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I just finished building a 4-bit adder on a breadboard and I am experiencing some quite odd/erratic behaviour and I am wondering why. The actual circuit/logic works as intended, meaning the 4-bit adder does what it is supposed to do. The weird behaviour I am experiencing is with the LEDs seemingly arbitrarily turn on/off depending on how close my fingers are to also seemingly arbitrary parts of the circuit.

I am using 74HC86 for XOR gates, 74HC08 for AND gates and 74HC32 for OR gates. The switches are SPST. A 4-bit adder requires 8 XOR gates, 8 AND gates and 4 OR gates. I've used all of the inputs of the logic gates so there shouldn't be any "floating inputs" coming from the logic gates? I have used pull-down resistors on the switches too. Why is this erratic behaviour occuring?

Do I have damaged ICs?

See video to see what happens (YouTube)

I made another post yesterday regarding the same problem. I got a ton of good help and we came to the conclusion that it was most likely "floating inputs" from the CMOS logic gates causing problems. Yesterday I hadn't completed the build yet and I sure had tons of unconnected inputs coming from the logic gates causing problems.

I thought if I just finished the build I surely wouldn't end up with any unconnected inputs because the build requires use of all of them. I finished it today and as stated before, I have no unconnected inputs so where am I experiencing "floating inputs"?

Logic gates are in order (left to right) XOR-AND-XOR-AND-OR

Top left resistors: 10kOhm

Top right resistors: 1kOhm

Bottom right resistors: 10kOhm

**Bottom right is where I use my Arduino to connect 5V to the breadboard and of course Arduino to ground.

enter image description here

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    \$\begingroup\$ The behavior you describe is indicative of open digital inputs on the 74HC series. Careful, floating digital inputs causes the chips to consume more power than they normally should. Fun fact: doing this with 4000 series CMOS would have destroyed them. \$\endgroup\$
    – rdtsc
    Commented Nov 11, 2020 at 14:15
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    \$\begingroup\$ You already have an unresolved question on this issue LEDs only light up when my fingers are near arbitrary parts of the circuit please do not re-post, rather edit your original question with any new findings. Splitting your issue over multiple posts just makes it hard for anyone to make sense of. Also, trying to ask such a question without a schematic of the system is absurd - how could you possibly expect anyone to help without one? \$\endgroup\$
    – Chris Stratton
    Commented Nov 11, 2020 at 15:59
  • \$\begingroup\$ Need a schematic for this design. A picture of the protoboard is too hard to decipher \$\endgroup\$
    – Voltage Spike ♦
    Commented Nov 18 at 16:53

1 Answer 1

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You do have floating inputs. You simply replaced the VCC wires to DIP switches with resistors. So again, when DIP switch is closed, the chip input is connected to VCC via resistor, which is fine. When DIP switch is open, the chip input is connectes to nowhere and floats.

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  • \$\begingroup\$ The black wires I had going from the DIP-switches to ground were replaced with the resistors you see on the image (top left). They are not connected to VCC bu to ground, or am I misinterpreting you? How do I ensure a state for the chip inputs when I have my DIP-switch open? What do I need to do now? \$\endgroup\$
    – NoName123
    Commented Nov 11, 2020 at 14:44
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    \$\begingroup\$ The resistors are there to set the default voltage at the chip input side when DIP switch is in OFF state, so that it does not float. So when the DIP switch is ON it will connect the to the other voltage directly. So it depends which way you want it, but the original circuit with LS chips connected the chip inputs to ground via DIP switch when DIP is ON. It also means the input pins were weakly pulled up when DIP switch was OFF. \$\endgroup\$
    – Justme
    Commented Nov 11, 2020 at 15:00
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    \$\begingroup\$ Pull-up and pull--down resistors must be connected to the IC inputs at all times - they must not be connected/disconnected by the switch. \$\endgroup\$
    – Peter Bennett
    Commented Nov 17 at 20:53

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