so much info on working with LEDs, just needing a solution. help??

[reurbo]

yes, i am new to working with electronics at the circuit board level & to me, until recently, just about every electrical solution for wiring was solved by a simple logic of connecting positive to negative leads, things working, & me moving on to something else.

now, i'm working on a project where i need to learn more besides just watts & volts. i'm upgrading a single IR LED on a, for lack of a better comparison, single-button remote that only sends 1 signal every time the circuit is closed to three 5mm high output IR LEDs i picked up from Radio Shack after reading several good reviews on them.
the given info for the LEDs are as follows:
Forward Voltage: 1.2V
Forward current: 100mA
Radiant power output (100ma): 16mW min.
Viewing angle to 1/2 intensity: 45degrees
Wavelength: 940nm

my questions are:
1) i know i need a resistor to regulate the voltage & current for the LEDs but when i use any online LED calculator, i don't know if i need to multiply the Voltage & Current fields by 3 (how many LEDs i'm using) or just input what a single LED lists.

2) how do i, or rather what do i change in the LED calculators if i want to double or even triple the brightness & range of the IR LEDs?... do i increase the Voltage or Current values in the calculator?
Side Note: the IR LEDS won't be staying on constantly; they will blink for a short double burst each time a push button is pressed.

3) the circuit board with the original IR LED has a resistor i haven't been able to find what it's specs are for it to know if it needs to be upgraded as well. it looks to have the bars Green, Black, Gold, Gold on it. any one know of a way? i've already tried using Google.


finally, the power source will be four 4AA battery holders (24V total) in a series with an inline fuse before connecting to the circuit board with the IR LEDs connected to it. the reason for so many batteries is that there is a second device that will be drawing power from the same source & don't want to have to replace batteries constantly.


any questions are welcome & i'll answer as best i can.
thank you all in advance for any progress that is made from helpful answers!

 

[carbonzit]

my questions are:
1) i know i need a resistor to regulate the voltage & current for the LEDs but when i use any online LED calculator, i don't know if i need to multiply the Voltage & Current fields by 3 (how many LEDs i'm using) or just input what a single LED lists.

I'll limit myself to just this one question. The answer is, it depends ... on whether your 3 LEDs are in series or in parallel. You didn't say in your posting.

Suggestion: it's better to put them in series, and use a single current-limiting resistor for the series string.

Add their forward voltages, then use that figure for your calculations.

Hint: you really don't need one of those on-line calculators. It's a very simple computation based on Ohm's law:

R = (Vtotal - Vled)/I​

Translation: the resistor value is equal to the (total supply voltage minus the LED's forward voltage) divided by the current.

Example:

Vtotal: 24V
Vled: 3.6V
I: 100mA

therefore R = (24-3.6) / 0.1 = 204Ω. Use the closest standard value (either 180Ω or 220Ω depending on whether you want to go a little over or a little under on the current). Remember: less resistance = more current.

But: that's only half the battle (calculating the resistance value). You also need to size the resistor in terms of how much power it will be dissipating.

P = E x I

(translation: power = voltage x current)

Since we have 20.4 (24-3.6) volts across the resistor, and since the current through the whole shebang is 100mA (0.1A), the power is 20.4 x 0.1 = 2W. So you're going to need a 2-watt resistor. (Doubt you'll be able to get it at Rat Shack; try going to a real electronics vendor. There's a "sticky" thread in the Electronics Chat forum with a list of suppliers.)

 

[Pommie]

Green, Black, Gold, Gold = 5.1Ω

To increase the brightness you increase current. However, 100mA is quite high for an LED and might be the pulsed current. Check the data sheet for the permitted pulse current and pulse length.

Mike.

 

[Boncuk]

P = E x I

(translation: power = voltage x current)

Since we have 20.4 (24-3.6) volts across the resistor, and since the current through the whole shebang is 100mA (0.1A), the power is 20.4 x 0.1 = 2W. So you're going to need a 2-watt resistor. (Doubt you'll be able to get it at Rat Shack; try going to a real electronics vendor. There's a "sticky" message in the Electronics Chat forum with a list of suppliers.)

I would not like to use a 2W rated resistor if the calculated power dissipation is already 2.04W. At that power it gets almost red hot.

If soldered to a PCB it makes it look crispy after some hours.

Better use a 5W resistor.

Boncuk

 

[Boncuk]

Green, Black, Gold, Gold = 5.1Ω

To increase the brightness you increase current. However, 100mA is quite high for an LED and might be the pulsed current. Check the data sheet for the permitted pulse current and pulse length.

Mike.

Very interesting. Have you ever seen a bright IR-LED?

 

[Pommie]

Very interesting. Have you ever seen a bright IR-LED?

Only through my video camera.

BTW, what would the term be if not brightness?

Mike.

 

[reurbo]

I'll limit myself to just this one question. The answer is, it depends ... on whether your 3 LEDs are in series or in parallel. You didn't say in your posting.

yes, though i was thinking it, i failed to list that i was planning on wiring them in a series, not in parallel.

Hint: you really don't need one of those on-line calculators. It's a very simple computation based on Ohm's law:

R = (Vtotal - Vled)/I​

Translation: the resistor value is equal to the (total supply voltage minus the LED's forward voltage) divided by the current.

Example:

Vtotal: 24V
Vled: 3.6V
I: 100mA

therefore R = (24-3.6) / 0.1 = 204Ω. Use the closest standard value (either 180Ω or 220Ω depending on whether you want to go a little over or a little under on the current). Remember: less resistance = more current.

But: that's only half the battle (calculating the resistance value). You also need to size the resistor in terms of how much power it will be dissipating.

P = E x I

(translation: power = voltage x current)

Since we have 20.4 (24-3.6) volts across the resistor, and since the current through the whole shebang is 100mA (0.1A), the power is 20.4 x 0.1 = 2W. So you're going to need a 2-watt resistor. (Doubt you'll be able to get it at Rat Shack; try going to a real electronics vendor. There's a "sticky" thread in the Electronics Chat forum with a list of suppliers.)

thank you for all of this..it is greatly appreciated & makes a lot more sense than the ramblings most sites tried showing the same thing.

Green, Black, Gold, Gold = 5.1Ω

To increase the brightness you increase current. However, 100mA is quite high for an LED and might be the pulsed current. Check the data sheet for the permitted pulse current and pulse length.

Mike.

thank you. & yes, after reading several sites list around 20mA as "common" for LEDs that produced visible light, 100mA did seem very high so i wasn't sure about it being increased to raise the signal range.

I would not like to use a 2W rated resistor if the calculated power dissipation is already 2.04W. At that power it gets almost red hot.

If soldered to a PCB it makes it look crispy after some hours.

Better use a 5W resistor.

Boncuk

thank you for that advice. i will be posting up a diagram of the full wiring layout asap to help give a better idea of what i'm building.
i'm trying to modify as little of the original PCBs as possible while still getting the functionality/feature i'm wanting out of this project.

 

[carbonzit]

I would not like to use a 2W rated resistor if the calculated power dissipation is already 2.04W. At that power it gets almost red hot.

I disagree. Since when can a 2-watt resistor not be used at 2 watts, more or less continuous?

It might get warm, true, but it's not going to get "almost red hot".

 

[audioguru]

A 2W resistor that is continuously dissipating 2W will be very hot. It will burn you, it will melt plastic and it might even char the circuit board.

RadioShack does not have a detailed datasheet for the IR LED so you must guess at the amount of current and the pulse width that will burn it out.
LEDs sold for much lower prices at a [b[real[/b] electronic parts distributor have detailed datasheets.

 

[crutschow]

If the unit is only on momentarily for a short period of time, as with a typical IR remote, then a 2W resistor should be fine.

 

[Boncuk]

I disagree. Since when can a 2-watt resistor not be used at 2 watts, more or less continuous?

It might get warm, true, but it's not going to get "almost red hot".

According to power dissipation calculation the value is already beyond 2W (2.04W).

I wouldn't even use a resistor like that for heating purposes.

Power resistors are best selected when they are rated double the calculated power dissipation.

Even with that power rating it's advisable to mount the resistor 2 to 3 mm above the PCB for free air circulation around it at a safe distance between the resistor and other components, especially electrolytic caps.

If they get heat caused by a hot part they will dry out faster than normally and malfunction.

Boncuk