enter image description hereI'm new to electronics, and it's my first time using a hex inverter. The following pins are connected:

  • 1(Input A)-GND or VCC
  • 2(Output A)-LED
  • 7-GND
  • 14-VCC

VCC is 5V, but whatever I do the output is high. It doesn't matter if I connect input a to 5v or 0v the LED still lights up. Can anyone help? Thanks in advance, Lukas

  • \$\begingroup\$ Well, it should not do that. Would you explain how you connect the input to 5V or GND, or how you connect the LED to output. Maybe there is a mistake. Take a photo of what you are trying to do. If you draw a diagram, the diagram may show what you intend to do, instead of what you actually do. \$\endgroup\$
    – Justme
    Sep 22 at 21:37
  • \$\begingroup\$ Thanks for the reply!! I have connected the input to 5V or GND simply using a wire, without any resistors. \$\endgroup\$
    – lukasm06
    Sep 22 at 21:45
  • 1
    \$\begingroup\$ That sounds correct but due to no available photo, there is no proof you are actually doing what you think you are doing. Also the wire may be broken or have bad contact so it does not work. Or you are using the wrong pins for input, output, or supplies. Everything is possible unless ruled out. Draw a diagram, or preferably, provide a photo of scenario that "does not work". For example, you don't mention about a resistor for LED, and don't mention if the LED is active high or low. If no resistor, the chip may have damaged due to overcurrent. Maybe it's the LED that's wired wrong? \$\endgroup\$
    – Justme
    Sep 22 at 21:51
  • 1
    \$\begingroup\$ You could try another gate in the chip as a further check. \$\endgroup\$
    – Barry
    Sep 22 at 21:53
  • 7
    \$\begingroup\$ It hurts to see a black wire at VCC and a red one at GND... \$\endgroup\$
    – Jens
    Sep 22 at 22:42

1 Answer 1

  1. The LED is in backwards or should be connected to 0V instead of 5V: The anode should be connected to 5V for a high on pin 1 to light the LED. If you connect the cathode to 0V then a 0V on pin 1 will light the LED.

  2. The red wire looks like it might accidentaly short to the resistor lead.

  3. You have not tied off the unused inputs. I can't say for sure that this is responsible for your problem but Note 3 in the datasheet says, "NOTE 3. All unused inputs of the device must be held at VCC or GND to ensure proper device operation. Refer to the TI application report: Implications of Slow or Floating CMOS Inputs, literature number SCBA004."

The 74LS04 can sink 8mA, but source 0.4 mA. It is better to connect the anode to 5V cathode to the resistor, then calculate the resitor value for <=8mA. These inverters can be paralleled to get higher current.

The 7404 can sink 16 mA per output.

The OP indicates that ensuring the unused inputs are tied off solved the problem. Also mentioned in a comment the problem changed from always on to always off.

This indicates a latch is created by cross-coupling through the power rails.

A decoupling capacitor from pin 14 directly across the chip to pin 7 must be used to decouple the internal inverters from each other.

A comment by Hearth reminded me of the time that we received counterfeit LS chips from a local shop. They tested as HCT chips.

So I zoomed in on the state of Texas logo on the chip in the OP’s photo. Looking closely reveals a mermaid folded to look like the state of Texas. This is a counterfeit that looks like a TI chip to a glancing eye.

The 74HCT04 data sheet may be a better fit. I had to measure the input characteristics to be certain.

The CMOS inputs will be more sensitive when not tied off, so further explaining the OP’s problem.

  • \$\begingroup\$ Thank you so much, it was due to the inputs that weren't tied off! You've really helped me out here, \$\endgroup\$
    – lukasm06
    Sep 23 at 6:49
  • \$\begingroup\$ An original 7404 would be hard to find these days; for higher current capacity, a 74HC04 would probably be the way to go--the HC series is even easier to find today than the LS series. \$\endgroup\$
    – Hearth
    Sep 23 at 12:28
  • \$\begingroup\$ So true @Hearth. And then there the counterfeiters. Look closely at the logo on the chip. The OP likely has a 74HC04. See my question update \$\endgroup\$
    – RussellH
    Sep 23 at 13:04

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