Question 11.5: Analysis of MOSFET Amplifier Determine the gate and drain-so......

Known Quantities: Drain, source, and gate resistors; drain supply voltage; MOSFET parameters.

Find: v_{GS} ; v_{DS} ; i_{D} .

Schematics, Diagrams, Circuits, and Given Data: R_1  =  R_2  =  1  M\Omega; R_D  =  6  k\Omega; R_S  =  6  k\Omega ; V_{DD}  =  10  V ; V_T  =  1  V; K = 0.5 mA/V² .

Assumptions: The MOSFET is operating in the saturation region. All currents are expressed in milliamperes and all resistors in kilohms.

Analysis: The gate voltage is computed by applying the voltage divider rule between resistors R_1 and R_2 (remember that no current flows into the transistor):

v_G  =  \frac{R_2}{R_1  +  R_2}V_{DD}  =  \frac{1}{2} V_{DD}  =  5  V

Assuming saturation region operation, we write

v_{GS}  =  v_G  −  v_S  =  v_G  −  R_S i_D  =  5  −  6i_D

The drain current can be computed from equation 11.3:

\begin{aligned} i_D  &=  K(v_{GS}  −  V_T )^2\left(1  +  \frac{v_{DS}}{V_A}\right) \\ &\cong  K(v_{GS}  −  V_T )^2 \end{aligned} \quad Saturation region        (11.3)

i_D  =  K (v_{GS}  −  V_T )^2  =  0.5 (5  −  6i_D  −  1)^2

leading to

36i^2_D  −  50i_D  +  16  =  0

with solutions

i_D  =  0.89  mA \quad\text{and}\quad i_D  =  0.5  mA

To determine which of the two solutions should be chosen, we compute the gate-source voltage for each. For i_D  =  0.89  mA , v_{GS}  =  5  –  6i_D  =  – 0.34  V. For i_D  =  0.5  mA , v_{GS}  =  5  –  6i_D  =  2  V . Since v_{GS} must be greater than V_T for the MOSFET to be in the saturation region, we select the solution

i_D  =  0.5  mA \quad v_{GS}  =  2  V

The corresponding drain voltage is therefore found to be

v_D  =  v_{DD}  −  R_D i_D  =  10  −  6i D  =  7  V

And therefore

v_{DS}  =  v_D  −  v_S  =  v_D  −  i_D R_S  =  7  −  3  =  4  V

Comments: Now that we have computed the desired voltages and current, we can verify that the conditions for operation in the saturation region are indeed satisfied: v_{GS}  =  2 \gt V_T and v_{GD}  =  v_{GS}  –  v_{DS}  =  2  –  4  =  – 2 \lt V_T . Since the inequalities are satisfied, the MOSFET is indeed operating in the saturation region .