Hello-

I was wondering if you guys can provide some insight and guidance on what voltage regulator I should choose for my project. I am using a rechargeable Lithium ion battery (9V 600mAH). I need my voltage to drop to around 5V but I will also be using the 9V supply to power another portion of my circuit. The two regulators I am deciding between are the LM317 and LM7805.

The questions I have are the following:

1) Which poses the least amount of danger of overheating and destroying my circuit long-term?
2) Which is most efficient?
3) Which regulator is best used to power an MCU (ATMega328)?

Thanks in advance for your replies!

 

1. You didn't specify current draw for the 5V supply, but both of the regulators are equivalent in most regards.
2. They are equally efficient. If you want to minimize power lost to heat, you need to use a buck regulator; but that could introduce noise that could affect other circuitry adversely.
3. Best is subjective.

 

@Fobio Design
I would not use either of those regulators. They are linear regulators and will waste 44% of your battery power. They are good regulators if you have the power to just throw away. I suggest that you go with a "simple switcher" regulator from Texas Instruments. With these kind of regulators you only loose 5% to 10% of your battery power. Look at an LM2576-5 or LM2596-5. These regulators will cost a little more then the LM7805 or LM317 but your battery will last 30% to 40% longer.

Edit: You can buy regulator modules/boards on EBay made with these parts pretty cheaply.
http://www.ebay.com/itm/LM2596-Powe...451145?hash=item20e9cc9089:g:VyUAAOSwVFlT12W7

 

Thank you for your help.
How can you calculate 44% wast of the battery power when I am using LM317 and LM7805 regulator?
I try to use voltage regulator to increase the battery operation hours as I can use for two hours 9v 600 ma/h battery. the device use about 300ma/h. When I use LM317 or LM7805 regulator, I can use about 4 hours, right?

 

...............
How can you calculate 44% wast of the battery power when I am using LM317 and LM7805 regulator?
I try to use voltage regulator to increase the battery operation hours as I can use for two hours 9v 600 ma/h battery. the device use about 300ma/h. When I use LM317 or LM7805 regulator, I can use about 4 hours, right?

No.
The power loss for any type of linear regulator is simply the voltage drop across the regulator times the current.
Thus the efficiency is simply the voltage drop divided by the input voltage or (9-5) / 9 = 44.4%.

And since the current in equals current out for a linear regulator, there is no gain in battery life for a given current load.
Think of a linear regulator as simply putting a variable resistor in series between the battery and the load, with the resistance being automatically adjusted to give the desired output voltage.
So to reduce the battery current, you need to use a switching buck regulator, as Les noted, which can be 90-95% efficient.
Its input (battery) current will thus be less than the load current.

 

thank you so much.