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I am trying to wire 7 LEDs together, and failing.

Lights: (Numbers are for image below, I have no clue what other info to put here, so I just made them links) They are 3V. Lights 3&4 Lights 1,2,5,6,7.

Resistors: I think this is a 100M ohms resistor. I think this is 100M ohms (It came marked as “5.1-9Vdc use”)

Battery: 9V With battery adapter. I have tried multiple batteries, so this is not the issue.

I am trying to wire these together, and read that parallel is the best way to do this. I kind of did a series/parallel mix.

I think the best way to explain this is to show an image, and than say what's not working, hopefully you can help me out :)

Key:

Red: Positive wire/power

Yellow: Negative wire/ground

Blue: Resistor (I think it is 100M ohms.)

White: LED (3V, 3&4 are white, rest (1,2,5,6,7) are warm white)

Pink: Battery (9V) and mess-up number

Diagram 1 Only LED #5 works here, and it's nice and bright. Others are off or super dim.

Diagram 2 LED #5 is full strength (The only one lighting up this bright out of all the diagrams below. I wish the rest could be this bright), and LED #3&4 are on but dim. The rest are off.

Diagram 3 Everything is on (Except for #5, but it's not plugged in), all are a little dim, but not much.

Again, any help would be appreciated, thanks to any ideas!

Also, I guess I should note that I've only been doing this for about 2 days, so sorry if the answer is obvious.

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  • \$\begingroup\$ You will want to check this website which is from one of the members of Stack Exchange here. LEDnique.com \$\endgroup\$
    – Miss Mulan
    May 5, 2021 at 0:12
  • \$\begingroup\$ 3x3V=9V only use 3 in series then another 3S in parallel with the 1st . It might last a day or two \$\endgroup\$
    – Tony Stewart EE75
    May 5, 2021 at 0:28
  • \$\begingroup\$ Ah, so if I have three sets of 3 lights in series and no resistor, my battery would last a day or two? If I were to have many sets (like 20) of 3 lights in series, how many hours (approx) would my battery last? Thanks! \$\endgroup\$
    – Greenreader9
    May 5, 2021 at 0:31
  • \$\begingroup\$ @Greenreader9 have you visited the link I provided you? \$\endgroup\$
    – Miss Mulan
    May 5, 2021 at 0:34
  • \$\begingroup\$ My answer on how to light up a series of 3 LEDs might also help: LED lighting for use in miniature tanks electronics.stackexchange.com/questions/563073/…. \$\endgroup\$
    – tlfong01
    May 5, 2021 at 1:40

1 Answer 1

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LEDs don't act like ordinary incandescent light bulbs. When lit, they have a fairly constant voltage drop that varies slightly with current. White LEDs have a typical forward voltage of about 3 volts.

The single #5 LED will draw enough current from the 9 volt battery to pull the voltage down to about 3 volts. The other LEDs which are connected as two LEDs in series, will require about 6 volts to light, but since #5 is holding the voltage down to ~3 V, they will not light.

When you remove LED #5, the voltage will rise to allow the other LEDs to light - but the 9 volt battery cannot deliver enough current to light them to full brightness.

LEDs should never be connected directly to a voltage source - they always require something to limit current (normally a resistor).

You should connect one of the resistors you show in series with each series pair of LEDs. That resistor is probably 100 Ohms - definitely not 100 MegOhms!

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  • \$\begingroup\$ Thanks. I think I get it. It won't let me edit my answer, so I cannot add images (I'll have to explain the best I can here). So your saying that the lights that are hooked in parellel are being restricted to 3V for both (when it should be 6V) causing them to be dimmer, and #5 is stealing all the power because it has access to the full 3V supplied by the resistor on the battery? So to solve this, I would need to put a resistor limiting to 6V on the double lights, and a 3V limiting one on the single ones, and none on the battery? What color would the 6V retricting ones be than? Thanks! \$\endgroup\$
    – Greenreader9
    May 5, 2021 at 0:09
  • \$\begingroup\$ For a single LED with a 9 volt battery, you should use a 390 Ohm resistor to limit the current to 15 mA. for two LEDs in series, you should use a 180 Ohm resistor. \$\endgroup\$
    – Peter Bennett
    May 5, 2021 at 0:36
  • \$\begingroup\$ You are connecting LED #5 which is a 3V or 4V LED directly to the 9V battery to blow it up. You are also connecting two series LEDs that need no more than 6V or 7V to a 9V battery to blow them up. The LEDs do not blow up if the battery is weak because it is old or is cheap. \$\endgroup\$
    – Audioguru
    May 5, 2021 at 0:37

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