Question 16: Semiconductors and Logic Gates

Question

The correct truth table for the given input data of the following logic gate is :

Accepted Answer

The given logic gate is,
Gate 1 is an AND gate for the inputs A and $\mathrm{B}, \Rightarrow \mathrm{Y}_1=\mathrm{A} \cdot \mathrm{B}$
Input
Output
A
B
Y
1
= A ÃÂ· B
1
1
1
0
0
0
1
0
0
1
1
1
Gate 2 is an OR gate for the inputs C and $\mathrm{D}, \Rightarrow \mathrm{Y}_2=\mathrm{C}+\mathrm{D}$
Input
Output
C
D
Y
2
= C + D
0
1
1
1
1
1
1
0
1
1
1
1
Gate 3 is a NOT gate for which the input is $Y_1$ (the output of gate 1), $Y_3=\overline{Y_1}$
Input
Output
Y
1
Y
3
=
Y
1
1
0
0
1
0
1
1
0
Gate 4 is an AND gate for inputs $Y_2$ and $Y_3, Y_4=Y_3 \cdot Y_2$
Input
Output
Y
2
Y
3
Y
4
= Y
2
+ Y
3
1
0
0
1
1
1
1
1
1
1
0
0
Gate 5 is a NOT gate which inverts the output of Gate 4 to give the final output, $Y=\overline{\mathrm{Y}_4}$
Input
Output
Y
4
Y =
Y
4
0
1
1
0
1
0
0
1
So, the final truth table is,
A
B
C
D
Output Y
1
1
0
1
1
0
0
1
1
0
1
0
1
0
0
1
1
1
1
1
Hence, the correct option is (A).

Answer

Inputs
Output
A
B
C
D
Y
1
1
0
1
1
0
0
1
1
0
1
0
1
0
0
1
1
1
1
1

Answer

Inputs
Output
A
B
C
D
Y
1
1
0
1
0
0
0
1
1
1
1
0
1
0
1
1
1
1
1
1

Answer

Inputs
Output
A
B
C
D
Y
1
1
0
1
1
0
0
1
1
0
1
0
1
0
1
1
1
1
1
0

Answer

Inputs
Output
A
B
C
D
Y
1
1
0
1
0
0
0
1
1
0
1
0
1
0
1
1
1
1
1
1

Semiconductors and Logic Gates

The correct truth table for the given input data of the following logic gate is :

Choose the correct answer: