Daedalus1632:
Hi all!
I'm kinda new to electronics, so I need your help to understand a thing.
I'm in the process of designing a standalone project that has to be mounted on my bike.
It will use an ATMega328, an RTC and +/- 12 LEDs.
I know that I have to drop the 12V provided by the battery to 5V. My inital idea was to use a LM7805, but I found out that with that Vin it would get very hot, so I thought I could use a second LM7809 (XV to 9V) to divide the stress. Is this a good solution?
In addition, I can't understand why I have to waste that energy that becomes heat. Isn't there any device that permits to use only the power you need?
Sorry for the probably idiot question but I'm quite ignorant on the subject.
Thank you all.
The 7805 is a linear regulator. This means that it "shaves off" the excess voltage off the top and gives you the difference (i.e. 5 volts). Since the current in a series circuit is the same at any point, the power dissipated by the 7805 is 12 volts - 5 volts = 7 volts multiplied by the current your load draws.
Since your load uses 5 volts and the regulator "throws away" 7 volts, you are actually wasting more power than you are using.
To prevent this, you need a switching regulator. A switching regulator works by turning on and off hundreds to thousands of times per second, and varying the ON time and the OFF time.
Imagine you had a 12 volt light bulb, a 12 volt battery and a switch. Turn the switch on, the light is 100% bright. The power wasted by the switch is virtually zero because almost no voltage is dropped across it. Now turn the switch off. The bulb is dark, and the power wasted by the switch is again zero because although the voltage across it is 12, the current is zero.
Now, imagine you can turn the switch on and off 1000 times per second and you turn it on for 1/2 of the time and off for 1/2 of the time. The light bulb is alternately getting 0 volts and 12 volts. Remember that in either case the power wasted by the switch is (for all intents and purposes) zero.
But, the bulb cannot go on and off 1000 times per second, so it assumes an average brightness of 50%. You dimmed the bulb, you made it act as if only 6 volts were being supplied to it and wasted ZERO power.
Now, replace the switch with a MOSFET or other high quality solid state switch, and drive it with a circuit that varies the ON and OFF time ratio. You can make the bulb go anywhere from off to full brightness with no power lost.
Lastly, take that same circuit and instead make it monitor the output voltage and constantly self-adjust the on/off duty cycle in order to maintain 5 volts on the output. This is a 5 volt switching regulator which converts the higher input voltage down to 5 volts while wasting virtually zero power.
Here's an example of a switching regulator: Pololu Step-Down Voltage Regulator D15V70F5S3
Cool, huh? That little bitty board can provide 7 amperes of output current because the switching regulator wastes virtually zero power and therefore does not get very hot.
Hope this helps and makes sense.
