Yes, your objective is correct, however you are using a resistive load to simulate a non-resistive, non-linear load, LEDs, you get lucky to get same result.
I call it a mess when using a current based switch in digital circuit -- looking at your "simulation" cicrcuit, your base current is rather large for a digital pin.
What luck? Current is current, the LEDs make for zero difference.
And the base current, approx 5mA, is well under-spec for a "digital pin" anyway that's sliced.
It's not unsound and it works.
Maybe it's "over-driven"; but that makes it "beta-proof" - no harm, no foul. Look what I have to work with here. smiley-tired_grin
Looks like it's just a matter of trial an error then.
Calculations are just too complex unless I try to use that SPICE simulator mjkzz talked about, which would be another project in itself.
It's not "trial and error".
Just do what I've told you - and get the wires right.
I don't know how better I can make the point other than to have you send me the parts and do it for you.
> > > > This person managed to get with the programme --
http://arduino.cc/forum/index.php/topic,156059.msg1178571.html#msg1178571
Well, the luck is that a non-linear device in this case behaved (or did it?) like your replacement resistor in that particular circuit, you might not be lucky to get the same result . . .
One example is, if the supply voltage is 3.3V and OP uses 3 three LEDs in series, your simulating circuit would still work, but OP's circuit won't if each LED has 1.25 forward voltage because there is not enough voltage to turn the LEDs on. Do you see what I am getting at? Of course, you could probably argue that you would use a different simulation circuit.
My point is, in general, it is not a good idea to use a linear device to simulate a non-linear device.