> Ledtronics LED 120 deg angle + current regulator for led
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fimdan |
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Class : Apprenti
Posts : 4
Registered on : 05/11/2007
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What do you think about this LED made by Ledtronics.
http://www.ledtronics.com/ds/SML10IR941T/Dsdc0053.pdf
It seems to have excellent 120 deg angles. What about electrical characteristics? I am asking cause I am not electronics expert I would rather not want to blow up these expensive LEDs.
To me 100mA for an LED seems to be rather high. I guess running it at lower current would not hurt. Also, is there a device that would allow me to manually control the amount of current (say between 10 - 100 mA). I would like to be able to adjust it for different LED intensity.
Also, sites like http://ledcalc.com/ give me different recommendations for the led setup. Can you point me to a tutorial how to calculate these numbers myself.
Thank you for making this site available to all of us!!
Daniel F. |
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wody |
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Class : Apprenti
Posts : 5
Registered on : 30/10/2007
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Hi Daniel,
according the datasheet 100mA is the maximum current! I would target some 20 to 40 mA.
The forward voltage is 1,9V (@50ma), so the calculation would be:
5V Supply - 1,9V = 3,1 V for the Resistor.
Ohm's formula is: R = U/I = 3,1V / 30mA = 103Ohm.
The next available resistor would be 120 Ohm (Gives you about 25mA).
If you want to regulate the brightness = current, just connect in series a fixed resistor with 68 Ohm (for 50mA) and a linear potentiometer with about 1kOhm.
By the way, the ledcalc.com site is ok. You enter the supply voltage, the forward voltage in the Voltage-drop-across-the-LED-field, and the current you want. All for 1 LED.
Hope that helps
Peter
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fimdan |
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Class : Apprenti
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Registered on : 05/11/2007
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Hello,
Thank you for clarifying some of these things for me.
I have decided to go with 9V battery and 4 LED setup. I will buy 3 different types of resistors (for 70, 50, 30mA current settings) . This way I can test the setup with three different values.
Also, I could not find a cheap potentiometer so I think I will dump the idea of adjustable intensity for now.
Now I have to wait for all the parts.
Thank again,
Daniel F.
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Kestrel |
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Class : Webmaster (admin)
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Registered on : 13/07/2007
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fimdan @ 05/11/2007 - 23h54 a dit:
Also, I could not find a cheap potentiometer so I think I will dump the idea of adjustable intensity for now.
You can buy inexpensive trimpots that are mini potentiometers.
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fimdan |
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Class : Apprenti
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Registered on : 05/11/2007
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Thanks
I am learning a lot here. Now those trimpots. Does 1KOhm trimpot allow me to control resistance between 0 and 1kOhms? So, if I put is behind a 68Ohm resistor I will be able to regulate total resistance from 68 - 1068 Ohms?
Am I getting it right?
Thanks
Daniel F.
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protonmw |
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Class : Apprenti
Posts : 15
Registered on : 25/09/2007
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Please take care of the maximum power that the potentiometer can handle!!!
P = U * I
And if you want to built the 4 leds in serial I wont do so because the battery changes his voltage with the time and with this little resistor you will get a big change in current! If you want to use a 9V battery I would built 2 leds in serial and the same thing again parallel to it. (I hope you understand my text   ) |
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tristan68 |
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Class : Electronicien en chef
Posts : 651
Registered on : 10/07/2007
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Hello fimdan
Please go for a 3led setup , 4 led is obsolete design.
@protonmw :
If you want to power your setup with a 9v battery , then the battery lifetime has an importance, and the battery voltage is high enough.
So instead of a parallel-serial (2 x 2leds and a resistor) design as you suggested, i would rather recommand a full serial setup (4 leds and a resistor)
cya
or
depending if you want to create a 3led or 4led setup |
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fimdan |
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Class : Apprenti
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Registered on : 05/11/2007
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Thanks
Regarding battery life. I assume as the battery is drained the voltage decreases.
Here is how I understand it. The formula says that r = v/i or i = v/r.
Now if v changes (decreases), and r stays constant, the value of I will also drop. So, the lower the voltage, the lower the current, and the lower the intensity of the LED.
For power. Formula is P = V x I. If battery voltage and current begin to fade, so will the power consumption.
So, if my math is correct, I will not set my head on fire. Or should I get a fireproof baseball cup??:)
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tristan68 |
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Class : Electronicien en chef
Posts : 651
Registered on : 10/07/2007
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no, no need for a fireproof cap 
and your math is correct
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