You've managed to take a 7492's output of 0-1-2-3-4-5-8-9-A-B-C-D and correct it to 0-1-2-3-4-5-6-7-8-9-reset, yet you can't make a simple one digit correction built onto that for the 12 hour? :shock: I'm assuming that instead, you used a 7490 to get 0-9 for the ones digit, and then a 7492 to get 0-5 for the tens digit. If so that makes a lot more sense, as there would be no correction logic necessary at this point.
Anyway, on to correcting a 7492's output to result in 12-01-02-03-04-05-06-07-08-09-10-11. Use a truth table and Karnaugh maps to transfer from your given bit patterns to the desired bit patterns.
Code:
Given | Desired
|
Q3 Q2 Q1 Q0 | Ten-HR 1-HR D4 D3 D2 D1 D0
----------------+------------------+------------------
0 0 0 0 0 | 1 2 | 1 0 0 1 0
1 0 0 0 1 | 0 1 | 0 0 0 0 1
2 0 0 1 0 | 0 2 | 0 0 0 1 0
3 0 0 1 1 | 0 3 | 0 0 0 1 1
4 0 1 0 0 | 0 4 | 0 0 1 0 0
5 0 1 0 1 | 0 5 | 0 0 1 0 1
8 1 0 0 0 | 0 6 | 0 0 1 1 0
9 1 0 0 1 | 0 7 | 0 0 1 1 1
A 1 0 1 0 | 0 8 | 0 1 0 0 0
B 1 0 1 1 | 0 9 | 0 1 0 0 1
C 1 1 0 0 | 1 0 | 1 0 0 0 0
D 1 1 0 1 | 1 1 | 1 0 0 0 1
You'll feed D3-D0 into the 7447 to get your hours digit output. As for the ten-hours output, it looks kind of funny to have a leading zero, so it's simple. It's either completely off, or it displays a 1. Just connect your D4 line straight to your 7 segment's B and C lines, so when D4 is on, a 1 displays, when D4 is off, nothing displays.
Another possibility is that you use a 7493 chip, which counts up to 16. When it starts out at 0, simply convert that to a 12. All of the rest of its outputs will be correct. Then when it reaches its own 12, have that trigger a reset so it starts the cycle all over again.