Find the correct combination of A, B, C and D inputs which can cause the LED to glow. [figure]
- 1000
- 0011
- 0100
- 1101
Answer
(4) 1101
Solution
A NOR gate with both inputs joined acts as a NOT gate.
Outputs of the first two gates are A and B.
NAND gate output: Y1 = A B = A + B (De Morgan's theorem)
Lower NOR gate output: Y2 = C + D
The LED (through the resistor) is connected from Y1 to Y2 with its p-side towards Y1. It glows only when it is forward biased:
Y1 = 1 and Y2 = 0
(A + B) = 1 and (C + D) = 1
Checking the options (ABCD):
1000: C + D = 0 Y2 = 1, LED off
0011: A + B = 0 Y1 = 0, LED off
0100: C + D = 0 Y2 = 1, LED off
1101: A + B = 1, C + D = 1 Y1 = 1, Y2 = 0, LED glows
Answer: (4)