10
\$\begingroup\$

As an absolute layperson in electronics I just found out that an LED only lights up when connected like in figure 1 but not when connected like in figure 2. And I just don't get why?

I know an LED is a diode and current can only move through it in one direction, but that's the case in both constructions. I don't see a difference in both constructions.

So why does it obviously make a difference?

figure 1 (battery not depicted): LED is on figure 2 (battery not depicted): LED is off

\$\endgroup\$
19
\$\begingroup\$

In the second arrangement, both legs of the LED are on the same horizontal row - In breadboards, each set of 5 holes on either side of the board is shorted together.

In other words, your LED's legs are shorted together, no current will flow through it.

See this image of how a breadboard's holes are typically connected internally:

Image (source)

The red thin lines indicate internal connections between the respective holes.

For a more detailed discussion on breadboards see this answer.


Specific to the Fritzing software the images in the question seem to be made in:

Note that as soon as any lead of a component is inserted into a hole in the on-screen breadboard, all the holes that are internally connected get a green indication - That shows that those holes are essentially electrically connected to the occupied hole.

\$\endgroup\$
  • \$\begingroup\$ Thanks Anindo, that makes perfect sense. I didn't think about the fact that the legs are shorted together. \$\endgroup\$ – JohnnyFromBF Aug 30 '13 at 7:03
0
\$\begingroup\$

What you have to understand is that horizontally placed objects on a breadboard does not mean you're connecting them in series. Rows on a breadboard implies that that's the direction of current. Placing an LED horizontally on a bread board is like connecting a wire from the positive to negative legs of an LED, it will short it out. Therefore, positioning LED's vertically on a breadboard removes this short.

TL;DR : always place electrical units in vertical progressions unless you want to simulate a parallel circuit.

\$\endgroup\$
  • 5
    \$\begingroup\$ I'm trying to figure out what value your answer added, exactly. \$\endgroup\$ – Anindo Ghosh Aug 30 '13 at 15:40
0
\$\begingroup\$

It's somehow clear in the second case that the LED is in parallel with a wire so the current prefers to move throgh the wire instead of the LED. In fact, the wire removes the LED.

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.