-1
\$\begingroup\$

I have two 1.5 Volt batteries, a Red 3V LED, and some cables.

The batteries I have connected in series like

+- +-

From my studying I believe I should now have 3 Volts ready when I need.

I connected two copper cables, 1 at the negative end and 1 at the positive end.

I believe the volts should push the current from the negative end of the battery towards the LED and then back out towards the positive end of the battery.

I noticed the length of the leg cable on one pin on the LED is longer than the other, I have tried two different LEDS both ways and the LED still does not light.

I have attached a picture below.

I have also tried inserting a 270 ohm resistor at the end of the negative cable of the battery but before the LED - This still doesnt allow the LED to light.

What am I doing wrong here? Is my understanding correct?

I have also tried lighting the LED with just one battery, it does not work either.

enter image description here

\$\endgroup\$
5
  • \$\begingroup\$ I suspect that you have poor (or no) contact between cells or between the wires and the cell terminals - bits of sticky tape do not make a good battery holder! \$\endgroup\$ Aug 12, 2015 at 0:57
  • \$\begingroup\$ if I remove the LED and touch the wires, the terminals get very hot, shouldnt this indicate that we have a full circuit? \$\endgroup\$
    – buntybudia
    Aug 12, 2015 at 0:58
  • \$\begingroup\$ Assuming it is a normal red LED, it should definitely glow. I don't think there is anything wrong with your conceptual understanding. The problem must be related to the setup. The polarity does matter. You have to connect the LED cathode to battery - terminal, and LED anode to battery + terminal. If you are not sure just try it both ways. \$\endgroup\$
    – mkeith
    Aug 12, 2015 at 1:02
  • \$\begingroup\$ Also remember that the LED may be manufactured with a longer lead indication the anode (+), BUT one could easily snip the leads to make them shorter. If you look at the LED closely you will see that the LED isn't perfectly round, there is a flat edge. This flat edge indicates the cathode (-). Make sure polarity is correct. And it is always good practice to place a resistor in series with the LED. Cheers! \$\endgroup\$
    – Josh Jobin
    Aug 12, 2015 at 1:52
  • \$\begingroup\$ @PeterBennett Your first comment was right, it was the poor contact, Thanks \$\endgroup\$
    – buntybudia
    Aug 12, 2015 at 2:43

3 Answers 3

0
\$\begingroup\$

Everything seems fine to me, these are few things that might be wrong:
1) Batteries are bad, maybe not completely empty, but empty enough to not give needed voltage. Try different ones
2) LEDs are dead, they may burn if you connect them without resistor, so its better to connect at least any resistance(if you don't have specific values like 330 Ohms, just put anything around it). But if they are dead - they are dead, you need different one then.
3) Check if you have any conductive stuff around batteries that may connect them directly to each other(between two poles), this will make a short circuit and current will not flow through LED.
4) Bad contact with LED leads.

Not much can be wrong here since the circuit is simple. If you have multimeter you can also check the voltage on the batteries and beep the LED to know if its dead or not(and find cathode and anode). If you don't have multimeter... buy one, life is hundred times easier if you have multimeter :)

\$\endgroup\$
2
  • \$\begingroup\$ Make sure you have the LED the correct way round - it will only work one way. \$\endgroup\$ Aug 12, 2015 at 1:41
  • \$\begingroup\$ The problem was 4), The leads wasnt pressed up against the terminals enough, Applying force made the LED light up. \$\endgroup\$
    – buntybudia
    Aug 12, 2015 at 2:42
0
\$\begingroup\$

To an approximation, LEDs are a constant-voltage drain. In other words, if you supply less than their voltage drop, you get nothing. If you try to supply more, though, their current goes through the roof and you blow them up. Even more fun, that voltage drop varies from component to component as well as over temperature. This means that in some circumstances your batteries would do nothing; in others your batteries would kill the LED. In summary, supplying an LED from a raw constant-voltage supply is in general a bad idea.

The standard way to drive an LED from a constant-voltage source is to use a dropping resistor, chosen by taking the difference between the source voltage and the LED voltage, and dividing by the desired current. This gives you a good, stable current to drive the LED, even under somewhat varying conditions.

\$\endgroup\$
0
\$\begingroup\$

Check the datasheet for the LED. To put the LED into conduction the supply voltage must exceed the forward voltage of the LED (Vf_max). This is specified in the datasheet. Without enough potential to forward bias the LED, it will not conduct. To illuminate, you should have enough current (also in the datasheet). I don't see a resistor in your setup either (or some other current control), so look into that

\$\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.