# Wiring LED Bulbs in Parallel

#### Paranoia08

##### New Member
Hi all... need help desperately...

I am wiring 108 x 3mm led bulb(white) in parallel.
Voltage drop of white led bulb should be appx 4 volts.
Desired led current should be appx 20mA.

I am currently using 4 x 1.5v AA batteries as source voltage

What kind of resistors should I get?

I used the resistor calculator and it showed me on "the wizard thinks 1/4W resistors/100ohms". Over at another website, it showed me "1/8 resistors/120ohms".

Please advice me on how much should I have for Supply/Source voltage and resistors.

Thanks a million in advance....

Resistors calculator link :
http://led.linear1.org/led.wiz
http://ledcalc.com/#

#### Limsgp

##### New Member
To do a detailed calculation, you need to specify the internal resistance of the batteries.

Actually, it may not even require a resistor. Since the total EMF is 4.5V and the LED needs "around" 4V, it should work fine.

Anyway, there's some numbers

Assuming the batteries have zero or negligible internal resistance.

Resistance of 1 LED = 4/20mA = 200ohms (Assume)

Total effective resistance of 108 LED = 200/108 = 1.85ohms

The resistor should be 0.23 ohms.

Voltage across LED = 4.5V * 1.85/(1.85+0.23) = 4V

note: The current thru the resistor is 0.02*108 = 2.16A
The resistor should be rated at least = 2.16 * 2.16 * 0.23 = 1.08W

Also, make sure the battery can deliver 2.16A. Carbon-Zinc Battery may not work, Alkaline maybe, maybe not.. Eneloop type might be better.

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#### proteonXPR

##### New Member
the current drawn by the LEDs is determined by its resistance across the supply

so if you have 3 parallel, it will draw more current, with lower resistance than 1 single.

you have a multi meter to measure? if you can measure the "cold" resistance?

then you can put in a resistor in series with them of a value that is half of this led resistance

so all in all you get a potential divider, 2/3 of voltage going to the LED which is 4V, and 2V through the resistor.

if you want to be safe, you can work with 4x rechargeable batteries first, they are 4x 1.2v = 4.8V

#### Paranoia08

##### New Member
First of all, thanks for the advices..

I actually called up the shop which I purchased the led bulbs from and got some info.

My led bulbs are white.

Led Voltage: 3V
mA: 20
Max rating: 50 ohm

He recommended me to have a "supply voltage" of 18v(12 x AA batt), if not they won't be bright.

Also recommended me resistor : 300 30ohm(I am not sure what he meant here, he was quite busy though..)

Really need such high voltage???

#### hacknet

##### New Member
you cannot model LEDs as resistances. its essentially a diode, ie, if you are using a 6v source, you probably want to do the maths this way

6-3=3v
3v/(3*20ma) = 50ohms in series.

thats all, you probably could get away with 3 batteries in series. i think the LEDs have about a 2.x v drop, you might even get away with 2 batteries.

if you are out for efficiency, youd probably want to look into led controllers.

#### hacknet

##### New Member
oh, yes i forgot, it be better if you had separate dropping resistors for each led in parallel.

#### allenleonhart

##### Deregistered
huh? shldnt he be getting a fuse instead? i can imagine the current flow gonna be quite large if its in parallel.

#### iceman

##### New Member
huh? shldnt he be getting a fuse instead? i can imagine the current flow gonna be quite large if its in parallel.
wire 108 LEDs in parallel. => 108 * 20 mA = 2.16A.
I wonder the quantity of AA batteries that is required. :bigeyes:

#### Limsgp

##### New Member
Really need such high voltage???
You high current supply, not high voltage.

Connect 3 eneloop in series to get 3.6V. Use 2 sets of these in parallel if the current from 1 set is not enough.

Not need any additional resistors or whatever.

That's the simplest and effective. Just try, should work.

#### allenleonhart

##### Deregistered
You high current supply, not high voltage.

Connect 3 eneloop in series to get 3.6V. Use 2 sets of these in parallel if the current from 1 set is not enough.

Not need any additional resistors or whatever.

That's the simplest and effective. Just try, should work.
and yes. i think this way works better. at least the math seems to add up.

#### hacknet

##### New Member
oh, i read wrongly, i didnt see that you were trying to wired up 108 leds. honestly, i doubt a 3mm led can dump 20ma @ 3v, thats a heap of power often left to the larger varieties. well, in this case, it would be sound to stack them in pairs and then parallel the whole lot. so you'd get 54 pairs in parallel. same mathematics, depending on voltage drop of each, recalculate the R.

oh yes, do expect them to all have different brightness, its something sadly intrinsic in LEDs.

#### ConnorMcLeod

##### New Member
Also recommended me resistor : 300 30ohm(I am not sure what he meant here, he was quite busy though..)
he probably means 330ohms. who would wire up 300 x 30ohm resistors (in any form)!?

#### Paranoia08

##### New Member
how about the source voltage?

anyone can advice me on how many batteries would I need to get the brightness, yet not going to blow the bulbs out?

#### Gizmore

##### New Member
somebody calculated for you that you probably need 2A of current in order to drive the LED at 20mA.

Even if you can wire enough batteries to achieve 2A, it will probably be drained in a matter of minutes.

You best option is to get an adapter that runs off AC.