Chapter 3, Problem 3.1EP

An NMOS transistor with $V_{TN}=1\text{V}$ has a drain current $i_D=0.8\text{mA}$ when ${\upsilon }_{GS}=3\text{V}$ and ${\upsilon }_{DS}=4.5\text{V}$ Calculate the drain current when: (a) ${\upsilon }_{GS}=2\text{V}$ , ${\upsilon }_{DS}=4.5\text{V}$ ; and (b) ${\upsilon }_{GS}=3\text{V}$ , ${\upsilon }_{DS}=1\text{V}$ . (Ans. (a) 0.2 mA (b) 0.6 mA)

To determine

The value of drain current for the given conditions.

Answer to Problem 3.1EP

  $i_D=0.2\text{mA}$

Explanation of Solution

Given:

The value of drain current is $i_D=0.8\text{mA when }{v}_{GS}=3\text{V and }{v}_{DS}=4.5\text{V}$ .

  $V_{TN}=1\text{V}$

New $v_{GS}=2\text{V,}{v}_{DS}=4.5\text{V}$

Calculation:

Consider NMOS which is also called n-channel enhancement MOSFET and assuming that the transistor is biased in saturation region, that is, $\left(v_{GS}>v_{TN}\right)$ , the drain current is given by

  $i_D=K_n{\left(v_{GS}-V_{TN}\right)}^2$

The value of conduction parameter $K_n$ is given by

  $\begin{array}{l}{K}_n=\frac{i_D}{{\left(v_{GS}-V_{TN}\right)}^2}\\ K_n=\frac{0.8\times {10}^{-3}}{{\left(3-1\right)}^2}\\ K_n=0.2{\text{mA/V}}^{\text{2}}\end{array}$

As,

  $\begin{array}{l}{v}_{DS\left(sat\right)}=v_{GS}-V_{TN}\\ v_{DS\left(sat\right)}=2-1\\ v_{DS\left(sat\right)}=1\text{V}\end{array}$

Comparing $v_{DS}$ and $v_{DS\left(sat\right)}$

  $v_{DS}>v_{DS\left(sat\right)}\left(\because 4.5>1\right)$

Therefore, the transistor is in saturation. Substituting the values in the equation,

  $\begin{array}{l}{i}_D=K_n{\left(v_{GS}-V_{TN}\right)}^2\\ i_D=0.2\times {10}^{-3}{\left(2-1\right)}^2\\ i_D=0.2\text{mA}\end{array}$

To determine

The value of drain current for given values.

Answer to Problem 3.1EP

  $i_D=0.6\text{mA}$

Explanation of Solution

Given:

The value of drain current is $i_D=0.8\text{mA when }{v}_{GS}=3\text{V and }{v}_{DS}=4.5\text{V}$ .

  $V_{TN}=1\text{V}$

New $v_{GS}=3\text{V,}{v}_{DS}=1\text{V}$

Calculation:

As,

  $\begin{array}{l}{v}_{DS\left(sat\right)}=v_{GS}-V_{TN}\\ v_{DS\left(sat\right)}=3-1\\ v_{DS\left(sat\right)}=2\text{V}\end{array}$

Comparing $v_{DS}$ and $v_{DS\left(sat\right)}$

  $v_{DS}<v_{DS\left(sat\right)}\left(\because 1<2\right)$

Therefore, the transistor is in non-saturation $\left(v_{DS}<v_{DS\left(sat\right)}\right)$

Drain current is given by

  $i_D=K_n\left[2\left(v_{GS}-V_{TN}\right)v_{DS}-v_{DS}^2\right]$

Substituting the values in the equation,

  $\begin{array}{l}{i}_D=0.2\times {10}^{-3}\left[2\left(3-1\right)1-1^2\right]\\ i_D=0.2\times {10}^{-3}\left[6-2-1\right]\\ i_D=0.6\text{mA}\end{array}$