Hi guys,

I wish someone could give me an advice with the following:

I wanted to replace a 240x64 LCD module, model PG24064-E (EL backlight) and ordered an LED backlight one, model ERM24064FS-1. Only upon arrival I have realised that it's a 22 pin version and the previous one is a 20 pin version. I have checked the data sheets and pins 1-20 are the same on both units and the last two pins on the new one are for powering the LED backlight. Now how could I use this with a 20 pin connector? Can I just solder in a 20 pin header and solder a piece of wire between the module power supply the LED power supply + to + and GND to GND? They both take 5V DC..
I know this might be a very silly question, but I don't actually know much about electronics, I just want to replace the LCD in my device. Your help is much appreciated guys! Cheers!

 

You are correct in being careful. I can't give you a definite answer to your question, but I think that you may need a current limiting resistor in series with one of the LED leads. If not provided, too much current could be allowed, which would quickly burn out the LED providing the backlight.

Look in the datasheet for the display to find the current required by the backlight LED and post it here. (It will be measured in milliamperes, abbreviated mA.) Then, someone here will help you calculate the resistance value of the current limiting resistor needed.

 

Have a look at the attached diagram, especially in the lower left corner, and see if that helps.

 

Okay, thank you tracecom! Here's the value you're talking about, I hope I deciphered the data sheet correctly:

Absolute maximum supply current for LED backlight (symbol ILED): 160 mA TYP.

So if I get your gist, in case I don't know the current of the 5V supply in my device that used to power the previous module, I should use a resistor after I split from that supply to the LED power input so I definitely don't fry the backlight right? But if I measure the supply current before soldering anything and it's below 160 mA, it should be fine.
I hope someone can calculate for me quickly the parameter of the resistor required.

Can I also deduct that the backlight will be the brightest when the current is reaching the limit of 160 mA and if it's too bright I shall also use a current limiting resistor of a lower value in order to dim it?

Cheers!

 

Have a look at the attached diagram, especially in the lower left corner, and see if that helps.

Thank you, unfortunately I can only guess but I would say it either shows that the LED power input already has some resistors in the circuit to keep the current between 120-160 mA and protect the backlight from frying or it it indicates what resistors shall be used when the supply current is outside these values. I hope it's the first one, because otherwise I'm still kinda lost I'm afraid.

 

You can take it in steps. Start with a 33Ω series resistor and observe the brightness. Reduce that to 22, 15 and 10Ω and note the increasing brightness.

Btw, you can use 33Ω, and another in parallel to give 16Ω and one more for 11Ω.

 

Thank you, unfortunately I can only guess but I would say it either shows that the LED power input already has some resistors in the circuit to keep the current between 120-160 mA and protect the backlight from frying or it it indicates what resistors shall be used when the supply current is outside these values. I hope it's the first one, because otherwise I'm still kinda lost I'm afraid.

It's the latter; you have to supply the resistors. MrChips has given you good direction; just be sure that if you go down to 11Ω, your resistor should be at least 1W, and 2W would be better.

Oh, one more thing. Too high a resistance will not hurt the LED; it will just not light as bright. But too low a resistance (less than 10 ohms) can damage the LED. I suspect that a 33 ohm resistor will give you enough brightness, and in that case, a 1 watt resistor would be fine.

 

Can I also deduct that the backlight will be the brightest when the current is reaching the limit of 160 mA

Correct.

if it's too bright I shall also use a current limiting resistor of a lower value in order to dim it?

No, just the reverse. The higher the resistance, the dimmer the light.

 

Thank you guys for pointing me in the right direction. I'm going to go with the 33Ω 1W resistor in series and if not bright enough, I'll try other resistors down to a 10Ω (1 or 2 Watt, depending on how much room I have).
This was educational for me, I'm grateful! Cheers!

 

Thank you guys for pointing me in the right direction. I'm going to go with the 33Ω 1W resistor in series and if not bright enough, I'll try other resistors down to a 10Ω (1 or 2 Watt, depending on how much room I have).
This was educational for me, I'm grateful! Cheers!

You are welcome. You might go ahead and buy 2 or 3 of the 33 ohm 1 watt resistors, and add them in parallel. In such an arrangement the reistance goes down, but the wattage is additive. That is, two 1 watt 33 ohm resistors would have a resistance of 16.5 ohms and a wattage rating of 2 watts. Three in parallel would have a resistance of 11 ohms and a wattage rating of 3 watts.

Good luck.

 

Most modern displays already have the backlight resistors onboard (so will run direct from +5v), you can see the resistors attached to the PCB tracks that go to the backlight LEDs.

However it can still be good practice to add additional resistance in series to reduce power consumption. If the LCD backlight is specced at 160mA you might find it will still have good visibility at 50-80 mA.

I'd start with a 100 ohm resistor and an ammeter, then reduce the resistor value as needed until you are happy with the display and are using as little current as possible. Even if you are not running from batteries that reduces heat and increases the life of the LEDs and LCD liquid.

It helps if the LCD is displaying something to properly judge the backlight brightness needed.