This set of Linear Integrated Circuit Multiple Choice Questions & Answers (MCQs) focuses 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
Explanation: 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) None of the mentioned
Explanation: Output current, Io = Vin /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
Explanation: In all the applications mentioned above, the input voltage Vin is converted into an output current of Vin/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
Explanation: 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
Explanation: 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
Explanation: 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
Explanation: Given Vin=1v ,R1=10+2kΩ ≅2kΩ
Io = Vin/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
Explanation: A combination of an ammeter and a full wave rectifier can be employed in the feedback loop to form an ac voltmeter.
9. For a full wave rectification, in a low voltage ac voltmeter, the meter current can be expressed as
a) Io = (1.9×Vin)/R1
b) Io = (3.9×Vin)/R1
c) Io = (0.9×Vin)/R1
d) Io = (2.9×Vin)/R1
Explanation: For full wave rectification, meter current is expressed as Io = 0.9xVin/R1.
Sanfoundry Global Education & Learning Series – Linear Integrated Circuit.