Electrical and Electronic Principles: Sensor Circuit Design, Logic Design, and A.C. Load Supply

This article discusses the design of sensor circuits, logic circuits, and A.C. load supply in Electrical and Electronic Principles. It includes Boolean expressions, Logisim circuits, and testing.

Running head: ELECTRICAL AND ELECTRONIC PRINCIPLES
EAT119 Electrical and Electronic Principles
Name of the Student
University of Sunderland
Author Note

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Student ID: 499304
Hence, ‘uvw’ = 499 and ‘xyz’ = 304.
Task A: Sensor Circuit Design
The sensor circuit design is given below.
Given that, Vs=( V DD
4 ) ( 2 ( ‘ uvw ’
999 ) +1 ) = ( 5
4 ) ( 2 ( 499
999 ) +1 ) = 2.499 Volts.
1) The output voltage V A is high when V+ is more than V- and the value of the
V A =V DD=5 volts, otherwise V A = 0 volts.
Now, applying voltage division rule supply voltage VS is divided across VR1 and VR2.
Or, VR1 = 5 – VS = 5-2.499 = 2.501 Volts.
VR2 = 2.499 Volts.

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Now, it is given that the potential divider takes a current of 0.5 mA.
Hence, 0.5∗10−3*R1 = 2.501 => R1 = 2.501∗103
0.5 = 5002 Ω ~ 5 kΩ
Now, 0.5∗10−3*R2 = 2.499 => R2 = 2.499∗103
0.5 = 4.997 kΩ.
2) Now, using two E12 resistors in the range given in the question the resistors R1 and R2
need to be designed.
R1 = series combination of (4.7*10^3) Ω and (3.3*10^3) Ω gives approximately equal to 5
kΩ.
R2 = series combination of (4.7*10^3) Ω and (1.8*10^3) Ω gives approximately equal to
4.997 kΩ.
Using the two E12 resistors for each of the resistors R1 and R2 approximately gives the same
current limit of 0.5 mA.
Task B: Logic Design
Water level can be in either of the three positions of Sensor A, Sensor B and Sensor C. So, at
most any of the three voltage signals V A, V B and V C can be high in one time, others must be

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low. Hence, the faulty signal occurs when the three voltages V A, V B and V C are high
simultaneously or any two of them are high simultaneously. In that case the Faulty signal VF
will go high(showing that fault has been detected) otherwise it will be low.
Hence, the Boolean expression of the faulty signal V F will be,
V F=V A V B V C + V A V B+V B V C +V A V C
Logisim circuit:

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Testing:

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Hence, from the testing it is clear that the fault signals become high when two or more than
two of the voltages V A, V B and V C are high.

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Task C: A.C. Load Supply
As given, I L=(0.09 ( ‘ xyz ’
999 )+0.01 ) = (0.09 ( 304
999 )+0.01 )= 0.037 Amps.
Now, in the saturation condition current free flows from the Collector to Emitter region.
Hence, in saturation the emitter current will be 0.037 Amps. The base current (IB) is low as
considered to be 10% of the collector current = 0.037* (10/100) = 0.004 Amps.
Now, the applied voltage across resistor RB is same as the faulty voltage VF = 5 Volts.
So, V(RB) = IB*RB => 5 = 0.004* RB => RB = 5/0.004 = 1250 Ω = 1.25 kΩ.
So, the value of RB required such that the NPN transistor work in saturation is equal to 1.25
kΩ.
Now, the diode is connected anti-parallelly across the relay coil to make the load current IL
unidirectional i.e. in case the current flow reverses due to Fault voltage VF getting higher than
supplied voltage VDD the current will not flow through the coil instead will flow through the
diode which will in turn reduce the difference between VF and VDD. Also, the diode ensures
that the voltage induced in the AC supplied load remains the same.