250+ TOP MCQs on Instrumentation Amplifier – 2 & Answers

Basic Linear Integrated Circuit Questions and Answerson “Instrumentation Amplifier – 2”.

1. The temperature of a thermistor increases, when the value of its resistance
A. Remain constant
B. Increase
C. Decrease
D. Depends on the heating material
Answer: C
Clarification: Thermistor is a semiconductor that behaves as resistor, with a negative temperature coefficient of resistance. As the temperature of thermistor increases, its resistance decreases.

2.
Consider the entire resistors in the bridge circuit are equal. The resistance and change in resistance are given as 3kΩ and 30kΩ. Calculate the output voltage of differential instrumentation amplifier?
A. 4.95v
B. 1.65v
C. 8.25v
D. 14.85v
Answer: B
Clarification: The output voltage of the circuit is V_o =-(R_F/R₁)×(△R/R)×V_dc
= (5.5kΩ/100Ω)×(30kΩ/3kΩ)×3 = 1.65v.

3. Consider a thermistor having the following specifications: R_F=150kΩ at a reference temperature of 35^oC and temperature coefficient of resistance = 25^oC. Determine the change in resistance at 100^oC.
A. -1.625MΩ
B. 9.75MΩ
C. 4.78MΩ
D. None of the mentioned
Answer: A
Clarification: Thermistor has negative temperature coefficient of resistance. Therefore, △R=-(25kΩ/^oC )×(100^oC-35^oC. = -1625kΩ .
△R=-1.625MΩ.

4. Consider the given bridge circuit, find the voltage across the output terminal, V_ab.

A. V_ab = 4.9v
B. V_ab = -5.6v
C. V_ab =1.2v
D. V_ab =-8.2v
Answer: A
Clarification: According to the voltage divider rule,
V_a =( R_a×V_dc)/[R_a+(R_T+△R)] = (1kΩ×5v)/(1kΩ+75kΩ) = 0.065v
V_b = ( R_b×V_dc)/(R_b+R_c) = (50kΩ×5v)/(50kΩ+250Ω) = 4.975v
The voltage across the output terminal of the bridge, V_ab = V_a– V_b = 4.9v.

5. Express the equation for transducer bridge, if all the resistor values are equal
A. v=-(△R×V_dc)/(2×R+△R)
B. v=-(△R×V_dc)/2×(R+△R)
C. v=-V_dc/[2×(2×R+△R)].
D. v=-(△R×V_dc)/ [2×(2×R+△R)].
Answer: D
Clarification: If the R_a=R_b=R_c=R_T=R(Equal), then the output voltage across the bridge terminals of the transducer bridge is v=-(△R×V_dc)/ [2×(2×R+△R)].

6. Which type of thermistor is chosen for temperature measurement and control?
A. High temperature coefficient of resistance
B. Low temperature coefficient of resistance
C. Positive temperature coefficient of resistance
D. None of the mentioned
Answer: A
Clarification: Thermistors with a high temperature coefficient resistance are more sensitive to temperature change and are therefore well suited to temperature measurement and control.

7. Photo conductive cell changes it resistance with
A. Change in temperature
B. Material composition
C. Incident radiant energy
D. Change in elasticity
Answer: C
Clarification: Photoconductive cell is a type of transducer that changes its resistance or varies its resistance with an incident radiant energy with light.

8. What will be the resistance of a photoconductive cell in darkness?
A. 1000-3000Ω
B. 100MΩ
C. 250-500Ω
D. None of the mentioned
Answer: D
Clarification: The resistance of the photoconductive cell in darkness is typically in the order of 100kΩ.

9. Which material is used for photoconductive cells?
A. Germanium
B. Cadmium sulphide
C. Lithium
D. Phosphorous
Answer: B
Clarification: The conductivity in cadmium sulphide is a function of incident radiant energy. So, it is used for photoconductive cell.

10. Name the resistive transducer that varies its resistance on application of external stress?
A. Photocells
B. Light dependent
C. Stain gage
D. None of the mentioned
Answer: C
Clarification: Strain gage is a type of resistive transducer whose resistance changes due to elongation or compression when external stress is applied.

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