Rf Amplifier Schematic

how do one select Inductor value at the collector of a RF pre-amp circuit..?

From a long time i have a simple doubt but not able to solve it and have no one to help me..
But in RF amplifier ,I am not able to choose the inductor value @ the collector of the transistor..

Is is selected by first determining how much voltage drop does the inductor needs to drop at the collector and then selecting the value of Inductor by it’s Inductive reactance of the amplifying Frequency…?

can u please please explain me how u select it value..
Take this schematic as example please and show how much value for the Inductor u would use.. and how would u calculate..
thank u

http://images.elektroda.net/74_1292775322.jpg

* A good starting objective would be to make Xʟ 10 to 20 times
signal load on Q1 output. As the 50Ω resistor provides this load,
Xʟ should be 500Ω to 1kΩ.

The reason for making Xʟ of the L much greater than Output Load
resistance (impedance) is, to effectively block any signal variations
occurring at the collector node going to the + supply line which is ac
grounded. As the 50Ω path from the collector to the next stage/output,
has much less resistance (impedance), over 90% of the signal current
will use this path.

The Xʟ of 700 nH at 95 MHz = 2πfL = 2π* 95 * 10^6 * 700 * 10^-9 = 417Ω

Signal current has two paths, one 50Ω, the other 417Ω.
Most will go through the lower 50Ω resistance. L is doing a good job here, OK.

Why is L in circuit? As well above reason, it allows dc Ic quiescent to be
supplied to the collector as static set-up conditions required.
So L lets dc through, but blocks signal frequency.

There is no point in having L much higher than shown here. It would also
involve extra cost, weight, pcb space/volume, increased self-capacitance
leading to deterioration of Xʟ.

A greater than around 100MHz, components behave a little odd due to
inherent stray capacitance and inductance.

Please request clarification, if required.

Low Distortion Wideband RF Amplifier Module