**How to Calculate the Value of Resistor for different types of LED’s Circuits**

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 = **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 as well.

**By: ****Engr Wasim Khan****Copyrigt @ http://electricaltechnology.org/**

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Its very usefully for Electrical Engineer

there are some errors in current value substitution in equations..<br />please correct it<br />

Dear Tony Stark. please mention it. Thanks

There 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<br />(18 – (2 x 6) / 20mA = 300 Ω)

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

Dear,<br />I Think here is also need of correction "Resistor Power Rating = IF2 x Resistor Value<br />= (20mA) 2 x 200Ω = 1.28 W" It might be 80mW <br />

For the single resistor and parallel LEDS: the 20mA is incorrect for the total current through the resistor. It should be 20mA x 4 = 80mA. Thus the minimum power for the resistor is 1.28W

quantum physics determines the forward voltage drops of leds (and any diode)<br />depending on the materials in the PN junction<br />some ir diodes as low as 1.5v<br />UV diodes as high as 4.5V<br />a lot of the white ones are florescent lamps a UV diode exciting phosphors <br /><br />If you parallel different colors red will take all the current and blue wont light at all <br /><br />if you have

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 <br />I work with LED Lamp circuits & it is very importan to me to know this <br /><br />hope you will help me <br /><br /><br />TY <br />Senarath <br />senarathdon@yahoo.com

HI,<br />I 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 <br /><br /><br />Thank you <br /><br />Senarath<br />senarathmdon@yahoo.com

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

Thank you very much.<br />It's very helpful for me.

I was in need of this formula. God bless you.<br />thanks a lot.<br /><br />Sohail Ahmad sambapk@yahoo.com

I have other solution to solve the prlboem.I used mesh analysis and others tools. I’ll explain how to get the results.We have tres values: See the image here:It’s not necessary resolve three equations by the mesh cause we know two of them. We only need do our mesh 1 equation:I puted two nodes on the circuits to have more information about the currents that we have.Node A = 0Node BWe have then:We could be more exactly in our equations to find Ix and I1Solving that:In this way, with one of our equation we could fin the other value:To find the , we could do a simple KVL across the mesh 2.

A genuine question as I’m trying a torch conversion. Why is the sum IF² ( = 400mA ) x Resistor value ,when the Total draw is 80mA ( 20mA x 4)?

Should it be 80mA x 350 = 0.28w = 280mw.

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

Resistor Power Rating = IF2x Resistor Value

= (20mA) 2 x 350 Ω = 0.14 = 140mW””

Yes!… for 4 number of LED’s Parallel Circuit, it would be 80mA x 350 = 0.28w = 280mw.

There are 40 led of 3.2 v and 60 ma of each led,combination is 8 in series and 5 in parallel. Current for whole led PCB is. ???? On 220 v ac.explain how to find wattage and current

Hi

I want to connect 300 resistor in parallel, can it will be possible to do this. And how much voltage should I pass through this circuit.please help me,As soon as possible

Can you provide an example using a Tri-Cree star LED module that is 750 mA instead of 20 mA, and a 3.7v li-ion cell 5000 mAh rating… if Vf is 2.8v for each RED die on the star pad… then what type of resistor would be used to satisfy these requirements, Ohms and Watts???

Ideally, if using a RGB star, each die has its own resistor based on Vf of red, green, and blue dies, AND, the I= values for mA draw of each die type, i.e. 750 red, 1000 blue or green, etc.

What if ledsupply.com only wired all 3 red dies in PARALLEL with only 2 wires exiting the heatsink module? I can’t use 3 seperate resistors now.

I understand that three dies in a star in parallel would draw 3000 mA and would damage a 2500 mAh cell, yes? I’m using a 5000 mAh cell 3.7v li-ion so it should do ok.

Also, if you are using a n-MOSFET to control this circuit, separate from a microcontroller (arduino) how can you determine if enough voltage is left and current to power a 5v 500mA boost breakout circuit, that powers the 5v microcontroller?

I am probably wrong… but I chose to buy at RadioShack a 0.47 Ohm 5watt ceramic resistor looks big and white rectangle shape. It seems to work for my red Tri-Cree module where I soldered the neg pads in series and used seperate wires (3) on the pos pads. I connected the resistor to the 3 wires…

Is that close, hot, or COLD?

I was told to use a 1 Ohm 2watt resistor for a single die blue 3w led and 3.7v cell. Or 2 Ohm, 1 watt… first sounds right…

Red only handles 750 mA compared to blue which requires 1000 mA. So, would my choice 0.47 ohm resistor 5watt be a bad choice based on this logic? It would allow more current than the 1 Ohm.

Thanks for listening and hopefully for ANSWERING my questions for this project. A module is $30 investment so I don’t want to sizzle it. Not fun.

Dave Hartman

Learning about electronic circuitry