How do encoder and decoder inputs and outputs differ?

Encoder has 2^n inputs and n outputs; decoder has n inputs and 2^n outputs

Encoder has n inputs and 2^n outputs

Encoder has one input and many outputs

Which is a limitation of the simple 8-to-3 encoder?

It gives a different code for each input twice

2081 Baisakh · 2+5 marks

Encoder vs decoder; 8-to-3 encoder

Short answer

An encoder has 2^n inputs and n outputs, converting one active input to binary using OR gates; a decoder does the reverse. An 8-to-3 encoder uses OR gates: Y2=D4+D5+D6+D7, Y1=D2+D3+D6+D7, Y0=D1+D3+D5+D7. D0 inactive gives 000.

Differentiate between encoder and decoder. Design an 8 to 3 encoder with diagrams.

Encoder vs decoder

Point	Encoder	Decoder
Function	Converts one active line to a binary code	Converts a binary code to one active line
Inputs / outputs	$2^n$ inputs, $n$ outputs	$n$ inputs, $2^n$ outputs
Main gates	OR gates	AND gates
Example	8-to-3 (octal to binary)	3-to-8 (74138)
Use	Keyboards, interrupt encoding	Address decoding, display driving

8-to-3 encoder

An 8-to-3 (octal-to-binary) encoder has eight inputs $D_0$ – $D_7$ and three outputs $Y_2Y_1Y_0$ . When one input $D_k$ is 1, the output is the binary code of $k$ . Only one input is assumed active at a time.

Truth table

$D_7$	$D_6$	$D_5$	$D_4$	$D_3$	$D_2$	$D_1$	$D_0$	$Y_2$	$Y_1$	$Y_0$
0	0	0	0	0	0	0	1	0	0	0
0	0	0	0	0	0	1	0	0	0	1
0	0	0	0	0	1	0	0	0	1	0
0	0	0	0	1	0	0	0	0	1	1
0	0	0	1	0	0	0	0	1	0	0
0	0	1	0	0	0	0	0	1	0	1
0	1	0	0	0	0	0	0	1	1	0
1	0	0	0	0	0	0	0	1	1	1

Output equations. Each output is 1 for the inputs whose code has a 1 in that bit:

\begin{aligned} Y_2 &= D_4 + D_5 + D_6 + D_7 \\ Y_1 &= D_2 + D_3 + D_6 + D_7 \\ Y_0 &= D_1 + D_3 + D_5 + D_7 \end{aligned}

Logic diagram

 D4, D5, D6, D7 ---[OR 4]---> Y2
 D2, D3, D6, D7 ---[OR 4]---> Y1
 D1, D3, D5, D7 ---[OR 4]---> Y0
 (D0 not connected: it gives 000)

Example: $D_6 = 1$ → $Y_2 = 1$ , $Y_1 = 1$ , $Y_0 = 0$ , output $110 = 6$ .

Limitations: $D_0 = 1$ and "no input active" both give 000, and two active inputs give a wrong code (e.g. $D_3$ and $D_4$ give 111). A valid-output bit and a priority encoder (74148) remove these problems.

✓ Answer checked by Claude Opus 5.5 (high effort)

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All Digital Logic chapter 4 questions →

$D_7$	$D_6$	$D_5$	$D_4$	$D_3$	$D_2$	$D_1$	$D_0$	$Y_2$	$Y_1$	$Y_0$
0	0	0	0	0	0	0	1	0	0	0
0	0	0	0	0	0	1	0	0	0	1
0	0	0	0	0	1	0	0	0	1	0
0	0	0	0	1	0	0	0	0	1	1
0	0	0	1	0	0	0	0	1	0	0
0	0	1	0	0	0	0	0	1	0	1
0	1	0	0	0	0	0	0	1	1	0
1	0	0	0	0	0	0	0	1	1	1

$D_7$	$D_6$	$D_5$	$D_4$	$D_3$	$D_2$	$D_1$	$D_0$	$Y_2$	$Y_1$	$Y_0$
0	0	0	0	0	0	0	1	0	0	0
0	0	0	0	0	0	1	0	0	0	1
0	0	0	0	0	1	0	0	0	1	0
0	0	0	0	1	0	0	0	0	1	1
0	0	0	1	0	0	0	0	1	0	0
0	0	1	0	0	0	0	0	1	0	1
0	1	0	0	0	0	0	0	1	1	0
1	0	0	0	0	0	0	0	1	1	1

Encoder vs decoder; 8-to-3 encoder

Encoder vs decoder

8-to-3 encoder

Related questions

$D_7$	$D_6$	$D_5$	$D_4$	$D_3$	$D_2$	$D_1$	$D_0$	$Y_2$	$Y_1$	$Y_0$
0	0	0	0	0	0	0	1	0	0	0
0	0	0	0	0	0	1	0	0	0	1
0	0	0	0	0	1	0	0	0	1	0
0	0	0	0	1	0	0	0	0	1	1
0	0	0	1	0	0	0	0	1	0	0
0	0	1	0	0	0	0	0	1	0	1
0	1	0	0	0	0	0	0	1	1	0
1	0	0	0	0	0	0	0	1	1	1