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Why does my LCD backlight draw so much current?

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Hero999 said:
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.

Thanks for this explenation, much appreciated. However, the actual point I wanted clarification on remains unclear. I know what the numbers 11.06-5=6.06 represent.

The question is, why do we use the 'excess voltage', if I can use this term to find the power dissipated? Why cant we just use 11.06 and multiply that with 0.7?

mvs sarma,

I did say on a later post that with the backlight off, current output was 23mA. I would immagine this is all that would matter.

But anyway, the other device connected to this 7805 is an 8051 microcontroller.

peace :)
 
PS.

I have installed a clip-on heatsink which is working rather nicely. The fins of the heatsink are not heating up too much indicating adequate heat-loss to the surrounding.

The backlight looks just right when connected to +5V and GND directly. Does the LCD not have a current limiting resistor built-in?

Thanks to all for all the help. I hope you understand I cannot reply to every single response/suggestions individually due to shortage of time, but I have taken all the advise onboard.

Thanks again.
 
AceOfHearts said:
The backlight looks just right when connected to +5V and GND directly. Does the LCD not have a current limiting resistor built-in?

Usually not! - because it depends on the voltage you're feeding it from. As suggested previously in this thread, measure the current the backlight is taking, if it's too high you MUST add a series resistor, otherwise it will blow fairly quickly.
 
AceOfHearts said:
the actual point I wanted clarification on remains unclear.
The question is, why do we use the 'excess voltage', if I can use this term to find the power dissipated? Why cant we just use 11.06 and multiply that with 0.7?
11.06V x 0.7A is the total power dissipated by the regulator plus dissipated by the LEDs in the display.
5.0V x 0.7A is dissipated by the LEDs and 6.06V x 0.7A is dissipated by the regulator.
 
Nigel Goodwin said:
Usually not! - because it depends on the voltage you're feeding it from. As suggested previously in this thread, measure the current the backlight is taking, if it's too high you MUST add a series resistor, otherwise it will blow fairly quickly.

I came across LCD displays 16*1 and 16*2 with on board smd resistors on the led circutry and solderable loops for controling the current. Instead we used to prefer external one for the safty purpose only.
 
AceOfHearts said:
PS.

I have installed a clip-on heatsink which is working rather nicely. The fins of the heatsink are not heating up too much indicating adequate heat-loss to the surrounding.

The backlight looks just right when connected to +5V and GND directly. Does the LCD not have a current limiting resistor built-in?

Thanks to all for all the help. I hope you understand I cannot reply to every single response/suggestions individually due to shortage of time, but I have taken all the advise onboard.

Thanks again.

Thanks i got your message in earlier posts. cool working of LM7805 alone is not sufficient perhaps. LCD back lit alone taking so much current is rather unhealthy for the device's life.
we have too to reduce it to around 100 to 120 mA toteal load with back lit.

In cidentally which is make and model no of the LCD that is claing so much backlit crt-- according to few spcs i went thro, even the blue color backlit has a limit at 220mA.

try by a series resistor on the LED+ line with 125mA approximate and see whether the illumination( for an yellow-green color) is nominally sufficient. at this current the LCD may live longer. No pot is needed once the value of the resistor is known.

they are taking voltage drop across the LEDs as 4.4V or so, i may interpret that this value is with 2 Yellow LEDs in seires and 4 or 5 such combinations in parallel. thus we may calculate

(5-(say)4)/.12mA= 11:eek:hm:

please try a 11 or 12 ohms 0.5 watts resistor and things would be safe.
 
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