# LED power consumption

#### christiannielsen

Joined Jun 30, 2019
330
I have a white LED I want to illuminate with a forward voltage of only 2,6v and a forward current of 0,02A.

1. If I am going to connect 39 of these white LED's in parallel how much power do they consump? (0,02A x 39) x 2,6v = 2 watt ?

2. The voltage drop over the current limiting resistor doesn't that also have a power consumption?

3. If I use a 9 volt power source will a 2 watt 320 ohm resistor be the right resistor to use for a current limiting resistor? (I know about the disadvantages in connecting LED's in parallel)

Thanks

#### Yaakov

Joined Jan 27, 2019
3,464
Your load is going to be 780mA because each LED needs 20mA to operate. To drop the voltage from 9V to 2.6V is going to take a very low resistance very high wattage resistor.

You need a switching supply, not a resistor.

• wayneh

#### christiannielsen

Joined Jun 30, 2019
330
I need to learn from my questions before I look into other solutions. Otherwise I don't get it.

Joined Feb 20, 2016
3,906
Try running groups of 3 LEDs in series with one resistor per group.
2.6x3=7.8V drop.
9-7.8=1.2V
resistor = 1.2/0.2=60ohms.
A standard 1/2W resistor will be plenty.

Of course, the voltage drop across each LED may well be different from the quoted 2.6V.
LEDs are current operated devices and running them relying on the assumed forward voltage can lead to some complications.
But, put one LED in series with a 390R resistor from the 9V and measure the LED drop to make sure.
It is not a real good idea to run LEDs at their max current level or a length of time if you want them to last.

• Yaakov

#### christiannielsen

Joined Jun 30, 2019
330
I really need the answers to my questions otherwise I don't understand the calculations.

So when I pick and calculate how big a resistor should be in wattage I shall use the power consumption of the voltage drop over the resistor and not the power consumption over the LED's ?

watt = 6.4 volt x (0.02 A x 39) = 5 watt

Is this correct?

watt = 2.6 volt x (0.02 A x 39) = 2 watt

And this is wrong?

Joined Feb 20, 2016
3,906
The resistor wattage is the voltage dropped over the resistor x the current through it.
So, if you run all the LEDs in parallel, then the total current will be 0.02A x 39 = 0.78A
Then, the voltage dropped across the resistor is 9 - 2.6 = 6.4V
Now, 0.78A x 6.4V = 5Watts. (close enough).
The resistor will be around 8.2 Ohms.
You need to use the resistor voltage, not the LED voltage in the calculation.
Running all the LEDs with only one resistor is a recipe for trouble.
If some LEDS become disconnected, the whole 780mA will flow through the other LEDS and they will fail.
Go with the groups of three LEDs and one resistor per group as above.

• christiannielsen

#### Yaakov

Joined Jan 27, 2019
3,464
I really need the answers to my questions otherwise I don't understand the calculations.

So when I pick and calculate how big a resistor should be in wattage I shall use the power consumption of the voltage drop over the resistor and not the power consumption over the LED's ?

watt = 6.4 volt x (0.02 A x 39) = 5 watt

Is this correct?

watt = 2.6 volt x (0.02 A x 39) = 2 watt

And this is wrong?
You seem to be ignoring the term “39” in your equation.

#### BobTPH

Joined Jun 5, 2013
3,652
1. You need a resistor for each LED or string of LEDs in series, or they will not get equal power.

2. You want to put as many in series as your supply voltage can support. This minimizes the wasted power.

For a nine volt supply, and 2.6V forward voltage, that would be 3 in series. 13 of these for your 39 LEDs.

There is a resistor for each string of three. You calculate by dividing the excess voltage by the current. So:

R = (9 - 2.6 x 3) / 0.02 = 60 Ohms.

The total power is 13 x 9 x 0.02 = 2.34W.

The power in each resistor is 1.2 x 0.02 = 0.024W. The total wasted power is 13 times that, or 0.3W

Bob

• #### KeepItSimpleStupid

Joined Mar 4, 2014
5,065
If you want the intensities to track, bin the LEDS by Vf. Put all of the similar Vf's in one string. You may have to adjust the resistance. One way is you say use 3 resistors with traces you cut to add that resistance. Say 60, 2 , 5, 10 ohms, but probablly closer.

#### christiannielsen

Joined Jun 30, 2019
330
If I go with the suggestion in #8 and I have 42 rows with 39 LED's each (1638 in total) how much will the power consumption be in total?

2,34 watt x 42 rows = 98,28 watt?

Is that a 11A power supply? sounds like a lot...

#### Audioguru again

Joined Oct 21, 2019
3,505
A white LED is 3.0V or more, not 2.6V. Google shows hundreds of graphs showing it.
Then 3 white LEDs in series might not light-up from a 9V supply.
Here is one graph:

#### christiannielsen

Joined Jun 30, 2019
330
A white LED is 3.0V or more, not 2.6V. Google shows hundreds of graphs showing it.
Then 3 white LEDs in series might not light-up from a 9V supply.
Here is one graph:
I have white led's and they illuminate sufficiently for my purpose at 2.6v.

I am going to use other colors too, for now I am only trying to learn the calculations.

#### Yaakov

Joined Jan 27, 2019
3,464
I have white led's and they illuminate sufficiently for my purpose at 2.6v.

I am going to use other colors too, for now I am only trying to learn the calculations.
I don't know where you are getting 20mA from. I have an example white LED and it does light at 2.6V though very dimly. I measured 4mA at 2.6V.

#### Audioguru again

Joined Oct 21, 2019
3,505
A white LED is a blue LED with a yellow-looking phosphor on top. A blue LED is also 3.0V or more.
Maybe your LEDs are bright pink or bright yellow?

#### Yaakov

Joined Jan 27, 2019
3,464
A white LED is a blue LED with a yellow-looking phosphor on top. A blue LED is also 3.0V or more.
Maybe your LEDs are bright pink or bright yellow?
No, it’s a warm white LED blue with phosphor. I can effectively forward bias it as low as 2.48V [EDIT: That should be 2.35V] and detect light, but the output is exceedingly small as you’d expect. It get down to .4μA consumption at that voltage. At 3V it uses 19.8mA as one would expect.

Last edited:

#### Yaakov

Joined Jan 27, 2019
3,464

#### christiannielsen

Joined Jun 30, 2019
330
If I go with the suggestion in #8 and I have 42 rows with 39 LED's each (1638 in total) how much will the power consumption be in total?

2,34 watt x 42 rows = 98,28 watt?

Is that a 11A power supply? sounds like a lot...

#### BobTPH

Joined Jun 5, 2013
3,652
If your LEDs take 20mA at 3.0V why are you using 20mA as the current at 2.6V? It might be as little as 2mA at 2.6V, which makes my calculation in post #8 way wrong. If you intend to run them at 2.6V you need to measure that current and redo the calculation.

Bob

• christiannielsen

#### christiannielsen

Joined Jun 30, 2019
330
My power bench supply and multimeter is not in sync. So lets say 2.7v at 2mA.
Then the LED illuminates so much that I actually can't look at it.

#### Yaakov

Joined Jan 27, 2019
3,464
My power bench supply and multimeter is not in sync. So lets say 2.7v at 2mA.
Then the LED illuminates so much that I actually can't look at it.
If you are using the bench supplies readout to decide the voltage you are going to be way off. Also, no LED will be blindingly bright at 2mA.