# LED

Discussion in 'The Projects Forum' started by whycanot, Jul 25, 2011.

1. ### whycanot Thread Starter Member

May 22, 2011
54
0
I connected 1 Infrared LED Transmitter with a 1.5V battery
It lighted up, but after a few minutes, it overheat n spoil
I want to know how much resistance should connect in series between the Infrared LED and the 1.5V battery

2. ### SgtWookie Expert

Jul 17, 2007
22,194
1,761
It would help a great deal to know what the particular diode's specifications are.

However, a Vf of 1.2v with 20mA current would be pretty typical.

You calculate the limiting resistor as:
Rlimit >= (Vsupply - Vf_LED) / Desired_Current
and then use a resistor that is >= the result.

A table of standard resistance values is here: http://www.logwell.com/tech/components/resistor_values.html
Bookmark/favorite that page.
E12 and E24 values are pretty commonly available E48 and higher are more expensive and usually need to be ordered.

If you can't find a value close enough, you can always use pairs of resistors in series or parallel to get much closer.
Here is a resistor calculator: http://www.qsl.net/in3otd/parallr.html
Very handy.

OK, let's say you have 1.5v as your supply, your IR LED needs 20mA current and will have a Vf of 1.2v.
Rlimit >= ( 1.5v - 1.2v) / 20mA = 0.3 / 0.02 = 15 Ohms.
Looking at the table of standard resistance values, you can see that 15 Ohms (shown as 150 Ohms; this is a decade table where you can multiply or divide the shown values by powers of ten) is a standard value of resistance.

If you want the LED to stay on longer, you could use two batteries in parallel.
If you wanted the LED to maintain it's brightness for a longer period of time, you could use two batteries in series, and a different resistor because your source voltage increased.

Two batteries in series would be 1.5v+1.5v=3v.
Then you would need to calculate:
Rlimit >= ( 3v - 1.2v) / 20mA = 1.8 / 0.02 = 90 Ohms.
Looking back at our standard values of resistors, we can see that (910/10) = 91 Ohms is a standard E24 value of resistance; very close to 90 Ohms.
But what if we didn't have a 91 Ohm resistor? What values could we use?

Going to that resistor calculator and asking for a 90 Ohm resistor from E12 resistors results in:
Code ( (Unknown Language)):
1. 68  +   22  =   90      (0 %)
2. 180 ||  180 =   90      (0 %)
3. 82  +   8.2 =   90.2        (0.222 %)
4. 150 ||  220 =   89.189      (-0.901 %)
5. 100 ||  820 =   89.13       (-0.966 %)
6. 56  +   33  =   89      (-1.111 %)
7. 120 ||  390 =   91.765      (1.961 %)
8. 47  +   47  =   94      (4.444 %)
9. 82  +   0   =   82      (-8.889 %)
The "nn || nn" means the resistors are wired in parallel, "nn + nn" means wired in series.
Almost all of them would be a reasonable choice except for the very last entry; too much current.

whycanot likes this.

Jul 7, 2009
1,585
141
The usual way is to find out the specified voltage drop of the device (e.g., read the specifications sheet) and its operating current. This may be given as a range or a maximum value may be given; use the highest value. Subtract the voltage drop from 1.5 volts; this is the voltage that must be dropped across the resistor -- call it V0. Calculate the value of the needed resistor from V0/i, where i is the rated current of the device. Use volts and amperes for units and the answer will be in ohms.

The best way is to find out the rated current of the device, then use a constant current DC power supply to set this current through the device. Read the voltage drop across the device and proceed as above.

If you answer that you don't have the spec sheet, then that's kinda like using a medical drug without reading the label...

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4. ### whycanot Thread Starter Member

May 22, 2011
54
0
the attached photo is my IR LED~
i dun have it's spec~

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283.2 KB
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5. ### whycanot Thread Starter Member

May 22, 2011
54
0
so you means that connect 1.5V with IRLED and 150ohms together in series then can work dy? the IR LED is shown on my previous reply~

6. ### RRITESH KAKKAR Senior Member

Jun 29, 2010
2,831
91
IR led require a switching pulse ( may be 38khz which is good for its app.) and connetcing directly to supply will d o nothing..!!

now i think you can connect the resistance in series with it..

If you want to see is Glowing use a digital camera like Mobile,etc.... not with naked eyes..!!

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7. ### wayneh Expert

Sep 9, 2010
13,440
4,274
It won't hurt to try. If it does not work, you'll know because the LED will not light. You can tell this by viewing it with your cellphone's camera, or maybe most digital cameras. You can test the camera with any working remote control you have - you should see a bit of light from the emitter LED if your camera can "see" IR.

150Ω may be too large a value, but will protect your LED. Only go to smaller values (120, 100, etc.) as you determine that the current is not too high. Be aware that a fresh battery will deliver more current than a used one. You want to choose a resistor that will protect your LED even with a fresh battery.

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8. ### Wendy Moderator

Mar 24, 2008
20,993
2,731
You can't tell the specs just by looking at it. You need to look where you bought it. Just from what you are saying I'm willing to bet you didn't have any resistor on it, this is almost always a mistake. LEDs require resistors. The only time this is not true is if you are using a very small watch battery, and that will work only because they have an effective resistance inside.

Use the 150Ω Wookie suggested.

LEDs, 555s, Flashers, and Light Chasers

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9. ### mcgyvr AAC Fanatic!

Oct 15, 2009
5,063
1,099
IF your device is a typical led with a Vf of 1.2V and requires 20mA then you would use a 15 ohm (not 150 ohm) resistor in series.
The formula is (Supply voltage - led VF)/current = required resistance
So (1.5-1.2)/.020 = 15 ohms

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10. ### wayneh Expert

Sep 9, 2010
13,440
4,274
That sounds better, and is what Wookie recommended. The OP missed his comment on the table and (Bill and) I didn't catch it.

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11. ### SgtWookie Expert

Jul 17, 2007
22,194
1,761
I posted 15 Ohms as the value of the limiting resistor.
150 ohms got confused in there, because that is what is shown in the standard decade resistance table, and I don't think that our OP understands that the table represents the values in each decade, not absolute values.

It depends upon the OP's application. They may be using it for IR illumination, not for signalling.

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12. ### RRITESH KAKKAR Senior Member

Jun 29, 2010
2,831
91
What is IR illumination...?

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13. ### nerdegutta Moderator

Dec 15, 2009
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14. ### whycanot Thread Starter Member

May 22, 2011
54
0
thanks for the help ... really appreciate it .... thanks guys !