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schematic

simulate this circuit – Schematic created using CircuitLab

I want to wire 6 LEDs together 2.75" apart; I'm using 2V red LEDs. I've also already calculated the resistors I need.

I'm thinking of making 2 separate circuits of 3 LEDS with each having their own 9v battery, if that makes it easier to do rather than 6 LEDS on 1 9v.

What exactly are the steps to get this done using a switch in between as well.

As of now my plan is: ( I Tried drawing a diagram here)

Wire postive end of 9v battery to a switch, then from there connect the switch to the first LED, using copper wire, i'd solder the wire to each consecutive LED, until I have all 6 connected. The from the last LED I'd connect the negative end to a resistor, and finally add a wire connecting the resistor to the negative side of the battery. Would this work?

schematic

simulate this circuit

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  • \$\begingroup\$ Please post a schematic. It would work better than a verbal description. \$\endgroup\$ – Nick Alexeev Oct 18 '13 at 5:11
  • \$\begingroup\$ Check this post for reference. \$\endgroup\$ – Chetan Bhargava Oct 18 '13 at 5:22
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    \$\begingroup\$ You've got half your LEDs hooked up backwards, so they are reverse biased and block current flow. \$\endgroup\$ – Kaz Oct 18 '13 at 5:55
  • \$\begingroup\$ Apart from obvious errors , the ideal LED config for 9V bat is 4Series (4S=8V) + 50 Ohms for 15mA to 3mA when battery is near dead at 8V. Here is a simulation with different values. tinyurl.com/y86vtcbu Avoid > -5V on any LED to prevent failure. \$\endgroup\$ – Sunnyskyguy EE75 Aug 8 '17 at 18:23
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Here is how you can wire your LEDs. The resistors used are 150 ohms and 1/4 watt. This is assuming that you have 20 ma flowing through each LED.


Wiring Diagram:

wiring



LED Polarity:

Polarity

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  • \$\begingroup\$ Where can I add a switch on that? at the end between the 2 resistors? would the 9v battery be able to power all 6 LEDS on its own? what does the last little circle on the right end of your diagram between the resistors represent? or is it nothing lol? \$\endgroup\$ – Roda Medhat Oct 18 '13 at 5:35
  • \$\begingroup\$ @RodaMedhat Add Switch between +9v in the wiring diagram and the battery +9v. \$\endgroup\$ – Chetan Bhargava Oct 18 '13 at 5:38
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    \$\begingroup\$ That's a very nice image illustrating the parts of an LED. \$\endgroup\$ – Magic Smoke Mar 18 '15 at 19:01
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The LEDs have a rating current which can be found in datasheet or asking from provider or etc. when you say 2volt LED, it means that its rating voltage is 2and also it almost locks on 2V.

as Vazquez said you can not put 6 LEDs in series against 9v. One way is tu put two 3LEDs in parallel that each 3LEDs need 6v. Now we subtract 6 from 9 volt source voltages. We'll have 3V.

now with this 3V we use ohm's law and find the appropriate resistor:

R = 3 / (rating current)

for example if the rating current is 20mA you'll need 3/20mA = 150 ohm

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No. You have a total Vf of 12V, but a supply of only 9V. You will need to create 2 parallel chains, and use resistors to limit the current through each chain.

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  • \$\begingroup\$ Sorry I'm still learning this stuff. So are you suggesting I create a 2nd circuit with another battery? so it's 3 LEDS to one 9v battery? And so other than my power supply, does the rest of my plan make sense? location of toggle switch and using copper wire to connect LED's? \$\endgroup\$ – Roda Medhat Oct 18 '13 at 5:12
  • \$\begingroup\$ They can't use separate batteries if they're in parallel. \$\endgroup\$ – Ignacio Vazquez-Abrams Oct 18 '13 at 5:20
  • \$\begingroup\$ No. You should connect two sets of < three LEDs and a resistor in series > in parallel, with the switch in series with that (and we would usually put the resistor at the positive terminal of the battery, but it will work either way.) \$\endgroup\$ – Peter Bennett Oct 18 '13 at 5:21
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The schematic you showed won't do what you want. Notice that every other LED is reversed. LEDs need to be forward biased to make light.

Another issue you are ignoring is that the voltage of all the LEDs in series is more than your battery. You want to light 6 LEDs, and you say these drop 2 V each. With all of them in series, they would need around 12 V.

A simple solution is to arrange the LEDs in two strings of three each. In each string, there are 3 LEDs in series. That means a string will require around 6 V.

Let's say you want 20 mA thru each LED, which also means thru each string. With 9 V, there is 3 extra voltage to drop in addition to the 6 V dropped by the LEDs. This can be done by a resistor. You know the voltage across it will be 3 V and the current thru it 20 mA. By Ohm's law, the resistance should be (3 V)/(20 mA) = 150 Ω. Each string should have its own resistor.

Run two strings in parallel from the 9 V battery. Each string will draw 20 mA, so both together will draw 40 mA. A 9 V battery can handle that for a while.

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