Roll No.: _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ Amrita Vishwa Vidyapeetham Amrita School of Engineering, Coimbatore B.Tech. Degree Examinations – August 2022 Fourth Semester (Electronics and Communication Engineering) 19ECE212 Linear Integrated Circuits Time : Three hours Maximum : 100 Marks CO Course Outcomes CO01 CO02 CO03 CO04 Able to understand the construction of electronic systems using operational amplifiers Able to understand the various specification parameters of op-amp Able to design and analyze linear and non-linear circuits with op-amps Able to design and analyze sinusoidal oscillator circuits Instructions: 1. Answer all questions in the order in which they appear. 2. All circuit diagrams / figures are to be drawn neatly and labelled, using pencil. 3. All components used may be considered to be ideal, unless specifically mentioned otherwise. 4. The following may be assumed: ππ = 25 ππ. 5. Any opamps used, may be considered to be powered off + 15 V supplies, unless mentioned otherwise. 6. Any variables / parameters used, if not explained, may be considered to have the same significance as those used in Microelectronic Circuits – Theory and Applications, by Sedra & Smith, 5th ed. 1. Fill in the blanks with the most appropriate answer. Only the answer needs to be noted down in the answer sheet. [15] [CO1, CO2,CO3,CO4] [BTL2] (i) For an inverting amplifier, keeping the feedback resistance a constant, if the input resistance is to decrease, the gain will -----------. (ii) For a difference amplifier of unity gain, with the resistances = R, the differential mode input resistance is ------------. (iii) If the NE555 timer IC were to be powered by a 15 V source, the voltage reference levels would be ------------- V and ------------ V. (iv) A comparator is designed to operate in the ----------- region of the Voltage Transfer Characteristic (VTC). (v) A Schmitt Trigger provides noise immunity by means of the -----------------, provided by the circuit. (vi) The time period of an astable multivibrator with a duty cycle of 25%, a feedback factor of 75% and a time constant of 1 ms, is -------------- s. (vii) A unity gain amplifier was found to have a bandwidth of 1.5 MHz. If the same amplifier were to be re-configured for a bandwidth of 50 kHz, its closed loop voltage gain would be ---------. (viii) The input offset current of an operational amplifier is defined as πππ = ----------. (ix) If an amplifier with gain π΄ and negative feedback π, is to oscillate, the condition to be satisfied is that --------------. Page 1/6 (x) The input resistance of an instrumentation amplifier, with ideal opamps, is ----------. (xi) The typical output resistance of the ππ΄741 is ----------------------. (xii) In a typical inverting amplifier using an opamp, a ------------- is sampled and a ------- is mixed at the input. (xiii) The Schmitt Trigger has ------------- stable state(s). (xiv) Employing negative feedback, if the output voltage were to be sampled, the output impedance -----------. (xv) The output impedance of a BJT based current mirror is due to the -----------------. 2. Match the entries in Column A against those in Column B. Write down only your choices in the answer sheet. [10] [CO1, CO2,CO3,CO4] [BTL2] A (i) (ii) B Inverting amplifier (a) (b) (c) (d) (e) (f) (g) (h) (i) (j) (k) (l) (m) (n) (o) vi = 1 V Fig. 1 (iii) (iv) (v) (vi) (vii) (viii) (ix) (x) 3. Non-inverting Comparator Schmitt Trigger Slew Rate of an opamp Negative feedback RC Phase Shift Oscillator Log Amplifier Superdiode Astable Multivibrator ππ = 1 2π√6π πΆ 15 π; π£π > ππ π£π = ⌊ −15 π π£π < ππ Prevents opamp from saturating Bistable Multivibrator Increases the bandwidth Expansion Compression Increases the output impedance Relaxation Oscillator Voltage Sampling, Voltage mixing vo = |vi| π£π = π£π Large Signal Parameter −15 π; π£π > ππ π£π = ⌊ 15 π π£π < ππ Voltage Sampling, Current mixing Choose the correct answer from amongst those given. Please note down clearly and legibly only your choice. For example, if your answer to question 3(i) is “None of these”, write down “(e)”. [15] [CO1, CO2,CO3,CO4] [BTL2] (i) For a difference amplifier with inputs V1 and V2, if the output vo is given by the expression, π£π = 10π1 − 9.99π2 , then the CMRR of the amplifier (in dB) is (a) 40 (b) 50 (c) 60 (d) 80 (e) None of these (ii) Given a non-inverting Schmitt Trigger of upper and lower voltage levels of 2 V and 1.8 V respectively, the reference voltage (VR) required would be (in V) (a) 1.95 (b) 2.05 (c) 0.87 (d) 1.85 (e) None of these (iii) The 1N4007 diode with a cut-in voltage (Vγ ) of 0.6 V, is used R in a super-diode configuration, such that the output, π£π = vo R + 0.1πππ(2 ∗ 0; π£π < 0. If the input to the circuit were π£π = [0.2 π ∗ 100π‘)]π, the average value of the output would be (in V) v vi o 10 kο 90 kο (a) 0.2 (b) 0.4 (c) 0.127 (d) 0.064 (e) None of these Fig. 2 v i 1 k ο 9 k ο Page 2/6 (iv) If the resistor R in Fig. 2 is provided for the purpose of input bias current compensation, its value is likely to be (in kΩ) (a) 1 (b) 9 (c) 10 (d) 0.9 (e) None of these (v) In a common-emitter amplifier, if an emitter resistance of 100 Ω is added, the input resistance of the amplifier (a) Decreases (b) Increases (c) is not affected (d) changes, depending on the nature of the BJT (e) None of these (vi) If the time period for which the output of an astable multivibrator remains high is indicated by TON, then TON depends on (i) the time constant, RC; (ii) the feedback network; (iii) the supply voltage levels, + VCC; (iv) the voltages to and from which the capacitor charges and discharges. Which of these four statements are true ? (a) (i), (ii) and (iii) (b) (ii), (iii) and (iv) (c) (i), (iii) and (iv) (d) (i), (ii), (iii) and (iv) (e) None of these. 5.6 π; 1 π < π£π < 3 π (vii) A circuit has the VTC given by π£π = [ ]. If the input to the circuit −5.6 π; πππ ππ€βπππ vi is given by π£π = 5 ∗ πππ(2 ∗ π ∗ 0.1π‘), the output vo at t = 4.2 s is (in V) (a) 0 (b) 5.6 (c) -5.6 (d) vi (e) None of these (viii) In an astable multivibrator, when the output levels change between + 3.3 V, the capacitor of 10 nF charges between + 1 V, through a resistor of 10 kΩ. Then, the time period of the output waveform would be (in μs) (a) 62.6 (b) 125.1 (c) 100.3 (d) 52.4 (e) 55.3 (ix) In a current amplifier, employing negative feedback, the output impedance ------, while the input impedance --------. (a) Decreases, Decreases (b) Increases, Increases (c) Increases, Decreases (d) Decreases, Increases (e) None of these When the inputs to a difference amplifier were 20 mV and -20 mV, its’ output was found to be 500 mV and 490 mV, when the inputs were 1.02 V and 0.98 V. Its’ CMRR is (in dB) (a) 62 (b) 42 (c) 71 (d) 56 (e) None of these (xi) For a monostable multivibrator, the frequency of the output signal is related to that of the trigger as (a) They are equal (b) The trigger frequency is greater than that of the output (c) The trigger frequency is lesser than that of the output (d) The trigger frequency is not related to that of the output (e) None of the above −3π£ ; π£π > 0 (xii) The VTC of a circuit is given by the expression, π£π = [ 0; π πππ ππ€βπππ . The circuit is (x) likely to be a(an) (a) Inverting amplifier (b) Window Comparator (c) Schmitt Trigger (d) Precision Full Wave Rectifier (e) None of these (xiii) The slew rate of the ππ΄741 operational amplifier is typically ------------ (in π/ππ ). (a) 0.5 (b) 1.2 (c) 0.2 (d) 0.3 (e) 0.8 Page 3/6 (xiv) If the differential gain of a difference amplifier with a CMRR of 80 dB is 10, the common mode gain would be ----------------. (a) 10−3 (b) 10−2 (c) 103 (d) 102 (e) None of these (xv) Pick the odd man out: ππ΄741, ππ07, ππΏ064, ππΏ071, πΏπ311. (a) ππ΄741 (b) ππ07 (c) ππΏ064 (d) ππΏ071 (e) πΏπ311 For questions 4 - 13, note down your answer to the question. There will be NO partial credit. [10*2 = 20 marks] [CO1, CO2, CO3,CO4] [BTL3] 4. For the circuit shown in Fig. 3, with π1 = 1 π; π2 = 2 π; π = 10 πΩ and ππ = 10 π, neatly plot its’ Voltage Transfer Characteristic. Determine the current flowing through the 10 πΩ resistor, when π£π = 2.5 π. 5. In Fig. 4, π 1 = π 3 = 10 πΩ; π 2 = 100 πΩ and π 4 = 99 πΩ. If π1 = 1.1 π and π2 = 1 π, determine π£π (rounded off to two decimal places). 6. If, for the circuit shown in Fig. 5, π = 1 πΩ, π£π = πππ(ππ‘), π = 100 π»π§, determine the output π£π , at t = 2.5 ms. Fig. 3 7. Fig. 5 Fig. 4 A non-inverting amplifier of gain 10 is found to have a bandwidth 150 kHz. If the amplifier is now re-configured to have a gain of 25, what will be the bandwidth ? Fig. 6 Fig. 7 Fig. 8 8. (i) Obtain an expression for the output π£π of the circuit shown in Fig. 6. It is given that π = 1 πΩ; πΆ = 1 ππΉ. (ii) If π1 = 1 π; π£π = 2 π, then, determine π£π , at time π‘ = 1 ππ . 9. (i) For the circuit of Fig. 7, obtain an expression for the output π£π . (ii) If π 1 = π 2 = 10 πΩ, π1 = 0.5 π and π£π = 0.5π‘, what is the output π£π at π‘ = 1 π ? 10. Consider the circuit given in Fig. 8. π 1 = 5 πΩ; π 2 = 10 πΩ; π 3 = 65 πΩ; VR = 0 π and πΆ = 100 ππΉ. Plot the output waveform and determine its period. Page 4/6 11. Fig. 11 Fig. 10 Fig. 9 The current (πΌπ· ) – voltage (ππ· ) relationship of the diode in the circuit of Fig. 9 is given by ππ· πΌπ· = πΌπ π ππ , where πΌπ = 1 ππ΄. If π 1 = 1 πΩ, ππ = 0.5 π and π£π = 1 π, determine π£π (rounded off to one decimal place). 12. The opamp of the circuit in Fig. 10 has an open-loop gain of 40. If π 1 = 1 πΩ; π 2 = 9 πΩ and π£π = 0.2 π, determine π£π . 13. If the output π£π of the circuit of Fig. 11 is 0.42 V, with π 1 = 100 Ω and π 2 = 560 πΩ, then, the input offset voltage of the opamp is ------------------. Each question from 14 to 17 carries 5 marks and should begin on a fresh page. There WILL be partial credit. [4*5 = 20 marks] [BTL3] 14. Consider the circuit shown in Fig. 12. If the opamp can be considered to be ideal, with the zener diodes (π·1 & π·2 ) having a zener voltage VZ = 3.3 V and a forward cut-in voltage Vγ = 0.7 V, plot the voltage transfer characteristic of the circuit. Mark the cardinal points. [CO3] vi 2 kο 9 kο vo D1 2 kο D2 1V Fig. 12 15. Plot the steady-state waveforms at VP, VN and vo for the circuit shown in Fig. 8. Clearly mark all salient voltages and time intervals. It is given that π 1 = 1 πΩ; π 2 = 9 πΩ; π 3 = 10 πΩ; ππ = 1 π and πΆ = 10 ππΉ. It is known that the output levels of the opamp are +13 V and -12 V. [CO3] 16. (i) Identify the circuit shown in Fig. 13. (ii) The current source is such 1 ππ΄ < πΌπ΄ < 10 ππ΄. It Fig. 13 was found that when ππ = 1 π, πΌπ΄ = 1ππ΄, the output π£π = 0.5 π. Determine the output for πΌπ΄ = 10 ππ΄. [CO3] Page 5/6 17. The circuit of Fig. 14 is required to oscillate at a frequency of 10 kHz. (i) If πΆ = 10 ππΉ, what should be the value of R ? (ii) If π 1 = 10 πΩ, what should be the minimum value of π 2 , for oscillations to be sustained ? [CO4] Fig. 14 Questions 18 and 19 each carry 10 marks and should begin on a fresh page. There WILL be partial credit. [BTL4] 18. (i) Design a circuit to produce a single positive pulse of duration 100 ππ . The circuit is normally in the OFF state. (ii) If provided with a pulse train of period 10 ms, with a duty cycle of 10%, develop a circuit to provide the necessary input to shift the circuit of (i) to the unstable state. Justify any assumptions that you make. [CO1, CO3] 19. A differential pressure transducer provides a differential output voltage (ππ ) given by the expression ππ = (100 + 0.3 πΏπ) mV, where πΏπ is the differential pressure to be measured, in mmHg. If πΏπ varies between 100 and 350 mmHg, design a circuit which will provide maximum common mode suppression and an output linear with πΏπ, with a sensitivity of 10 mV/mmHg. Minimum component spread is to be ensured. An alarm is to be raised when the pressure rises above 300 mmHg – a red LED, with a cut-in voltage of 1.8 V is to light up to indicate this status. If the maximum current through the LED is to be 10 mA, design an appropriate circuit for this purpose, so that the output voltage will not be greater than 5 V. Justify the use of any component and any assumptions that you make. You are provided with ideal components and power supplies of + 10 V. [CO1, CO2, CO3] Course Outcome / Bloom’s Taxonomy Level (BTL) Mark Distribution Table CO CO01 CO02 CO03 CO04 Marks 37 18 37 8 Average BTL BTL 1 2 3 4 Marks 0 40 40 20 2.8 Page 6/6
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