AceOfHearts
New Member
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