This set of Linear Integrated Circuit Multiple Choice Questions & Answers (MCQs) focuses on “Current to Voltage Converter”.
1. The output current equation for MC1408 digital to analog converter would be
a) Io= -(Vref/R1)×[(D7/2)+(D6/4)+(D5/8)+(D4/16)+(D3/32)+(D2/64)+(D1/128)+(D0/256)].
b) Io= (Vref/2R1)×[(D7/2)+(D6/4)+(D5/8)+(D4/16)+(D3/32)+(D2/64)+(D1/128)+(D0/256)].
c) Io= (Vref/R1)×[(D7/2)+(D6/4)+(D5/8)+(D4/16)+(D3/32)+(D2/64)+(D1/128)+(D0/256)].
d) Io= -(Vref/2R1)×[(D7/2)+(D6/4)+(D5/8)+(D4/16)+(D3/32)+(D2/64)+(D1/128)+(D0/256)].
Explanation: MC1408 is a combination of a DAC and current to voltage converter. If the binary signal is input to MC1408 DAC then the output current would be,
Io= (Vref/R1)×[ (D7/2)+(D6/4)+(D5/8)+(D4/16)+(D3/32)+(D2/64)+(D1/128)+(D0/256)].
2. Determine the maximum value of output current of the DAC in MC1408?
Explanation: The output current of DAC is the maximum when all the inputs are logic 1.
Therefore, Io= (Vref/R1)×(1/2+1/4+1/8+1/16+1/32+1/64+1/128/+1/256)=(0.996)×(Vref/R1).
Explanation: 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 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
Explanation: 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 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 active transducer in photosensitive devices?
a) Photovoltaic cell
b) Photo diode
c) Photo sensor
d) All of the mentioned
Explanation: A photovoltaic cell is semiconductor junction device that convert radiation energy into electrical energy and hence it does not require external voltage.
6. If the input applied to DAC using current to voltage converter is 10110100, determine the reference voltage (Assume Io= 2mA and R1=1.2kΩ)
Explanation: Io=Vref/R1×[(D7/2)+(D6/4)+(D5/8)+(D4/16)+(D4/32)+(D4/64)] Vref =(Io×R1)/ (1/2+0+1/8+1/16+0+1/64+0+0)=(2mA×1.2kΩ)/0.703 .
=> Vref = 3.41v.
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
Explanation: 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
Explanation: For full wave rectification, meter current is expressed as Io = 0.9xVin/R1.
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.