Category: Linear Integrated Circuits Objective Questions

Linear Integrated Circuit Multiple Choice Questions on “IC Chip Size and Circuit Complexity”.

1. How many gates per chip are used in first generation Integrated Circuits?
A. 3-30
B. 30-300
C. 300-3000
D. More than 3000

Answer: A
Clarification: The first generation ICs belongs to small scale integration, which consists of 3-30 gates per chip (approximately).

2. Find the chip area for a Medium Scale Integration IC?
A. 8 mm³
B. 4 mm²
C. 64 mm³
D. 16 mm²

Answer: D
Clarification: The approximate length and breadth of Medium Scale Integration would be 4 mm. Therefore, its area is given as = length × breadth = 4mm × 4mm = 16mm^2.

3. The number of transistors used in Very Large Scale Integration is
A. 10⁷ transistors/chip
B. 10⁶ – 10⁷ transistors/chip
C. 20³ – 10⁵ transistors/chip
D. 10² – 20³ transistors/chip

Answer: C
Clarification: Very Large Scale Integration (VLSI) ICs are fabricated using more than 3000 gates/chip, which is equivalent to 20,000 – 1,00,00,00 transistors/chip.

4. What type of integration is chosen to fabricate Integrated Circuits like Counters, multiplexers and Adders?
A. Small Scale Integration (SSI)
B. Medium Scale Integration (MSI)
C. Large Scale Integration (LSI)
D. Very Large Scale Integration (VLSI)

Answer: B
Clarification: Fabrication of ICs like counter, multiplexers and Adders requires 30-300 gates per chip. Therefore, Medium Scale Integration is best suitable.

5. Determine the chip area for Large Scale Integration ICs.
A. 1,00,000 mil²
B. 10,000 mil²
C. 1,60,000 mil²
D. 16,000 mil²

Answer: C
Clarification: The chip area for a Large Scale Integration IC is 1 cm².
=> Area of LSI = 10mm × 10mm = 1cm × 1 cm = 1cm².
=> 1,60,000mil² (1cm=400mil).

6. Ultra Large Scale Integration are used in fabrication of
A. 8-bit microprocessors, RAM, ROM
B. 16 and 32- bit microprocessors
C. Special processors and Smart sensors
D. All of the mentioned

Answer: C
Clarification: Ultra Large Scale Integration have nearly 10⁶ – 10⁷ transistors/chip. Hence, it is possible to fabricate smart sensors and special processor.

7. The concept of Integrated circuits was introduced at the beginning of 1960 by
A. Texas instrument and Fairchild Semiconductor
B. Bell telephone laboratories and Fair child Semiconductor
C. Fairchild Semiconductor
D. Texas instrument and Bell telephone Laboratories

Answer: A
Clarification: The concept of Integrated circuits was introduced by Texas instrument and Fairchild Semiconductor, whereas Bell telephone laboratories developed the concept of transistors.

8. Which process is used to produce small circuits of micron range on silicon wafer?
A. Photo etching
B. Coordinatograph
C. Photolithography
D. Ion implantation

Answer: C
Clarification: It is possible to fabricate as many as 10,000 transistors on a 1cmX1cm chip, using photolithography process.

9. Mention the technique used in photolithography process
A. X-ray lithographic technique
B. Ultraviolet lithographic technique
C. Electron beam lithographic technique
D. All of the mentioned

Answer: D
Clarification: All these techniques are used to produce device dimension as small as 2µm or even down to sub micron range (<1µm).

Linear Integrated Circuit Multiple Choice Questions on “Open-Loop Voltage Gain as a Function of Frequency – 1”.

1. Determine the output voltage for an op-amp with single break frequency.

A. V_O ={ jX_C /[R_o+(iX_C)]} × AV_id
B. V_O = = AV_id / [1+ j2πfR_oC].
C. V_O = AV_id /(R_o+j2πfC.
D. V_O = V_id / [R_o-(j2πfR_oC.].
Answer: B
Clarification: The output voltage for an op-amp with single break frequency,
V_O = {(-jX_C) / [(R_o)-(jX_C)} ×AV_id
∵ -j=1/j & X_C =1/2πfC
=> V_O = {(1/j2ΠfC./[R_o + (1/ j2πfC.] } × AV_id = AV_id / [1+ j2πfR_oC].

2. Compute the break frequency of an op-amp, if the output resistance=10kΩ and capacitor connected to the output =0.1µF.
A. 159.2Hz
B. 6.28Hz
C. 318.4Hz
D. 1000Hz
Answer: A
Clarification: Break frequency of the op-amp is given as f_o = 1/(2πR_oC.= 1/ (2π×10kΩ×0.1µF) = 1/ (6.28×10^-3) = 159.2Hz.

3. The open loop voltage gain as a function of frequency is defined as
A. A_OL(f) = V_O/V_in
B. A_OL(f) = V_O/V_id
C. A_OL(f) = V_O/V_f
D. All of the mentioned
Answer: B
Clarification: The open loop voltage gain as a function of frequency is defined as ratio of output voltage to the difference of input voltages.

4. Which among the following factor remain fixed for an op-amp?
A. Open loop voltage gain
B. Gain of the op-amp
C. Operating frequency
D. Break frequency of the op-amp
Answer: D
Clarification: Break frequency f_o depends on the value of capacitors and on output resistance. Therefore, f_o is fixed for any op-amp.

5. Find the gain magnitude and phase angle of the op-amp using the specifications:
f= 50Hz; f_o=5Hz ; A=140000.
A. A_OL(f)= 22.92dB , Φ(f) = – 89.99^o
B. A_OL(f)= 66dB , Φ(f) = – 90^o
C. A_OL(f)= 26dB, Φ(f) = – 89.99^o
D. A_OL(f)= 20dB , Φ(f) = – 84.29^o
Answer: A
Clarification: The open loop gain magnitude |A_OL(f)|= 20log[A/√[1+ f/f_o)²] = 20logA-20 log[A/√ [1+(f/f_o)²] = 20log(140000)- 20log[√(1+(50,000/5)²)]
A_OL(f) dB= 102.922-80 = 22.92dB.
Phase angle, φ(f) = -tan^-1(f/f_o) = -tan^-1(50000/5) = -89.99^o.

6. Consider an op-amp where the inverting input voltage =3.7mv, non-inverting input voltage=6.25mv and open loop voltage gain =142dB. Find the output voltage.
A. 0.21v
B. 0.45v
C. 0.78v
D. 0.36v
Answer: D
Clarification: Open loop voltage gain, A_oL(f) = V_o/V_id
V_O = A_OL(f) × (V_in1-V_in2) = 142 dB×(6.25-3.7) = 142×2.55 = 0.36v.

7. Express the open loop gain of the op-amp in complex form?
A. A/√ [1+(f/f_o)²
B. 20log{A/√[1+(f/f_o)²}
C. A/[1+j(f/f_o)].
D. None of the mentioned
Answer: C
Clarification: The open loop gain of the op-amp A_OL(f) is a complex quantity and is expressed as A_OL(f) = A/[1+ j(f/f_o)] . The remaining equations are expressed in polar form.

8. Determine the difference between two A_OL(f) at 50Hz and 500Hz frequency? (Consider the op-amp to be 741C.
A. 40dB
B. 30dB
C. 20dB
D. 10dB
Answer: C
Clarification: A_OL(f) dB= 20log[√ [1+ (f/f_o)²]
At f= 50 Hz,
A_OL(f) dB = 20log(200000)- 20log(√(1+(50/5)²) = 106.02-20.04 ≅ 86dB
At f= 500Hz
A_OL(f) dB =20log(200000)-20log(√(1+(500/5)²) = 106.02-40 ≅ 66dB
Therefore, the difference between A_OL(f)dB = 86-66 = 20dB.

9. At what frequency, the phase shift between input &output voltage will be zero?
A. -40Hz
B. 0Hz
C. -22Hz
D. 20Hz
Answer: B
Clarification: At 0Hz the phase shift between input and output voltage is zero.
At f=0Hz, φ(f) = – tan^-1 (f/f_o) = -tan^-1(0/5) = 0^o

10. At what frequency A_OL(f)=A?
A. 50Hz
B. 10Hz
C. 5Hz
D. 0Hz
Answer: D
Clarification: For any frequency less than break frequency (f_o =5Hz) the gain is approximately constant and is equal to A. For example, f_o =0Hz,
Then A_OL(f) dB= 20log(200000-20log[√1+(0/5)²)] = 106dB. Where A =20,000 ≅ 106dB.

11. What happen when the frequency increases?
A. A_OL(f) continues to drop
B. A increases
C. f_o –> 0Hz
D. None of the mentioned
Answer: A
Clarification: The open loop voltage gain as a function of frequency is given as A_OL(f) = A/ [√ 1+ (f/f_o)] at frequency above f_o, the denominator value increases, causing the gain, A_OL(f) to decrease. Thus, as frequency increases, the gain A_OL(f) continuous to drop.

12. What will be the absolute value of phase shift, if the frequency keeps increasing?
A. Increase towards 45^o
B. Decrease towards 45^o
C. Increase towards 90^o
D. Decrease towards 90^o
Answer: C
Clarification: For any frequency above break frequency, the absolute value of phase shift increases towards 90^o with increase in frequency.

.

Linear Integrated Circuit Multiple Choice Questions on “Current to Voltage Converter”.

1. The output current equation for MC1408 digital to analog converter would be
A. I_o= -(V_ref/R₁)×[(D₇/2)+(D₆/4)+(D₅/8)+(D₄/16)+(D₃/32)+(D₂/64)+(D₁/128)+(D₀/256)].
B. I_o= (V_ref/2R₁)×[(D₇/2)+(D₆/4)+(D₅/8)+(D₄/16)+(D₃/32)+(D₂/64)+(D₁/128)+(D₀/256)].
C. I_o= (V_ref/R₁)×[(D₇/2)+(D₆/4)+(D₅/8)+(D₄/16)+(D₃/32)+(D₂/64)+(D₁/128)+(D₀/256)].
D. I_o= -(V_ref/2R₁)×[(D₇/2)+(D₆/4)+(D₅/8)+(D₄/16)+(D₃/32)+(D₂/64)+(D₁/128)+(D₀/256)].
Answer: C
Clarification: MC1408 is a combination of a DAC and current to voltage converter. If the binary signal is input to MC1408 DAC then the output current would be,
I_o= (V_ref/R₁)×[ (D₇/2)+(D₆/4)+(D₅/8)+(D₄/16)+(D₃/32)+(D₂/64)+(D₁/128)+(D₀/256)].

2. Determine the maximum value of output current of the DAC in MC1408?
A. 0.773×(V_ref/R₁)
B. 0.448×(V_ref/R₁)
C. 0.996×(V_ref/R₁)
D. 0.224×(V_ref/R₁)
Answer: C
Clarification: The output current of DAC is the maximum when all the inputs are logic 1.
Therefore, I_o= (V_ref/R₁)×(1/2+1/4+1/8+1/16+1/32+1/64+1/128/+1/256)=(0.996)×(V_ref/R₁).

3. Determine the range or the output voltage?

A. 0 – 2.51v
B. 0 – 2.22v
C. 0 – 3.74v
D. 0 – 4.93v
Answer: D
Clarification: When all binary input D₀ through D₇ are logic 0, the current I_o =0.
∴ The minimum value of V_o =0v.
When all the inputs are at logic 1, I_o = (V_ref/R₁) × (1/2+1/4+1/8+1/16+1/32+1/64+1/128/+1/256) = (3/2kΩ) × (0.996) =1.494mA.
Hence, the maximum value of output voltage is V_o= I_o×R_F = 1.494×3.3kΩ =4.93v. Thus, the output voltage range is from 0 to 4.93v.

4. Calculate the change in the output voltage if the photocell is exposed to light of 0.61lux from a dark condition. Specification: Assume that the op-amp is initially nulled, Minimum dark resistance = 100kΩ and resistance when illuminated (at 0.61lux) = 1.5kΩ.

A. V_o –> 23v to 50v
B. V_o –> 0v to 33.11v
C. V_o –> -1.653v to 8.987v
D. V_o –> -0.176v to -11.73v
Answer: D
Clarification: The resistance R_T in darkness is 100kΩ. The minimum output voltage in darkness is V_{o min} = -(V_dc×R_F)/ R_T = -(3.2v×5.5kΩ)/100kΩ = -0.176v.
When photocell is illuminated, its resistance R_T =1.5kΩ. Therefore, the maximum output voltage is V_{o max} = -(V_dc×R_F)/ R_T = -(3.2v×5.5kΩ)/1.5kΩ =-11.73v.
Thus, V_o varies from -0.176v to -11.73 as the photocell is exposed to light from a dark condition.

5. Which cell can be used instead of a photocell to obtain active transducer in photosensitive devices?
A. Photovoltaic cell
B. Photo diode
C. Photo sensor
D. All of the mentioned
Answer: A
Clarification: A photovoltaic cell is semiconductor junction device that convert radiation energy into electrical energy and hence it does not require external voltage.

6. If the input applied to DAC using current to voltage converter is 10110100, determine the reference voltage (Assume I_o= 2mA and R₁=1.2kΩ)
A. 53.1v
B. 3.41v
C. 9.21v
D. 67.34v
Answer: B
Clarification: I_o=V_ref/R₁×[(D₇/2)+(D₆/4)+(D₅/8)+(D₄/16)+(D₄/32)+(D₄/64)]
V_ref =(I_o×R₁)/ (1/2+0+1/8+1/16+0+1/64+0+0)=(2mA×1.2kΩ)/0.703 .
=> V_ref = 3.41v.

7. The current to voltage converter photosensitive device can be used as
A. Light intensity meter
B. Light radiating meter
C. Light deposition meter
D. None of the mentioned
Answer: A
Clarification: The photosensitive device can be used as a light intensity meter by connecting a meter at the output that is calibrated for light intensity.

8. For a full wave rectification, in a low voltage ac voltmeter, the meter current can be expressed as
A. I_o = (1.9×V_in)/R₁
B. I_o = (3.9×V_in)/R₁
C. I_o = (0.9×V_in)/R₁
D. I_o = (2.9×V_in)/R₁
Answer: C
Clarification: For full wave rectification, meter current is expressed as I_o = 0.9xV_in/R₁.

.

Linear Integrated Circuit Multiple Choice Questions on “Basic Principles of Sine Wave Oscillator – 2”.

1. What will be the phase shift of feedback circuit in RC phase shift oscillator?
A. 360^o phase shift
B. 180^o phase shift
C. 90^o phase shift
D. 60^o phase shift

Answer: B
Clarification: The RC feedback network provide 180^o phase shift and amplifier used in RC phase shift oscillator provide 180^o phase shift (op-amp is used in the inverting mode) to obtain a total phase shift of 360^o.

2. How many RC stages are used in the RC phase shift oscillator?
A. Six
B. Two
C. Four
D. Three

Answer: D
Clarification: The RC stage forms the feedback network of oscillator. It consist of three identical RC stages, each of 60^o phase shift so as to provide a total phase shift of 180^o.

3. Calculate the frequency of oscillation for RC phase shift oscillator having the value of R and C as 35Ω and 3.7µF respectively.
A. 1230 Hz
B. 204 Hz
C. 502Hz
D. 673 Hz

Answer: C
Clarification: The frequency of oscillation of RC phase shift oscillator is,
f_o=1/(2πRC√6) = 1/(2×3.14×√6×3.7µF×35Ω)
=> f_o= 1/ 1.9921×10^-3 = 502Hz.

4. What must be done to ensure that oscillation will not die out in RC phase shift oscillator?
A. Gain of amplifier is kept greater than 29
B. Gain of amplifier is kept greater than 1
C. Gain of amplifier is kept less than 29
D. Gain of amplifier is kept less than 1

Answer: A
Clarification: For a sustained oscillation in RC phase shift oscillator the gain of the inverting op-amp should be at least 29. Therefore, gain is kept greater than 29 to ensure the variation in circuit parameter will not make |Aß|<1, otherwise oscillation will die out.

5. Calculate the feedback voltage from phase shit network.

A. V_f = ( V_oR³S³C³) / (1+ 6SRC+5S²C²R²+S³R³C³)
B. V_f = ( V_oR³S³C³) / (1+ 5SRC+6S²C²R²+S³R³C³)
C. V_f = ( V_oR³S³C³) / (1+ 5SRC+5S²C²R²+S³R³C³)
D. V_f = ( V_oR²S³C³) / (1+ 6SRC+5S²C²R²+S³R³C³)

Answer: B
Clarification: Applying KVL equation to the circuit, we get
=> I₁(R+(1/SC.)-I₂=V_o -> Equ1
=> -I₁R+ I₂(2R+(1/SC.)- I₃R=0 -> Equ2
=> 0- I₂R+ I₃(2R+(1/SC.=6 -> Equ3
WKT, V_f = I₃× 2R,
Solving Equ 1, 2, and 3 for I₃
=> I₃= ( V_oR²S³C³)/ (1+ 5SRC+6S²C²R²+S³R³C³)
=> V_f = I₃× 2R
=( V_oR³S³C³) / (1+ 5SRC+6S²C²R²+S³R³C³).

6. Which type of op-amp is avoided for high frequencies?
A. LM318
B. Op-amp 741
C. LF 351
D. None of the mentioned

Answer: B
Clarification: Op-amp741 is generally used for low frequencies < 1 kHz.

7. Find out the constant values of α and ß in phase shift oscillator.
A. α = √6, ß = -1/29
B. α = 6, ß = -1/29
C. α = √6, ß = 1/29
D. α = 6, ß = -1/29

Answer: A
Clarification: From phase shift network, we obtain
ß= 1/(1-5 α²)+j α(6- α²) -> Equ1
For Aß=1, ß should be real and the imaginary terms must be zero
α(6- α²) =0
=> α=√6
Now substituting α²=6 in Equ 1, we get
ß=-1/29 (Negative sign indicates that the feedback network produces a phase shift of 180^o).

8. A phase shift oscillator is designed to oscillate at 155Hz. Determine the value of R_f. (Take C=0.30µF)
A. 399Ω
B. 3.98MΩ
C. 13.9kΩ
D. 403kΩ

Answer: D
Clarification: R = 1/(2πC√6×f_o)
=> R= 1/7.153×10^-4= 1398=13.9kΩ.

9. The value of feedback resistor in phase shift oscillator is 180kΩ. Find its input resistance?
A. 52kΩ
B. 151kΩ
C. 209kΩ
D. 6.2kΩ

Answer: C
Clarification: To obtain sustained oscillation in phase shift oscillator.
=> |A|=29 or |R_f / R₁|=29
=> R₁|= R_f / 29 = 180kΩ/29= 6.21kΩ,

10. Determine the frequency of oscillation (f_o) in phase shift oscillator?
A. f_o = √6/ωRC
B. f_o = 0.56/ωRC
C. f_o = 0.065/ωRC
D. f_o = 6/ωRC

Answer: C
Clarification: The frequency of oscillation of phase shift oscillator is given as
f_o = 1/(2π×RC×√6) = 1/15.38×RC
=> f_o = 0.065/RC.

11. The condition for zero phase shift in wein bridge oscillator is achieved by
A. Connecting feedback to non-inverting input terminal of op-amp
B. Balancing the bridge
C. Applying parallel combination of RC to the feedback network
D. All of the mentioned

Answer: B
Clarification: In wein bridge oscillator, the feedback signal in the circuit is connected to the non-inverting input of op-amp. So, feedback network does not provide any feedback and the condition of zero phase shift around the circuit is achieved by balancing the bridge.

Linear Integrated Circuit Multiple Choice Questions on “555 Timer as an Astable Multivibrator”.

1. Free running frequency of Astable multivibrator?
A. f=1.45/(R_A+2R_B)C
B. f=1.45(R_A+2R_B)C
C. f=1.45C/(R_A+2R_B)
D. f=1.45 R_A/( R_A+R_B)

Answer: A
Clarification: The frequency of the Astable multivibrator is T=0.69(R_A+2R_B)C.
Therefore, f = 1/T =1.45/(R_A+2R_B)C.

2. Find the charging and discharging time of 0.5µF capacitor.

A. Charging time=2ms; Discharging time=5ms
B. Charging time=5ms; Discharging time=2ms
C. Charging time=3ms; Discharging time=5ms
D. Charging time=5ms; Discharging time=3ms

Answer: B
Clarification: The time required to charge the capacitor is t_High=0.69(R_A+R_B)C =0.69(10kΩ+5kΩ)x0.5µF =5ms.
The time required to discharge the capacitor is t_Low=0.69xR_C =0.69x5kΩx0.5µF=2ms.

3. Astable multivibrator operating at 150Hz has a discharge time of 2.5m. Find the duty cycle of the circuit.
A. 50%
B. 75%
C. 95.99%
D. 37.5%

Answer: D
Clarification: Given f=150Hz.Therefore,T=1/f =1/150 =6.67ms.
∴ Duty cycle, D%=(t_Low/T) x 100% = (2.5ms/6.67ms)x100% = 37.5%.

4. Determine the frequency and duty cycle of a rectangular wave generator.

A. Frequency=63.7kHz; Duty cycle=50%
B. Frequency=53.7kHz; Duty cycle=55%
C. Frequency=43.7kHz; Duty cycle=50%
D. Frequency=60kHz; Duty cycle=55%

Answer: B
Clarification: Frequency=1.45/(R_A+R_B)C .
Where R_A=100Ω+50Ω=150Ω,
R_B=100Ω+20Ω=120Ω.
=>∴f=1.45/((150+120)x0.1µF) = 53703Hz = 53.7kHz.
Duty cycle, D% = [R_B/(R_A+R_B)] x 100% = 120Ω/(150Ω +120Ω) x 100% = 0.55×100% = 55%.

5. How to achieve 50% duty cycle in adjustable rectangular wave generator? (Assume R₁ –> Resistor connected between supply and discharge and R₂ –> Resistor connected between discharge and trigger input.)
A. R₁2
B. R₁ > R₂
C. R₁ = R₂
D. R₁ ≥ R₂

Answer: C
Clarification: The equation of duty cycle, D = R₂/(R₁ + R₂). If R₁ is made equal to R₂ then 50% duty cycle is achieved.

6. How to obtain symmetrical waveform in Astable multivibrator?
A. Use clocked RS flip-flop
B. Use clocked JK flip-flop
C. Use clocked D-flip-flop
D. Use clocked T-flip-flop

Answer: b
Clarification: Symmetrical square wave can be obtained by adding a clocked JK flip-flop to the output of Astable multivibrator. The clocked flip-flop acts as a binary divider to the times output and produces 50% duty cycle without any restriction on the choice of resistors.

7. Determine the output frequency of the circuit.

A. 1450Hz
B. 1333Hz
C. 1871Hz
D. 1700Hz

Answer: C
Clarification: The output frequency of the frequency shift keying generator is
f=1.45/[(R_A||R_C)+2(R_B)]xC = 1.45/[(2.3kΩ||2.3kΩ) + (2×3.3kΩ)] x 0.1µF = 1.45/{[(2.3×2.3)/(2.3+2.3)] + 6.6kΩ}x0.1µF = 1.45/(7.75×10^-4) = 1870.9 ≅ 1871Hz.

8. How does a monostable multivibrator used as frequency divider?
A. Using square wave generator
B. Using triangular wave generator
C. Using sawtooth wave generator
D. Using sine wave generator

Answer: A
Clarification: Monostable multivibrator can be used as a frequency divider when a continuously triggered monostable circuit is triggered using a square wave generator. Provided the timing interval is adjusted to be longer than the period of triggering square wave input signal.