In the following circuit, the correct relation between output (Y) and inputs A and B will be :

Solution

Physics sir · 3 steps

Split the circuit in two: the diodes make an AND at X, and the transistor flips it.

1. Diode stage. Each diode has its cathode at the input and its anode at X, and X is pulled up to 10 V through R. If either input is at 0, that diode is forward biased and drags X down to about 0. X goes high only when both A and B are high, so X = AB.
2. Transistor stage. X drives the base of an npn transistor in common-emitter mode, with Y taken from the collector. X high puts the transistor in saturation, so Y falls to about 0. X low cuts it off, so no current flows in R_C and Y sits at 5 V. This stage is a NOT gate, so Y = X̅.

   A	B	X (diode AND)	Y (transistor NOT)
   0	0	0	1
   0	1	0	1
   1	0	0	1
   1	1	1	0

3. So. Y is 0 only when both inputs are 1. That is a NAND gate: Y = A̅B̅, option (c).

The idea to keep. Diodes with their anodes tied together and pulled up to the supply form an AND gate. Pointing the other way, with a pull-down resistor, they form an OR gate. A transistor with output at the collector always inverts.

Answer from the NTA final answer key. Solution by PrepWiser.