In a BCD-to-decimal decoder, why are the input codes 1010 to 1111 treated as don't-cares?

They are not valid BCD digits and never occur

They activate all outputs at once

What happens for the invalid input 1111 in the simplified decoder, and how does the 7442 avoid it?

D7 and D9 both go active; the 7442 uses full decoding so all outputs stay inactive

Only D9 goes active; the 7442 adds a clock

No output is active; the 7442 uses a latch

D0 goes active; the 7442 uses a ROM

Why is a decoder classed as a combinational circuit rather than a sequential one?

Its output depends only on the present inputs and it has no memory

It uses flip-flops but no clock

Its output depends on past states

It needs feedback from output to input

2073 Shrawan · 2+6 marks

Combinational vs sequential; BCD-to-decimal decoder

Short answer

Combinational circuits have no memory and output depends only on current inputs; sequential circuits use memory elements and past state. A BCD-to-decimal decoder activates one of ten outputs for valid 4-bit BCD inputs (0–9), with don't-care conditions for invalid codes 10–15 simplifying logic.

Differentiate between combinational and sequential circuits. Explain BCD-to-Decimal decoder circuit with suitable diagram.

Combinational vs sequential circuits

Point	Combinational	Sequential
Output depends on	Present inputs only	Present inputs and past state
Memory	None	Has memory (flip-flops)
Feedback	No	Yes
Clock	Not needed	Usually needed (synchronous)
Building blocks	Gates	Gates + flip-flops
Examples	Adder, MUX, decoder	Counter, register, FSM

BCD-to-decimal decoder

A BCD-to-decimal (4-line to 10-line) decoder accepts a 4-bit BCD code $ABCD$ ( $A$ = MSB) and makes exactly one of ten outputs $D_0$ – $D_9$ active, showing the decimal digit. The 7442 is the TTL IC (active-low outputs).

Truth table

ABCD	Active output	ABCD	Active output
0000	$D_0$	0101	$D_5$
0001	$D_1$	0110	$D_6$
0010	$D_2$	0111	$D_7$
0011	$D_3$	1000	$D_8$
0100	$D_4$	1001	$D_9$

Codes 1010–1111 are invalid in BCD, so they are don't-cares.

Output expressions. With full decoding each output is a 4-variable minterm, e.g. $D_5 = A'BC'D$ . Using the don't-cares (K-map grouping), shorter expressions are obtained:

Output	Full decoding	Simplified
$D_0$	$A'B'C'D'$	$A'B'C'D'$
$D_1$	$A'B'C'D$	$A'B'C'D$
$D_2$	$A'B'CD'$	$B'CD'$
$D_3$	$A'B'CD$	$B'CD$
$D_4$	$A'BC'D'$	$BC'D'$
$D_5$	$A'BC'D$	$BC'D$
$D_6$	$A'BCD'$	$BCD'$
$D_7$	$A'BCD$	$BCD$
$D_8$	$AB'C'D'$	$AD'$
$D_9$	$AB'C'D$	$AD$

For example, $D_8$ : 1000 can be grouped with don't-cares 1010, 1100, 1110, giving $AD'$ .

Logic diagram

 A  B  C  D
 |  |  |  |---[NOT]-- D'
 |  |  |------[NOT]-- C'
 |  |---------[NOT]-- B'
 |------------[NOT]-- A'

 A',B',C',D' --[AND4]--> D0
 A',B',C',D  --[AND4]--> D1
 B',C,D'     --[AND3]--> D2
 B',C,D      --[AND3]--> D3
 B,C',D'     --[AND3]--> D4
 B,C',D      --[AND3]--> D5
 B,C,D'      --[AND3]--> D6
 B,C,D       --[AND3]--> D7
 A,D'        --[AND2]--> D8
 A,D         --[AND2]--> D9

Working: for input $0111$ , only the gate $BCD$ has all inputs 1, so $D_7 = 1$ and the others are 0. The simplified version gives false outputs for invalid codes (e.g. 1111 activates $D_7$ and $D_9$ ); if invalid codes must be rejected, full decoding is used, as in the 7442 which keeps all outputs inactive for 1010–1111. Uses: driving decimal indicators (Nixie tubes, LEDs) and selecting one of ten devices.

✓ Answer checked by Claude Opus 5.5 (high effort)

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