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2-to-4 Decoder

Enable input, active high and active low outputs.

Mohith N
Updated: 7 April 2026
4 min read

Address decoding in 8085 microprocessor systems uses a 2-to-4 decoder to assert chip-select lines for four separate memory banks from just two address bits. The 74LS139 dual 2-to-4 decoder handles this in one IC.

2-to-4 Decoder74LS139A0A1ENY0Y1Y2Y3Truth Table (active-low outputs)EN A1 A0 | Y0 Y1 Y2 Y3 1 x x | 1 1 1 1 (disabled) 0 0 0 | 0 1 1 1 0 0 1 | 1 0 1 1 | 0 1 = Y1 low
Figure 1: 2-to-4 decoder (74LS139) with truth table showing active-low output behavior

Core Concept

A decoder converts an n-bit binary input into 2^n unique output lines. The 2-to-4 decoder maps each of the four input combinations (00, 01, 10, 11) to one of four outputs. Unlike a DEMUX, there is no separate data line—the output pattern depends solely on the address inputs.

The 74LS139 contains two independent 2-to-4 decoders. Outputs are active-low: the asserted output goes to logic 0 while others remain at 1. Propagation delay is 22 ns at 5 V. Fan-out is 20 standard TTL loads. Supply voltage: 4.75 V to 5.25 V.

Two 74LS139 decoders can be cascaded with one address bit used as an enable for each, yielding a 3-to-8 decoder. This cascading approach appears frequently in 8085 memory interface questions at Anna University.

Boolean Expression

Each output is a minterm of the two address inputs. Y0 = A1'·A0', Y1 = A1'·A0, Y2 = A1·A0', Y3 = A1·A0. For active-low outputs (74LS139): Y0_bar = NOT(A1'·A0') = A1 + A0 by De Morgan's law. Only one minterm evaluates to 1 at a time, so only one output is low at any given address.

Example
Given:
  A1 = 1, A0 = 0, EN = 0 (enabled)

Formula:
  Y0 = A1'.A0'
  Y1 = A1'.A0
  Y2 = A1 .A0'
  Y3 = A1 .A0

Step by step:
  A1' = NOT(1) = 0
  A0' = NOT(0) = 1

  Y0 = 0 . 1 = 0  → active-low output = 1 (inactive)
  Y1 = 0 . 0 = 0  → active-low output = 1 (inactive)
  Y2 = 1 . 1 = 1  → active-low output = 0 (ACTIVE)
  Y3 = 1 . 0 = 0  → active-low output = 1 (inactive)

Final Answer:
  Y2 is asserted (logic 0 on active-low output)
  Address 10 (decimal 2) selects Y2
Exam Tip: The 74LS139 has active-low outputs. Questions that ask "which output is HIGH when A1A0=10" expect you to say none—Y2 goes LOW. Always check whether the question asks for active-high or active-low behavior. Anna University papers from 2018–2022 frequently test this distinction with the 74138/139 family.

Key Properties

  • IC: 74LS139 (dual 2-to-4), 74HC139 (CMOS version, 2 V to 6 V)
  • Outputs: active-low on 74LS139; invert externally for active-high
  • Propagation delay: 22 ns (74LS139), 7 ns (74HC139)
  • Fan-out: 20 standard TTL loads (LS family)
  • Supply: 4.75–5.25 V (LS), 2–6 V (HC)
  • Enable: active-low; when EN=1, all outputs are deasserted (held high)
  • One IC decodes four unique binary addresses simultaneously

Quick Revision

  • 2-to-4 decoder: 2 inputs → 4 outputs, one active at a time
  • Each output = one minterm of the input variables
  • 74LS139: active-low outputs, dual decoder in one 16-pin DIP
  • No data input: output depends only on address, unlike DEMUX
  • Used for chip select in memory interfacing, instruction decode in simple CPUs
  • Cascade two 74LS139 to make 3-to-8 using one address bit as enable
  • Exam trap: treating active-low outputs as active-high and answering that Y2=1 when A1A0=10

2-to-4 Decoder Quiz

Test your understanding of decoder output logic, enable input, and active-low behavior.

Question 1 of 3

Q1.A 2-to-4 decoder with active-low enable (E_bar) and active-low outputs: when E_bar = 0 and inputs A1=1, A0=0, which output line is asserted (goes LOW)?