You could go with 9V and replace the LM317 with a simple resistor (assuming Vf =3.6 then R2 would be 18Ω). I only use the LM317 because I like it, but it isn't that special overall.

There are also other ways to regulate current that drop a lot less voltage. Take the circuit below, reverse the polarity, use PNP transistors, and it would do the same thing as the LM317 shown. The emitter resistors would be 7Ω or so.

Generally transistors in these designs are not critical, the 2N2222 and 2N2907 (its compliment) are as generic as they come, since you're in the UK go with whatever is closest. Same thing with the diodes shown, almost any small signal or power diode will work, in other words, junkbox parts.

This circuit will drop a minimum of 0.8 volts. You have two LEDs that are 3.6V worst case, so that would be 7.2V + .8V, or 8V. If you're using a 9V battery this will be sucked dry in less than an hour with that kind of current though.

The resistor R2 is 12.5Ω, you could use a 12Ω or a 13Ω, or a 13Ω in parallel with a 330Ω. Don't forget, these resistors are usually ±5%. You can always make custom values by using parallel resistors.

 

Bill,
You're assuming that the LED's Vf will be the highest specified Vf, or 3.6v. If a range is given (like Marko did of 3.2v to 3.6v), I assume the midpoint for the typical Vf; in this case 3.4v. This is how I calculated 21 Ohms, with 22 being the recommended value.

If you assume the maximum Vf, those LEDs with a lower Vf will receive considerably more current.

As far as the transistor arrangement, he'd probably be better off to stick with what you already drew up, rather than march off in a different direction trying to homebrew current limiter circuits. The LM317 & R1 is a good and simple current limiter. Besides, if he'll be using the 12v supply, he'll have plenty of headroom for it. The LM317 will dissipate about 350mW, and the 12.5 Ohm resistor around 125mW.

For the 12.5 Ohm resistor, here's a list of serial (+) and parallel (||) combinations that could be used, and how close they are percentage-wise:
11 + 1.5 = 12.5 (0 %)
8.2 + 4.3 = 12.5 (0 %)
15 || 75 = 12.5 (0 %)
13 || 330 = 12.507 (0.058 %)
20 || 33 = 12.453 (-0.377 %)
16 || 56 = 12.444 (-0.444 %)
10 + 2.4 = 12.4 (-0.8 %)
7.5 + 5.1 = 12.6 (0.8 %)
9.1 + 3.3 = 12.4 (-0.8 %)
6.8 + 5.6 = 12.4 (-0.8 %)
18 || 43 = 12.689 (1.508 %)
22 || 30 = 12.692 (1.538 %)
24 || 27 = 12.706 (1.647 %)
13 + 0 = 13 (4 %)

These pairs were calculated using IN3OTD's online calculator, here:
http://www.qsl.net/in3otd/parallr.html
A very handy page to have bookmarked.

The ULN2804A driver IC is what I would use for the LED current sink. It replaces 16 components with ONE, reduces the number of solder connections necessary in the driver by nearly 1/2 (from 32 to 17), reduces the output load on the 4017 from 2.8mA down to 1mA, and costs under $1 US.

 

Beats the price of individual transistors. How are you getting the parallel characters (||)?

 

Beats the price of individual transistors. How are you getting the parallel characters (||)?

The vertical bar character | is on the backslash (\) key; so shift+backslash. Usually right next to the backspace key.

If you haven't experimented with the ULN2003/ULN2803 ULN2004/ULN2804, you should pick up a few, as they are really handy Darlington drivers.

The ones ending in xxx3 are primarily for 5v (TTL) input levels.
The ones ending in xxx4 are for 6v-15v (CMOS) input levels.
ULN20xx are 7-channel. Handy for driving large 7-segment displays. Usually cheaper than the ULN28xx drivers.
ULN28xx are 8-channel. Handy for lots of things. You can use just one of them to drive a pair of unipolar stepper motors. They already have built-in protection diodes.

Get the Allegro datasheets for them; it has handy charts for saturation voltage vs sink current.

 

OK, you're right. I'm always looking for a better solution. In this case a current regulator will compensate for whatever he has.