I have constructed a simple fade in and fade out circuit similar to given on this website https://roboindia.com/tutorials/arduino-analog-output-led-fade Now I am connecting a wire from 5V power pins to +ve side line, but i don't know why I am doing this because when I remove this wire or connect it to 3.5V pin the led is still running fine. Also as side lines are connected horizontally there is no other further connection given to the +ve wire. Can you tell me why I am joining the wire?
-
1You should include the diagram in the questions, so that people can see which wire you're talking about without following the links.– Dmitry GrigoryevCommented May 21, 2019 at 13:43
Add a comment
|
1 Answer
This is indeed not needed. As you can see in the diagram, the horizontal line which you connected the 5 V pin to.
The horizontal line of holes just above the red horizontal line) are connected. However, since it only connected to the 5V pin of the Arduino, the all pins on that horizontal line have 5 V, but since there is nothing connected to it (thus no closed circuit), the effect is zero.
The reason why the connection is made (the wire from Arduino to 5V), is that it is 'default' to connect both GND and 5V (or 3.3V) as a starting point. It is easy to forget later and than no current will flow when you connect something to the +5 V horizontal line of pins.
Btw, the 3.5 V pin you mention, is 3.3 V, not 3.5 V.
answered May 20, 2019 at 12:07
