Electrical – Current calculations

This is actually rather simple, given your assumption. We can apply kirchoff's voltage law, which states that the sum of the voltages around a loop equals zero. $$V_{in} - V_R - V_D = 0$$ With \$V_{R}\$ The voltage over the resistor, \$V_{D}\$ voltage over the diode and \$V_{in}\$ the voltage you apply to the circuit. Solving for \$V_{R}\$: $$V_R = V_{in} - V_D = V_{in} - 0.7$$ Simply fill this in in Ohm's law: $$I = V_R / R = (V_{in} - 0.7)/R$$ Edit : This formula is only valid when \$V_{in} > 0.7V\$ as Warren said in the comments. A voltage lower than that would not induce any negative current. (Assuming an ideal diode)

Yes. Your application of the formula is correct. I can provide part numbers for better LEDs than that. "Soon" ... (just stumbled out of bed to complete a report)

At about 15 mW a 'good' modern LED will give >= 150 l/W or about 2.25 lumen. That's enough to illuminate 1+ sheet of A3 paper to good readable level with colour well discerned. If illuminating a small enclosed space it should be more than you need.

The formula (from above) is

$$ P = {V_{in}^2 \over L \cdot 10^-6} $$

or

$$ L = {V_{in}^2 \over P \cdot 10^{-6}} $$

\$L\$ in henries, or remove \$10^{-6}\$ and use \$L\$ in μH.

With this sort of IC the brightness varies approximately with the square of the input voltage so you get a power ratio of 9:4 for voltage changing from 3V to 2V.

• A 2:1 difference is hardly noticeable to the eye if you look at one light at a time.
• Look at two side by side and you can distinguish down to the 20% - 50% difference range.
• "Wall wash" with two LEDs side by side and you can typically see differences down to 10%.

LED: Look at Cree ML-E to start. There are much better now. More on that soonish.