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

This article covers the design of a sensor circuit, logic design, and A.C. load supply in Electrical and Electronic Principles. It includes calculations, truth tables, and Boolean expressions.

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

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Given student ID number: 289967
Hence, ‘uvw’ = 289 and ‘xyz’ = 967.
Task A: Sensor Circuit Design
The Op-Amp with the sensor circuit is given below.
Given that, Vs=( V DD
4 ) ( 2 ( ‘ uvw ’
999 ) +1 ) = ( 5
4 ) ( 2 ( 289
999 ) +1
) = 1.97 Volts.
1) Given that,
VA = VDD if V+ > V-
= 0 V if V+ < V-
Now, applying voltage division rule supply voltage VDD is divided across VR1 and VR2.
Or, VR1 = 5 – 1.97 = 5-1.97 = 3.03 Volts.

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VR2 = 1.97 Volts.
The potential divider has a current of 0.5 milliamps.
So, 0.5∗10−3*R1 = 3.03 => R1 = 3.03∗103
0.5 = 6060 Ω ~ 6.06 kΩ.
Now, 0.5∗10−3*R2 = 1.97 => R2 = 1.97∗103
0.5 = 3940 Ω ~ 3.94 kΩ.
2) Now, the resistor R2 can be modelled as the series combination of two E12 resistors
of the range given below.
3.9*10^3 + 3.9*10 = 3939 Ω ~ 3.94 kΩ.
Now, the resistor R1 can be modelled using two E12 resistors in series as given below.
5.6*10^3 + 4.7*10 = 6070 ~ 6.06 kΩ.
Task B: Logic Design
The water level cannot be in position A, B and C simultaneously, the level must be in a single
position which is either A, B or C. Hence, the faulty condition occurs when V A , V B , V C are
high at a time or any two of them are high at a time. When the water level is in a position of

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either A, B or C, the sensors senses and the voltage of either V A , V B∨V C is applied which is
equal to supply voltage VDD = 5 Volts. The faulty signal VF becomes high when either 3
voltages V A , V B , V C are high simultaneously or any two of them are high simultaneously.
Otherwise, the faulty signal voltage VF remains zero.
The Boolean expression is
V F=V A V B V C + V A V B+V B V C +V A V C
Truth table:
V A V B V C V F
0 0 0 0
0 0 1 0
0 1 0 0
0 1 1 1
1 0 0 0
1 0 1 1
1 1 0 1
1 1 1 1

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Task C: A.C. Load Supply
Given that, I L=
(0.09 ( ‘ xyz ’
999 )+0.01 ) = (0.09 ( 967
999 )+0.01 )= 0.097 Amps.
Now, current freely flows from collector to the emitter region in the saturation condition. So,
in saturation the current through the emitter will be 0.097 Amps. The base current (IB) is very
low and considered to be 10% of the collector current = 0.097* (10/100) = 0.0097 Amps.
Now, the V(RB) = VF = 5 Volts.
So, Voltage across (RB) = IB*RB => 5 = 0.0097* RB => RB = 5/0.0097 = 515.5 Ω = 0.515 kΩ.
So, the value of RB required such that the NPN transistor work in saturation is equal to 0.515
kΩ.
Now, the diode is connected anti-paralleled direction to the relay coil so that load current IL
do not flows in the reverse direction when VF becomes more than the supply voltage VDD.
The current will instead go through the diode not through the coil. The diode also blocks the

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flow of current from VDD to transistor ensuring that the current flows from AC supplied load
only.