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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×ππππ /πππ . If the intrinsic carrier concentration is 1.5×πππ /πππ .Then calculate the hole concentration. Given that electron concentration n= 2.15×1010 /ππ3 intrinsic carrier concentration ππ =1.5×108 /ππ3 according to the law of mass action n×p = ππ 2 hole concentration p = ππ 2 n = 108 2 ) ππ3 2.15×1010 /ππ3 (1.5× p = 1.046 ×106 /ππ3 b) The intrinsic carrier density at 300K is 2.5×ππππ /πππ , in silicon. For p-type silicon dopped to 1.25×ππππ atoms/πππ 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. π πΏ = 12π£β0.3π£β0.7π£ 2 mA π πΏ = 12π£β1π£ 2 mA = 11π£ 2 mA = 5.5kΞ©. d) What is the ripple, 3.5 π½πΉπ΄πΊ on average of 120 V? Given that R.M.S value of ac component ππ ππ = 3.5π Average value of output voltage πππ or πππ£π =120V ᡧ ππ ππ Ripple factor ( ) = πππ = 3.5 π 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 πππΏ = 36v Full load voltage ππΉπΏ = 30v Voltage regulation = = = πππΏ β ππΉπΏ πππΏ 36π£β30π£ 36v 6π 36 V =0.1667 f) If αΌ is 0.98, π°π© is 100 ΞΌA ,π°πͺπΆ is 6 ΞΌA .then find out π°πͺ . Ξ²= Ξ²= αΌ ( 1βαΌ ) = 0.98 ( 1β0.98 ) = 98 2 =49. πΌπΆ πΌπ΅ ο° πΌπΆ = Ξ² πΌπ΅ = (49)(100 ΞΌA) = 4.9mA. ο° πΌπΈ = πΌπΆ + πΌπ΅ = (4900+100) ΞΌA = 5000 ΞΌA =5mA. For common base transistor configuration: π°πͺ πππ πππππππ = π°πͺ ππππππππ + π°πͺ ππππππππ Where asπ°πͺ ππππππππ = αΌπΌπΈ and π°πͺ ππππππππ = πΌπΆπ΅π πΌπΆ = αΌπΌπΈ + πΌπΆπ΅π = (0.98×5mA)+(6 ΞΌA) =4.906mA. g) Differentiate between fixed bias and self bias. Fixed Bias Self Bias 1. S=Ξ² + 1. 1. S<(Ξ² + 1). It has poor stability due to large stability factor. Stability is improved i.e. better than fixed biasing. 2.Absence of π πΈ 2. presence of π πΈ 3.Due to large value of s as Ξ² is greater ,circuit 3.Due to less value of s which increases the is unstable. stability. h) βFET is known as voltage controlled deviceβ justify the statement. It is called as voltage controlled device because the output drain current (πΌπ· ) is controlled by input gate to source voltage (ππΊπ ). i) What do you mean by intrinsic standoff ratio of UJT? The Greek letter Ξ· (eta) is called the intrinsic stand-off ratio of the device and is defined by Ξ·=π π΅1 π π΅1 + π π΅2 | πΌπΈ = 0 = π π΅1 π π΅B 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 π¬π located in the energy band of silicon, at 300K with p=ππππ ππβπ and π΅π½ =1.40×ππππ . p= ππ exp [ ] β(πΈπΉ β πΈπ ) πΎπ p=ππ = 1014 ππβ3 and ππ =1.40×1019 ο° exp ο° [ [ ]= (πΈπΉ β πΈπ ) πΎπ (πΈπΉ β πΈπ ) πΎπ ] =ln( ππ p ππ p ) ο° (πΈπΉ β πΈπ ) = KT ln( ππ p ) 1.40×1019 ο° (πΈπΉ β πΈπ ) = (26πππ)ln( 14 β3 ) 10 ππ ο° (πΈπΉ β πΈπ ) = 0.31eV. Hence πΈπΉ is above πΈπ by 0.31eV b) Define mobility, conductivity and diffusion? Mobility :It is defined as the ratio between the particle drift velocity (π½π ) per unit electric field (E) . Mobility = π½π E Conductivity :reciprocal of resistivity. It is the measure of semiconductorβs 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 Ξ©-cm and hall coefficient of 100 πππ /πππππππ .assuming only electrons as carriers determine the electron density and mobility. Given that resistivity = 0.1 Ξ©-cm hall coefficient π π» = 100 ππ3 /πππ’ππππ electronic charge (e) = 1.6×10β19 πππ’ππππ. ο° Conductivity = 1 resistivity ο° Electron density = 1 π π» = = 1 0.1 Ξ©βcm 1 100 ππ3 /πππ’ππππ =10 S/cm. = 0.01coulomb/ππ3 . ο° Mobility = (Conductivity)×(π π» ) =(10)(100) =(1000 S ππ2 )/ coulomb. b) Neatly draw and explain the V-I characteristics of a tunnel diode? ο· ο· ο· ο· ο· Here forward current (πΌπΉ ) increases sharply as applied voltage (ππΉ ) increases. πΌπΉ increases upto the point βAβ on the curve i.e, the peak voltage. As the F/B is increases beyond this point ,πΌπΉ decreases and continuous to drop until the point B is reached. This is known as βValley Voltage ππ£ β. 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. π½π π½π = π΅π π΅π =>π½π = (π΅π΅π)π½π = ( π 1 )(230)=46v = π½π 5 (π½π ) Hence the peak value of the secondary voltage (π½π )πππ = πππ β2 (π½π )πππ =>(π½π )πππ = β2 46 = β2 (π½π )πππ = 32.5269v 2 A.C input power = πΌπππ × (πΉπ³ + πΉπ ) = π·ππ(π.π) = π·ππ(π.π) = πΌπππ₯ 2 π ( ×πΉπ³ = ( πππ― ππππ )2 × (π½π ) πππ πΉπ³ )2 × πΌπππ₯ 2 π × (πΉπ³ + πΉπ) πΉπ³ 2 ππππ 2 =>π·ππ(π.π) = 1.3225watt Rectification efficiency (Ξ·) = π·π π(π/π) π·ππ(π/π) = πΌππ 2 × πΉπ³ πΌπππ 2 × (πΉπ³ + πΉπ ) = ( 2 πΌ )2 × πΉπ³ 3.14 max πΌπππ₯ 2 × (πΉπ³ + πΉπ ) π =>Ξ· = πβπΉπ³ (π©π’βπ©π’)×(πΉπ³ + πΉπ ) =>Ξ· = π (π©π’βπ©π’)×(π + πΉπ /πΉπ³ ) Assuming diode is ideal πΉπ³ β« πΉπ =>Ξ· = 0.πππ (π + πΉπ /πΉπ³ ) × 100% = 0.812 × 100% = 81.2% We know Rectification efficiency (Ξ·) = Dc power π·π π(π/π) = π·π π(π/π) π·ππ(π/π) π·ππ(π/π) × = 0.812 0.812 = 0.812 × π. πππππ°πππ π·π π(π/π) = 1.0738watt Form factor (πΎπΉ ) = π°πππ π°πππ π° ( πππ ) = β2 ππ°πππ ( pi ) = π π©π’β2 = 1.11 Ripple factor (ᡧ ) = β((πΎπΉ )2 ) β 1 = β(1.11 β 1.11) β 1 = 0.482 b) With a neat sketch, explain the zener regulator circuit. ο· ο· ο· ο· ο· ο· 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 β voltage regulator β or β zener regulator β. It is required to provide a constant voltage across load resistancs (π πΏ ) where the input voltage may be varying over a range. Zener diode is reverse bias and as long as input voltage πππ not less than ππ§ . 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 ΞΌf .The connected load is of 8kΞ© determine, 1) ripple voltage and 2) form factor. Given c= 45 ΞΌf , π πΏ = 8kΞ©, f = 50Hz As per formula Ripple factor (ᡧ ) = πππ(πππ ) πππ Ripple factor (ᡧ ) = = ππ 2 β3πΌππ π πΏ 1 4β3f Cπ πΏ = 0.00801875 Maximum voltage across the load We know that πππ = ππΏ πππ₯ - ( πΌππ π πΏ + ( πΌππ (8000 + πΌππ 4fC (ππΏ )πππ₯ = (ππ )πππ₯ = 50v πΌππ 4fC ) ) = ππΏ πππ₯ 1 4β50β45β 10β6 ) = 50 πΌππ = 0.00616438 Amp = 6.164 mA πππ = πΌππ * π πΏ = 49.31v Ripple voltage ππ = (πππ )πππ = ππ 2β3 πΌππ 2fC =1.3697v = 0.3953 Ripple factor (ᡧ ) = πππ(πππ ) πππ = 0.3953v 49.312v = 0.0080163 Ripple factor (ᡧ ) =β((form factor)2 ) β 1 ᡧ = β((πΎπΉ )2 ) + 1 πΎπΉ = β((ᡧ)2 ) + 1 πΎπΉ = β((0.0080163)2 ) + 1 πΎπΉ = β1.00006426 = 1. b) Differentiate between Half wave and full wave rectifier. Half waverectifier S.No Particulars 1. 3. Number of diodes Transformer required πΌππ 4. πΌπππ 2. 5. 6. Peak Inverse Voltage Ξ· Half waverectifier 1 Not essential Full wave rectifier Center tapped full wave rectifier 2 Essential Bridge type full wave rectifier 4 Not essential πΌππ΄π π 2πΌππ΄π π 2πΌππ΄π π πΌππ΄π 2 πΌππ΄π πΌππ΄π β2 β2 πππ΄π 2πππ΄π πππ΄π 40.6% 81.2% 81.2% 6. a) In a voltage divider biasing circuit of BJT, R1 =47kΞ© ,R2=82k Ξ© ,Rc=2.2kΞ© ,Re=1.2kΞ© ,and Ξ² =352,Vcc=12V.Draw the load line for the above biasing circuit and locate the Q-point that is Ic and ππΆπΈ . Given that R1 =47kΞ© , R2=82k Ξ© , Rc=2.2kΞ© , Re=1.2kΞ© ,and Ξ² =352, Vcc=12V. Exact Analysis: on applying thevenins theorem, we have π ππ» = (π 1 ||π 2 ) = (47kΞ© ||82k Ξ©) =29.875kΞ© πππ» = πππ × π 2 (π 1 +π 2) = 7.627V On applying the K.V.L to the input loop equation we have πππ» β π ππ» πΌπ΅ β ππ΅πΈ β π πΈ πΌπΈ = 0 ο° πππ» β π ππ» πΌπ΅ β ππ΅πΈ β π πΈ πΌπ΅ (1 + π½) = 0 ο° πΌπ΅ = πππ» βππ΅πΈ π ππ» +(1+π½)π πΈ = 0.0153mA = 15.3159ΞΌA . ο° πΌπΆ = πΌπ΅ π½ = 5.391mA On applying the K.V.L to the output loop equation we have ππΆπΆ β π πΆ πΌπΆ β ππΆπΈ β π πΈ πΌπΈ = 0 ο° ππΆπΈ = ππΆπΆ β πΌπΆ (π πΆ + π πΈ ) ο° ππΆπΈ = (β6.33)π As βve signifies that collector to emitter is reverse bias . Again from the output loop equation we know that ππΆπΆ β π πΆ πΌπΆ β ππΆπΈ β π πΈ πΌπΈ = 0 ππΆπΈ = ππΆπΆ β πΌπΆ (π πΆ + π πΈ ) πΌπΆ(π ππ‘) = ππΆπΆ (π πΆ +π πΈ) | ππΆπΈ = 0 | ππΆπΈ = ππΆπΆ πΌπΆ = 0 Point A (0,(π ππΆπΆ πΆ +π πΈ) ) Therefore Point A = (0, 3.529mA) Point B (ππΆπΆ , 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. ο· ο· ο· ο· ο· ο· ο· ο· ο· ο· ο· ο· Upto the peak point P, the diode is reverse bias and hence, the region to the left of the peak point is called βcutoff regionβ. 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 βsaturation regionβ. 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 βUJT Relaxation Oscillatorβ. It consists of a UJT and a capacitor which is charged through π πΈ as the supply voltage ππ΅π΅ is switched on. The voltage across the capacitor increases exponentially and ππ = ππ 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 ππ =0 , the device is cutoff and capacitor πΆπΈ starts to charge again. This cycle is repeated continuously generating a sawtooth waveform across πΆπΈ . 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: ο· ο· ο· ο· 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. π= π πΏ ππΏ3β2 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 π πΏ 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 βonβ 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: ο ο ο ο ο ο ο ο ο Identifying the terminals of diode Identifying the terminals of transistor Measuring voltage Finding knee voltage Whether the transistor is PNP or NPN Ξ² value of transistor To find exact resistor values Measuring current Depending upon the resistance values we can identify the terminals of transistor.