7805 getting hot when using LCD backlight

Hi guys,
I'm building a small project that has a small 16x2 character Winstar LCD (WH1602B). For this project I'm using a "UNO Standalone board" I've built. In this project I'm using 12V and 5V so I'm using a 7805 regulator for the ATMEGA328 and rest of the circuit.

The 7805 is getting really hot when the display is connected (better, when the backlight is turned on). I just saw that the LCD led array expect 4.2V and I'm using instead the full 5V.
Led array seems to expect 100mA (datasheet here).
I think I have to add a 20ohm resistor before the 15th pin of the LCD to draw some voltage...

Will this be enough not to burn the 7805?
Thanks a lot.

Hi.

You can also look for the datasheet of the 780x.
You would find that this regulator has to "burn the excess voltage" (more politely described as "dissipate") to get to the 5 volts.
So it has to dissipate more than it would offer at its output ( 7 to 5 ratio).

Have a look in the displays datasheet to see whether it has resistors for the backlight on board.
If it doesn't you absolutely need to provide that.
Else you might very likely burn the regulator as well as the LEDs.

If you need to use the setup you've got right now, attach a heatsink to the 7805, so it can get rid of the heat.
You could also use an array of 780x'es so each one dissipates a bit (heatsinks still recommended).
But do read the datasheet, you need to use a bunch of capacitors to ensure stability.

I just saw that the LCD led array expect 4.2V and I'm using instead the full 5V.
Led array seems to expect 100mA (datasheet here).
I think I have to add a 20ohm resistor before the 15th pin of the LCD to draw some voltage...

You do not understand how LEDs are rated. Unlike resistors you do not apply some specific voltage and then get a resulting current.

Assuming the values mentioned above, it is your responsibility to use some means of current limiting to limit the LED current to 100 mA regardless of what power supply voltage you are using. If you succeed in limiting the current to 100 mA then you can expect to measure 4.2v across the LED terminals.

Don

For around five bucks, use a Buck converter. For a little more you can even get one with adjustable current limiting.

Thnx everyone for your answers, but

Have a look in the displays datasheet to see whether it has resistors for the backlight on board.
If it doesn't you absolutely need to provide that.
Else you might very likely burn the regulator as well as the LEDs.

I checked the datasheet, but I'm not so proof to understand it completely.
It says

PIN 15: 4.2V for LED (RA = 0 ohm) / Negative Voltage Output
LED Forward Voltage: Typical 4.2V, Maximum 4.6V
LED Forward Current: Array 100mA

What should I do?

Whenever you see a zero Ohm resistor, forget its name "resistor".
It is a jumper wire disguised as a resistor, to make the end result look better.

So there is no resistor on board, you need to add one.
You are powering the board with 5 volts, need to drop 0.8 volts to get to that 4.2 and you need to limit the current to 100 Ohms or less.

So the resistor would have 0.8 volts over it, current through them is 0.1 A.
U=I*R; R=U/I; 0.8/0.1; R = 8 Ohms.
You will have a hard time finding an 8 Ohm resistor, 9.1 Ohms is the next higher standard value (but will still not be easy to find).
The current and brightness will be a bit smaller, but the parts will survive a bit longer.

MAS3:
Whenever you see a zero Ohm resistor, forget its name "resistor".
It is a jumper wire disguised as a resistor, to make the end result look better.

So there is no resistor on board, you need to add one.
You are powering the board with 5 volts, need to drop 0.8 volts to get to that 4.2 and you need to limit the current to 100 Ohms or less.

Thanks a lot MAS3!
Good answer. I do not know why I made 5-4.2=1.8 in all my calculations, that's why was coming the 20ohm resistor. But tell me, could this be a reason to make the 7805 so hot?

wgh000:
I checked the datasheet, but I'm not so proof to understand it completely.

That's because it actually tells you nothing!

It says that the backlight could be any of three different things, or nothing. Obviously you do have one of them, but it is not entirely obvious which. I think you should post a perfectly focused photo of the back of the display (preferably not as an attachment, either).

You do have a heatsink on the 7805, do you not? Is it powered from 12V or something higher?


MAS3:
Else you might very likely burn the regulator as well as the LEDs.

I thought 7805 regulators were supposed to implement thermal shutdown and were thus short-circuit proof?

Might means not will.
Because OP didn't mention any heatsink, i pointed that out too and had to assume it wasn't used.
I don't know how well the thermal shutdown works and if it works as well without a heatsink.

Paul__B:
You do have a heatsink on the 7805, do you not? Is it powered from 12V or something higher?

No, I do not have heatsink, I thought I can have less Watt to dissipate lowering the current used.
7V*200mA should give around 0.3W isn't it?
7V is the drop of voltage, 200mA twice the current used by the led (I've got the display itself and also the ATMEGA running...)

If you use a big enough heat sink, or a 'buck converter, or anything else to keep the regulator from shutting down then all you will achieve is getting the backlight to be really really bright for a very short amount of time, then dark forever.

Don

I will wait for that picture of the actual display (and wiring) as we still have no idea what is going on.

7*0.2=1.4 last time I looked.
It should be heat sinked anyway.
You can not parallel 7805s. One will supply all the current until it
limits than another will begin to take the load.
There is a way to get LM317s to share current.
Dwight

Adding ( 100ma) current limiting resistor to drop .8V is little overkill (FULL 5V MAY shorted the life of the back-light LED to 100 years) , but the 7805 will still dissipate close to 1.5W so it will NOT run any cooler. Heatsink is must,thermally protected regulator or not , 1.5 W needs to go somewhere.
Why do you need 12V supply to start with?

Why not use a 7.5VDC supply?

http://www.dipmicro.com/store/DCA-07510

Or better yet, a 5V supply?
http://www.dipmicro.com/store/DCA-0510

I skip linear regulators when I can and use 5V wallwart switching regulators instead.
They also have short circuit protection built in, in case you short the supply out by accident.

In this project I need both 5 and 12V. So I decided to hold a regulator and not to have different power supply.

The display is the one I published in the first post, the datasheet say everything about it.
Pin 1 and 16 ground, Pin 2 and 15 5V (15 should be 4.2v with 100mA).
Connection as manuals say, leaving the first 4 data bus empty. Check any CristalLibrary example to see it.

So it seems the unique solution is to have a heatsink over the 7805, is it?
To low temperature I just have not to use backlight or reduce a bit the current (lowering the backlight).
Otherwise i could split the load, using 2 different 7805? Could this be an Idra?

Adding ( 100ma) current limiting resistor to drop .8V is little overkill ...

You don't understand how LEDs work either.

  • The specified LED forward voltage is a characteristic, not a rating.
  • The current limiting resistor limits current not voltage.
  • IF you limit the current properly then the resulting forward voltage will be near the specified value.

(FULL 5V MAY shorted the life of the back-light LED to 100 years).

The voltage applied to the LED-resistor combination is irrelevant.

  • If you do not somehow limit the LED current you will shorten the life of the LED, probably closer to 100 mSec than 100 years.
  • If you do limit the current properly then the resulting forward LED voltage will be less than the applied voltage, typically around 4.2v and at most 4.8v for this LED.

@wgh000

In order to calculate the desired resistance you start out with the desired current, the power supply voltage, and the typical LED forward voltage drop.

Use the resulting calculated resistance value as a guide to select the actual resistor. Choose a resistance larger rather than smaller than your calculation. Remember that the typical LED forward voltage drop you used in your calculation is probably not exactly what your particular device will exhibit. The actual voltage, however, should be less than the maximum value specified on the data sheet.

You really don't need to use the maximum allowable current for your backlight. You will probably find that the difference in brightness is indistinguishable (to the human eye) if you use a considerably lower current.

If you run the LED from your 12v supply you will lessen the power dissipated in the 5v regulator and you may eliminate the need for a heat sink.

Don't forget to check the power dissipated in the series resistor. This is particularly important if you choose to run the backlight from your 12v supply.

Don

wgh000:
Otherwise i could split the load, using 2 different 7805? Could this be an Idra?

If this idea originates from my remark to have 2 780x´es share the load, that is not what i meant.
I saw someone telling that it is not a good idea to put 2 of these in parallel too, which is something i absolutely agree with.
What i meant to say, was to use a 7808 for a first stage, followed by that 7805.
This way each has to dissipate a lot less.
Still a heatsink is recommended for both of them.

Good to know it is not good to place two regulator in parallel.

I can opt to place a 100ohm resistor and attach the backlight to 12v: what do tou say?
(12-4.2)/100mA = 78ohm
Using a 100 ohm i will have 78mA, is it?
And i will remove the load from the 7805.

(12 - 4.2) / 100 = 78mA

This is just regular Ohms Law. i.e. V / R = I

You can calculate power as V × I or V × V / R
i.e. 0.61 watts

David.