. It consists of a UJT and a capacitor which is charged through R E as the supply voltage V BB is switched on. The voltage across the capacitor increases exponentially and V c V p and UJT starts conducting. After the peak voltage of a UJT is reached, it provides negative resistance to the discharge path which is useful in working of relaxation oscillator. As the V c 0, the device is cutoff and capacitor C E starts to charge again. This cycle is repeated continuously generating a sawtooth waveform across C E.
7. a) N-type semiconductor: A semiconductor in which electrical conduction is due chiefly to the movement of electrons. b)led: Light Emitting Diode is nothing but a P-N junction diode which emits light when it is forward bias. In all semiconductors or P-N junction diode some of energy is radiated as heat and photons. Here light is generated by recombination of electrons and holes where by the excess energy is transferred to a emitted photon. The brightness of emitted light is transferred to an emitted photon. c)avalanche Breakdown: It is due to thermally generated charge carriers. It is lightly doped. It occurs more than 5volts. Deplection layer will be wider. d) L-filter: The output of rectifier circuit contains dc and ac components. In order to get the pure form of dc we are using one of the filter circuit i.e, L-filter circuit.an inductor filter in which the rectified output signal will be the input for this filter circuit. The property of inductor will allows to pass dc and blocks ac. ν R L ωl3 2 PART-C 8. a) Write down how to measure the average value of output voltage and ripple factor by help of a digital multimeter in the laboratory. Let us assume a rectifier circuit, here to find the output voltage first we are setting the digital multimeter in voltage measurement mode and measuring the output of rectifier circuit by placing the multimeter probes (red probe and black probe) across the load resistance R L side of the rectifier circuit. b)how to measure the knee voltage and identify the terminals of an isolated diode as well as the terminals of an isolated BJT only by the help of a digital multimeter. For Diode: The small diode symbol as the bottom option of the rotating dialwhen set in this position and hooked up, the diode should be in the on state and the display will provide an indication of the forward-bias voltage such as 0.67 V (for Si). For Transistor:Digital multimeter provides the level of hfe using the lead sockets appearing at the bottom left of the dial. The choice of pnp or npn and the availability of two emitter
connections to handle the sequence of leads as connected to the casing. In the diode testing mode it can be used to check the p-n junctions of a transistor. With the collector open the base-to-emitter junction should result in a low voltage of about 0.7 V with the red (positive) lead connected to the base and the black (negative) lead connected to the emitter. A reversal of the leads should result in to represent the reverse-biased junction. Similarly, with the emitter open, the forward- and reverse-bias states of the base-to-collector junction can be checked. c) Draw the pictorial diagram of a digital multimeter with proper labelling. What are the parameters that can be measures by help of the digital multimeter Parameters that we can measure by using digital multimeter are: Identifying the terminals of diode Identifying the terminals of transistor Measuring voltage Finding knee voltage Whether the transistor is PNP or NPN β value of transistor To find exact resistor values Measuring current Depending upon the resistance values we can identify the terminals of transistor.