I'm struggling to implement the HD74HC32 (quadruple 2-input positive-OR gate) package:

My breadboard looks as following:


  • 1.8 volt power supply, sufficient according to the HD74HC32 datasheet
  • 1A is located at the left bottom
  • Vcc connected to [+] and GND connected to [-]
  • 1Y connected to LED Anode
  • LED cathode connected to [-]

I tried every possible combination, with and without resistors, but I cannot get the gate to work. The LED even stays on while 1A and 1B are disconnected.

How do I get this OR gate to work as expected?

The following question looks alike, but I just can't get it to work :(
Connecting AND Gate Chip to an Integrated Circuit

My reputation is too low to embed images, that's why linked directly, sorry!

  • \$\begingroup\$ Can you link the datasheet? Why do you use 1.8V? \$\endgroup\$
    – jippie
    Commented Jun 7, 2013 at 21:16
  • \$\begingroup\$ Thanks for your edit and comment! The HD74HC32 datasheet. I choose the 1.8 volt because I ran out of proper resistors for the LED to work with 5 volt. I'm just "playing around" with logic gates for the first time! \$\endgroup\$
    – Anne
    Commented Jun 7, 2013 at 21:38

1 Answer 1


On CMOS IC's like the one you are using, unused input pins float high. This means that if nothing is connected to them, the circuit will act as if they have a Vcc input applied to them. To resolve this, connect inputs you want to have a value of '0' to ground.

  • \$\begingroup\$ Thanks four your quick and clear answer, greatly appreciated! Works like a charm :) \$\endgroup\$
    – Anne
    Commented Jun 7, 2013 at 21:35
  • 1
    \$\begingroup\$ No, they don't float high, they simply float. All 8 inputs to this chip should be driven (or tied) high or low at all times. \$\endgroup\$
    – Dave Tweed
    Commented Jun 7, 2013 at 21:36

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.