# Linear Integrated Circuit Questions and Answers – Instrumentation Amplifier – 2

This set of Basic Linear Integrated Circuit Questions and Answers focuses on “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

Explanation: 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

Explanation: The output voltage of the circuit is Vo =-(RF/R1)×(△R/R)×Vdc
= (5.5kΩ/100Ω)×(30kΩ/3kΩ)×3 = 1.65v.

3. Consider a thermistor having the following specifications: RF=150kΩ at a reference temperature of 35oC and temperature coefficient of resistance = 25oC. Determine the change in resistance at 100oC.
a) -1.625MΩ
b) 9.75MΩ
c) 4.78MΩ
d) None of the mentioned

Explanation: Thermistor has negative temperature coefficient of resistance. Therefore, △R=-(25kΩ/oC )×(100oC-35oC) = -1625kΩ .
△R=-1.625MΩ.

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

a) Vab = 4.9v
b) Vab = -5.6v
c) Vab =1.2v
d) Vab =-8.2v

Explanation: According to the voltage divider rule,
Va =( Ra×Vdc)/[Ra+(RT+△R)] = (1kΩ×5v)/(1kΩ+75kΩ) = 0.065v
Vb = ( Rb×Vdc)/(Rb+Rc) = (50kΩ×5v)/(50kΩ+250Ω) = 4.975v
The voltage across the output terminal of the bridge, Vab = Va– Vb = 4.9v.

5. Express the equation for transducer bridge, if all the resistor values are equal
a) v=-(△R×Vdc)/(2×R+△R)
b) v=-(△R×Vdc)/2×(R+△R)
c) v=-Vdc/[2×(2×R+△R)].
d) v=-(△R×Vdc)/ [2×(2×R+△R)].

Explanation: If the Ra=Rb=Rc=RT=R(Equal), then the output voltage across the bridge terminals of the transducer bridge is v=-(△R×Vdc)/ [2×(2×R+△R)].
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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

Explanation: 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
d) Change in elasticity

Explanation: 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

Explanation: 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
c) Lithium
d) Phosphorous

Explanation: 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

Explanation: 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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