How much gain for optical open-air receiver

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.

It's not obvious to me if you want to calibrate your instrument to measure the size of your particles, or to measure the concentration of particles in a volume. I'm assuming you want to measure particle concentration, but most of this applies to either case.

I'm not familiar with the measurement you're describing, but the obvious way to calibrate it is to buy samples of powders of known particle size, disperse them in a known volume, and measure the response of the detector. A web page at the University of Manchester supports this method:

"In practise, theoretical instrument response curves are not used for sizing of particles, rather a series of calibration particles of known size are used to determine the response of the instrument over its full size range. A curve is fitted to these measurements to allow sizing of particles which fall between the calibration points."

If the particles you're measuring are not very tightly distributed in size (say they vary by +/- 50% in diameter), then you probably don't need to worry about the ripples in the Mie scattering curve you showed. Just use an average value of the response over the range of sizes you expect in your sample. To calibrate the instrument you'd need a reference sample with a comparable spread of particle sizes.

Generally with a measurement like this there's so many factors that vary between experiments (maybe the shape of the particles, the amount of turbulence in the sample volume, or other sources of reflected light) that a factory calibration from the instrument vendor isn't likely to be very precise. More likely you'll need to start with a reference measurement in your lab, and then make further measurements relative to that.