Clarification on LED millicandela and light output

[Rajinder]

Hi all,
I have two LEDs which have the following specifications

LED1 LED2
VF 3.0V 3.2V
IF 30mA 25mA
Viewing Angle 20 15
Luminuos intensity (mcd) 4500 8500
Reverse Voltage 5V 5V
Reverse Current 10uA 10uA

The LEDs are powered from 5V with suitable resistor. LED1 is used at the mment at 15mA, so 5-3.0V / 15mA = 133R. I am trying to get LED2 so it gives off the same brightness or light output as LED1. LED2 has a higher milli candela i.e. 8500. But LED1 has a better viewing angle. Do i need to convert the mcd milli candela readings to lumens to get the light output? Or do i need to increase the series resistor in LED2 to get the same light output? Or am i missing something out which I may have overlooked?

I would be very grateful for your help.
Best regards,
rajinder

 

[KJ6EAD]

Do i need to convert the mcd milli candela readings to lumens to get the light output? Or do i need to increase the series resistor in LED2 to get the same light output? Or am i missing something out which I may have overlooked?

Yes, if you want a numerically accurate match and assume the specifications are correct. Maybe. Almost certainly.

If you're looking for a visual match, power both equally then use a potentiometer on the brightest one to determine the correct resistance for matching. This is done by eye either direct viewing or preferably while projecting the light onto a white sheet of paper.

 

[Herschel Peeler]

Hi all,
I have two LEDs which have the following specifications

LED1 LED2
VF 3.0V 3.2V
IF 30mA 25mA
Viewing Angle 20 15
Luminuos intensity (mcd) 4500 8500
Reverse Voltage 5V 5V
Reverse Current 10uA 10uA

The LEDs are powered from 5V with suitable resistor. LED1 is used at the mment at 15mA, so 5-3.0V / 15mA = 133R. I am trying to get LED2 so it gives off the same brightness or light output as LED1. LED2 has a higher milli candela i.e. 8500. But LED1 has a better viewing angle. Do i need to convert the mcd milli candela readings to lumens to get the light output? Or do i need to increase the series resistor in LED2 to get the same light output? Or am i missing something out which I may have overlooked?

I would be very grateful for your help.
Best regards,
rajinder

Calculations are a theory only. Forget the math and adjust the resistor until you get the brightness you want.

 

[KMoffett]

If the two LEDs have different view angles, they can only have their perceived output matched from one position.

Ken

 

[Rajinder]

yes, i am after a visual match. the LEDs are used in an escape light for an alarm unit. So will light up the exit point to evacuate the building. They have different viewing angles.
So it is best to power what i have at the moment i.e. LED1. Then use LED2 in its place with a POT until I get the same output Visually? This would mean checking it just for the brightness. Should i check this at a certain distance?

 

[KMoffett]

The perceived brightness will drop off as an LED is viewed from any angle other that straight-on. The LED with a narrower viewing angle with appear dimmer as you move away for a straight-on view. So by using an LED with a narrow view angle, you cannot maintain a constant perceived brightness. Maybe you can better describe the environment. You may need multiple LEDs projecting at different angles. Or LEDs lighting a diffuser surface that will appear as a more constant brightness at wide angles.

Ken

 

[Rajinder]

Thanks all for your replies, much appreciated. Unfortunately i can't add multiple LEDs the second LED is supposed to be a drop in replacement (+ resistor change) for the first one.

 

[BobK]

It is odd that such a narrow viewing angle LED would be used for this purpose.

As KMoffet already said, you cannot make the second LED have the same visual effect from all angles as the first one. That said, I don't think it is going to make much difference in how well it performs if you simply run it at the nominal Vf and If.

Bob

 

[Rajinder]

Yes it is. There are 4 in total to give the required brightness. I just found that out today. Thanks for all the help.
How should I compare the outputs, I was thinking of mounting them on the ceiling in a dark room. Then visually checking that there are no big discrepancies in the output light transmitted.

 

[KMoffett]

To be semi-scientific, you might try the oil spot photometry method of comparative light output:
http://www.exploratorium.edu/snacks/oil-spot-photometer

Ken

 

[Herschel Peeler]

To be semi-scientific, you might try the oil spot photometry method of comparative light output:
http://www.exploratorium.edu/snacks/oil-spot-photometer

Ken

Fascinating!