$2.5 \times 10^{-3} A/V^2$
This question asks us to calculate the device parameter 'K' for a MOSFET, given specific operating conditions. MOSFETs (Metal-Oxide-Semiconductor Field-Effect Transistors) are fundamental semiconductor devices where the current flow is controlled by an input voltage. The relationship between the input voltage and the output current isn't always straightforward (it's non-linear), and this parameter 'K' is crucial in defining that behavior.
The behavior of a MOSFET, particularly the output current ($I_D$), is related to the voltage applied between the gate and source ($V_{GS}$) and the threshold voltage ($V_{th}$). The device parameter 'K' quantifies this relationship. Based on the provided options and the typical formulas, we can infer the relationship used here is:
$ I_D = K (V_{GS} - V_{th})^2 $
Where:
Note: A common formula includes a factor of 1/2, but deriving the provided answer requires using the formula above.
Let's list the values provided in the question:
First, we need to convert the output current from milliamperes (mA) to amperes (A):
$ I_D = 10 \text{ mA} = 10 \times 10^{-3} \text{ A} $
To find 'K', we need to rearrange the formula:
$ K = \frac{I_D}{(V_{GS} - V_{th})^2} $
Now, substitute the given values into the rearranged formula:
$ V_{GS} - V_{th} = 6 \text{ V} - 4 \text{ V} = 2 \text{ V} $
$ (V_{GS} - V_{th})^2 = (2 \text{ V})^2 = 4 \text{ V}^2 $
$ K = \frac{10 \times 10^{-3} \text{ A}}{4 \text{ V}^2} $
$ K = 2.5 \times 10^{-3} \text{ A/V}^2 $
Therefore, the calculated value for the MOSFET parameter 'K' is $2.5 \times 10^{-3} \text{ A/V}^2$. This matches the first option.
An n-channel MOSFET is connected as shown in the Figure.
Assume $V_{TH} = 1\text{ V}$, $V_{DD} = 5\text{ V}$, and $\mu C_{ox}\left(\frac{W}{L}\right) = 2\text{ mA V}^{-2}$ and neglect channel length modulation effects.
The gate voltage ($V_G$) of the n-channel MOSFET (in Volt) is __.
(rounded off to two decimal places)

MOSFET can be used as a ____________.