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Recently bought some TI SN74HC14 Hex Inverter IC's to learn and am having weird issues with them. I've looked at many examples online and have tried wiring them the same with the same passive values. Here's one that I tried: http://www.learningaboutelectronics.com/Articles/7414-schmitt-trigger-oscillator-circuit.php

I hooked the output to my oscilloscope and messed around. It seemed like when I turn on the power from my supply to 5V, the output goes high for a period, then the output goes low for a period, then the output goes to a 'middle noisy state' (similar to what you see if you have take out the capacitor). I tried different capacitor values which would extend the pause period but no matter what it would stop working after the first cycle. Its like the capacitor would start low (bring the output high), get charged (bring the output low), and then never discharge again (like having no capacitor there at all).

As a random last ditch effort I tried going down to 2-3V, and the output is what you'd expect - a square wave. Success!!... kind of...

The chip is rated for up to 7V so i'm not sure why this would be happening. I would assume there must be a way to get this to work everywhere within the IC's spec voltage. My circuit is the exact same as the one in the link posted, with the output going to just an LED and to my oscilloscope (i've also tried it just going to my oscilloscope with the same results). Anyone have any ideas?

Thanks,

Andrew

Edit: 1/23/17 Thanks for the comments so far. To give more information based on your comments, I have added a 0.1 uF decoupling cap (I also tried other values). I've tested the circuit with and without the LED, and the resistor on the LED is 330 Ohms (also tested other values). Also you're right about 7V not being the max, the datasheet says 2V-6V operating voltage, I think I just typed the wrong number.

Edit: After trying stuff...

I tried bumping up the LED resistor from 330 ohms to 1-2k, which let me bring the power supply voltage up a few tenths of a volt before the output became unstable. I then tried grounding the unused outputs, and as I ground them one by one it allows me to increase the supply voltage up to the usable 2-6V range! Looks like the issue was mostly just grounding the unused pins. Thanks!

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    \$\begingroup\$ You don't mention the value of the current-limiting resistor in series with your LED. What is it? Is the power supply to the IC decoupled? \$\endgroup\$ – replete Jan 23 '17 at 1:00
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    \$\begingroup\$ "The chip is rated for up to 7V" - no, it isn't! SN74HC14 Datasheet section 6.1 Absolute Maximum Ratings:- "These are stress ratings only, and functional operation of the device at these or any other conditions beyond those indicated under Recommended Operating Conditions is not implied". \$\endgroup\$ – Bruce Abbott Jan 23 '17 at 4:38
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    \$\begingroup\$ As was already said, that circuit is awful since it has no decoupling of the device supply. Add a 100nF capacitor close to pin 14 of the device (Vcc) with a short connection to ground (pin 7) \$\endgroup\$ – Claudio Avi Chami Jan 23 '17 at 4:59
  • \$\begingroup\$ I know this is late, but it is a better idea to ground the unused inputs and leave the outputs floating. \$\endgroup\$ – evildemonic Mar 9 '18 at 22:37
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Check the function without LED. It's well possible that your LED has too low resistance current limiting resistor which prevents the full output swing. This theory is compatible with the improved operation at lower supply voltages.The resistance in series with the LED should be at least 2 kOhm.

Addendum due the downvotes and comments:

At +5V supply voltage 74HC series output current sink or source capablity is rated to be max 5 mA. Loading that heavily or more is well visible as narrowed output voltage swing. Let's assume the led has 1,5 V forward drop. This leaves 3,5 V to series resistor. 2 kOhm series resistor determines the LED current to be = 1,75 mA. This is about a third part of the rating, surely on the safe side. If we want to stretch to the limit, then 680 Ohm series resistor is absolute minimum, but the oscillation frequency is difficult to pretend, it can be remarkably lower than without the output load.

As written in comments, a capacitor over the supply voltage, with short wires to the IC (=decoupling) is needed to keep the supply voltage stable. This is a basic practice in logic circuits. No decoupling implies parasitic oscillations (=unquessably complex instability).

Another basic practice is to short unused logic inputs to high or low. Open MOS inputs easily collect stray signals that drive the IC to have half-state logic elements which are unstable and draw plenty of current. On breadboard all IC pins are connected to metal strips that are substantial capacitors. That increases the stray-effect remarkably.

Reducing the supply voltage makes a logic IC slower, but it still can operate. Slowdown is caused by the increase of internal resistances which makes all time constats longer. A slow IC is less prone to parasitic oscillations because its frequency bandwidth is low when compared to normal.

By the questioner noticed "noise" can be high frequency parasitic oscillation. It can be quite chaotic, far from pure sinewave, but some short term periodicity should be seen, if the oscilloscope is set to have 0,1us/div.

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    \$\begingroup\$ 2k is far too high. \$\endgroup\$ – Leon Heller Jan 23 '17 at 5:25
  • \$\begingroup\$ @LeonHeller - Only if you're willing to operate beyond ratings. At Vcc 5 volts and output current 4 mA, the NXP version is only rated for a Voh of 3.7 volts. \$\endgroup\$ – WhatRoughBeast Jan 23 '17 at 6:22
  • \$\begingroup\$ It will be OK if if the output is buffered by another inverter, won't it? I assumed that is what he did. \$\endgroup\$ – Leon Heller Jan 23 '17 at 8:02
  • \$\begingroup\$ I tried bumping up the LED resistor from 330 ohms to 1-2k, which let me bring the power supply voltage up a few tenths of a volt before the output became unstable. I then tried grounding the unused outputs, and as I ground them one by one it allows me to increase the supply voltage up to the usable 2-6V range! Looks like the issue was mostly just grounding the unused pins. Thanks! \$\endgroup\$ – Andrew Southworth Jan 24 '17 at 3:42

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