Chapter 3, Problem 3.2TYU

The NMOS devices described in Exercise TYU 3.1 have parameters $W=20\mu \text{m}$ , $L=0.8\mu \text{m}$ , to $t_{\text{ox}}=200\text{Å}$ , ${\mu }_n=500{\text{cm}}^{\text{2}}\text{/V}-\text{s}$ , and $\lambda \text{=0}$ . (a) Calculate the conduction parameter $K_n$ for each device. (b) Calculate the drain current for each bias condition listed in Exercise TYU 3.1. (Ans. (a) $K_n=1.08{\text{mA/V}}^{\text{2}}$ ; (b) $i_D=0.518\text{mA,}0.691\text{mA}$ , and $0.691\text{mA}$ ; $i_D=2.59\text{mA,5}\text{.83}\text{mA}$ , and $\text{11}\text{.1}\text{mA}$ )

To determine

To find:Conduction parameter for the given NMOS devices.

Answer to Problem 3.2TYU

The conduction parameter for the NMOS devices is $1.08{\text{ mA/V}}^2$

Explanation of Solution

Given Information

Two MOSFETS, first one being an n-channel enhancement mode MOSFET with threshold voltage $V_{TN}=1.2\text{ V}$ and $v_{GS}=2\text{ V}$ being operated with three different values of drain to source voltages $v_{DS}=0.4\text{ V}$ , $v_{DS}=1\text{ V}$ and $v_{DS}=5\text{ V}$ , the second one being an n-channel depletion mode MOSFET with threshold voltage $V_{TN}=-1.2\text{ V}$ and operating under same gate to source and drain to source voltages.

The device parameters for both the MOSFETS are

  $W=20\mu \text{m}$ , $L=0.8\mu \text{m}$ $t_{o\text{x}}=200\stackrel{\text{o}}{\text{A}}$ ${\mu }_n=500{\text{ cm}}^2/\text{V-s}$ and $\lambda =0\text{}$ .

Calculation:

The conduction parameter of an n-channel MOSFET, $K_n$ is expressed in terms of the device parameters as,

  $K_n=\frac{{\mu }_n{C}_{o\text{x}}W}{L}$

  $C_{o\text{x}}=\frac{{\in }_{o\text{x}}}{t_{o\text{x}}}$

Here, ${\mu }_n$ is the mobility of electrons ininversion layer.

  $W$ and $L$ represents the channel width and length respectively.

  $C_{o\text{x}}$ represents the capacitance of the oxide layer in the MOSFET.

The oxide permittivity for silicon devices is ${\in }_{o\text{x}}=\left(3.9\right)\left(8.85\times {10}^{-14}\right)\text{F/cm}$ .

Substituting the given parameters and the permittivity, the conduction parameter can be obtained as,

  $\begin{array}{c}{K}_n=\frac{{\mu }_n{\in }_{o\text{x}}W}{t_{o\text{x}}L}\\ =\frac{500\times 3.9\times 8.85\times {10}^{-14}\times 20\times {10}^{-4}}{2\times 200\times {10}^{-8}\times 0.8\times {10}^{-4}}\\ =1.078{\text{ mA/V}}^2\\ =1.08{\text{ mA/V}}^2\end{array}$

To determine

To find: The drain currents for the two NMOS devices with each of the bias conditions.

Answer to Problem 3.2TYU

The drain current of the enhancement MOSFET for each bias conditions in the order of the given drain to source voltages is,

Case 1: $0.518\text{ mA}$

Case 2: $0.6912\text{ mA}$

Case 3: $0.6912\text{ mA}$

The drain current of the depletion mode MOSFET for each bias conditions in the order of the given drain to source voltages is,

Case 1: $2.59\text{ mA}$

Case 2: $5.83\text{ mA}$

Case 3: $11.1\text{ mA}$

Explanation of Solution

Given Information

Two MOSFETS, first one being an n-channel enhancement mode MOSFET with threshold voltage $V_{TN}=1.2\text{ V}$ and $v_{GS}=2\text{ V}$ being operated with three different values of drain to source voltages $v_{DS}=0.4\text{ V}$ , $v_{DS}=1\text{ V}$ and $v_{DS}=5\text{ V}$ , the second one being a n-channel depletion mode MOSFET with threshold voltage $V_{TN}=-1.2\text{ V}$ and operating under same gate to source and drain to source voltages.

The device parameters for both the MOSFETS are $W=20\mu \text{m}$ , $L=0.8\mu \text{m}$ $t_{o\text{x}}=200\stackrel{\text{o}}{\text{A}}$ ${\mu }_n=500{\text{ cm}}^2/\text{V-s}$ and $\lambda =0\text{}$ .

Calculation

Consider the first n-channel device which is enhancement type MOSFET with threshold voltage $V_{TN}=1.2\text{ V}$ and $v_{GS}=2\text{ V}$ .

The drain current of the device is given by different expression based on the bias conditions .

For non-saturation condition, the expression is,

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

For saturation condition, the expression is

  $i_D=K_n\left({\left(v_{GS}-V_{TN}\right)}^2\left(1+\lambda v_{DS}\right)\right)$

Here, the parameter $\lambda$ represents channel length modulation parameter. It is given to be zero for the device. The saturation value of drain-source voltage is

  $\begin{array}{c}{v}_{DS\left(sat\right)}=v_{GS}-V_{TN}\\ =2-1.2\\ =0.8\text{ V}\end{array}$

Now, consider each of the biasing conditions.

Case 1 : For the bias voltage of $v_{DS}=0.4\text{ V}$ , this voltage is less than the saturation value $v_{DS\left(sat\right)}$ . Thus the drain current is given by,

  $\begin{array}{c}{i}_D=1.08\times {10}^{-3}\left(2\left(0.8\right)\times 0.4-{0.4}^2\right)\\ =0.518\text{ mA}\end{array}$

Case 2 : For the bias voltage of $v_{DS}=1\text{ V}$ , this voltage is greater than the saturation value $v_{DS\left(sat\right)}$ . The device is in saturation region and the drain current is given by,

  $\begin{array}{c}{i}_D=1.08\times {10}^{-3}{\left(0.8\right)}^2\\ =0.6912\text{ mA}\end{array}$

Case 3 : For the bias voltage of $v_{DS}=5\text{ V}$ , this voltage is greater than the saturation value $v_{DS\left(sat\right)}$ . The device is in saturation region and the drain current is given by,

  $\begin{array}{c}{i}_D=1.08\times {10}^{-3}{\left(0.8\right)}^2\\ =0.6912\text{ mA}\end{array}$

Now, consider the second n-channel device which is depletion type MOSFET with threshold voltage $V_{TN}=-1.2\text{ V}$ and $v_{GS}=2\text{ V}$ .

The drain current of the device is given by different expression based on the bias conditions .

For non-saturation condition, the expression is,

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

For saturation condition, the expression is

  $i_D=K_n\left({\left(v_{GS}-V_{TN}\right)}^2\left(1+\lambda v_{DS}\right)\right)$

Here, the parameter $\lambda$ represents channel length modulation parameter. It is given to be zero for the device. The saturation value of drain-source voltage is

  $\begin{array}{c}{v}_{DS\left(sat\right)}=v_{GS}-V_{TN}\\ =2-(-1.2)\\ =3.2\text{ V}\end{array}$

Now, consider each of the biasing conditions.

Case 1 : For the bias voltage of $v_{DS}=0.4\text{ V}$ , this voltage is less than the saturation value $v_{DS\left(sat\right)}$ . Thus the drain current is given by,

  $\begin{array}{c}{i}_D=1.08\times {10}^{-3}\left(2\left(3.2\right)\times 0.4-{0.4}^2\right)\\ =2.59\text{ mA}\end{array}$

Case 2 : For the bias voltage of $v_{DS}=1\text{ V}$ , this voltage is less than the saturation value $v_{DS\left(sat\right)}$ . The device is in nonsaturation region and the drain current is given by,

  $\begin{array}{c}{i}_D=1.08\times {10}^{-3}\left(2\left(3.2\right)\times 1-1^2\right)\\ =5.83\text{ mA}\end{array}$

Case 3 : For the bias voltage of $v_{DS}=5\text{ V}$ , this voltage is greater than the saturation value $v_{DS\left(sat\right)}$ . The device is in saturation region and the drain current is given by,

  $\begin{array}{c}{i}_D=1.08\times {10}^{-3}{\left(3.2\right)}^2\\ =11.06\text{ mA}\\ \text{=11}\text{.1 mA}\end{array}$