Consider the circuit shown in the figure below. The transistor pa

SOLUTION

Concept:

For V_DS ≥ V_GS - V_TN, the MOSFET operates in saturation with the current given by:

I_D(sat) = K_n (V_GS - V_TN)²

\({K_n} = \frac{{W{\mu _n}{C_{ox}}}}{{2L}}\) 

The above current equation can be written as:

\({V_{GS}} - {V_{TN}} = \sqrt {\frac{{{I_D}}}{{{K_n}}}}\)     ...1)

Application:

The given circuit is redrawn as:

Using virtual ground concept, we can write:

V⁺ = V^- = 0 V

For the MOSFET, the drain current will be the same at drain and source terminal i.e.

\({I_D} = \frac{{{V_ - } - \left( { - 10} \right)}}{{10}}\)

\({I_D} = \frac{{10}}{{10}} = 1\;A\)

Since the MOSFET is operating in saturation, this is the saturation current, i.e.

I_D(sat) = 1 A

With I_D = 1A and K_n = 0.25, Equation (1) can be written as:

\({V_{GS}} - {V_{TN}} = \sqrt {\frac{1}{{0.25}}}\)

V_GS - V_TN = 2

Also, since V_DS = V_D - V_S, we can write:

V_DS = V_DD – V^-

For a MOSFET to operate in saturation, the following condition must be satisfied:

V_DS ≥ V_GS - V_TN, i.e.

V_DD – V^- ≥ V_GS - V_TN

V_DD – 0 ≥ 2

V_DD ≥ 2

∴ The minimum value of V_DD such that the MOSFET remains in saturation is