OK, so the voltage is 21.6V, you want a current of 0.28A, and you have 6 LEDs with a Vf of (say) 2.2V.

You need a resistor of (21.6 - (6 * 2.2)) / 0.28 ohms. That's 30 ohms. A 39 ohm resistor should be OK.

The power dissipated in the resistor is I²R watts. This is 0.28 * 0.28 * 30 = 2.35W. I would get a 5W 39 ohm resistor.

In addition I would measure the actual current to make sure it's OK, and I would ensure those LEDs are on heatsinks.

The amount of power that will be dissipated in the resistor will be the same for a given current. The fact that they get to a higher surface temperature doesn't mean there's more power. A 22 ohm resistor in this case will reduce the current to about 0.38A. At this current it will dissipate about 3.2W. A 22 ohm 100W resistor would still dissipate the same power, but might hardly feel any warmer. A 22 ohm 1/4W resistor would burst into flame. A 3W resistor might get to a surface temperature in excess of 250°C (resistors can be specced for very high surface temperatures).

Placing two 22 ohm resistors in series would drop the current to about 190mA (0.19A) and the total dissipation would be 1.6W (or 0.8W per resistor). If these resistors are 3W resistors, and you think they're too hot when dissipating about 1/4 of their rated power then one of the following is true:

- You're too sensitive
- A higher current is flowing than I have calculated
- The resistors are a style requiring a heatsink that you haven't provided (unusual for such low power devices)

If it turns out to be (2), then maybe the Vf is less than 2.2V (I assumed 2.2V for my calculations) or your voltage is more than 21.6V, or you've connected things up wrong (resistors or LEDs in parallel?)

The best way to measure Vf is to connect the string up and measure the voltage across the resistor. This allows you to calculate the current, and knowing the power supply voltage you can then determine the voltage across the LEDs, and by division, the voltage across each LED. If you find that you get the following:

power supply voltage = 2V

resistance 38 ohms

voltage across the resistor = 9.1V

We can calculate:

current = 9.1 / 38 = 0.24 A

voltage across the LEDs = 22 - 9.1 = 12.9V

Voltage across a single LED = 12.9 / 6 = 2.15V

From that we know that Vf is about 2.15V at 240mA, so if we were aiming for 200mA, we should use a resistor of the value:

(22 - (6 * 2.15)) / 0.2 = 45.5 ohms (we would probably use 47 ohms).

I hope this helps a little.