1
\$\begingroup\$

I am having some issues with debouncing a switch with an RC filter and 74HC14 Schmitt-trigger inverter IC. I am using a standard circuit that is linked in the datasheet. The circuit in question here and here.

I am using the normal switch debounce circuit. The output LED is partially on when the button is not pressed, and the Schmitt-trigger has an output voltage of around 2V. When the button is pressed, the output voltages goes to about 2.7V. However, if I remove the capacitor the LED stays off until the button is pressed.

The circuits in question: screenshot of the circuits mentioned

My question is why is the Schmitt-trigger output never 0V? The IC has a push-pull output, so the output does not need a pullup/pulldown.

Edit: when the capacitor is removed, the output has a normal range of 0V-5V. Also, if it matters, all unused inputs are tied directly to ground, just like the datasheet states.

\$\endgroup\$
9
  • 1
    \$\begingroup\$ There's something wrong somewhere. Bad (low resistance) capacitor or missing ground somewhere. \$\endgroup\$
    – Transistor
    Commented May 29 at 23:29
  • \$\begingroup\$ What is your Vcc? What are you using for the debounce capacitor? \$\endgroup\$ Commented May 29 at 23:38
  • \$\begingroup\$ @MathKeepsMeBusy VCC is 5V and the debounce cap is C2 - 100nF \$\endgroup\$ Commented May 29 at 23:53
  • 1
    \$\begingroup\$ Try a new chip. That one might be damaged from static. \$\endgroup\$
    – MOSFET
    Commented May 29 at 23:54
  • \$\begingroup\$ @MOSFET the chip works, I am using another gate with no issue. \$\endgroup\$ Commented May 30 at 0:01

1 Answer 1

0
\$\begingroup\$

In my case, @Transistor was correct, in that the capacitor was bad. I switched it out with another, and the circuit works properly.

\$\endgroup\$
1
  • \$\begingroup\$ Good work. You should be able to "Accept" your own answer after a day or so. (I can't remember how long.) \$\endgroup\$
    – Transistor
    Commented May 30 at 12:22

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.