In an attempt to make the clock I'm building even more confusing to those who are already confused by it, I'm trying to implement a 7-segment display which shows the current time converted into hexadecimal.

I already have the LED clock display circuit planned out using a 555 timer and a bunch of cascading 7490 counters, and was thinking of ways I may be able to accomplish this. I want the actual time to be converted into a decimal number, so 23:59:59 would equal 235,959, or 399B7 in hex.

My original thought was to feed each counter into a multiplier depending on its place in the decimal format, and then feed each of the products into an adder. So, for 23:59:59, I would have (2*100000)+(3*10000)+(5*1000)+(9*100)+(5*10)+9. 08:14:06 would be (8*10000)+(1*1000)+(4*100)+6

This is kind of where I get stuck. I'm not sure how I would achieve this multiplication without using some sort of programmable device, and if I could, how I could send the large number converted to hex to the display.

I poked around on TI's website and looked at Dadda and Wallace Trees, but I don't believe it'd be feasible to construct one for such large numbers.

Any thoughts?

Thanks.

EDIT: Was just thinking, would it be possible to send my these values into a DAC, then run that through a multiplier with the the right amount of gain, then run that through a ADC? (haven't gotten to that lab in class yet, so dunno if it would work or not, just a thought)

 

Why not just set the counters to count in binary? Each 4-bit counter will then give you a hexadecimal value directly. That will be different from what you want but it would still be fairly confusing and a lot easier to implement. Thus 23:59:59 would be 17:3b:3b in hex.

I have a binary clock that has a matrix of LEDs to indicate the time. Each row has binary values vertically increasing, e.g. 1st row is 1, 2nd row is 2, 3rd row is 4, 4th row is eight. The left two columns are hours, the middle two columns are minutes, and the last two columns are seconds. The LEDs are lit such that you add the total lit in each column to get the time. Thus none lit in the left column and the bottom two lit in the next column would be 3 o'clock.