Currently I am messing around with the HC74 4049 Hex Inverter, which has the following pinout:

enter image description here

I began by connecting one input (marked by the lowest brown stripe on the breadboard) to the first input pin (1A). Then, I used another two jumpers to position the corrosponding inverted signal (1Y!) right next to the actual input signal. As a matter of fact, the whole wiring works and the input 1A is inverted, indicated by how the LED (which is connected to 1Y!) lights up when 1A is low.

The problem, however, is that as soon as I move my (human) body away from the breadboard, the LED gradually grows darker and eventually 1Y! turns low, even though 1A is low as well. In contrast, as I approach the chip, the LED becomes lighter.

enter image description here

Then, as I connected a second input and output, the same body movements caused the LED (no matter whether connected to 1Y! or 2Y!) to grow lighter and darker even quicker. In addition, the outputs start to bug horribly.

enter image description here

Does anyone know what might cause the Hex Inverter to malfunction like this? What did I do wrong?

Unfortunately, I did not find a single entry on the web addressing this issue. Is it possible that the problem is due to certain interdepencendies between the wires (taking into account that the problem worsens the more cables are connected)?

  • 5
    \$\begingroup\$ Try searching google for "open CMOS input". \$\endgroup\$ Jul 23 at 22:18
  • 2
    \$\begingroup\$ @KevinWhite likely has the cause. We like schematics rather than breadboard photos (following wires is error-prone). Other potential problems: (1) a resistor in series with LED seems missing. (2) Vdd-to-GND bypass capacitor missing. \$\endgroup\$
    – glen_geek
    Jul 23 at 22:24
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    \$\begingroup\$ CMOS inputs have extremely high impedance and are very sensitive to any coupling. Humans are somewhat conductive bags of saltwater. Depending on how you power the circuit, there is likey enough capacitive coupling between you and the CMOS input that it detects mains hum. How do you power the circuit? \$\endgroup\$
    – Justme
    Jul 23 at 22:47
  • \$\begingroup\$ The messy long wires and rows of contacts on a breadboard are antennas for electrical, radio, TV and other kinds of interference. Whenever we talk about an electronic circuit we always say its supply voltage. You did not. \$\endgroup\$
    – Audioguru
    Jul 24 at 0:41
  • \$\begingroup\$ I think a schematic is essential. For the first picture i don’t understand the LED position and the green wire routing. Are you shorting vcc with that blanc wire (LED)? \$\endgroup\$
    – RemyHx
    Jul 24 at 1:14

1 Answer 1


Does anyone know what might cause the Hex Inverter to malfunction like this?

It's functioning perfectly as expected. It is however being mis-used.

The inputs to CMOS logic are very high impedance, essentially open circuits.

In normal logic use, you would connect all inputs to ground or rail via a relatively low resistance connection, for instance a 100k resistor.

With open inputs, what you have is a charge detector. With tens of volts of mains pickup capacitively coupled to your body, and possibly thousands of volts of static on it, you can see it won't take much stray capacitance, and changes in that capacitance as you move around, to impress a few volts onto a wire connected to a high impedance CMOS input pin, which will then cause the gate to operate.

CMOS can draw quite high currents when the inputs are not at ground or rail, it's possible to overheat a CMOS IC by leaving its inputs floating.


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