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I have a digital clock connected to an MM74C915 7 segment to BCD encoder, which goes into a 74LS42 BCD to 1 of 10 decoder. (Think like a CD4017 but BCD)

Need to somehow convert the 1 of 10 output into a single pulse, everytime the decoder increments.

What I'm using it for:

I have an LM8361 digital clock IC, connected to 7 segments with a ring of 60 LEDs that are driven by a CD4017 from 1Hz pulses.

I need to reset the 4017 every minute to keep the 60 LED ring in sync/sequence with the clock.

Thats where the 7 segment to BCD encoder and BCD to 1 of 10 decoder comes in. It's on the minutes digit. Need to get 1 quick pulse per minute when the digit on the clock changes, from the counter chain as a single pulse (one shot). I've tried or gates, but that just results in a constant minute long output.

What can I connect to the 74LS42 to achieve this? Please note, I am trying to avoid using microcontrollers if at all possible. schematic of the clock. Section in question is top right.

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  • \$\begingroup\$ The 74LS42 is a BCD to decimal (one-of-10) decoder, not a counter. \$\endgroup\$ Commented Apr 1 at 1:14
  • \$\begingroup\$ @PeterBennett when connected to the MM74C915, it does count. But is there a way to get a single pulse everytime the 74ls42 advances? Ie, serially 1010101010 \$\endgroup\$ Commented Apr 1 at 1:26
  • \$\begingroup\$ The 74LS42 is not a counter - it just decodes the BCD count on its input to provide a single output depending on the input count. You might get what you want from the A input bit on pin 15. Look for the 74LS42 datasheet to see how it operates. \$\endgroup\$ Commented Apr 1 at 2:44
  • \$\begingroup\$ Are you trying to do this in the most complicated way possible? Forget about the 7-segment to binary and binary to decimal conversions, and just do change detection directly on the 7-segment data. You don't even need to look at all 7 bits -- I believe that there is a subset of 3-4 where at least one of them changes on each count. \$\endgroup\$ Commented Apr 1 at 3:55
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    \$\begingroup\$ In fact, just looking at segments d and e would be enough. \$\endgroup\$ Commented Apr 1 at 3:58

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