How to power Lumex LCD backlight?

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Fred.Amoson

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I recently picked up a Lumex 4 line LCD with a backlight. The LCD works fine, but I can't figure out how to power the backlight. The datasheet for the LCD is pretty lame and didn't mention anything about the backlight. Powering it directly from the 5V is not what I want to do since I am pretty sure it is an LED backlight, but if I put a resistor in series it doesn't light very much. Does anyone know the proper way to light the backlight? What voltage or resistor value I should be using? My LCD is LCM-S01604DSF if that helps.

Thanks
 
hi,
If you are driving the backlight LED from +5V, note its has a Vfwd drop of 4.2V
and requires 260mA to operate..

So 5v - 4.2v = 0.8v/0.26 = 3R

I would suggest initially you try a 10R..
 
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Thanks alot!

Where did you get that datasheet?

Also, why do LCD backlights use so much current? I would have expected the current draw to be much lower.
 
Fred.Amoson said:
Thanks alot!

Where did you get that datasheet?

Also, why do LCD backlights use so much current? I would have expected the current draw to be much lower.

hi,
The backlight has a number of fairly bright LED's in series.

I couldnt relocate the address for the datasheet, so I have attached it.

Lets know how it goes, others may have the same problem..
 
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The backlight probably consists of several strings of leds. There are probably 2-3 leds in each series, and probably a couple of strings in parallel to light the screen. Each string consumes a certain amout of current. Typical leds range from about 20mA all the way up to 150mA and higher in some cases. For example, let's say you had a series of 2 leds with a 2.1V drop on each led. This would give you 4.2V drop for the series. If these leds took 50mA and you had 4 strings total, that would make 200mA. (just for example)

To determine the correct resistor value, just do the math as ericgibbs said.

You are supplying 5 volts. You know the forward voltage is 4.2V, and the current is 260mA. This leaves 0.8 volts to be dissipated on the resistor. R=V/I, so R = 0.8V/.260mA = 3.076 ohms (just repeating you ericgibbs)

Also important to keep in mind is the power of the resistor. In this case, the power the resistor will have to dissipate is 0.8V * 0.260mA = 0.208 watts.

Using a 1/4 watt resistor should be fine, but I would use a 1/2 watter to be safe.


//edit - sorry eric did not see your post
 
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