Electronic – Can the voltage drop of an LED driven at a constant current be higher than the supply voltage

An LM317-based solution is probably the simpliest one if heatsink and extra power consumption is not a problem. It can also be turned into an advantage if you need to warm up water in your aquarium.

A better solution is to use a single-chip PWM controller, a power MOSFET and an inductor to drive the LEDs. A UC3843 and an IRF540 may suit you well.

I would never power LEDs in parallel without a series resistor in each branch to balance the currents between the branches, especially if the LEDs are intended to be powered close to their maximum current. If you don’t try the balance the currents, a branch may get slightly more current than the others, which will make the LEDs in the branch slightly hotter, changing the U-I characteristics such that the branch will get more current, and you have thermal runaway. I think a 1Ω resistor in each branch should be enough.

The pulse current is the maximum current allowed in the LED for a short time (for example if one wants to flash the led for a still picture camera). The max current should be the maximum current the LED can accept, probably with perfect thermal dissipation. I’d rather not use currents much higher than the nominal current.

Your calculations for resistor values and power dissipations look fine to me.

Edit: it is not fine. First, if the maximum current in your LEDs is \$4 \times 90\textrm{mA} = 360\textrm{mA}\$ you should certainly not design a current regulator for a higher current, or you will burn your LEDs. You should rather design it for a lower current to ensure you won’t burn them. I’d go for \$4 \times 60\textrm{mA} = 240\textrm{mA}\$. Then, you’d get \$\textrm{R5} \parallel \textrm{R6} = \frac{0.7\textrm{V}}{0.24\textrm{A}} = 2.9\Omega\$, with \$\textrm{R5} \parallel \textrm{R6} = \frac{\textrm{R5} \times \textrm{R6}}{\textrm{R5} + \textrm{R6}}\$. If you choose \$\textrm{R5} = \textrm{R6}\$ (which is sane), you have \$\textrm{R5} \parallel \textrm{R6} = \frac{\textrm{R5}}{2} = \frac{\textrm{R6}}{2}\$, hence \$\textrm{R5} = \textrm{R6} = 5.8\Omega\$.

Your circuit is more a current limiter than a current regulator. It works because when current gets (too) high, the Vbe of T1 gets high, and then T1 will reduce the Vgs voltage of Q1, which become more resistive and will reduce the current. R7 is useful so that T1 can reduce the voltage. Without it, you might just burn T1 if WHITE_GPIO was connected to a low-impedence voltage source.

I have no idea about the use of R21.