Connecting LED's to a receptacle

IF I HAVE 3.6VOLT(DC Forward Voltage) LEDS

AND I WANT TO CONNECT 50 OF THEM

WILL THAT MAKE 180VOLTS HOW CAN I

USE A WALL OUTLET WITH ONLY 120VOLTS?

IS THIS POSSIBLE? PLEASE help .

 

Do some searches online for "how to wire LEDs". There are different ways they can be wired but for a long string they are commonly wired in groups. Within the groups they are in series. Then the groups are wired in parallel to the power supply. How many LED's in each group depends on the system's voltage but 12 and 24 volts is common.

 

Welcome to the forums!

The DC draw of your lights is not the same as their AC load requirements and volts are not equivalent to watts. What AC wattage does each set draw at 120V?

 

You add up the load wattages. The voltage stays the same.

 

I changed the thread title to better reflect your question.

You can't realistically generate 180v so you would size your series strings to match the available line voltage.

In other words.... if you put 25 LED's in series you would need a 90 volt source so you could introduce a resistor into the string and make up two 25 LED strings.

Or you could make up series strings of 35 LED's which you should be able to connect directly to the line.

 

Specifications:

LED Emitter: 3W

- Output Lumens: 180-210 Lumens

- DC Forward Voltage (VF) : 3.6-3.8Vdc

- DC Forward Currect (IF) : 700mA

- Color Temp: 6000~6500K ( White)

- Beam Angle: 120 degrees

- LifeSpan Time : > 50,000 hours



can someone please tell me how many number of ohms resistors i
need please help

 

What is your power source? Then we can tell you the resistor you need. If you want to hook this up to normal house power you can't. Your house has 120 volts AC. LED's need DC.

 

Part Number: 24MD-DL
Weight: 45.41 lbs
Warranty: 1 YR
12 Volt
550 CCA
Application: Marine
BCI Cold Cranking Amps (CCA): 550
BCI Cold Cranking Amps Value: Tested to BCI Standard
BCI Designation: 24
Battery Height: 225 mm - 8 7/8 in
Battery Length: 260 mm - 10 1/4 in
Battery Width: 173 mm - 6 13/16 in
Core Charge Applicable: Yes
Cranking Amps (CA): 685
EN / DIN Designation: Not Applicable
JIS Designation: Not Applicable
Positive Terminal Side: Left
Post Location: Top Post
Reserve Capacity (min): 140
Voltage: 12
MSDS Information: Click to View

 

Can anyone answere me as soon as possible

 

I changed your original title as you were asking about a wall receptacle.
In your last post you left information about a 12volt battery.

In you LED spec post where you asked about what resistor to use..... you posted that each LED uses 700ma of current. That is a lot of current for an LED. You are talking about 35 amps of load.

We aren't mind read readers. If we don't have all the information and what you are trying to accomplish it's extremely hard to help you. I have a hunch you are working with some expensive LED fixtures based on their output.

 

Source voltage 12v
diode forward voltage 3.6
diode forward current (mA) .7
number of LEDs in your array 50
View output as: ASCII schematic wiring diagram


LED series parallel array wizard


in this webpage i put the detail in the boxed

and got a schematic of what it would look like.

:No Beer 4U:

 

mighty mouse can u answer me

 

Diode forward current is 700 ma. if what you posted previously is correct.
That would be .7amps. each.

According to that wizard the resistors would not be able to handle the load as well as you would be wasting a ton of energy in resistor heat since you would need massive resistors.

 

LED series parallel array wizard

Source voltage 12v
diode forward voltage 3.6
diode forward current (mA) .7
number of LEDs in your array 50
View output as: ASCII schematic wiring diagram

I have put these information in the box

 

In post # 6 you put down 700ma of forward current.
Is it 700ma or .7 ma ?

It makes a big difference.

 

.700 MA

3.6Vdc

LEDs

I want to power them with 12volt or 24v

 

Put the numbers in the wizard like I've posted and click on wiring diagram. However you will need some large wattage resistors and a lot of your power is going to be wasted as heat.

If you want to use 24v..... put that in instead of 12v.

 

its 700 MA

not .7

Im waiting for any type of information that would help

 

Thnk for the correction

A resistor of 1.8 ohms for each 3 LEDs? Is this correct.

 

Using a 1.8 ohm resistor would be a poor design for a 12v battery. Batteries never stay at just 12v. They can range from 11 to 13.5 volts, and those changes can reduce by half or double the power being provided to each diode. If you were using a regulated power supply that remained at precisely 12v, then each diode would be getting 667 ma. But it would still be a poor design. For a battery powered circuit, a 1.8 ohm resistor with 3 diodes is wrong. 2 diodes in series with a larger resistor would be better.

When you design a power source for LEDs you base the circuit on the current the diodes needs. In the manufacturers specifications they will have a curve for voltage and current and that is where you start.

Bud

 

It is more efficient to make strings of LEDs larger to get the voltage drop closer to the source voltage. But if you get too close to the source voltage than a resistor to control the current has little room to work with source voltage variations. You may want to invest in an active circuit current control.

It sounds like you have a lot of money tied up in those LEDs so you want to be sure you don't overcurrent them.