Hi,

Its a bit strange. I have two 7805, one for two small DC motors and one for other devices including my LCD and its backlight (I'm building a robot).

Strangely, when my backlight is on, the 7805 it is connected to (not the DC motor's one) becomes very hot, and I have to turn it back off again.

This is so strange because there is no noticable heat in the 7805 powering the DC motors but the 7805 for LCD backlight becomes really hot...

I have tried inluding series resistance in with the backlight (using a potentiometer) but getting the same problem.

I can always get a heatsink for that 7805, but its just too strange a simple backlight could cause so much power dissipation in that 7805?

I have looked for short circuits etc. Nothing found.

 

Have you measured the current that the 7805 is passing?

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What is the input voltage to the 7805 regulator. If it is much higher than 8V then heat is produced when there is also some current.

How many LEDs in the backlight? 10 LEDs at 25mA each is 250mA.

If the input to the 7805 is 12V then it heats with (12 V - 5V) x 0.25A= 1.75W and it will be very hot without a heatsink.

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unless the 7805 is somewhere it can surprise someone with its hotness, don't worry about it. it has a built in safety feature that will reduce the current output if the ic overheats.

if you have some piece of metal to screw it to, you'll make things easier, otherwise, just tuck it someplace where it's not easy to touch, or put a HOT label on it.

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No.
It doesn't reduce the current, it shuts off. When it cools enough then it starts working and heating up again. The thermal stress of heating and cooling over and over will kill it.

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Uncle $crooge

 

Ace, what is the current rating of the backlight. the LCD data sheet should have it.

what is the input voltage of the backlight's 7805?

My guess is that between your backlight current draw and Vin-Vout you are dissipating at least 750 milliWatts. That's getting up there for your package.

Note that the 7805 is spec'd up to 125C (iirc). even 100C won't kill it. It wouldn't hurt to measure the temperature. All things considered, it's better to run cool as possible, though.

 

You may have to limit the current by an external series resistor for the Backlit path alone. generally a 10 or 6.8 ohm 0.3 watt resistor would so.

Though the internal facility is there for strapping the series resistors in some diaplays- but they are generallt SMD. so an external is safer.

Sarma

 

The chip inside the regulator IC is a lot hotter than the surface of the IC's package.

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Uncle $crooge

 

Thanks for all the help.

Well, the input voltage is 11.06V.

I have checked the current output from the 7805 last night, and if I can recall, with the backlight on, it was outputting just more than 0.7A!

With it off, it was outputting something like 23mA.

I dont know how many LEDs it has, it does not say on the datasheet (which is just a single sheet, or do I have the actual full datasheet?!)

I will buy a heatsink for the 7805.

My method of placing series resistance was to connect a potentiometer between the -ve terminal (ground) of the LCD, then I turned it (reduced the resistance) to get the LCD just ON (quite dim). But at decent backlight level, which is when it looks its best, the 7805 becomes really hot.

I think I just need a heatsink to dissipate all that heat.

 

Your 7805 is dissipating (11.06V - 5V) x 0.7A= 4.24W. If the ambient temperature is 30 degrees C then its chip is at 111 degrees C if it is in free flowing air. It is close to shutting off.
Get a heatsink for it.

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Uncle $crooge

 

That sounds like the contrast control, not the back light. Just put a small resistor in series with the back light.

 

Hi thanks.

Can you kindly explain why you did 11.06-5[V] ?

 

No, I am using the same method for backlight. This should be the same thing right?

I am connecting the GND of the backlight to the middle terminal of the potentiometer, which is gives me varied voltage as I adjust the resistance.

If I put a normal resistor there, we are getting a voltage drop between the resistor and the LCD is seeng 5V - V. drop accros resistor. Using the potentiometer this way should achieve the same thing right?

Or have I got it wrong?

 

audioguru, you sometimes need to explain your calculations more clearly.

He was calculating the power dissipated by the LM7805 and therefore the temperature rise.

Voltage drop = 11.06 - 5 = 6.06V
Power dissipation = 6.06 * 0.7 = 4.242W

The thermal resistance of an unheatsinked 7805's die to ambient in free flowing air is about 19˚C/W.

In free flowing air, the temperature rise will be:
Δt = 4.242*19 = 80.6˚C

If the abient air temperature is 30˚C the 7805's die temperature will be:
t = 30 + 80.6 = 110.6˚C

The LM7805 will shut down if its die gets hotter than 125˚C. Your LM7805 is operating close to, or above its upper limits so it's no surprise it's not working very well.

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why not draw the backlit current from unregulated 12 v by increasing the series resistor value to get the current as per the manufacturers spec? this will make 7805 COOL

Sarma