Electrical theory 1.3f - Series LED connections

We’ll design a rainbow light emitting LED string. Note a resistor, relay contact or switch can be connected either side of the LED.

Case 1

The circuit is shown below. The LED voltages and the physical connections for the LEDs and resistor are also shown. The last LED is a white (light emitting) LED.

 

Step 1

Initially we'll set the voltage across the current limiting resistor at 1.0V.

 

 

You remember for a series circuit, the voltages are added together.

So the total of voltages is 2.0 + 2.1 + 3.4 + 3.2  + 3.3 + 1.0 = 15V 

Note: this is the minimum acceptable supply voltage for good light output! The manufacturers' data was used for LED voltages.

 

Step 2

We’ll assume we have a 15Vdc supply and will operate the LEDs at .020A current (20mA).

To calculate the value of limiting resistance (R ), use Ohm’s Law as follows:

 

R = voltage/ current or V/I

R = 1.0/ .020 = 50 Ohms; the closest standard value is 47 Ohms

Step 3 (the resistance value changed slightly, so recalculate current - optional step)

Using Ohm's Law in different format: I= (V supply - VLED total) /R

Revised current = (15 – (2.0 + 2.1 + 3.4 + 3.2  + 3.3) )/ 47 = .021A  This current is within reliable tolerance limits.

Step 4
Resistor power rating: P = V*I = 1.0 * 0.021 = .021W; we can use a 1/4W size which is easy to buy.

Alternate equation: P = I squared * R = .021 * .021 * 47 = .021W

Note: some LED videos show a LED connected directly to a small battery without a resistor. The video circuit uses the resistance of the small battery to limit current. It has the disadvantage of discharging the battery quicker. You can't do this with a car battery because the LED will melt!

Case 2

Here’s what to do if you only have a 24Vdc power supply. (If this circuit was connected to 24V it would quickly melt because a current around 0.2A will flow!) There are 2 options as solutions:

 

Option 1: put more LEDs into the circuit and repeat above process (Case 1);

Option 2: we only want 5 LEDs so we increase the size of the resistor as calculated below:

 

Total LED voltage = 2.0 + 2.1 + 3.4 + 3.2  + 3.3  = 14.0V
To calculate the voltage across the resistor, subtract the total LED voltage from the supply voltage:

V resistor = V supply – V total LEDs = 24.0 -14.0= 10.0V

Now we want .02A flowing through the LEDs which also flows through the resistor (it’s a series circuit). So use Ohm’s Law to calculate the resistance value:  

R = voltage/ current or V/I

R = 10.0/ .020 = 500 Ohms; the closest standard value is 470 Ohms

Since 470 is close to 500, there's no real need to recalculate the current - it's within safe limits.

Resistor power rating: P = V*I = 10 * 0.02 = .20W;

We could use a 1/4W size but it will get too hot. So use a resistor about 4 times the wattage to keep it cool, say a 1W size. This will only cost an extra 10c! (Note resistor wattage sizes: 1/4W, 1/2W, 1W, 2W, 5W, 10W)

Did you find designing a series LED circuit relatively easy?

KEY EQUATIONS:

Resistance in Ohms = (V supply - total LED typical voltages)/ required current

Power rating of resistor = resistance * current * current * resistor cooling factor (approx =4)

Here's a website LED calculator to use: http://led.linear1.org/led.wiz
Note: this program only doubles the resistor wattage; I recommend doubling this wattage again to ensure the resistor runs cool and does not become a fire hazard.

NEXT >> Parallel connected LEDs