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GATE 2016 EC – Question 53

Digital Circuits · Combinatorial Circuits · 2 marks · Multiple choice

The functionality implemented by the circuit below is

Four data inputs P, Q, R, S each feed a tristate buffer. The outputs of the four buffers are joined together to form the output Y. A 2:4 decoder with select inputs $C_1$ and $C_0$ and $Enable = 1$ drives the enable of each buffer, with one decoder output per buffer.
  1. 2-to-1 multiplexer
  2. 4-to-1 multiplexer
  3. 7-to-1 multiplexer
  4. 6-to-1 multiplexer

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Correct answer: (B) 4-to-1 multiplexer

Explanation

The enabled 2-to-4 decoder turns on exactly one of its four outputs according to $C_1 C_0$. Only the tristate buffer connected to that output drives the common output line, so Y equals one of P, Q, R or S chosen by $C_1 C_0$. That is a 4-to-1 multiplexer.