Without using an Arduino and microcontrollers, is this the most efficient way to implement two seven-segment displays to show 2-digit numbers (0-30) with the help of BCD decoders?

$$ d_1=d_2=0\\d_3=AC\ +\ AB \\d_4=A^\prime BD+A^\prime BC+AB^\prime C^\prime +BCD\\d_5=A^\prime BC^\prime D^\prime +AB^\prime C^\prime D+ABCD^\prime\\d_6=A^\prime B^\prime C+A^\prime CD+AC^\prime D^\prime+ ABC^\prime\\d_7=A^\prime B^\prime D+B^\prime CD+A^\prime BCD^\prime+ AB^\prime C^\prime D^\prime+ ABC^\prime D\\d_8=E $$

where \$d_i\$ represents the input to the two BCD decoders connected to the two seven-segment displays and \$ABCDE\$ are the 5-bit binary inputs that must be displayed by two seven-segment displays.

The minterms are derived from the truth table for example:

$$ \text{ ABCDE = 11101 (29)}\to \text{ 0010 (2) and 1001 (9) } $$ which will then be fed to the two BCD decoders that is: \$0010\$ to BCD deocder 1 and \$1001\$ to BCD decoder 2 so that the two seven-segment displays would show 29.


I feel that there is a more efficient way to implement this, since the truth table somehow exhibits a "pattern".

  • \$\begingroup\$ Comments are not for extended discussion; this conversation has been moved to chat. \$\endgroup\$
    – Voltage Spike
    Jan 4, 2022 at 6:04

1 Answer 1


is this the most efficient way...

I'd use a 74185A decoder chip.

It's still available and here's the clincher part of the data sheet showing the circuit (up to 6 bits binary in and 2-blocks of data out to feed the 7-seg display decoders): -

enter image description here

So, if by efficient you mean it mops up all the logic gates, then this looks like the route to go.

  • \$\begingroup\$ Logically this is very efficient, but beware of the current consumption - 100mA max! \$\endgroup\$
    – user131342
    Jan 3, 2022 at 17:14
  • \$\begingroup\$ @Hearth that comment is best placed under the question but, it's already been mentioned as a comment there. \$\endgroup\$
    – Andy aka
    Jan 3, 2022 at 17:56

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