0
\$\begingroup\$

In my project I'm working on right now, I'm trying to use a Schmitt trigger inverter to help debounce input signals. Below is my overall circuit I am trying to implement:

enter image description here The Schmitt trigger inverter I'm using is the following: https://assets.nexperia.com/documents/data-sheet/74HC_HCT14.pdf

The transistor I'm using is: https://www.diodes.com/assets/Datasheets/ZXTN25100DG.pdf

By default, 5V is going into the inverter and 0V is coming out which is what I expect. However when I press the button to connect the inverter to ground, I read that 0V is going in, but only about 1.3V is being outputted. Where have I gone wrong with the circuit?

\$\endgroup\$
7
  • \$\begingroup\$ Well, which one is it, the HC or HCT, the datasheet is for both models? And how is it connected to power supply? What load is it driving? \$\endgroup\$
    – Justme
    Feb 8, 2021 at 5:32
  • \$\begingroup\$ It is the HC, apologies. I've connected 5V to the inverter's VCC and common ground to the GND pin. I am unsure what you mean by what load it is driving? I have the output of the inverter wired to an Arduino Mega as well as a BJT transistor. \$\endgroup\$ Feb 8, 2021 at 5:58
  • \$\begingroup\$ That is exactly what I mean. You don't tell what you have connected and how so we don't know why your output is wrong. And now that you do mention it, I suspect you are driving the BJT in a way that does not allow the output voltage to fully rise. Please add a diagram how the output is connected to Arduino input, and to the BJT, including all resistors that are in the circuit, and even mention if the Arduino is powered on or off, and is the IO pin properly configured as input or is it accidentally an output. Problem is not the HC14 output, the problem is what inputs it drives. \$\endgroup\$
    – Justme
    Feb 8, 2021 at 6:07
  • \$\begingroup\$ Just updated with a diagram of the overall circuit I'm trying to implement. I currently do not have the Arduino connected, so it technically shouldn't be affecting this. I implemented a way for my solenoid to be activate without the use of the Arduino which is why I think I should be able to use the circuit without an Arduino and is how I've been testing so far. \$\endgroup\$ Feb 8, 2021 at 7:14
  • \$\begingroup\$ Are you absolutely sure about the connections, transistor pinout, and resistor values? The only thing wrong in the schematic is the diode orientation, but in real life, we don't know how you build it and if there are some errors. \$\endgroup\$
    – Justme
    Feb 8, 2021 at 7:42

1 Answer 1

1
\$\begingroup\$

Sounds like you forgot to provide a 5V power supply for the chip.

Or the chip is damaged.

\$\endgroup\$
8
  • \$\begingroup\$ I did provide 5V power and ground for the chip. Sorry it's not clear on the diagram. I am not sure how the chip would be damaged. Could soldering the chip on have damaged it? I soldered the legs on fine I believe and when testing it the chip did not pop, so I did not overload it. \$\endgroup\$ Feb 8, 2021 at 5:58
  • \$\begingroup\$ Soldering is extremely unlikely to damage the chip, however probing with power applied (or ESD, especially with power applied) can cause the demise of the chip. It will often (not always) feel very hot as a result (it should never feel warm). \$\endgroup\$ Feb 8, 2021 at 6:00
  • 1
    \$\begingroup\$ If connected to nothing else it should output almost exactly the supply voltage when high. I would spend a lot of effort inspecting (the art, a blank board, and the finished article) before picking up a soldering iron. \$\endgroup\$ Feb 8, 2021 at 8:02
  • 1
    \$\begingroup\$ If it outputs 1.3V when connected to nothing else (with in=0 and VCC present) it has probably been destroyed, possibly by ESD. (There are 5 more on the same chip : have you tried another one?) \$\endgroup\$
    – user16324
    Feb 8, 2021 at 12:03
  • 1
    \$\begingroup\$ Incidentally, I've seen very, very rarely hairline whisker shorts on a PCB that were roughly in the right resistance range to cause that. Unlikely with tested boards or from a high-volume supplier but it's possible. \$\endgroup\$ Feb 8, 2021 at 12:06

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.