logo
  • Home
  • Rank
  • Tests
  • About
  • Training
  • Programming
  • CS
  • IT
  • IS
  • ECE
  • EEE
  • EE
  • Civil
  • Mechanical
  • Chemical
  • Metallurgy
  • Instrumentation
  • Aeronautical
  • Aerospace
  • Biotechnology
  • Agriculture
  • MCA
  • BCA
  • Internship
  • Contact

Linear Integrated Circuits Multiple Choice Questions | MCQs | Quiz

Linear Integrated Circuits Interview Questions and Answers
Practice Linear Integrated Circuits questions and answers for interviews, campus placements, online tests, aptitude tests, quizzes and competitive exams.

Get Started

•   Differential Amplifier
•   Op-Amp Internal Circuit - 1
•   Op-Amp Internal Circuit - 2
•   Op-Amp Internal Circuit - 3
•   Op-Amp Internal Circuit - 4
•   Integrated Circuits - 1
•   Integrated Circuits - 2
•   IC Chip Size & Circuit
•   Basic Planar Process - 1
•   Basic Planar Process - 2
•   Basic Planar Process - 3
•   Active & Passive IC - 1
•   Active & Passive IC - 2
•   Active & Passive IC - 3
•   Thin Film Technology
•   FET Fabrication
•   IC Package Type - 1
•   IC Package Type - 2
•   Data Sheets Interpreting - 1
•   Data Sheets Interpreting - 2
•   Ideal Operational Amplifier
•   Open Loop Op-Amp
•   Feedback Configurations
•   Voltage Series Amplifier - 1
•   Voltage Series Amplifier - 2
•   Voltage Shunt Amplifier
•   Multiple Op-Amp - 1
•   Multiple Op-Amp - 2
•   Input Offset Voltage - 1
•   Input Offset Voltage - 2
•   Input Offset Voltage - 3
•   Input Bias Current
•   Input Offset Current
•   Thermal Drift
•   Offset Voltage Effect
•   Noise
•   Rejection Ratio
•   Frequency Response
•   Non-Compensating Op-Amp
•   Open-Loop Voltage Gain-1
•   Open-Loop Voltage Gain-2
•   Circuit Stability
•   Slew Rate - 1
•   Slew Rate - 2
•   DC & AC Amplifiers
•   AC Amplifiers
•   Peaking Amplifier
•   Averaging Amplifier - 1
•   Averaging Amplifier - 2
•   Instrumentation Amplifier - 1
•   Instrumentation Amplifier - 2
•   Instrumentation Amplifier - 3
•   Op-Amp Equivalent Circuit
•   Voltage-Current Converter-1
•   Voltage-Current Converter-2
•   Current-Voltage Converter
•   Input Impedance Circuit
•   Log & Antilog Amplifier
•   Multiplier & Divider - 1
•   Multiplier & Divider - 2
•   Integrator - 1
•   Integrator - 2
•   Differentiator
•   Active Filters - 1
•   Active Filters - 2
•   1st Order Butterworth Filter
•   2nd Order Butterworth Filter
•   Higher Order Filters
•   Band Pass Filters
•   All Pass Filters
•   Sine Wave Oscillator - 1
•   Sine Wave Oscillator - 2
•   Sine Wave Oscillator - 3
•   Sine Wave Oscillator - 4
•   Square Wave Generator
•   Sawtooth Wave Generator
•   Comparator
•   Schmitt Trigger
•   Precision Type Comparator
•   Basic DAC Techniques - 1
•   Basic DAC Techniques - 2
•   A to D Converter - 1
•   A to D Converter - 2
•   DAC / ADC Specification
•   Clippers & Clampers
•   Peak Detector & Sampling
•   Monostable Multivibrator
•   Astable Multivibrator
•   Phase-Locked Loops
•   Phase Detector
•   Controlled Oscillator
•   PLL Applications
•   IC Voltage Regulator
•   Switching Regulator
•   Monolithic Phase Loop
•   Power Amplifiers

Best Reference Books

Linear Integrated Circuits Books
« Prev Page
Next Page »

Linear Integrated Circuit Questions and Answers – Operational Amplifier Internal Circuit – 2

Posted on August 27, 2017 by staff10

This set of Linear Integrated Circuit Multiple Choice Questions & Answers (MCQs) focuses on “Operational Amplifier Internal Circuit – 2”.

1. How are the arbitrary signal represented, that are applied to the input of transistor? (Assume common mode signal and differential mode signal to be VCM & VDM respectively).
a) Sum of VCM & VDM
b) Difference of VCM & VDM
c) Sum and Difference of VCM & VDM
d) None of the mentioned
View Answer

Answer: c
Explanation: In practical situation, arbitrary signal are signal are represented as Sum and Difference of common mode signal and differential mode signal.
advertisement

2. How the differential mode gain is expressed using ‘h’ parameter for a single ended output?
a) – hfeRC/hie
b) 1/2×(hfeRC)/hie
c) – 1/2×hfeRC
d) None of the mentioned
View Answer

Answer: b
Explanation: Formula for differential mode gain using ‘h’ parameter model for a single ended output.

3. Find Common Mode Rejection Ration, given gm =16MΩ-1, RE=25kΩ
a) 58 db
b) 40 db
c) 63 db
d) 89 db
View Answer

Answer: a
Explanation: Formula for Common Mode Rejection Ration, CMRR= 1+2gmRE,
⇒ CMRR = 1+(2×16MΩ-1×25kΩ)
= 801 = 20log⁡801 = 58.07 db.

4. In differential amplifier the input are given as V1=30sin⁡Π(50t)+10sin⁡Π(25t) , V2=30sin⁡Π(50t)-10 sin⁡Π(25t), β0 =200,RE =1kΩ and RC = 15kΩ. Find the output voltages V01, V02 & gm=4MΩ-1
a) V01=-60[10 sin⁡Π(25t) ]-6.637[30sin⁡Π(50t) ], V02=60[10 sin⁡Π(25t) ]-6.637[30sin⁡Π(50t) ].
b) V01=-6.637[10 sin⁡Π(25t) ]-60[30sin⁡Π(50t) ], V02=6.637[10 sin⁡Π(25t) ]-60[30sin⁡Π(50t) ].
c) V01=-60[30 sin⁡Π(50t) ]-6.637[10sin⁡Π(25t) ], V02=60[30 sin⁡Π(50t) ]-6.637[10sin⁡Π(25t) ].
d) V01=-6.637[30 sin⁡Π(50t) ]-60[10sin⁡Π(25t) ], V02=6.637[30 sin⁡Π(50t) ]-60[10sin⁡Π(25t) ].
View Answer

Answer: a
Explanation: Differential mode gain, ADM = -gm RC,
⇒ ADM = -4MΩ-1×15kΩ = 60
⇒ rΠ=β0/gm =200/4MΩ-1 =50kΩ
Common mode gain, ACM=-βo×RC/rΠ+(βO+1)×RE
⇒ ACM =-200×15kΩ/50kΩ+2(1+200)×1kΩ=-6.637
Common mode signal, VCM=(V1+V2)/2= 30sin⁡Π(50t)
Differential mode signal, VDM=(V1-V2)/2= 10 sin⁡Π(25t)
Output voltages are given as
⇒ V01=ADM)× VDM)+ ACM× VCM
= -60[10 sin⁡Π(25t)]-6.637[30sin⁡Π(50t)],
⇒ V02=-ADM× VDM+ ACM× VCM
= 60[10 sin⁡Π(25t)]-6.637[30sin⁡Π(50t)].

5. If the value of Common Mode Rejection Ratio and Common Mode Gain are 40db and -0.12 respectively, then determine the value of differential mode gain
a) 0.036
b) -1.2
c) 4.8
d) 12
View Answer

Answer: d
Explanation: Common mode rejection ratio, CMRR =log-1×(40/20) = 100
⇒ CMRR =(∣ADM∣/ ∣ACM∣)
⇒ ∣ADM∣ =100×0.12 = 12.
advertisement

6. To increase the value of CMRR, which circuit is used to replace the emitter resistance Re in differential amplifier?
a) Constant current bias
b) Resistor in parallel with Re
c) Resistor in series with Re
d) Diode in parallel with Re
View Answer

Answer: a
Explanation: Constant current bias offers extremely large resistor under AC condition and thus provide high CMRR value.

7. What is the purpose of diode in differential amplifier with constant current circuit?
a) Total current independent on temperature
b) Diode is dependent of temperature
c) Transistor is depend on temperature
d) None of the mentioned
View Answer

Answer: a
Explanation: The base emitter voltage of transistor (VBE) in constant current circuit by 2.5mv/oc, thus diode also has same temperature. Hence two variations cancel each other and total current IQ become in depend of temperature.

8. How to improve CMRR value
a) Increase common mode gain
b) Decrease common mode gain
c) Increase Differential mode gain
d) Decrease differential mode gain
View Answer

Answer: b
Explanation: For a large CMRR value, ACM should be small as possible.

9. Define total current (IQ) equation in differential amplifier with constant current bias current
a) IQ=1/R3×(VEE/R1+R2)
b) IQ =(VEE×R2)/(R1+R2)
c) IQ=1/R3×(VEE×R2/R1+R2)
d) IQ)=R3×(VEE/R1+R2)
View Answer

Answer : c
Explanation: The equation for total current is obtained by applying Kirchhoff’s Voltage Law to constant current circuit in differential amplifier.

10. Constant current source in differential amplifier is also called as
a) Current Mirror
b) Current Source
c) Current Repeaters
d) All of the mentioned
View Answer

Answer: a
Explanation: The output current is reflection or mirror of the reference input current. Therefore, the constant current source circuit referred as Current Mirror.
advertisement

11. When will be the mirror effect valid
a) β≫1
b) β=1
c) β<1
d) β≠1
View Answer

Answer: a
Explanation: If value of β is used in the equation, IC=β/(β+2)×Iref. It almost become unity and the output current become equal to reference current.

12. Calculate the value of reference current and input resistor for current mirror with IC=1.2μA & VCC=12v. Assume β=50.
a) 1.248mA, 9kΩ
b) 1.248mA, 9.6kΩ
c) 1.248mA, 9.2kΩ
d) 1.2mA, 9.6kΩ
View Answer

Answer: a
Explanation: We know that collector current, IC=β/(β+2)×Iref,
⇒ Iref=(β+2)/β×IC= (50+2)/50× 1.2μA = 1.248mA
⇒ Iref=(VCC-VBE)/R1
⇒ R1=(12v-07v)/1.248mA = 9.05kΩ.

Sanfoundry Global Education & Learning Series – Linear Integrated Circuits.

To practice all areas of Linear Integrated Circuits, here is complete set of 1000+ Multiple Choice Questions and Answers.

« Prev Page - Linear Integrated Circuit Questions and Answers – Operational Amplifier Internal Circuit – 1
» Next Page - Linear Integrated Circuit Questions and Answers – Operational Amplifier Internal Circuit – 3

« Linear Integrated Circuit Questions and Answers – Operational Amplifier Internal Circuit – 1
Linear Integrated Circuit Questions and Answers – Operational Amplifier Internal Circuit – 3 »
advertisement

Deep Dive @ Sanfoundry:

  1. Optical Communications Questions and Answers
  2. Network Theory Questions and Answers
  3. Signals & Systems Questions and Answers
  4. Electronic Devices and Circuits Questions and Answers
  5. Electrical Engineering Questions and Answers
  6. Linear Integrated Circuits Questions and Answers
  7. Linear Integrated Circuit Questions and Answers – Voltage Series Feedback Amplifier – 1
  8. Linear Integrated Circuit Questions and Answers – Voltage Shunt Feedback Amplifier
  9. Linear Integrated Circuit Questions and Answers – Instrumentation Amplifier – 1
  10. Linear Integrated Circuit Questions and Answers – Input Offset Voltage – 3
Manish Bhojasia, a technology veteran with 20+ years @ Cisco & Wipro, is Founder and CTO at Sanfoundry. He is Linux Kernel Developer and SAN Architect and is passionate about competency developments in these areas. He lives in Bangalore and delivers focused training sessions to IT professionals in Linux Kernel, Linux Debugging, Linux Device Drivers, Linux Networking, Linux Storage & Cluster Administration, Advanced C Programming, SAN Storage Technologies, SCSI Internals and Storage Protocols such as iSCSI & Fiber Channel. Stay connected with him below:
LinkedIn | Facebook | Twitter | Google+

Best Careers

Developer Tracks
SAN Developer
Linux Kernel Developer
Linux Driver Developer
Linux Network Developer

Live Training Photos
Mentoring
Software Productivity
GDB Assignment
Sanfoundry is No. 1 choice for Deep Hands-ON Trainings in SAN, Linux & C, Kernel Programming. Our Founder has trained employees of almost all Top Companies in India such as VMware, Citrix, Oracle, Motorola, Ericsson, Aricent, HP, Intuit, Microsoft, Cisco, SAP Labs, Siemens, Symantec, Redhat, Chelsio, Cavium, ST-Micro, Samsung, LG-Soft, Wipro, TCS, HCL, IBM, Accenture, HSBC, Mphasis, Tata-Elxsi, Tata VSNL, Mindtree, Cognizant and Startups.

Best Trainings

SAN I - Technology
SAN II - Admin
Linux Fundamentals
Advanced C Training
Linux-C Debugging
System Programming
Network Programming
Linux Threads
Kernel Programming
Kernel Debugging
Linux Device Drivers

Best Reference Books

Computer Science Books
Algorithm & Programming Books
Electronics Engineering Books
Electrical Engineering Books
Chemical Engineering Books
Civil Engineering Books
Mechanical Engineering Books
Industrial Engineering Books
Instrumentation Engg Books
Metallurgical Engineering Books
All Stream Best Books

Questions and Answers

1000 C Questions & Answers
1000 C++ Questions & Answers
1000 C# Questions & Answers
1000 Java Questions & Answers
1000 Linux Questions & Answers
1000 Python Questions
1000 PHP Questions & Answers
1000 Hadoop Questions
Cloud Computing Questions
Computer Science Questions
All Stream Questions & Answers

India Internships

Computer Science Internships
Instrumentation Internships
Electronics Internships
Electrical Internships
Mechanical Internships
Industrial Internships
Systems Internships
Chemical Internships
Civil Internships
IT Internships
All Stream Internships

About Sanfoundry

About Us
Copyright
Terms
Privacy Policy
Jobs
Bangalore Training
Online Training
Developers Track
Mentoring Sessions
Contact Us
Sitemap
© 2011 Sanfoundry. All Rights Reserved.