Count number of hours elapsed

Question

It's been a long while since I played with circuits, I'd appreciate a sanity check of the following design; I'm trying to count the number of hours since the reset button was last pressed. Ideally it'd last a long time just powered by batteries, but if I ever get it to work I'll be happy!

(Circuit diagram below)

I calculated my NE555 astable values by brute forcing the equation on Wikipedia for 'common values' of resistors and capacitors which were closest to 3600 seconds (I've added the ruby code below for those interested). Do they seem sensible?

• C: 100mF
• R1: 51KΩ
• R2: 470Ω
• CTL capacitor: 10µF (pulled this out of thin air…)
• This should set the duty cycle to 1.00057 hours.

I'm planning on using the following components - is there anything else I should be taking into account?

• 1x low power NE555 (TLC555)
• 1x quad 2-input NAND (SN7400N)
• 1x dual binary ripple counter (74HC393)
• 2x BCD to 7 segment decoders (4511)
• 1x 2 digit 7 segment display (Common Cathode)
• The above capacitors and resistors - I have literally no idea which ones to pick from the ridiculously vast array out there…

(I realised the NAND logic may not be clear - it took me a while to figure out how to implement an AND + OR combo with NANDs! - so I pulled it out beneath the circuit diagram)

Thanks!

Circuit diagram

Reset Logic gate layout

---

Ruby code for brute-forcing the closest NE555 component values:

require "pp"
capacitor = [1, 1.5, 2.2, 3.3, 4.7, 6.8]
resistor = [10, 11, 12, 13, 15, 16, 18, 20, 22, 24, 27, 30, 33, 36, 39, 43, 47, 51, 56, 62, 68, 75, 82, 91]
powers = [0.001, 0.1, 1, 10, 100, 1_000, 10_000, 100_000, 1_000_000]

target = 3600
tolerance = 1

best = []

ln2 = Math.log(2)

powers.each do |c_p|
  capacitor.each do |c_v|
    c = c_v * c_p
    powers.each do |r1_p|
      resistor.each do |r1_v|
        r1 = r1_v * r1_p
        powers.each do |r2_p|
          resistor.each do |r2_v|
            r2 = r2_v * r2_p

            value = ln2 * c * (r1 + 2*r2)

            proximity = (target - value).abs

            if proximity <= tolerance
              best.push(
                proximity: proximity,
                value: value,
                c: c,
                r1: r1,
                r2: r2
              )
            end
          end
        end
      end
    end
  end
end

best.sort! do |a, b|
  a[:proximity] <=> b[:proximity]
end

pp best.take(5)

Answer 1

A few comments to start with:-

• The 4511 needs unused inputs tied to an appropriate logic level
• The 4511 needs series resistors on each of the segment outputs to the common-cathode display
• You are using museum-quality TTL, so you should have a pull-up on the RESET input and pull it down to ground with the switch.
• One day (Edit: or one hour) is not a reasonable period to get from a 555 (even a CMOS one). Keep it to 1-10 seconds. The leakage on a 0.1F capacitor would make a mockery of your precise calculations. You can count down a higher frequency if you like. A 74HC4040 prescaler would allow a ~1 second clock.
• You'll never get better than maybe 5% without adjustment on a 555, and even if you trim it, tenths of a percent. That may not bother you, but the count might be off by an hour in ten or twenty without trimming. Think of how many weeks(Edit: days) it would take to trim a timer with a 1-day (Edit: or one hour) cycle!
• I don't know what a 7469 is.. something like a 74390 dual decade counter perhaps?
• 10uF is unnecessarily high for the control bypass- 100nF is more than enough.
• I didn't check your logic (the semi-pictorial is a bit painful to read) but it looks plausible as a plugboard starting point.