How to calculate the value of resistor for LED’s (with 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
LED Symbol, Construction and Lead Identification
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.
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How to calculate the value of resistor for LED’s
This is the Simplest LED Series circuit ever. Here, the supply voltage is 6V, LED Forward Voltage (VF) is 1.3 Volt and Forward Current (IF) is 10mA.
Now the Value of resistor (which we will connect in Series with LED) for this circuit would be:
Resistor Value = (Vsupply- VF)/ IF
= (6 -1.3) / 10mA = 470 Ω
Current draw = 20mA
Resistor Power rating formula for this circuit
Resistor Power Rating = IF2x 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.
  • IF = Forward Current of LED: 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.
  • If you are still unable to find the LED forward current, than assume it 20mA because a typical LED’s run on 20mA.
  • VF = Forward Voltage of LED: 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.
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 (VF) of LED’s is 2V and the forward Current (IF) is 20mA each.
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how to calculate the value of resistor for series circuit
Resistor Value (LED’s in Series) = (Vsupply – (VF x No. of LED’s)) / IF
Here, Total forward voltage (VF) of 6 LED’s = 2 x 6 = 12V and forward Current (IF) 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:
= (Vsupply – (VF x No. of LED’s)) / IF
= (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 = IF2x 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
Formula for finding the value of resistor(s) to connect LED’s in Parallel With Common Resistor


In this circuit, we have connected LED’s in parallel with common resistor. Supply Voltage is 18V, The Forward Voltage (VF) of LED’s is 2V and the forward Current (IF) is 20mA each.
Resistor Value (LED’s in parallel With Common Resistor)
= (Vsupply – VF)/ (IF x No. of LED’s)
Here, Total forward Current (IF) of 4 LED’s = 20mA x 4 = 0.08A, and forward Voltage (VF) 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:
= (Vsupply – VF)/ (IF 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 = IF2x 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) 
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Formula for finding the value of resistor(s) for connecting LED’s in Parallel With Separate resistor


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 (VF) of LED’s is 2V and the forward Current (IF) is 20mA each.
Resistor Value (LED’s in parallel with separate Resistor)
= (Vsupply – VF)/ IF
Here, Total forward voltage (VF) of LED’s = 2 and forward Current (IF) 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 1st simple circuit above, you will find that these are same)
Now, the value of resistor (for parallel Circuit with separate resistors) would be:
= (Vsupply – VF)/ IF
= (9 – 2) / 20mA = 350 Ω
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
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/

About Electrical Technology

Electrical Engineer. Writing for www.electricaltechnology,org You can contact us via Email or on Facebook. follow us on Also, Follow

14 comments

  1. Its very usefully for Electrical Engineer

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

  3. 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 &amp; advice. There Should Be as this<br />(18 – (2 x 6) / 20mA = 300 Ω)

  4. 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

  5. Senarath Seneviratne

    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 &amp; capasitors <br />I work with LED Lamp circuits &amp; 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

  6. Senarath Seneviratne

    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 &amp; the Paralel capasitor valu <br /><br /><br />Thank you <br /><br />Senarath<br />senarathmdon@yahoo.com

  7. Thank you very much.<br />It&#39;s very helpful for me.

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

  9. 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.

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