GOVERNMENT OF KARNATAKA KARNATAKA STATE PRE-UNIVERSITY EDUCATION EXAMINATION BOARD II YEAR PUC EXAMINATION MARCH-0 SCHEME OF VALUATION Subject Code: 0 Subject: Qn. PART - A 0. Which is the largest of three transistor currents? Emitter current OR I E. 0. Name the transistor amplifier that has current gain less than unity? Common base amplifier OR CB amplifier. 03. What kind of negative feedback increases input impedance and decreases output impedance? Voltage series feedback. 0. When does a comparator give zero output? When both the inputs are at equal voltage. 05. Why are oscillators damped in a tank circuit? Due to loss of energy in tank circuit. 06. Define single hop distance. The distance from the transmitting antenna measured along the earth surface where the first hop takes place is called single hop distance. 07. What is the value of modulation index if a carrier wave of amplitude 6V is amplitude modulated by audio signal of amplitude V? m a = V m /V c = /6 = 0.66 08. Write the Boolean expression for two input XNOR gate. Y = A B OR Y = A B + AB 09. Convert (3) 0 into excess-3 code. Decimal Nr = 3 + 3 + 3 + 3 Excess-3 code = 000 00 00 0. Expand PSTN. Public Switched Telephone Network PART - B. Obtain the relation between α and β. We have I E = I C + I B () Divide equation() by I C We get I E /I C = I C /I C + I B /I C /α = + /β Therefore α = β/( + β). A transistor is connected in CE mode. When collector to emitter voltage increases from V to 5V the collector current increases from 5 ma to 5.5 ma. Calculate the dynamic output resistance.
Qn.. r 0 = V CE / I C = (5 )/(5.5 5) x 0-3 = 3/ 0.5 x 0-3 = 6000 Ω OR 6 k Ω 3. An amplifier has a voltage gain of 50. Express the gain in decibels. Gain in db = 0 log 0 A = 0 log 0 50 = 3.5. What are the characteristics of CC amplifier? High/very high input impedance Low output impedance Voltage gain < (or unity) High current gain O/p signal is in phase with input signal Any two of the above, each carry mark 5. Distinguish between Open loop gain and closed loop gain? Open loop gain - Gain without feedback Closed loop gain - Gain with feedback 6. Mention any four characteristics of ideal Op-Amp. The open loop voltage gain is infinity (Av = ). The input impedance is infinity (Zi = ). The output impedance is zero (Z 0 = 0). The bandwidth is infinite (BW = ). The common mode rejection ratio is infinity (CMRR = ). The slew rate is infinity (SR = ). Perfect balance i.e., the output voltage is zero when both the inputs are equal. Characteristics do not drift with temperature. Any four of the above, each carry / mark 7. State Barkhausen criteria for sustained oscillations. The loop gain A V β should be unity OR A V β = OR + A V β = 0. The overall phase shift of the feedback circuit and the amplifier must be 0 or integral multiple of π. 8. What are ground waves and sky waves? Ground Waves: These are the radio waves travels along the earth surface. Sky waves: These are the radio waves, sent towards the sky, they get reflected back to earth from the ionosphere. 9. A 0 khz audio signal is used to frequency modulate a 00 MHz carrier causing carrier deviation of 75 khz. Determine modulation index m f = δ/f m = (75 x 0 3 )/(0 x 0 3 ) = 7.5 0. Using X-OR gate, convert the gray code 00 into binary code. Draw the logic diagram.
0.. Draw the block diagram of JK Master-slave flip-flop. JK Master Slave flip-flop. Construct OR and AND gates using NAND gates. OR gate using NAND gates AND gate using NAND gates PART - C 3. Using the following data, calculate the theoretical and experimental values of voltage gain for an Op-Amp inverting amplifier. Input voltage = 0.75 volts. Trial R i in kω R f in kω V 0 in volts Voltage gain Theoretical Practical. 8. -.7.7 0 -.6 3
3. Theoretical voltage gain : A VT = - R f /R i Practical voltage gain : A VP = V 0 /V i Trial A VT = - 8.k/.k = - 3.7 A VP = -.7/0.75 = - 3.6 Trial A VT = - 0k/.7k = -.3 A VP = -.6/0.75 = -.3 OR 3. The following readings were recorded in a CE amplifier experiment. Calculate the voltage gain. Input voltage = 50 mv. Frequency 00 500 k 5k to 500k 700k in Hz 00k Output voltage.5.5 5. 3.5 in volts Voltage gain Voltage gain: A V = V 0 /V i At 00 Hz A V = /50x0-03 = 0 At 500 Hz A V =.5/50x0-03 = 50 At khz A V =.5/50x0-03 = 90 At 5kHz to 00 khz A V = 5/50x0-03 = 00 At 500 khz A V =./50x0-03 = 8 At 700 khz A V = 3.5/50x0-03 = 70. An amplifier has a midband gain of 00. If the lower cut-off frequency and upper cut-off frequency are 500 Hz and 300 khz, calculate the bandwidth and gain at cut-off frequencies. Bandwidth = f f = 300 x 0 3 500 = 99.5 khz Gain at cut-off frequency = A/ = 00/ =. 5. Using re-model, derive the expression for voltage gain and input impedance of CE amplifier
5 Where R BB = R R and R AC = R C R L re-model equivalent circuit υ i = i b βre υ 0 = - i c (R C R L ) Voltage gain: A υ = υ 0 /υ i A υ = - i c (R C R L )/i b βre = - (R C R L )/re A υ = - R C / re - may also be considered Input impedance: Z i = (R R ) βre 6. With a block diagram, derive the expression for voltage gain of negative feedback amplifier. Voltage Series negative feedback Amplifier Gain of the amplifier without feedback, A = V 0 /V i () Gain of the amplifier with feedback, A f = V 0 /V s () V i = V s - V f (3) 5
6. β = V f /V 0.() V 0 = AV i from eqn () = A(V s V f ) from eqn (3) = A(V s βv 0 ) from eqn () V 0 = AV s - AβV 0 V 0 ( + Aβ) = AV s (for proper steps marks) A f = V 0 /V s = A/( + Aβ) 7. Calculate the output voltage for the circuit shown below For non-inverting amplifier: V 0 = ( + R f /R i )V i V 0 = ( + 0k/.7k)0.5 =.56 V For inverting adder: V 0 = - [(R f /R )xv + (R f /R )xv ] V 0 = - [(0k/6.8k)x.56 + (0k/3.3k)x0.75] = -.56 V 8. A Colpitt s oscillator generates a frequency of 500 khz. The capacitors to be used are C = 00 pf and C = 0 pf. Find the value of inductance. f = CT = L = π L = LC T 00 π X 0 X 3 (500 X0 ).57 mh Where 0 X (00 + 0) X0 (9.09 OR 0 X 0 ) L = C C C = (C C ) T + = 9.09 x0 F (for substitution and simplications marks) π f C T 6
Qn. 9. Derive an expression for modulation index in terms of V max and V min. Draw the modulated wave. Amplitude Modulated wave From figure, Modulation index, m a = V m /V C Vmax Vmin V m = V C = Vmax V m Vmax Vmin V C = Vmax Vmax + Vmin V C = m Vmax Vmin = a Vmax + Vmin..() from equation() (proper steps - marks) 30. Explain two-input DTL NAND gate with a circuit. Write its truth table Truth table Inputs Output A B Y(Vout) 0 0 0 0 0 DTL NAND Gate Ckt mark Table - mark Working When A = 0, B = 0, then D and D conducts, Q off, therefore Y(V out ) = A = 0, B =, then D conduct and D doesn t conducts, Q off, Y(V out ) = A =, B = 0, then D doesn t conducts and D conduct, Q off, Y(V out ) = A =, B =, then D and D doesn t conducts, Q on, therefore Y(V out ) = 0 7
Qn. 3. Using K-map, simplify the following Boolean expression Y = f(a, B, C, D) = m(0,,, 5, 6, 8, 9,, 3, ). Draw the logic diagram for the simplified expression using basic gates PART - D 3. Describe an experiment to study the frequency response of CC amplifier 6 Aim: To conduct an experiment to draw frequency response of CC amplifier Equipment & components: CC amplifier circuit, DC power supply, signal generator, CRO, etc., Observations: Voltage gain = V 0 /V i CC Amplifier Note: Suitable components values must be considered Frequency Input voltage, Vi Output voltage, V0 Gain = V 0 /V i 8
3 Frequency response Procedure: Connect signal generator to input terminals. Connect CRO across input & output terminals. Keep suitable input voltage, observe & record output voltage. Do the experiment for different frequencies. Determine gain for each frequency using the formula, A V = V 0 /V i Plot a graph of voltage gain verses frequency (Proper procedure must be considered) Result: Frequency response of CC amplifier is drawn experimentally OR 3. Describe an experiment to study RS flip-flop using NAND gates 6 Aim: To conduct & to study RS flip-flop using NAND gates Equipment & components: Digital IC trainer, IC 700, connecting wires etc., Pin diagram of IC 700 Pin diagram - mark RS flip-flop using NAND gates Circuit diagram - mark Note: Pin number connections in proper order must be considered. 9
Inputs Truth table Outputs S R Q Q Conditions 0 0 Last state Last state No change 0 0 Reset 0 0 Set Forbidden Procedure: Connect pin 7 to Gnd and pin to +V CC. Circuit connections are made to form RS flip-flop. Verify the truth table for various input conditions. (Proper procedure must be considered) Result: RS flip-flop using NAND gate is constructed and truth table is verified. 33a. With the circuit diagram, describe the procedure to draw output characteristics in CB mode. CB Amplifier An arrangement to draw characteristics of CB amplifier is shown in figure. Suitable meters and power supplies are used to carryout the experiment. Output Characteristics: To draw output characteristics keep input current I E constant. By varying output voltage V CB note down corresponding output current I C. Do experiment for different I E. Plot a graph of I C verses V CB for constant I E. (Proper procedure must be considered) 0
33a. Output Characteristics of CB amplifier 33b. Mention the steps involved to obtain dc equivalent circuit Reduce all AC sources to zero Open all capacitors 3. Explain the working of RC coupled amplifier with a circuit. Draw its frequency response. Mention its one advantage. 6 RC Coupled amplifier Circuit diagram of RC coupled amplifier is shown in figure. One stage of CE amplifier is coupled to next stage by coupling capacitors. Working: Coupling capacitor allows ac signals but blocks dc signals. The first CE amplifier produces a phase shift of 80 0 and second stage also produces a phase shift 80 0. Therefore the output V 0 is in same phase with input V i. Frequency response of RC coupled amplifier:
3 Advantage of RC coupled amplifier: Inexpensive. Wide frequency response. Less frequency distortion. Gain is large. Anyone advantage 35a. What is a differentiator? Derive an expression for its output voltage. Differentiator: The Op-amp circuit whose output is proportional to the derivative of the input signal is called differentiator. Op-Amp differentiator From the figure i C = if dq/dt = (0 V 0 )/R C[dV i /dt] = V 0 /R Therefore, V 0 = -RC[dV i /dt] (Proper steps must be considered) 35b. Draw the circuit diagram of Op-Amp subtractor. Write the expression for output voltage when all the resistors are equal. Op-Amp Subtractor Expression for output when Rf = Ri = R = R = R V 0 = V - V
Qn. 36a. What is piezoelectric effect? Draw the circuit of crystal oscillator. Mention any one advantage. Piezoelectric effect: When an AC voltage is applied between the faces of quartz crystal it vibrates at the frequency of the applied voltage. Consequently, if the crystal is vibrated mechanically an AC voltage is generated. Crystal Oscillator: Advantage of xtal oscillator: Frequency stability. High quality factor. Long life. Any one advantage 36b. An RC phase shift oscillator produces oscillations of frequency 00 Hz with C = 0. µf. Calculate the value of R. OR f = 0.065/RC f = π R C f C 6 OR OR R = 0.065/fC R = R = π x 3. x 00 6 x 0.x 0 = 3.8 kω 37a. The equation of FM wave is V FM = 0(sin8 x 0 8 t + 0 sin800t). Find: (i) Carrier frequency (ii) Modulting frequency (iii) Modulation index & (iv) Frequency deviation. The standard form of FM equation is V FM = V C sin(ω c t + m f sinω m t)..() But the given expression is V FM = 0(sin8 x 0 8 t + 0 sin800t)...() - 06 x 6 3
37a. To solve this problem, equation () need tobe equated to equation (). This is not possible because equation () is not of the standard FM equation form. Note: If the quastion number (37a) is written by the student, full () marks must be awarded. 37b. Express Y = A+ B C into standard cononical SOP form. Y = A+ B C Y = A( B+ B)( C+ C) + ( A+ A) B C Y = ABC+ ABC + ABC + A B C + A B C 38a. Explain linear diode detector with a circuit and waveforms. Linear diode detector Working: Diode D rectifies AM wave. Capacitor C provides low reactance to carrier and high reactance to signal. Resistor R provides discharging path to C. Capacitor C adds zero level to detected signal. 38b. Draw the block diagram of AM transmitter. Block diagram of AM Transmitter 39a. What is full adder? Draw block diagram of full adder using two half adders and OR gate. Write its truth table.
39a. Full adder: Full adder adds three binary digits at a time. Full adder using two half adders Truth Table of full adder Inputs Outputs A B C in Carry Sum 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 39b. Draw the block diagram of digital computer and label the blocks. Block diagram of digital computer 5
Qn. 0a. Draw the block diagram of monochrome TV transmitter. Block diagram of monochrome TV transmitter 0b. What are the advantages of E-mail? Advantage of E-mail: Any information (text, audio, video) can be send or receive. Transmission is immediate and to any distance. Any data size can be sent. A data can be send to any number of recipients. The information sent/received can be directly used for further processing. Economical. (OR Any two acceptable advantage each carry mark) ***** s * h * i * v * a * s * h * a *n * k * a * r ***** 6