And gate without input, output lights up the led???

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babaliaris

Joined Nov 19, 2019
160
IMG_20191126_181933.jpg

I just wanted to see if the chip works, so I made the circuit above. I just connected the chip to the 5VDC supply and took the first output and connected it to a led through a 100Ω resistor. The cathode of the led goes to the ground. I believe the led should not light up.

The resistor and the ground of the chip don't touch each other,
 

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dl324

Joined Mar 30, 2015
16,845
You're not supposed to leave CMOS inputs floating. The state is indeterminate.

When discussing circuits, a schematic is more readable than a breadboarded circuit. The only time a breadboard is more helpful than a schematic is when you suspect a wiring error.

You can demand more current from a gate than it is spec'ed to provide, but the output voltage isn't guaranteed to be in the specified range for valid logic levels.
 

Tonyr1084

Joined Sep 24, 2015
7,852
Simply grounding either of the two inputs to the AND gate SHOULD drive the output low and extinguish the LED. You can ground BOTH inputs if you like; but as you know, an AND gate requires two high inputs to give a high output. Leaving them floating allows them to act as radio antenna and can pick up stray signals and thus modulate the output at a given frequency. Your LED appears to be on steady but could be flashing as slow as 28 times per second to many kiloHertz or higher.

I was looking at the data sheet for a like device and it didn't say whether its inputs are FET type transistors, which I suspect they may be. One thing I did see clearly stated was that all inputs must be tied to either "0" or "1" logic levels. When the gate is steady (held high or low) it consumes very little power. But when it is switching it is consuming a lot more power. Therefore, leaving inputs to flap in the breeze will drive the current higher. Your chip is working hard even though you've only applied power and a single LED.
 
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