JEE MainPhysicsSemiconductors
A logic circuit has two independent branches. The first branch produces an output Y₁ using a NAND gate with inputs A and B. The second branch produces an output Y₂ using an XOR gate with inputs C and D. An ideal diode and a resistor are connected in series between the terminals Y₁ and Y₂ . The p-side of the diode is connected to Y₁ and its n-side is connected to the resistor, which is in turn connected to Y₂ . For wh
Options
- AA=1, B=1, C=1, D=0
- BA=1, B=1, C=0, D=0
- CA=1, B=0, C=1, D=0
- DA=0, B=1, C=1, D=1
Correct answer
D. A=0, B=1, C=1, D=1
Step-by-step solution
For current to flow through the resistor, the ideal diode must be forward-biased. This requires the potential at the p-side to be strictly greater than the potential at the n-side. Therefore, we need Y₁ = 1 (HIGH) and Y₂ = 0 (LOW). The output Y₁ is given by the NAND gate: Y₁ = A B . For Y₁ = 1 , we must have A B = 1 , which means inputs A and B cannot be both 1. The output Y₂ is given by the XOR gate: Y₂ = C D . For Y₂ = 0 , the inputs C and D must be identical (either both 0 or both 1). Let us check the given opti