JEE Main 2025 · 8 April · evening · Single correct · +4 / −1
The output voltage in the following circuit is (Consider ideal diode case)
- (a) 0 V
- (b) +5 V
- (c) −5 V
- (d) 10 V
Answer and solution
Answer (a) 0 V
Solution
Physics sir · 3 steps
Both diodes point away from the node, so the one going to ground turns on and pins the node at 0 V.
- Read the diodes. In both D₁ and D₂ the triangle points away from the output node. So each anode is at the node, and the cathodes go to +5 V and to ground.
- Pull-up tries first. The resistor tries to raise the node toward +5 V. As soon as the node rises above 0 V, D₂ is forward biased (anode higher than its grounded cathode).
Diode Anode (node) Cathode State D₂ 0 V 0 V (ground) ON, ideal, so zero drop D₁ 0 V +5 V reverse biased, OFF - So. The ideal diode D₂ clamps the node to ground, and a current 5/R flows through the resistor into ground. D₁ stays off, so Vₒᵤₜ = 0 V, option (a).
The idea to keep. In a diode-resistor logic circuit, the diode whose input is lowest conducts and fixes the output. This is a diode AND gate: one input at 0 drags the output to 0.
Answer from the NTA final answer key. Solution by PrepWiser.