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7493 counter

  • 10-03-2011 1:27am
    #1
    thecatspjs
    Closed Accounts Posts: 519 ✭✭✭


    This seems like the best place for this thread but if the mods think I'll get more feedback please move it.

    For my electronics class I've had to build a circuit containing a 555 timer, binary counter, decoder and 7 segment display. Everything is working fine except for the reset function on the binary counter.

    I'm having problems troubleshooting so I was wondering if anyone could give me any tips.

    The counter is set so that it only counts to 8 (ie. resets at 9), so I have QA and QD connected to R1 and R2. Both reset pins should be tied to high to reset the counter but it's not working for me. Should the resets be disconnected from QA and QD first? (I would have tried this already but my breadboard is a bit crammed so I want to be sure before I take it all apart!!)

    I hope someone knows what I'm on about, cheers.


Comments

  • #2
    Capt'n Midnight
    Moderators, Recreation & Hobbies Moderators, Science, Health & Environment Moderators, Technology & Internet Moderators Posts: 92,625 Mod ✭✭✭✭Capt'n Midnight


    http://www.kpsec.freeuk.com/components/74series.htm#7493
    The 74393 contains two separate 4-bit (0 to 15) counters, one on each side of the IC. They are ripple counters so beware that glitches may occur in logic systems connected to their outputs due to the slight delay before the later outputs respond to a clock pulse.

    For normal operation at least one reset0 input should be low, making both high resets the counter to zero (0000, QA-QD low). Note that the 7490 has a pair of reset9 inputs on pins 6 and 7, these reset the counter to nine (1001) so at least one of them must be low for counting to occur.

    Counting to less than the maximum (9 or 15) can be achieved by connecting the appropriate output(s) to the two reset0 inputs. If only one reset input is required the two inputs can be connected together. For example: to count 0 to 8 connect QA (1) and QD (8) to the reset inputs.

    if you are using 74HC93 or similar then you may need to add tie down resistors ( 10K whatever) to inputs in case the it doesn't sink enough current ( cmos has high input impedence) but that's only if all else fails


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