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Decoder; 3×8 decoder from two 2×4 decoders

Short answer

A decoder converts an n-bit input into one active output among 2^n lines. A 3×8 decoder uses two 2×4 decoders with enable: inputs B and C connect to both decoders; A=0 enables decoder1 (D0–D3), A=1 enables decoder2 (D4–D7), using an inverter on A.

What is meant by a decoder? Construct 3×8 decoder using two 2×4 decoders and additional gates if required.

A decoder is a combinational circuit that converts an nn-bit binary input code into one of 2n2^n unique output lines; for each input combination exactly one output is active (it generates all minterms).

3×8 decoder using two 2×4 decoders

Let the inputs be AA (MSB), BB, CC (LSB). Each 2×4 decoder has an enable input EE (active high).

  • BB and CC go to the inputs of both 2×4 decoders.
  • AA selects which decoder works: A′A' (through a NOT gate) drives EE of decoder 1 and AA drives EE of decoder 2.
  • When A=0A = 0, decoder 1 is enabled and gives D0D_0–D3D_3; when A=1A = 1, decoder 2 gives D4D_4–D7D_7.
                 +---------+
B,C --------+--->| 2x4 #1  |--> D0..D3
            |  +>| E       |
            |  | +---------+
            |  |
A --+-[NOT]-|--+
    |       |    +---------+
    |       +--->| 2x4 #2  |--> D4..D7
    +----------->| E       |
                 +---------+
ABCActive output
000D0D_0
001D1D_1
010D2D_2
011D3D_3
100D4D_4
101D5D_5
110D6D_6
111D7D_7

Only one extra gate (an inverter) is needed. With an overall enable EE, use E⋅A′E\cdot A' and E⋅AE\cdot A (two AND gates) for the two enables.

✓ Answer checked by Claude Opus 5.5 (high effort)

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