I have always used USB power but it's time to take my project and box it. This means I now need an external power source. I plan on using a 9v battery. My problem is all of the components are rated for less. Can I just use resistors to step it down? (i.e. on 5.0v PIC @ 220uA should I just use a 10k resistor?)

Or, should I use a Voltage Regulator, such as one of the one's below?

1) http://www.radioshack.com/product/index.jsp?productId=2062599

2) http://www.sparkfun.com/products/107


Here are the parts I am using right now:

PIC16F690 2.0-5.5v 220 uA
7-Segment LED 2.0v 30mA (Currently using 100 ohm resistor)
MQ-3 Sensor 5.0v 100mA
3 LED's @ 3.3v typical, 20mA (Currently using 82 ohm resistors)


Also, any advice on how to use a voltage regulator if you think I should go that route? I read capacitors maybe are needed? Why I am not sure.

 

If the load current changes, you can not depend on resistors to keep the voltage in the right range. That's why voltage regulators were invented. The 317 and the 7805 are linear regulators that will waste about 44% of the power because they just "burn" the excess voltage as heat, but they are simple and cheap. The "other" kind is called a switching regulator.

I'm adding a datasheet, but I'm getting tired so this is a partial answer.

ps, the capacitors on linear regulators are to stop them from oscillating.

 

If you went to four "AA" alkaline batteries instead, you would have 6v available. If you have a fairly constant current flow from the batteries, you might just use a diode to drop the voltage down to ~5.3v.

If you use four rechargeable (NoMH or NiCD) AA batteries, you'll have ~4.8v available; you wouldn't need a regulator or step-down.

 

Can I just use resistors to step it down? (i.e. on 5.0v PIC @ 220uA should I just use a 10k resistor?)

No,instead use voltage regulator like 7805.
Here is the datasheet : http://www.datasheetcatalog.org/datasheet/fairchild/LM7805.pdf on page 21 you will find Figure 8,its a typical way to connect a 7805.

Your circuit will be powered by battery so it might be helpfull to say this....
7805 is a linear regulator which burns out the difference between the input and output voltage as heat,and when this difference is much large like 9 -5 = 4 ,the loss will be also much greater,so they are extremely inefficient but they are cheap and they are easy to use i.e.. less complex circuit.A rule of thumb is if your linear regulator is wasting less than 0.5 watts then its ok to use a linear regulator.

On the other hand you can use switching regulators which are very efficient,not so much cheap and their circuit is also complex.

Good Luck

 

No,instead use voltage regulator like 7805.

No.
A little 9V battery voltage quickly drops to 7V then an old ordinary 7805 will not work. Use a modern "low dropout" and smaller 5V regulator that still works perfectly when the 9V battery drops to 5.5V.

 

If this - http://www.sparkfun.com/datasheets/Sensors/MQ-3.pdf - is the sensor you are using, the datasheet indicates that it needs a closely regulated supply voltage (5.0 +/- .1v), which means you need a voltage regulator.

 

Most definitely NOT the LM7805, maybe the LM78L05(TO-92). Either one, however will suck up power and the Zener might just be able to regulate within the tolerance of the IC.

 

Although not recommended, a PIC will be perfectly happy on 6V for years. I did this for a school project and it's still going. The *absolute* maximum is 7V, although they can take 9V for a couple seconds... (accidentally plugged it in briefly - no idea how I may have reduced the lifespan of the PIC - but it got hot!)

If you want regulation, Microchip make a few PIC's with HV regulators in them which act as a kind of dynamic zener diode. They usually have part numbers with "HV" in them. They also make some older PICs which are directly rated for 6V, but I don't think those are in production any more.

 

No.
A little 9V battery voltage quickly drops to 7V then an old ordinary 7805 will not work. Use a modern "low dropout" and smaller 5V regulator that still works perfectly when the 9V battery drops to 5.5V.

I have suggested linear regulator instead of resistors that he was going to use,and we already have said about inefficient of linear regulators so switching regulators have been suggested which I think are better than LDO Linear regulator.

Paradoxz if you don't want to use switching regulators,do it like SgtWookie suggested.

 

There are literally hundreds of "micropower low-dropout voltage regulators" now on the market. Most will work from 5v-16v and only waste some microamps, and be perfect for powering a 5v PIC from a 9v battery.

Just think of them like the "perfect" form of a 7805 regulator;
http://www.micrel.com/page.do?page=/product-info/products/mic5209.shtml

and don't be afraid of the surface mount packages, the SOT-223 is large enough, and has 0.1" pin spacing just like a 7805 and can easily be soldered onto the bottom of stripboard or your PCB instead of a 7805.