As further issues arise I see myself stuck once more with a dysfunctional circuit. Whenever I press the clock button I get all of the leds on except maybe one or two, with some flashing during the button press. I have added a debounce circuit, but still to avail.


simulate this circuit – Schematic created using CircuitLab

Debounce Circuit: http://www.labbookpages.co.uk/electronics/debounce.html#what Datasheet for 74161N: http://www.jameco.com/Jameco/Products/ProdDS/49664.pdf

  • \$\begingroup\$ 100uF ! :O Really? That's probably causing as much trouble as no debounce was. I'd reduce that down to something in the 100nF to 1uF range. \$\endgroup\$ – brhans Feb 5 '16 at 2:03
  • \$\begingroup\$ Oh, Ill change that and see what happens. \$\endgroup\$ – tjpc3 The Redstoner Feb 5 '16 at 2:06
  • \$\begingroup\$ 100nf didn't work and I don't have a 1uf capacitor \$\endgroup\$ – tjpc3 The Redstoner Feb 5 '16 at 2:12
  • \$\begingroup\$ Try increasing R5 (and keep the 100nF) \$\endgroup\$ – brhans Feb 5 '16 at 2:34

Several issues.

1) You claim to have copied the circuit you linked, but you didn't - you've failed to include the Schmitt trigger, typically a 7414. Without this buffering, your switch should not work on a button push, only on a release.

2) Your reset, pin 1, needs a 1k pullup from pin 1 to +5.

3) Pins 7, 9, and 10 should be tied together and pulled to +5 through another 1k resistor.

As a matter of principle, ALL unused inputs of a logic IC need to be connected to something. This is not always true of 7400 circuits, but it will be critical when you graduate to CMOS, such as 74HC chips. Get into the habit now.

Finally, you should not count on TTL outputs driving high, as you have shown. TTL works much better pulling loads down. So your outputs should be connected like


simulate this circuit – Schematic created using CircuitLab

You may object that this will provide LEDs which are on when the output is low, and off when the outputs are high, which is confusing. True enough. In that case, invert the signals going to your LEDs. Since you already need one section of a 7414, and there are 6 gates per IC, you simply go


simulate this circuit

  • \$\begingroup\$ Okay I wired 7,9, and 10 together and then connected them to a 1K resistor to 5V. Then I used a pullup resistor for the clear switch. I switched the LEDs around and connected their input to 5V and their ground to the output pins. I'm still getting the same effect. (inverted of course because of the LEDs) \$\endgroup\$ – tjpc3 The Redstoner Feb 5 '16 at 23:25
  • \$\begingroup\$ Also, have you added a 0.1 uF ceramic cap from pins 8 to 16, as close to the IC as possible? \$\endgroup\$ – WhatRoughBeast Feb 6 '16 at 1:01
  • \$\begingroup\$ Thanks! I added a 0.1uf ceramic capacitor across 5V and ground and it started to count properly! \$\endgroup\$ – tjpc3 The Redstoner Feb 6 '16 at 1:43
  • \$\begingroup\$ Ah. It's always the simple things. For a circuit board with a ground plane, the usual guideline is at least one cap per 5 ICs. For breadboards, use a cap for each IC. You can often get away with less, but as you've just learned, insufficient decoupling can have weird symptoms. \$\endgroup\$ – WhatRoughBeast Feb 6 '16 at 1:59

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