University:
Kentucky State University
Course:
PHY 212 | General Physics II
Academic year:

2021
Views:

80

Pages:

21
Author:

Kundulu8mbj

Electronics Lab Questions Solve the answers below Question No.1 1-The main difference between clipper and clamper is the presence of: Capacitor 2-The transistor has Vc=4V ,Vb=3V and Ve=5V , then the transistor is biased in Vb< Vc Reverse Vb< Ve Reverse Cut off mode 3- The term BJT is short for bipolar junction transistor 4- Compared to HWR,the FWR has Full wave Rectifier has: low ripple voltage=VP/2RFC =1/2*( low ripple voltage half wave) low ripple voltage 5- A certain transistor has IC = 15 mA and IB = 167 μA; DC is Β=IC/IB =15mA/167 μA= 90 6- is an electronic circuit that changes the DC level of a signal to the desired level without changing the shape of the applied signal Clamper 7- The circuit which changes the shape of AC input signal is called Clipper 8- The transistor has Vc=2V ,Vb=3V and Ve=1V , then the transistor is biased in Vb>Vc Forword Vb> Ve Forword Saturation mode 9- Id=3mA ,Vd=0.66V and when Id=3.5mA , Vd=0.7V, the Dynamic or AC Resistance rac= = Δv/ Δi =0.7- 0.66 /3.5 mA - 3 mA = 80 Ω 10- Which of the following diodes is reverse biased A 11-What are the two types of bipolar junction transistors npn and pnp 12- The transistor has Vc=4V ,Vb=3V and Ve=1V , then the transistor is biased in Vb Ve Forword Foreword Active mode 13- In which of the following figures, junction diode is forward biased B Question No.2 1- For the circuit shown below consider Vɣ=0.6 then fill the table 2) the RD at 3V RD=VD/IF=0.6/2.4*10^-3=250 3) the rD at 1V rD= Δv/ Δi=VD(3)-VD(0.9)/IF(3)-IF(0.9)=0.6-0.6/2.4-3=0 Question No.3 1) For the circuit shown below ,the Vo Avg is VP-Vɣ/π= 5-0.6/ π= 1.4 2) For the circuit the average power dissipated in the diode is V=IR 0.6=I*2000 I=0.3mA P=VI P=1.4*0.3Ma=0.42mW Question No.4 1) For the circuit shown below , What is the name of the circuit Clipper 2) the output wave form (Vɣ=0.7) A Question No.5 Question No.6 1) For the circuit shown below , What is the name of the circuit ? (Ideal diode) Clamper 2) the output wave form C Question No.7 1) For the circuit shown below (Ideal diode), the circuit represents full wave rectifier 2) The average value of the output signal is =2Vinp /π 2*5/ π = 3.18v 3) the output wave form B Question No.8 1) For the following circuit below fill the table below B Question No.9 1) For the circuit shown below assume (Ideal diode), the circuit represents half wave rectifier 2) the output wave form A 3) The average voltage V0 avg=Vin/ π 20/ π=6.366v 4) The average current in the diode Id avg= V0 avg/R load 6.366/1000= 6.36mA Question No.10 Question No.11 1) If β=200 , α equal= α= β/1+ β 200/1+200=65.66 2) If α=0.99 , β equal= β = α/1- α 0.99/1-0.99=99 3) If β=100 and Ic=30mA , IB equal IB = Ic/ β 30mA/100 = 300 μA Question No.12 1) V Average in full wave formula V Average =2(VP-2Vɣ)/π 2) V Average in half wave (Ideal diode) V Average =VP/π 3) V Average in half wave (offest diode) formula V Average =VP-Vɣ/π Question No.13 Question No.14 1) What type of diode circuit is used to add or restore a dc level to an electrical signal Clamper 2) An open circuit can have any voltage across its terminals, but the current is always 0A 3) Refer to the figure given below. Which diode arrangement will supply a negative output voltage? c 4) What best describes the circuit Half-wave rectifier 5) Which diode arrangement will supply a positive output voltage A 6) When a transistor is used as a switch, it is stable in which two distinct regions saturation and cutoff 7) Which of the following is true for an npn or pnp transistor IE = IB + IC 8) DC power should be connected to forward bias a diode as follows +anode, – cathode 9) The average value of a half-wave rectified voltage with a peak value of 200V is V Average =VP/π =200/ π=63.7 10) bride rectifier consists of 4diodes 11) Which of the following is not a necessary component in a clamper circuit Independent DC Supply Diode, Capacitor and Resistor are necessary to build a clamper circuit. An independent DC supply is required to bring an additional shift 12) For a sinusoidal input of 20 Vpeak to the given circuit, what is the peak value of the output waveform 25V Explanation: In the given circuit, the output becomes zero for v i less than -5 V. Hence, the peak value of the output is 25 V owing to the additive effect of V for vi. 13) For the given circuit for a 20 Vpeak sinusoidal input vi, what is the value of vi at which the clipping begins -5V Explanation: Considering the connection of diode, it is evident that the diode becomes reverse biased when vi < -5 V. Hence, clipping starts at -5 V. 14) The transistor has Vc=3V ,Vb=4V and Ve=5V , then the transistor is biased in Vb>Vc Forword Vb< Ve Reverse Reverse active mode 15) What is the peak-to-peak voltage value (VPP +10V-(-10V)=20V 16) is the conversion of alternating current (AC) to direct current (DC) Rectifier 17) The basic rectifier circuits are A+B+C 18) choose the correct answer A (NPN) - B (PNP) Question No.15 1) For normal operation of a pnp BJT, the base must be ______ with respect to the emitter and ______ with respect to the collector negative, positive 2) βDC is the ratio of IC to IE False 3) Refer to this figure. Determine the minimum value of IB that will produce saturation.Use VCE = 0.2V Apply kirchoff's voltage law in the output side 10=4.7*10^3*Ic+0.2 on solving Ic=2.085 ma. For common emmiter configuration we know Ic=b*Ib where b=200 So, Ib = 2.085/200 = 10.425μA 4) In this circuit βDC = 100 and VIN = 8 V. The value of RB that will produce saturation is: (Vce = 0.2 v) (vbe=0.7 v) Then apply kcl for output circuit. 20-0.2/2.5 = 7.92mA = ic ic = bib ib = ic/b = 7.92/100 = 0.0792. Then kvl for input side. 8 - 0.7-0.0792 * rb = 0, rb = 8-0.7/0.0792 = 92kohms Question No.16 1) Choose the correct answer below A B C D E F G function generator capacitor breadboard digital multimeter resistor transistor dual trace oscilloscope 2) These lines of five holes are known as nodes 3) ripple voltage of half wave rectifier formula Vr =(VP-Vɣ)/F*R*C 4) ripple voltage of full wave rectifier formula Vr =(VP-2Vɣ)/2*F*R*C 5) choose the right answer V avg full wave > V avg half wave 6) What is the peak voltage value 10 7) Transistor biasing represents ……………. Conditions d.c 8) Operating point represents Zero signal values of IC and VCE 9) Total emitter current is IB + IC 10) ripple factor of half wave rectifier formula %r = ( Vr / Vavg ) *100% 11) The ripple factor of a full-wave rectifier circuit compared to that of a half wave rectifier circuit without filter is less than half that for a half-wave rectifier circuit 12) Shunting the ac component away from the load is the task of a Filter 13) From the given I-V characteristics of a silicon diode, what is the approximate value of rav between marked points Explanation: rav is the average AC resistance, found between two marked points in the I-V graph. Here rav=0.1/14mA=7Ω. 14) What is the name of the figure below zener diode 15) In the image below, the diode characteristics diagram Answer the questions below A B reverse forward

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