This
tutorial will help you to find the proper value of resistor (or resistors) for
one or more LED’s to connect with battery.

If
you pick this topic, you will be able to:

Calculate
the value of resistors for different LED’s Circuit diagrams

Calculate
the Forward Current of LED’s

Calculate
the Forward Voltage for different LED’s Circuits

Connect
LED’s in Series with batter

Connect
LED’s in Parallel with battery

Connect
LED’s in Series-Parallel Circuits

**Update:**You can Also use this

**LED Resistor Calculator**for this purpose

**Typical LED Symbol, Construction and Lead Identification.**

*Click Image to enlarge*

Before
we go in detail, we will try to get ride on below simple circuit, so that the
other calculation will be easier to understand.

*Click Image to enlarge*

**This is the Simplest LED Series circuit ever**. Here, the supply voltage is 6V, LED Forward Voltage (V

_{F}) is 1.3 Volt and Forward Current (I

_{F}) is 10mA.

Now
the Value of resistor (which we will connect in Series with LED) for this
circuit would be:

**Resistor Value**= (V

_{supply}- V

_{F)}/ I

_{F}

= (6
-1.3) / 10mA =

**470 Ω**
Current
draw = 20mA

Resistor
Power rating formula for this circuit

Resistor Power Rating = I

_{F}^{2}x Resistor Value
= (10mA)

^{ 2}x 470**Ω = 0.047W = 47mW****But**This is the minimum required resistor value to ensure that resistor will not overheat, so its recommended that to double the power rating of resistor that you have calculated, therefore, choose 0.047W x 2 = 0.094W = 94mW resistor for this circuit.

Resistor
power rating (Value is doubled) = 0.094 W = (94 mW)

**Also keep in mind that:**

- It is too difficult to find the exact power rating resistors that you have calculated. Generally, Resistors come in 1/4 watt, 1/2 watt, 1 watt, 2 watt, 5 watt, and so on. Therefore, select the next higher value of power rating. For example, if you’re calculated value of resistor power rating is 0.789W = 789mW, then you would select 1W Resistor.
- It is too difficult to find the exact value of resistors that you have calculated. Generally, Resistors come in standard values. If you are not able to find the exact value of resistor that you have calculated, and then select the next coming value of resistor that you have calculated, For Example, if the calculated value is 313.5Ω, you would use the closest standard value, which is 330 Ω. if the closest value is not close enough, then you can make it by connecting resistors in series – parallel configuration.
**I**This is the amount of maximum current that LED can accept continuously. It is recommended that provide 80% of LED forward current rating for long life and stability. For example, if the rating current of LED is 30mA, then you should run this LED on 24mA. Value of current over this amount will shorten LED life or may start to smock and burn._{F}= Forward Current of LED:- If you are still unable to find the LED forward current, than assume it 20mA because a typical LED's run on 20mA.
**V**This is the forward voltage of LED i.e. the voltage drop when we supply the rated forward current. You can find this data on LED’s Packages, but is somewhere between 1.3V to 3.5V depending on type, color and brightness. If you are still unable to find the forward voltage, simply connect the LED through 200Ω with 6V battery. Now measure the voltage across LED. It will be 2V and this is the forward voltage._{F }= Forward Voltage of LED:

**Formula for finding the value of resistor(s) to connect LED’s in Series:**

Below
is another simple LED’s (LED’s Connected in Series) Circuit. In this circuit,
we have connected 6 LED’s in Series. Supply Voltage is 18V, The Forward Voltage
(V

_{F}) of LED’s is 2V and the forward Current (I_{F}) is 20mA each.*Click Image to enlarge*

Resistor Value (LED’s in Series) =
(V

_{supply}- (V_{F }x No. of LED’s)) / I_{F}
Here, Total forward voltage (V

_{F}) of 6 LED’s = 2 x 6 = 12V and forward Current (I_{F}) is same (i.e. 20mA)
(

**Note:**this is a series circuit, so current in series circuit in each point is same while voltages are additive)
Now, the value of resistor (for
Series Circuit) would be:

= (V

_{supply}- (V_{F }x No. of LED’s)) / I_{F}
= (18 – (2 x 6)) / 20mA

= (18-12) / 20mA =

= (18-12) / 20mA =

**300****Ω**
Total
Current draw = 20mA (This is series circuit, so currents are same)

Resistor Power Rating = I

_{F}^{2}x Resistor Value
= (20mA)

^{ 2}x 300**Ω = 0.12 = 120mW****But**This is the minimum required resistor value to ensure that resistor will not overheat, so its recommended that to double the power rating of resistor that you have calculated, therefore, choose 0.12W x 2 = 0.24W = 240mW resistor for this circuit.

Resistor
power rating (Value is doubled) = 0.24 W = (240 mW)

**Formula for finding the value of resistor(s) to connect LED’s in Parallel (With Common Resistor):**

*Click Image to enlarge*

In
this circuit, we have connected LED’s in parallel with common resistor. Supply
Voltage is 18V, The Forward Voltage (V

_{F}) of LED’s is 2V and the forward Current (I_{F}) is 20mA each.
Resistor Value (LED’s in parallel With
Common Resistor)

= (V

_{supply}- V_{F)}/ (I_{F}x No. of LED’s)
Here, Total forward Current (I

_{F}) of 4 LED’s = 20mA x 4 = 0.08A, and forward Voltage (V_{F}) is same (i.e. 2V)
(

**Note:**this is a parallel circuit, so voltage is parallel circuit is same in each point while currents are additive)
Now, the value of resistor (for parallel
Circuit with common resistor) would be:

= (V

_{supply}- V_{F)}/ (I_{F}x No. of LED’s)
= (18 – 2) / 0.08 =

**200****Ω**
Total
Current draw = 20mA x 4 = 80mA (This is parallel circuit, so currents are additive)

Resistor Power Rating = I

_{F}^{2}x Resistor Value
= (20mA)

^{ 2}x 200Ω = 0.08 W = 80mW**But**This is the minimum required resistor value to ensure that resistor will not overheat, so its recommended that to double the power rating of resistor that you have calculated, therefore, choose 1.28W x 2 = 2.56W resistor for this circuit.

Resistor power rating (Value
is doubled) = 2.56W (280 mW)

**Formula for finding the value of resistor(s) for connecting LED’s in Parallel (With Separate resistor)**

*Click Image to enlarge*

This
is another way to connect LED’s in parallel with separate resistors. In this
circuit, we have connected 4 LED’s in parallel with separate resistors. Supply
Voltage is 9V and the Forward Voltage (V

_{F}) of LED’s is 2V and the forward Current (I_{F}) is 20mA each.**Resistor Value**(LED’s in parallel with separate Resistor)

= (V

_{supply}- V_{F})/ I_{F}
Here, Total forward voltage (V

_{F}) of LED’s = 2 and forward Current (I_{F}) 20mA (i.e. 20mA)
(

**Note:**this is a parallel circuit, but we are finding the value of resistor for each section, not for whole circuit. So in each section, the circuit becomes in Series position (refer to the Series Circuit formula or the 1^{st}simple circuit above, you will find that these are same)
Now, the value of resistor (for
parallel Circuit with separate resistors) would be:

= (V

_{supply}- V_{F})/ I_{F}
= (9
– 2) / 20mA =

**350 Ω**
Total
Current draw = 20mA x 4 = 80mA (This is parallel circuit, so currents are
additive)

Resistor Power Rating = I

_{F}^{2}x Resistor Value
= (20mA)

^{ 2}x 350**Ω = 0.14 = 140mW****But**This is the minimum required resistor value to ensure that resistor will not overheat, so its recommended that to double the power rating of resistor that you have calculated, therefore, choose 0.14W x 2 = 0.28W = 280mW resistor for this circuit.

Resistor
power rating (Value is doubled) = 0.28 W (280 mW)

There
is another way (Series-Parallel Combination) to connect LED’s with battery; if
you understood this simple calculation then I’m sure that you can easily calculate
the value of resistors for Series-Parallel Combination LED’s connection
circuit.

**By:**

**Engr Wasim Khan**

**Copyrigt @ http://electricaltechnology.org/**

Its very usefully for Electrical Engineer

ReplyDeletethere are some errors in current value substitution in equations..

ReplyDeleteplease correct it

Dear Tony Stark. please mention it. Thanks

DeleteThere is need to correction here (18 – (12 x 6) / 20mA = 300 Ω) If the calculation will be as this there would be negative value -2700. Please check & advice. There Should Be as this

ReplyDelete(18 – (2 x 6) / 20mA = 300 Ω)

Ya sure..we had put 12 instead of 2. Thanks for correction

DeleteDear,

DeleteI Think here is also need of correction "Resistor Power Rating = IF2 x Resistor Value

= (20mA) 2 x 200Ω = 1.28 W" It might be 80mW

quantum physics determines the forward voltage drops of leds (and any diode)

ReplyDeletedepending on the materials in the PN junction

some ir diodes as low as 1.5v

UV diodes as high as 4.5V

a lot of the white ones are florescent lamps a UV diode exciting phosphors

If you parallel different colors red will take all the current and blue wont light at all

if you have a bag of assorted LEDs use a 20mA current source and measure the voltage drop directly.

Hi I need to exacly know how this works when this LEDs connected to main supply (230 V A/C ) how do I do the calculation for dropdown resistor & capasitors

ReplyDeleteI work with LED Lamp circuits & it is very importan to me to know this

hope you will help me

TY

Senarath

senarathdon@yahoo.com

HI,

ReplyDeleteI need to know how this calculation gose when LEDs are connected to main(230v A/C) how do I calculate the resisted valu & the Paralel capasitor valu

Thank you

Senarath

senarathmdon@yahoo.com

http://www.electricaltechnology.org/2013/03/230-v-ac-main-operated-led-powerful.html

DeleteThank you very much.

ReplyDeleteIt's very helpful for me.

I was in need of this formula. God bless you.

ReplyDeletethanks a lot.

Sohail Ahmad sambapk@yahoo.com