# LED Help

Discussion in 'General Electronics Chat' started by Lightfire, Jun 8, 2011.

1. ### Lightfire Thread Starter Well-Known Member

Oct 5, 2010
690
21
Hello folks,

This is my first project which compose only of Light-emitting diode.
Now, I want to ask if what resistor should I use for this project.

Now, refer to the schematic image I attached.

As you may see, every groups were connected in series (as somebody recommended) and all groups were connection in parallel.

As you may see again, every group has a different numbers of Light-emitting diode. The first group has a 9 Light-emitting diode connected in series, the second group has a 13 Light-emitting diode connected in series, the third group has an eighteen Light-emitting diodes connected in series, the fourth group has a 19 Light-emitting diode connected in series, the fifth group has a 13 Light-emitting diode connected in series, the last group has a nice Light-emitting diode connected in series.

For the battery, the battery that I will use is rated as 12 volts, 12 amperes-hour.

Any answers and help will be greatly appreciated.

Lightfire

• ###### aaa.PNG
File size:
17.5 KB
Views:
27
Last edited: Jun 8, 2011
2. ### Lightfire Thread Starter Well-Known Member

Oct 5, 2010
690
21
That would be great if a 9 volt battery will work just fine.

3. ### Jotto Member

Apr 1, 2011
159
17
what is the Vf of the diode, what is the Ir of the diode? only thing we have is the voltage.

4. ### Wendy Moderator

Mar 24, 2008
21,350
2,918
Did you read the link I gave? It covers LEDs in detail.

Each diode drops so much volts, this is called Vf. The total dropping voltage of each diode adds up. Red LEDs can be 2.0Vf to 2.5Vf. The answer I gave you here covered how to measure real Vf, and should give you clues how to use LEDs.

So three LEDs in series (if they were 2.5Vf) would add up to 7.5Vf, and a 9V battery powering these 3 LEDs would have 1.5V left over.

LEDs are current devices, so you must select a resistor that will take that 1.5V and only allow 10ma (0.01A) through.

Remember Ohm's Law?

1.5V / 0.01A = 150Ω

If you go to the next illustration in my LEDs, 555s, Flashers, and Light Chasers, you will see this illustration (the first on I showed you also came from this link).

In this illustration, I used 20ma.

I hate to break it to you, but reading is required if you want to learn this stuff, and you will need to learn some math, just like school.

5. ### Jotto Member

Apr 1, 2011
159
17
I would change the layout.
1. 9 diodes would be two rows, 5 220 ohms, 4 470 ohms, 20ma.
2. two rows 5 220 ohm, 1 row 3 680 ohm, 30ma
3. 3 rows 6 1 ohm, 30ma.
4. 3 rows 6 1 ohm, 1 row 1000 ohm, 30ma.
5. same as 2.
6. same as 1.

Vf used 2vdc, Ir 10ma, 12vdc Vcc.

6. ### mbohuntr Senior Member

Apr 6, 2009
431
32
Hey catapult,

To help a bit, A LED has a foward voltage rating, and a current limit. If you have a 5V source, and you want to light (ONE) LED with a Vf of 1.8volts, and limit it to 20mA, heres how. 5V-1.8Vf = 3.2V 3.2V/.02= 160ohms

Add the Vf of your leds in series and you can see you are higher than 12V !!! Take some out of series and re-post. Here is a calculator
I found on line. Do the math and see if it makes sense to you before you connect anything.
http://www.ledcalculator.net/

7. ### Lightfire Thread Starter Well-Known Member

Oct 5, 2010
690
21
Take a look at this, below.

But why? 2.5 divided by 0.02 (20 milliampere) equals 325? This should be 125.
Correct me if I were wrong, please.

Lightfire

8. ### Lightfire Thread Starter Well-Known Member

Oct 5, 2010
690
21
To Bill_Marsden
As mbohuntr,

Okay. The power source was rated as 9 volts and the Vf of the diode was rated at 2.5 and the diode ampere rating was rated as 0.02 (20 milliamperes). So the answer should be 180 ohms. Not 325 ohms. How it came 325 ohms? A bit confusing.

Lightfire

9. ### gerty AAC Fanatic!

Aug 30, 2007
1,231
341
I think where you are confused is that you are using the led voltage,2.5v, instead of the voltage across the resistor.
As Bill wrote:

10. ### mbohuntr Senior Member

Apr 6, 2009
431
32
9V-2.5V = 6.5V

6.5V/.02= 325ohms

Like these guys said,
1) Make sure you know the Vf of the LED, It is not always 2.5V, it depends on what you have.
2) Make sure of the max current the LED allows and stay below it.
3) Add the Vf of each series led and keep it below your source.
4) Subtract the total Vf from the source an divide that number by the desired current.

This gives you one string of series leds. Try this first. Then move into parallel strings when you get the hang of it.

11. ### debjit625 Well-Known Member

Apr 17, 2010
790
186
As gerty and mbohuntr said you need the voltage drop across the resistor rather then the voltage drop across the LED, here you go...

Ohm's law
V = I * R
So if we need R then it will be R = V/I
Now we know that I is 20mA or 0.02A
Now we need the V,this V is the voltage across the resistor not the voltage across LED, because the equation is supposed to be for the resistor not for the LED.

Now the power supply is of 9V and we know the LED will drop 2.5V, so what is left
(9 - 2.5 = 6.5) so their will be 6.5V across the resistor, now lets solve for R
R = 6.5/0.02 = 325 ohms

Like you know ohm’s law,it is also very important to know two other very important laws Kirchhoff’s Voltage Law and Kirchhoff’s Current Law.Give it a look from AAC’s e-book.

Good Luck

12. ### Wendy Moderator

Mar 24, 2008
21,350
2,918
The voltage drop across the LED is a constant. It doesn't change. You can't change it. The only thing you can change is the resistor. It is the resistor that sets the current. It can't change the voltage, but it can take the voltage left over from the LEDs and set their current. Look at the illustrations.

9V (battery) - 3 X 2.5V (LEDs) = 1.5V

This is the voltage the resistor has to work with.

13. ### SgtWookie Expert

Jul 17, 2007
22,202
1,792
9v "transistor" PP3 batteries are compact, but expensive, don't have much of a mAh rating (500mAH is typical for a fresh alkaline battery) and won't last long under a heavy load.

A 500mAh battery will supply 25mA for 20 hours before the battery voltage falls below the "discharged" threshold. That's not a lot of power.