A. 0 – 2.51v
B. 0 – 2.22v
C. 0 – 3.74v
D. 0 – 4.93v
Answer: D
When all binary input D0 through D7 are logic 0, the current Io =0.
∴ The minimum value of Vo =0v.
When all the inputs are at logic 1, Io = (Vref/R1) ×
(1/2+1/4+1/8+1/16+1/32+1/64+1/128/+1/256)
= (3/2kΩ) × (0.996) =1.494mA.
Hence, the maximum value of output voltage is
Vo= Io×RF = 1.494×3.3kΩ =4.93v.
Thus, the output voltage range is from 0 to 4.93v.
4. Calculate the change in the output voltage if the photocell is exposed to a light of 0.61lux from a dark condition. Specification: Assume that the op-amp is initially nulled, Minimum dark resistance = 100kΩ, and resistance when illuminated (at 0.61lux) = 1.5kΩ.
A. Vo –> 23v to 50v
B. Vo –> 0v to 33.11v
C. Vo –> -1.653v to 8.987v
D. Vo –> -0.176v to -11.73v
Answer: D
The resistance RT in darkness is 100kΩ.
The minimum output voltage in darkness is
Vo min = -(Vdc×RF)/ RT
= -(3.2v×5.5kΩ)/100kΩ = -0.176v.
When the photocell is illuminated, its resistance RT =1.5kΩ.
Therefore, the maximum output voltage is
Vo max = -(Vdc×RF)/ RT
= -(3.2v×5.5kΩ)/1.5kΩ =-11.73v.
Thus, Vo varies from -0.176v to -11.73 as the photocell is exposed to light from a dark condition.
5. Which cell can be used instead of a photocell to obtain an active transducer in photosensitive devices?
A. Photovoltaic cell
B. Photodiode
C. Photosensor
D. All of the mentioned
Answer: A
A photovoltaic cell is a semiconductor junction device that converts radiation energy into electrical energy and hence it does not require external voltage.
6. If the input applied to DAC using the current to voltage converter is 10110100, determine the reference voltage (Assume Io= 2mA and R1=1.2kΩ)
7. The current to voltage converter photosensitive device can be used as
A. Light intensity meter
B. Light radiating meter
C. Light deposition meter
D. None of the mentioned
Answer: A
The photosensitive device can be used as a light intensity meter by connecting a meter at the output that is calibrated for light intensity.
8. 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
Answer: C
For full-wave rectification, meter current is expressed as
Io = 0.9 ×Vin/R1.
9. Which of the following circuit has the highest input resistance?
A. Voltage follower
B. Inverting amplifier
C. Differential amplifier
D. None of the mentioned
Answer: A
The voltage follower has the highest positive input resistance of an op-amp circuit. For this reason, it is used to reduce voltage error caused by source loading.
