Question 11: Semiconductors and Logic Gates

Question

Find the correct combination of $\mathrm{A}, \mathrm{B}, \mathrm{C}$ and D inputs which can cause the LED to glow.

Accepted Answer

Gate 1 is a NOR gate with both inputs tied together to A. A NOR gate with shorted inputs acts as a NOT gate.
$$ Y_1=\overline{A+A}=\bar{A} $$
Inputs
Output, Y
1
=
A+ A
=
A
1
1
0
0
0
1
Gate 2 is also a NOR gate with both inputs tied to B , which acts as a NOT gate.
$$ Y_2=\overline{B+B}=\bar{B} $$
Inputs
Output, Y
2
=
B+ B
=
B
1
1
0
0
0
1
Gate 3 is a NAND gate receiving inputs $Y_1$ and $Y_2$.
$$ Y_3=\overline{Y_1 \cdot Y_2} $$
A
B
Y
1
=
A
Y
2
=
B
Y
1
ÃÂ· Y
2
$Y_3=\overline{Y_1 \cdot Y_2}$
1
1
0
0
0
1
1
0
0
1
0
1
0
1
1
0
0
1
0
0
1
1
1
0
Gate 4 is a NOR gate receiving inputs C and D .
$$ Y_4=\overline{C+D} $$
C
D
C + D
Y
4
=
C + D
1
1
1
0
1
0
1
0
0
1
1
0
0
0
0
1
For the LED to emit light, it must be forward-biased. The anode is connected to output of gate 1 and gate 2, i.e., $Y_3$. The cathode is connected to the output of gate 4 , i.e., $Y_4$.
To achieve forward bias in a digital logic circuit, the anode must be at a HIGH voltage (1) and the cathode must be at a LOW voltage (0). Therefore, the required conditions are:
$$ Y_3=1 * Y_4=0 $$
Options
A
B
Y
3
(Anode)
C
D
Y
4
(Cathode)
ONN/OFF
(1)
1
0
1
0
0
1
OFF
(2)
0
0
0
1
1
0
OFF
(3)
0
1
1
0
0
1
OFF
(4)
1
1
1
0
1
0
ONN
In case of 1101 , the anode is HIGH and the cathode is LOW. The LED is forward-biased and ON.
Therefore, the correct option is (D).

Answer

1000

Answer

0011

Answer

0100

Answer

1101

Semiconductors and Logic Gates

Find the correct combination of \mathrm{A}, \mathrm{B}, \mathrm{C} and D inputs which can cause the LED to glow.

Choose the correct answer: