Download ECE1250F14_HW4_4

1250
F 14
Homework 4-4
1.
Find vo for the op-amp circuit shown below. Assume the op-amp is ideal.
2.
In this problem, you will design a circuit to output a specific (negative) voltage at a
specific light level. The circuit schematic diagram is shown below. The CdS
(Cadmium Sulphide) photocell changes its resistance in response to the amount of
illumination hitting it. The goal of the design is for the circuit to output vo = –10 V
when the light level is 50 lux, which is living room lighting (in Australia!), according
to Wikipedia: http://en.wikipedia.org/wiki/Lux
A datasheet for the Advanced Photonics PDV-P9008 photocell we assume is used in
the design may be found here:
http://www.advancedphotonix.com/wp-content/uploads/PDV-P9008.pdf
According to the datasheet, the resistance of the photocell varies widely from one
photocell to the next. To simplify matters, assume the resistance of our photocell is
10 kΩ at 50 lux.
You are free to choose the values of R, Rs, and Rf. Complete the design by specifying
R, Rs, Rf. (Answers will vary, and the values of R and Rs can be almost arbitrary.)
Assume vs = 5 V, V+ = 15 V, and V– = –15 V.
Hint: Convert everything from the reference through Rs into a Thevenin equivalent
when you calculate vo, or use the circuit system design cards.
ANS:
1. vo = 15 V
2. The Thevenin equivalent through Rs is vTh = vs
The output voltage is vo = -vTh
Rf
R
RCdS + R
, RTh = R || RCdS + Rs .
. One solution would be R = 10 kΩ, giving vTh
RTh
= vs/2 = 5 V/2 = 2.5 V and RTh = 5 kΩ + Rs. Suppose we choose Rs = 10 kΩ so that
RTh = 15 kΩ. We need a gain of –4 for Rf /RTh, so we want Rf = 60 kΩ. We could
use a 100 kΩ potentiometer for Rf.