Electronic – Designing simple class a amplifier using mosfet

The K constant you refer to (more specifically \$K_n\$) is called the conduction parameter of the n-channel device.

\$K_n\$ is given by:

$$K_n = \frac{k_n'}{2}\cdot\frac{W}{L}$$

Where

$$k_n' = \mu_nC_{ox}$$

\$\mu_n\$ is the mobility of the electrons in the inversion layer and \$C_{ox}\$ is the oxide capacitance per unit area. According to Neamen the \$k_n'\$ parameter is called the "process conduction parameter" and is considered to be a constant for a fabrication technology. Therefore the ratio \$\frac{W}{L}\$ is the transistor design variable.

Neamen goes on to say that the design variable is used to design MOSFETS to produce specific current-voltage characteristics in MOSFET circuits.

EDIT:

Yes w refers to width and L to length. It relates to the geometry of the semiconductor.

Even if the graphs in the datasheet say 1 A at 5 V you should at a load to limit the current. FET characteristics can vary widely, just look at \$V_{GS(th)}\$: minimum 2V, maximum 4 V. For some FETs even a 3:1 range is given. Your 5 V may be 1 V above \$V_{GS(th)}\$, but it might as well be 3 V over it. The 1 A is typical, but \$R_{DS}\$ may already be low. Note that at 10 V this will be as low as 0.5 Ω. Now you have the full 12 V across the FET, and if \$R_{DS}\$ would go to even 2 Ω this would cause a dissipation of 72 W, which without any cooling is way too much for the poor thing.

BTW, even the 12 W of a 1 A current would be too much without cooling, though I wouldn't expect it to explode then.

So make sure you add a load so that the FET won't see the full 12 V across it.

Also, it is important to note that the value of the transconductance, G, specified in the data sheet is NOT the same as the K parameter for a mosfet. They are related by the following expression:

G = 2 * Sqrt( K * Id )