LEDengin RGBA direct drive wiring and resistor placement

[Cantankerous]

OK, Ive been doing a fair amount of searching on the forums and have come across some very useful info on serial wiring a multi-die LED. I believe I have this figured out, just one thing Im not too sure on. Im looking to make a purple saber using the RGBA LEDengin as the the R and B on this one make a nice shade of purple on their own, as opposed to the RGBW (which make more of a pink). I have read to wire the R and B dice in series (source+>red+, red->blue+, blue->source negative), but this was with a buckpuck as the driver. I only want to use a resistor. Ive added both Vf values together and plugged 'em into the calculator and got my resistor value (LEDs R=2.00min B=3.2min) for a 7.4V pack run at 1500mah as 1.5ohms, 3.4W. My questions is, will I pop the LED as im using the min values of two different Vf value LEDs in series to calculate a resistor? Also, will I need to put the resistor on the source+ or source- lead? I appreciate any help, as this is the last little bit im not clear on. If I blow the LED, the wife probably wont let me buy a replacement as money is an issue. Otherwise, id just use a puck. :)

Thanx

[DarkarNights]

I think your safe with that calculation, you'd be more likely to blow the LEDs by using the maximum numbers, not the minimums. That calculation tells you the right resistor to drop the voltage to the number you plug in. So using that resistor limits your forward voltage on the RB series to 5.2V, which sounds right for what you want to do. If I'm wrong someone please chime in.

[Cantankerous]

The resistor goes on the source+ lead, then?

[DarkarNights]

I would wire the resistor into the positive side of the circuit.