1
\$\begingroup\$

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.

![pic

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

\$\endgroup\$
1
  • \$\begingroup\$ Comments are not for extended discussion; this conversation has been moved to chat. \$\endgroup\$
    – Voltage Spike
    Jan 4 at 6:04

1 Answer 1

2
\$\begingroup\$

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.

\$\endgroup\$
2
  • \$\begingroup\$ Logically this is very efficient, but beware of the current consumption - 100mA max! \$\endgroup\$
    – henros
    Jan 3 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 at 17:56

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.