Continue to Site

Welcome to EDAboard.com

Welcome to our site! EDAboard.com is an international Electronics Discussion Forum focused on EDA software, circuits, schematics, books, theory, papers, asic, pld, 8051, DSP, Network, RF, Analog Design, PCB, Service Manuals... and a whole lot more! To participate you need to register. Registration is free. Click here to register now.

Created a sensor using only a nor integrated circuit, LED, button?

Status
Not open for further replies.

wildlandschool

Newbie level 4
Newbie level 4
Joined
Nov 15, 2016
Messages
5
Helped
0
Reputation
0
Reaction score
0
Trophy points
1
Activity points
31
We were doing a school project, attempting to create binary data storage. In the process of putting the circuit together, we discovered that moving closer or farther from the circuit affects the brightness of the LED. When we disconnected the button from the nor integrated circuit, the circuit got extremely hot and burnt out. We also recreated this same circuit on another separate bread board, and the same exact thing happened. If anyone has any idea what is happening, please let us know!

IMG_0004.jpg

 

Don't expect much help without a schematic.
 

Your input is almost certainly responding to static charge on your arm. This is typical. You can cure it by attaching a pullup or pulldown resistor to the input. You should tie all unused inputs to a supply rail.

The input also can pick up ambient 60-cycle hum, although that would not necessarily make the led glow brighter in every instance.

(Side note: My electronics lab kit is identical to yours, except mine is gray, made by Radio Shack in the mid 1990's. It was a gift from an electronics engineer and teacher. The breadboard provided me with a major step forward in my pursuit of electronics.)
 

other objects such as a meterstick (held from a long distance), scissors, or a pencil also create the same effect. Also, the circuit also randomly inverts (sometimes it gets brighter the closer we get, sometimes it inverts and becomes dimmer the closer we get).
 

Also why is your NOR circuit only have two connections?

A 2-input NOR gate (assuming the part has 2-input NOR gates) has 3 connections not 2. Leaving input pins floating is never a good idea.

- - - Updated - - -

Ah, I see BradtheRad, noticed the same thing :)

- - - Updated - - -

Also there doesn't appear to be any current limiting resistor on the LED.
Besides that once both NOR gate inputs are disconnected (removing the switch) the floating inputs to the gate could result in allowing all the transistors to partially turn on shorting Vcc to Gnd.
 

other objects such as a meterstick (held from a long distance), scissors, or a pencil also create the same effect. Also, the circuit also randomly inverts (sometimes it gets brighter the closer we get, sometimes it inverts and becomes dimmer the closer we get).

Your question has been already answered in post #4 and post #5.

Correct the circuit as suggested (pull-down resistor for the switch input, connect all unused logic inputs). The LED should be also connected with a current limiting resistor.
 

we aren't trying to figure out how to fix it, we are trying to figure out why the circuit is able to act as a sensor using only what is on the circuit. so, our question has not yet been answered. we are uploading a schematic now.

- - - Updated - - -

IMG_0009.jpg

here it is
 

we aren't trying to figure out how to fix it, we are trying to figure out why the circuit is able to act as a sensor using only what is on the circuit. so, our question has not yet been answered. we are uploading a schematic now.

It has been answered, by both BradtheRad and I...

If you leave the inputs floating on the NOR gate static charge of your hand or whatever else you move around the circuit can partially turn on the transistors that drive the output of the NOR gate. You don't need much gate voltage to partially turn on a CMOS transistor.

Look at a CMOS NOR gate circuit.
https://upload.wikimedia.org/wikipedia/commons/4/43/NOR_gate_%28CMOS_circuit%29.PNG

Maybe you should also learn about ESD, before you damage every electronic part in sight.

https://www.radio-electronics.com/info/circuits/esd/what-is-basics-electrostatic-discharge-tutorial.php
 

we are trying to figure out why the circuit is able to act as a sensor using only what is on the circuit
Yes, I was under the impression that this part of the question had been also answered.

Please notice that primarily the floating input pin12 is acting as "sensor", you can considerably reduce chip heating by grounding the other unused inputs.
 

Status
Not open for further replies.

Part and Inventory Search

Welcome to EDABoard.com

Sponsor

Back
Top