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
Answer:A
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.