P = V * I. In your case for the LED, P = 20 W, V = 11 V and therefor I = 1.82 A.
To drive the LED correctly you will need to limit the current to 1.82 A. (you will also need to provide adequate heat sinking as 20 W is a lot of power to dissipate and most of it will be in the form of heat!!!! .... but that is another matter)
Diodes, using a PN junction have forward voltage drop, in your case that drop is 11 V. When referring to the diode datasheet they will include a V I curve, this will relate how much current flows for a given voltage. When you examine this curve you will notice that the current heads toward infinite for the given voltage drop. If you allow this to happen, you diode will simply burn out.
The simplest method to limit the current is to use a resistor. V = I * R. For a given voltage drop, a fixed resistor will limit the current to a fixed amount.
Assuming you have a 18 V battery and an 11 V LED, the difference, 7 V will need to be dropped across the resistor. This is fixed. Also fixed is the desired current, of 1.82 A. Rearranging the equation you get R = V / I => R = 3.8 ohm. This resistor will need to be rated to a minimum of 12 W. That is a large resistor that will get hot!!!!
The brings up the second issue - heat. At a total of 32 W to dissipate, you will need some serious heatsinking.
You can probably do away with the resistor and replace it with a controlled current circuit based on transistor and or mosfets but I will leave that to someone else to answer. The same issue of heat dissipation may still be present depending on the circuit design!!!
I know you mentioned you have an 11 V battery and an 11 V LED. Unless the LED has built in current limiting circuit, or the battery has, it will be difficult to drive it at the same voltage (depending on the VI curve of the diode) without a boost converter.