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In article <16307@s.ms.uky.edu>, tindle@ms.uky.edu (Ken Tindle) writes...

>I have three binary lines, for eight possible states (yeah, oh wow).  How
>can one light seven LED's from this, with zero having all LED's off and
>seven lighting all seven of them, in bar graph fashion?  A dot display is
>easy, but bar is tougher.  As a final touch, one should tax each digital
>line with but one standard TTL load. 
> 

  Probably no better than your solution but here are 2 ways I can think
of.  

   1. Use resitors to binary weight your 3 lines and turn it into analogue.
  Then use one of the bar graph driver chips made for stereos. You need
  to find one that has linear levals not logrithmic.

   2. Use 74XX138 (or is it 139) one of eight decoder to get 1 of eight
  lines active low. drive LEDs as follows

                         +-|\
                           | >-----+----|<|----330ohm--5v
             line n+1------|/      |
                                   |
                        +----------+
                        |
                        +--|\
                           | >-----+----|<|----330ohm--5v
             line n--------|/      |
                                   |

      Takes three chips 1 74xx138 and 2 quad and gates. How many did
 it take you?



					Chuck@mitlns.mit.edu