Category: Linear Integrated Circuits Objective Questions

Linear Integrated Circuit Multiple Choice Questions on “Integrated Circuits, Types & Manufacturer’s Design – 1”.

1) What is the best choice of IC package used for experimental purpose?
A. DIP package
B. Metal can package
C. Flat pack
D. Transistor pack

Answer: A
Clarification: DIP package are used as it is easy to mount. The mounting does not require bending or soldering of the leads.

2. What is the general information specified in ordering an IC?
A. Temperature range
B. Device type
C. Package type
D. All of the mentioned

Answer: D
Clarification: Generally, in ordering an IC, all the three informations must be specified.

3. Find the ordering information for µA741TC.
A. Sprague 741 DIP with Industrial temperature range
B. Intersil 741 DIP with commercial temperature range
C. Fairchilds 741 DIP with commercial temperature range
D. Texas instrument 741 metal can with Industrial temperature range

Answer: C
Clarification: Here “µA” represents the identifying initials used by Fairchild,
T represents Mini DIP package and C represents Commercial temperature range.

4. How a Motorola IC with plastic DIP and commercial temperature range is ordered?
A. ICLxxxP -> 0^o to 75^oc
B. CAxxE -> -55^o to +125^oc
C. LMxxxxA -> -40^o to+85^oc
D. MCxxxP -> 0^o to 70^oc

Answer: D
Explantion: The ordering format for a typical Motorola IC is,
MCxxxx –> Device type
P –> Package type(Plastic DIP)
0^o to 70^oc –> Temperature range (Commercial).

5. What does the 1-2-3 numbering system used in National Semiconductor IC denotes
A. Validity in years
B. Temperature range
C. Package type
D. Ordering information

Answer: C
Clarification: In National linear ICs, a 1-2-3 numbering system is used to represent the temperature range.

6. How does a industrial temperature range device in National Semiconductor IC is represented?
A. LM305
B. LM101
C. LM201
D. All of the mentioned

Answer: C
Clarification: In LM201, the number 2 denotes an industrial temperature range device.

7. Use device identification method to find the IC of Fairchild chip manufactured in the year 1980.

8. Dual-In-Line pack is considered to be suitable for mounting because,
A. Easy to handle
B. Fits mounting hardware
C. Inexpensive
D. All of the mentioned

Answer: C
Clarification: DIP pack is easy to handle, fit standard mounting hardware and is inexpensive when moulded on plastic.

9. What is the use of notch and dot in DIP ICs?
A. Determine the pin configuration
B. Designed to represent device type
C. Represent property of IC
D. Find the pin number

Answer: D
Clarification: A notch and dot as viewed form top view is used to find the pin terminal. The terminals are numbered counter clockwise.

10. How an eight pin Dual-In-Line Package is shortly named
A. 8p DIP
B. Maxi DIP
C. Mini DIP
D. ES DIP

Answer: C
Clarification: An eight pin Dual-In-Line Package is called as Mini DIP as it is used for devices with minimum number of inputs and outputs.

Linear Integrated Circuit Multiple Choice Questions on “Open Loop Op-Amp Configuration”.

1. Open loop op-amp configuration has
A. Direct network between output and input terminals
B. No connection between output and feedback network
C. No connection between input and feedback network
D. All of the mentioned

Answer: A
Clarification: In an open loop configuration, the output signal is not fed back in any form as part of the input signal and the loop that would have been formed with feedback is open.

2. In which configuration does the op-amp function as a high gain amplifier?
A. Differential amplifier
B. Inverting amplifier
C. Non-inverting amplifier
D. All of the mentioned

Answer: D
Clarification: An op-amp functions as a high gain amplifier when connected in open loop configuration. These three are the open loop configuration of an op-amp.

3. How does the open loop op-amp configuration classified?
A. Based on the output obtained
B. Based on the input applied
C. Based on the amplification
D. Based on the feedback network

Answer: b
Clarification: Open loop configurations are classified according to the number of inputs used and the terminal to which the input is applied when a single input is used.

4. What will be the voltage drop across the source resistance of differential amplifier when connected in open loop configuration?
A. Zero
B. Infinity
C. One
D. Greater than one

Answer: A
Clarification: The source resistances are normally negligible compared to the input resistance. Therefore, the voltage drop across input resistors can be assumed to be zero.

5. The output voltage of an open-loop differential amplifier is equal to
A. Double the difference between the two input voltages
B. Product of voltage gain and individual input voltages
C. Product of voltage gain and the difference between the two input voltages
D. Double the voltage gain and the difference between two input voltages

Answer: C
Clarification: The output voltage is equal to the voltage gain times the difference between the two input voltages.

6. Calculate the output voltage for the given circuit.

A. V_o = 7v
B. V_o = 5.9v
C. V_o = 12v
D. V_o = 11.4v

Answer: C
Clarification: The output voltage, V_o = A*(V_in1-V_in2).(Since, R_in1 and R_in2 are negligible compared to input resistance in open loop differential amplifier).
=> V_o = 4*(12v-9v) = 12v.

7. Find the output of inverting amplifier?
A. V_o = AV_in
B. V_o = -AV_in
C. V_o = -A(V_in1– V_in2)
D. None of the mentioned

Answer: B
Clarification: In an inverting amplifier the input signal is amplified by gain A and is also inverted at the output. The negative sign indicates that the output voltage is of opposite polarity.

8. Determine the output voltage for the non-inverting amplifier input voltage 37µVpp sinewave. Assume that the output is a 741.
A. -7.44 Vpp sinewave
B. 74 Vpp sinewave
C. 7.4Vpp sinewave
D. 0.7 Vpp sinewave

Answer: C
Clarification: The output voltage for non-inverting amplifier V_o = A*V_in = 200000 * 37µ = 7.4 Vpp sinewave.

9. Find the non-inverting amplifier configuration from the given circuit diagram?

Answer: C
Clarification: In a non-inverting amplifier, the input is applied to the non-inverting input terminal and the inverting terminal is connected to ground.

10. What happen if any positive input signal is applied to open-loop configuration?
A. Output reaches saturation level
B. Output voltage swing’s peak to peak
C. Output will be a sine waveform
D. Output will be a non-sinusoidal waveform

Answer: A
Clarification: In open-loop configuration, due to very high gain of the op-amp, any input signal slightly greater than zero drives the output to saturation level.

11. Why open-loop op-amp configurations are not used in linear applications?
A. Output reaches positive saturation
B. Output reaches negative saturation
C. Output switches between positive and negative saturation
D. Output reaches both positive and negative saturation

Answer: C
Clarification: When operated in open loop, the output switches between positive and negative saturation levels. For this reason, open loop op-amp configurations are not used in linear applications.

Linear Integrated Circuit Multiple Choice Questions on “Frequency Response and Compensating Networks”.

1. Variation in the operating frequency of op-amp causes
A. Variation in gain amplifier
B. Variation in gain phase angle
C. Variation in gain amplitude and its phase angle
D. None of the mentioned
Answer: C
Clarification: The gain of the op-amp is a function of frequency. It will have a specific magnitude as well as a phase angle.

2. A graph of the magnitude of the gain versus frequency is called
A. Break frequency
B. Frequency response plot
C. Frequency stability plot
D. Transient response plot
Answer: B
Clarification: A frequency response plot is obtained by plotting the gain of the op-amp responding to different frequencies.

3. In the frequency response plot, the frequency is expressed in
A. Anti-logarithmic scale
B. Logarithmic scale
C. Linear scale
D. Exponential scale
Answer: B
Clarification: To accommodate large frequency ranges the frequency is assigned to a logarithmic scale.

4. Why the gain magnitude in frequency response plot is expressed in decibels (dB.
A. To obtain gain > 10⁵
B. To obtain gain < 10⁵
C. To obtain gain = 0
D. To obtain gain = ∞
Answer: A
Clarification: In frequency response plot, gain magnitude is assigned a linear scale and is expressed in decibels to accommodate very high gain ( ≅ of the order 10⁵ or higher).

5. Which technique is used to determine the stability of op-amp?
A. Frequency response plot
B. Transient response plot
C. Bode plot
D. All of the mentioned
Answer: C
Clarification: Although frequency response and bode plots indicate the effect of frequency variation on gain, the Bode plot is generally used for stability determination and network design.

6. How many types of plots can be obtained in the AC analysis of network using Bode plot?
A. Five
B. Four
C. Three
D. Two
Answer: D
Clarification: Two types of plots can be obtained using Bode plot. They are magnitude versus frequency and phase angle versus frequency plots.

7. What happens when the operating frequency of an op-amp increase?
A. Gain of the amplifier decrease
B. Phase shift between output and input signal decrease
C. Gain and phase shift of amplifier decreases
D. None of the mentioned
Answer: A
Clarification: When the operating frequency is increased the gain of the amplifier decrease. As it is linearly related to frequency, the phase shift is logarithmically related to frequency.

8. Which of the following causes change in gain and phase shift?
A. Internally integrated Resistor
B. Internally integrated inductors
C. Internally integrated Capacitor
D. All of the mentioned
Answer: C
Clarification: The change in function of frequency is attributed to the internally integrated capacitor as well as stray capacitor. These capacitors are due to the physical characteristic of semiconductor device.

9. Which plot is not provide by the manufactures?
A. Magnitude plot
B. Phase angle plot
C. Frequency response plot
D. None of the mentioned
Answer: B
Clarification: Phase angle plot are not generally provided because phase shift of later generation op-amp are less than 90^o even at cross over frequency.

10. Find out the non-compensating op-amp from the given circuit

Answer: C
Clarification: Non-compensating op-amp has external compensating components, that is , resistors and / or capacitors, are added at designated terminals. The mentioned op-amp has three compensating components: a resistor and two capacitors.

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Linear Integrated Circuit Multiple Choice Questions on “Voltage to Current Converter with Floating and Grounded Load – 1”.

1. Voltage to current converter is also called as
A. Current series positive feedback amplifier
B. Voltage series negative feedback amplifier
C. Current series negative feedback amplifier
D. Voltage series positive feedback amplifier
Answer: C
Clarification: Voltage to current converter is also called as current series negative feedback amplifier because the feedback voltage across internal resistor applied to the inverting terminal depends on the output current and is in series with the input difference voltage.

2. Given voltage to current converter with floating load. Determine the output current?

A. 3mA
B. 6mA
C. 4mA
D. 2mA
Answer: D
Clarification: Output current, I_o = V_in /R₁ = 10/5kΩ =2mA.

3. Which of the following application uses voltage to current converter?
A. Low voltage dc and ac voltmeter
B. Diode match finding
C. Light emitting diode
D. All of the mentioned
Answer: D
Clarification: In all the applications mentioned above, the input voltage V_in is converted into an output current of V_in/R₁ or the input voltage appear across resistor.

4. The op-amp in low voltage DC voltmeter cannot be nullified due to
A. D’Arsonaval meter movement
B. Offset voltage compensating network
C. Selection of switch
D. Gain of amplifier
Answer: A
Clarification: The op-amp sometimes cannot be nullified because the output is very sensitive to even slight variation in wiper position of D’Arsonaval meter movement (ammeter with a full scale deflection of 1mA..

5. What is the maximum input voltage that has to be selected to calibrate a dc voltmeter with a full scale voltage range of 1-13v.
A. ≤ ±14v
B. ≥ ±13v
C. ≤ ±15v
D. = ±14v
Answer: A
Clarification: The maximum input voltage has to be ≤ ±14v, to obtain the maximum full scale input voltage of 13v.

6. Higher input voltage can be measured in low voltage DC voltmeter using
A. Smaller resistance value
B. Higher resistance value
C. Random resistance value
D. All of the mentioned
Answer: B
Clarification: Higher resistance values are required to measure relatively higher input voltage. For example, if the range of switch is at x10 position in the low voltage dc voltmeter then, the corresponding resistance value would be 10kΩ. So, it requires a 10v input to get a full scale deflection (if 1v cause full scale deflection in the ammeter with a full scale deflection of 1mA..

7. In the diagram given below, determine the deflection of the ammeter with a full scale deflection of 1mA when the switch is at X2kΩ. Consider resistance of the offset voltage compensating network to be 10Ω.

A. Full scale deflection in the ammeter
B. Half scale deflection in the ammeter
C. Quarter scale deflection in the ammeter
D. No deflection occurs in the ammeter
Answer: B
Clarification: Given V_in=1v ,R₁=10+2kΩ ≅2kΩ
I_o = V_in/R₁= 1v/2kΩ =0.5mA. This means that 2v causes half scale deflection of the ammeter.

8. How to modify a low voltage DC voltmeter to low voltage ac voltmeter
A. Add a full wave rectifier in the feedback loop
B. Add a half wave rectifier in the feedback loop
B. Add a square wave rectifier in the feedback loop
B. Add a sine wave rectifier in the feedback loop
Answer: A
Clarification: A combination of an ammeter and a full wave rectifier can be employed in the feedback loop to form an ac voltmeter.

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Linear Integrated Circuit Multiple Choice Questions on “Band-Reject and All-Pass Filters”.

1. How many types of band elimination filters are present
A. Three
B. Two
C. Four
D. None of the mentioned

Answer: B
Clarification: Band-reject filters are also called as band elimination filters. They are classified into two types.
i) Wide band-reject filter and
ii) Narrow band-reject filter.

2. Find the wide band-reject filter

Answer: B
Clarification: A wide band-reject filter is made using a low pass filter, a high pass filter and a summing amplifier.

3. A narrow band-reject filter is commonly called as
A. Notch filter
B. Band step filter
C. Delay filter
D. All of the mentioned

Answer: A
Clarification: A narrow band-reject filter is also called as notch filter because of its higher quality factor, Q (>10).

4. Find the expression for notch-out frequency?
A. f_N = 2πRC
B. f_N = 2π/RC
C. f_N = 1/2π×√(R/C.
D. f_N = 1/2πRC

Answer: D
Clarification: The notch-out frequency is the frequency at which maximum attenuation occurs: it is given by f_N =1/2πRC.

5. The quality factor of passive twin T-network is increased by using
A. Inverting amplifier
B. Non-inverting amplifier
C. Voltage follower
D. Differential amplifier

Answer: C
Clarification: The passive twin T-network has a selectively low figure of merit. The Q of the network can be increased significantly, if it is used with the voltage follower.

6. Find out the application in which narrow band-reject filter can be used?
A. Embedded system
B. Biomedical instrument
C. Digital computer
D. None of the mentioned

Answer: B
Clarification: Notch filters or narrow band-reject filters are used in biomedical instruments for eliminating undesired frequencies.

7. Design 120Hzactive notch filter?

Answer: A
Clarification: Since C value n = 1/2πRC.
R= 1/(2πf_nC.= 1/(2π×120Hz×0.68×10^-6) = 1.95kΩ ≅2kΩ.
For R/2, parallel of two 2kΩ resistor
=>R/2 = 2kΩ||2kΩ =(2×2)/(2+2)=1kΩ.
For C/2 , parallel of two 0.68µF capacitor
C/2=> 0.68µF + 0.68µF= 1.36µF.

8. Find the application of area where all-pass filters are used?
A. Cathode ray oscilloscope
B. Television
C. Telephone wire
D. None of the mentioned

Answer: C
Clarification: When signals are transmitted in transmission lines like telephone wire, they undergo change in phase, all-pass filters are used to compensate these phase changes.

9. Determine the output voltage for all the all-pass filter and express it in complex form?
A. V_O =V_in/ [(1-j2πfRC. /(1+ j2πfRC.].
B. V_O =V_in× [(1+j2πfRC. /(1- j2πfRC.].
C. V_O =V_in ×[(1- j2πfRC. /(1+ j2πfRC.].
D. None of the mentioned

Answer: C
Clarification: The output voltage of all-pass filter is given as V_O =V_in× [(1-j2πfRC. /(1+j2πfRC.] .

10. Determine the input frequency for all-pass filter with phase angle as 62^o. Consider the value of resistor and capacitor are 3.3kΩ and 4.7µF.
A. Input frequency= -7.65Hz
B. Input frequency= -6.77Hz
C. Input frequency= -3.89Hz
D. Input frequency= -9.65Hz

Answer: D
Clarification: The phase angle is given as Φ = -2tan^-1×(2πfRC.
=> f=-tanΦ/4πRC =-tan(62^o)/(4π×3.3kΩ×4.7µF)= -1.88/0.1948 =-9.65Hz.

11. Determine the angle for given circuit diagram, if the frequency of input signal is 1khz
A. -45^o
B. -180^o
C. -270^o
D. -90^o

Answer: D
Clarification: Phase angle Φ=-2tan^-1×(2πfRC/1) = -2tan^-1×(2π×1kHz×16kΩ×0.01µF)
= -2tan^-1×(1.0048)=-90^o.

12. The voltage gain magnitude of all-pass filter is
A. Zero
B. One
C. Infinity
D. None of the mentioned

Answer: B
Clarification: The magnitude of voltage gain of all-pass filter |V_O /V_in| = √(1+(2π/RC.²) / √(1+(2 π/RC.²) =1.

13. What happens if the position of R and C are interchanged in the below circuit diagram?
A. V_in leads V_O
B. V_in lags V_O
C. V_O leads V_in
D. V_O leads V_in

Answer: C
Clarification: For the circuit given, the phase angle changes from 0 to 180^o as frequency is varied from 0 to ∞. If the positions of R and C are interchanged, the phase shift and band width input and output becomes positive. That is the output (V_O) leads input (V_in).

14. Choose the incorrect statement “In wide band-reject filter” .
A. Low cut-off frequency of low pass filter must be larger than the high cut-off frequency of the high pass filter.
B. Low cut-off frequency of high pass filter must be equal than the high cut-off frequency of the high pass filter.
C. Low cut-off frequency of high pass filter must be smaller than the high cut-off frequency of the low pass filter.
D. None of the mentioned

Answer: D
Clarification: In wide band-reject filter, low cut-off frequency of high pass filter must be larger than the high cut-off frequency of the low pass filter.

Linear Integrated Circuit Multiple Choice Questions on “Peak Detector, Sampling & Hold Circuit and Absolute Value Output Circuit “.

1. Which circuit can be used as a full wave rectifier?
A. Absolute vale output circuit
B. Positive clipper with two diodes
C. Negative clipper with two diodes
D. Peak clampers

Answer: A
Clarification: Absolute value output circuit produces an output signal that swings positively only, regardless of the polarity of the input signal; because of the nature of its output wave form, the circuit is used as full wave rectifier.

2. For the circuit shown below find the output voltage
A. V_o (+) = +10 v
B. V_o (+) = +12v
C. V_o (+) = +7v
D. None of the mentioned

Answer: B
Clarification: The voltage at the terminal V₁ = (V_p -V_d1) /2
V₁ = (12-0.7) /2 = 5.65 v (V_d1= voltage drop across diode=0.7)
Similarly, the voltage at the negative terminal V₂ = (V_o -V_d3 ) /2 = (V_o – 0.7) /2
Since V_id ≅ 0v , ∴ V₁ = V₂
V_o = (5.65 *2 ) + 0.7 = 12v.

4. What is the alternate method to measure the values of non-sinusoidal waveform other than ac voltmeter?
A. Clipper
B. Clamper
C. Peak detector
D. Comparator

Answer: C
Clarification: A conventional ac voltmeter is designed to measure rms value of the pure sine wave whereas, the peak value of the non-sinusoidal wave forms can be a peak detector.

5. State the condition needed to be satisfied by peak detector for proper operation of circuit.
A. CR_d ≤ T/10 and CR_L ≥ 10T
B. CR_d ≤ 10T and CR_L ≥ T/10
C. CR_d ≥ T/10 and CR_L ≤ 10T
D. CR_d ≥ 10T and CR_L ≤ T/10

Answer: A
Clarification: For proper operation of the circuit, charging and discharging time constant must satisfy the following: CR_d ≤ T/10 and CR_L ≥ to 10T.

6. The resistor in the peak detector are used to
A. To maintain proper operation
B. Protect op-amp from damage
C. To get shaped non-sinusoidal waveform
D. None of the mentioned

Answer: B
Clarification: The resistor is used to protect the op-amp against the excessive discharge current, especially when the power supply is switched off.

7. How the recovery time of the op-amp is reduced?
A. Diode is connected at the output of amplifier
B. Load resistor
C. Forward biased diode resistor
D. Discharge capacitor

Answer: A
Clarification: The diode connected at the output of op-amp conducts during negative half cycle of input voltage. Hence, prevent the op-amp from going into negative saturation. This in turn helps to reduce the recovery time of the op-amp.

8. How to detect the negative peaks of input signals in the peak detector given below?

A. Reversing D₁ diode
B. Reversing D₁ and D₂ diodes
C. Reversing D₂ diode
D. Charging the positions of D₁ and D₂

Answer: B
Clarification: The negative peaks of the input signal V_in can be detected by reversing diodes D₁ and D₂.

9. In the sample and hold circuit, the period during which the voltage across capacitor is equal to input voltage
A. Sample period
B. Hold period
C. Delay period
D. Charging period

Answer: A
Clarification: The time periods of the sample and hold control voltage during which the voltage across capacitor is equal to the input voltage are called sample period.

10. During which period the op-amps output of sample and hold circuits is processed?
A. Delay period
B. Sample and hold period
C. Sample period
D. Hold period

Answer: D
Clarification: Hold period is the period during which the voltage across the capacitor is constant and the output of the op-amp is processed or observed during hold periods.

11. Which IC is mostly preferred for sample and hold circuit?
A. µ771
B. IC741
C. LF398
D. µ351

Answer: C
Clarification: LF398 have significant reduction in size and improved performance and require only an external storage capacitor.

12. Sample and hold circuit are used in
A. Analog to Digital modulation
B. Digital to analog modulation
C. Pulse position modulation
D. All of the mentioned

Answer: D
Clarification: All types of modulation involve taking samples of an input signal and hold on to it last sample value until the input is sampled.