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

My breadboard looks as following:

Notes:

  • 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!

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  • \$\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

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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.

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  • \$\begingroup\$ Thanks four your quick and clear answer, greatly appreciated! Works like a charm :) \$\endgroup\$
    – Anne
    Commented Jun 7, 2013 at 21:35
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    \$\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

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