I'm not following what exactly this Minicircuits thing is, but it sounds like they thought you want to turn on a LED when RF is present, hence the detector. It seems you actually want to drive the LED with 20 MHz.
At that speed, it's a good idea to actively turn off the LED, not just on. I haven't tried this, but this double emitter follower might do what you need:
When the digital output is at 5V, there should be around 4.3V on the emitter of Q1, which should be enough to turn on the LED thru R1. If D1 needs about 2V, for example, then R1 of 47Ω allows about 50mA thru the LED. Of course you need to adjust this for your particular LED. Note that you can drive it at twice its rated average current since you'll be doing it for half the time.
When the digital output goes low, the emitter of Q2 will go to about 700mV. That's a lot less than what it takes to turn on the LED, and will actively remove some charge to turn off the LED quicker. A ordinary CMOS 5V logic gate should be able to drive this circuit. I don't know why you think you need some sort of amplifier in there.
Added:
The circuit you show will work to drive the LED since it can drive 0 to some maximum current thru the LED as a function of the control signal. However, the big question is how well it will work at 20 MHz. At that frequency you have to think about semiconductors being actively turned off, not just on. You have nothing to actively turn off the LED (that's what Q2 is for in my circuit). You do have resistors to ground on both transistor bases, but you have to think about the values carefully to make sure the transistors turn off fast enough.
You haven't said what the maximum LED current needs to be, so I can't tell whether you really need the gain of two transistors to make a controlled current sink. Unless the current is really high (100s or mA or more), the gain of a single transistor is likely enough and it will be easier to drive a single transistor effectively at 20 MHz.
Added 2:
You now say you want to run the diode in linear mode with a bias of 125mA and a signal level of +-75mA from that. Here is something that might work. I say "might" because there are too many unknowns, especially at 20 MHz. You will have to test and adjust according to what you find:
Q1 acts like a voltage-controlled current sink. R2 is adjusted to get the right bias current with no RF signal in. With 5Vpp AC added to the 5V bias on the base of Q1, the current should vary about over the range you want.
C2 is only for a bit better speed. I took a rough stab at a plausible value, but you'll have to experiment to see what works best in your setup. It will depend on how slow the transistor really is. Note that since this is running the LED in linear mode, there is nothing actively removing charges from the junction when lowering the current. Actual light output will therefore probably lag decreasing current a bit. How much depends on things we don't know at this point. C2 will make the current lead the input voltage a bit in a attempt to compensate for the slowness of the diode and the transistor.
In simplest term no, you dont need to handle it differently
but
Laser diodes are usually much more sensitive than regular LEDs. Due to this you will usually see them driven by circuits with capacitors for voltage filtering/smoothing, signal diodes for polarity protection and some form of current limiting, either via a simple resistor or some other means.
So your diode might work but be aware that it might have a low life expectancy.
Resources:
Laser driver circuit
http://laserpointerforums.com/f42/diy-homemade-laser-diode-driver-26339.html
Best Answer
A laser diode is usually a three terminal device: a common point, a supply pin for power to the laser diode itself, and a photodiode output for feedback. The device you have looks like it has either a built-in controller or is running in straight open-loop (uncontrolled) mode.
If you keep the current within the Safe Operating Area (SOA) for the laser diode, you won't hurt anything, but you end up with an uncontrolled laser. It's not as scary as it sounds, but you probably aren't lasing or, if you are, you have no idea what your actual laser power output level is, which can be dangerous to eyeballs.
Laser diodes are designed to be operated in closed-loop mode where the power supply senses the output level of the laser through that photodiode and adjusts its output current on the fly in order to keep the device laser diode operating in the region where it is actually a laser and also to keep the power output level where it is designed to be. The link you provided isn't a laser power supply in that sense; it's a simple open-loop power supply which is probably better at maintaining a specific voltage/current level, but not actually maintaining a specific laser output power level.