Semiconductor Electronics Part 3

Concept:The circuit uses two diodes with their cathodes connected to inputs A and B, and a pull-up resistor R to V_dc. It functions as an AND gate where the output C is HIGH (5 V) only when both inputs are HIGH (5 V), otherwise it is LOW (0 V).Explanation:We assume positive logic: LOW = 0 V (logic 0), HIGH = 5 V (logic 1).Supply voltage $V_{dc} = 5$ V is connected to output C through resistor R.Diodes $D_1$ and $D_2$ are connected with their cathodes (n-side) to inputs A and B respectively. They act as switches that conduct when forward-biased (cathode lower than anode).Case 1: $A = 0$, $B = 0$ (both LOW).Both diodes are forward-biased (anode at $+5$ V, cathode at $0$ V). They conduct, pulling output C to approximately 0 V. Output C = 0.Case 2: $A = 0$, $B = 1$ (one LOW, one HIGH).Diode $D_1$ (cathode at 0 V) is forward-biased, conducting current to ground. This forces output C to near 0 V. Diode $D_2$ (cathode at 5 V) is reverse-biased (off). Output C = 0.Case 3: $A = 1$, $B = 0$ (one HIGH, one LOW).Diode $D_2$ is forward-biased, conducting to ground at B, so output C stays at 0 V. Diode $D_1$ is reverse-biased. Output C = 0.Case 4: $A = 1$, $B = 1$ (both HIGH).Both cathodes are at 5 V, same as anodes. Diodes are reverse-biased (no current). No voltage drop across R, so output C is at supply voltage 5 V. Output C = 1.Truth table:The output is HIGH only when both inputs are HIGH. This is the characteristic of an AND gate.Answer:The circuit functions as an AND gate. Therefore, the correct option is A.