LM317 Driver, capacitor type?

I am currently building a setup that includes a current regulator and voltage regulator using 2x LM317. I have a few questions before I start however. I am using the following setup on my driver board: http://laserpointerforums.com/f42/diy-homemade-laser-diode-driver-26339.html

1. I see in the pics it is a ceramic disk style capacitor. Will an electrolytic work here? My local Ratshack isn't very well stocked, and they don't seem to have any ceramic capacitors above .1uf. I'm hoping I don't need to order one online and have to wait another week to build my laser.

2. In the circuit for the voltage regulator, I see the common diagram includes a 1uf (I believe) capacitor on the output end. Again, will electrolytic work in this case? Its about all thats available locally past 0.1uf.

3. Do I need both capacitors, or one or the other? I'm fairly new to electronics and this is my first project. I'm not entirely sure if I need both, just one, or none at all.

4. I'm assuming the flow of current in my setup should go from power source -> voltage regulator -> current regulator -> LD,
correct?

EDIT:
5. Thought of one more question. I see both the current control board and voltage regulator setup include pots. How much power is actually flowing thru the pot in each one of these cases? The reason I ask is I'm wondering if I can use a small trimmer style pot rated for 0.25W. My laser should be pulling about 350mW, but correct me if I'm wrong, that amount of current shouldn't be flowing thru the pots. From my understanding the pots are just used for adjustment purposes, and they aren't flowing the full current of the circuit.

I would omit the capacitors. They are usually unnecessary as the LM317 is sufficient to protect the diode. Also the capacitor can contribute to accidentally frying your diode if you forget to discharge it before connecting it up. Happens all the time.

If you must use the capacitor, connect it in parallel between ground and the input of the LM317, not at the output that will go to your laser diode like it shows in that diagram. If you use an electrolytic, make sure you have the polarity correct (there's a stripe that shows a negative sign on them; connect that to ground) otherwise the capacitor can explode.

Finally, why are you using 2x LM317s? Just using the current-regulating LM317 will be enough in general. It will convert excess voltage to heat as needed. With two LM317s, and each LM317 having a drop-out voltage of around 3V (i.e. the input must be 3V higher than the output) means that having two in series will require 6V on top of the laser diode's forward voltage drop just to power the circuit. For a 3V laser diode that is 9V you need at the input, of which 6V is essentially wasted.

Thanks for the quick reply, thats just what I needed to know.

I'm using 2 LM317's because I wanted to be able to control the voltage. I plan on powering it with 9v's as of now, but would like to go to rechargeable solution in the future. I assumed since I was going to be using a 9v for power I'd need voltage regulation to get it down to the 2.5v required by the diode. Is it safe to feed 9v of power directly to the current regulator only without killing the diode?

Better yet, would there be a better source of power then 9v's? I was thinking about 6v watch style battery's, but I hate those because I never have one available to replace it when I need it. I would like something around a 6-9v rechargeable battery pack, that I could simply wire in a connector for a charger and I would never have to remove the battery pack at all. I just figured a 9v would be the best bang for the buck since size of the project box isn't really an issue.

9V batteries are terrible to use. They're essentially 6x AAAA batteries in series. They don't last very long, and can't source very much current (relatively). Those 9V batteries are expensive too. You can use 4x AA batteries instead of the 9V battery to produce 6V for your setup. It would last longer, and not heat the LM317 chip as much. Plus those AA batteries can source more current than a 9V battery. You can also use 2x LiPo CR123A batteries. They produce ~4.2V each fully charged, 3.6V discharged, but are rechargeable and better suited for powering your lasers.

The current regulator can feed directly from 9V. It'll just turn the voltage difference into heat, but only on the regulator, not on the diode. The output diode will set the voltage that the voltage regulator puts out. Still, it is better if you can reduce the voltage at the input of the regulator so that it doesn't need to heat up. For that, put some rectifier diodes in series with the LM317's input to drop the voltage by the input diode's forward voltage. Rectifier diodes are diodes like the 1N400x (x being 1 to 7) series of diodes that can handle 1A of current and work well; even a 1N4148 can work if you're using less than 200mA.

I know you probably don't want to buy them, but the parts are very cheap if you buy like 100 or so at a time. You can use them for test loads or other things (like your project above). I bought 2000 of the 1N4148s for $10 at Mouser.com; the 1N4001s are more expensive but are very nice for test loads, and not too much either ($6 for 100). For very high currents (3A) get some 1N540x diodes. Every time I order something from there I get some more bulk parts so that I have them on-hand for other things. Stuff like common resistors, capacitors, and cheap transistors are always useful for projects like laser controls, etc., and you can get them cheaply if you're buying larger quantities of them. I would buy LEDs from eBay or DealExtreme though.

I like the looks of the CR123A battery setup, I believe I'll go that route instead. Thanks!

I don't mind buying my parts online at all, its just that the laser diode itself will arrive in the mail today and I'd like to get something running today. I'll definatly pick up a selection of the more common parts as soon as I get some more funds after this project.

One more question. On the current regulator, how can I determine the resistor I need for specific current outputs? If I were going to replace the pot/variable resistor with a single resistor, how would I determine what resistance I need? I see the formula for the voltage regulator setup on the package of the LM317, but nothing about it as a current regulator. I'm guessing I want around ~350ma for a 808nm 300mW diode? Im waiting to see if the diode comes with a data sheet, if not I'm kind of in the dark.

EDIT: Google'd for a calculator and it looks like for 350mA output I want a single 3.6ohm 1/2w resistor. Does that sound right?

Break out the LM317 datasheet and look at the example circuits that are provided. That's the best way to learn. That's where the whole current-regulating laser circuit was derived from.

Basically the LM317 work on only the following principles: between the Input and Output pins is a drop of 3V (the drop-out voltage), and between the Output and the ADJ pin is 1.25V. The output -> adj pin voltage is enforced by the regulator and can be used with Ohm's law to force a specific current, or a specific voltage at the output (ex: put a resistor there and the use I = V / R to calculate the current).