Chapter 10, Problem 10.6TYU

The bias voltages of the MOSFET current source in Figure 10.17 are $V^{+}=3V$ and $V^{-}=-3V$ . The transistor parameters are $V_{TN}=0.5V,{k^{\prime }}_n=80\mu A/V^2$ , and $\lambda =0.02V^{-1}$ . The transistor width-to-length ratios are ${\left(W/L\right)}_3=3,{\left(W/L\right)}_1=12$ , and ${\left(W/L\right)}_2=6$ . Determine: (a) $I_{REF}$ , (b) $I_O$ at $V_{DS2}=2V$ , and (e) $I_O$ at $V_{DS2}=4V$ . (Ans. (a) $I_{REF}=1.33mA$ , (b) $I_O=0.6936mA$ , (c) $I_O=0.7203mA$ )

To determine

The reference current $I_{REF}$ .

Answer to Problem 10.6TYU

  $I_{REF}=1.39\text{mA}$

Explanation of Solution

Given Information:

  $\begin{array}{l}{V}^{+}=3\text{V}\\ V^{-}=-3\text{V}\end{array}$

  $\begin{array}{l} V_{TN}=0.5V\\ K{\text{'}}_n=80\mu A/V^2\\ \lambda =0.02V^{-1}\\ $ {(W/L)}_1=12$$ {(W/L)}_2=6$$ {(W/L)}_3=3$\end{array}$

Calculation:

  

According to the circuit the reference current will be,

  $I_{REF}=\frac{K{\text{'}}_n}{2}{\left(\frac{W}{L}\right)}_1{(V_{GS1}-V_{TN1})}^2(1+\lambda V_{DS1})$

Substitute given values in above expression,

  $\begin{array}{}$ I_{REF}=\frac{80\times {10}^{-6}}{2}\left(12\right){(V_{GS1}-0.5)}^2(1+(0.02)V_{DS1})$$ I_{REF}=480\times {10}^{-6}{(V_{GS1}-0.5)}^2\left[1+(0.02)V_{DS1}\right]\left(1\right)$\end{array}$

The gate- to- source voltage $V_{GS1}$ will be,

  $V_{GS1}=\frac{\sqrt{\frac{{\left(W/L\right)}_3}{{\left(W/L\right)}_1}}}{1+\sqrt{\frac{{\left(W/L\right)}_3}{{\left(W/L\right)}_1}}}(V^{+}-V^{-})+\frac{1-\sqrt{\frac{{\left(W/L\right)}_3}{{\left(W/L\right)}_1}}}{1+\sqrt{\frac{{\left(W/L\right)}_3}{{\left(W/L\right)}_1}}}{V}_{TN}$

Substitute given values in above expression,

  $V_{GS1}=\frac{\sqrt{\frac{3}{12}}}{1+\sqrt{\frac{3}{12}}}(3-(-3)+\frac{1-\sqrt{\frac{3}{12}}}{1+\sqrt{\frac{3}{12}}}(0.5)$

  $\begin{array}{l}{V}_{GS1}=\frac{0.5}{1.5}(6)+\frac{0.5}{1.5}(0.5)\\ V_{GS1}=2+0.167\\ V_{GS1}=2.167V\end{array}$

As $V_{DS1}=V_{GS1}=2.167V$ so substitute the values in equation (1),

  $\begin{array}{l}$ I_{REF}=480\times {10}^{-6}{(2.167-0.5)}^2\left[1+(0.02)(2.167)\right]$I_{REF}=1.39mA\end{array}$

To determine

The load current $I_O$ for the given value of $V_{DS2}$ .

Answer to Problem 10.6TYU

  $I_O=0.6936mA$

Explanation of Solution

Given Information:

  $\begin{array}{l}{V}^{+}=3V\\ V^{-}=-3V\end{array}$

  $\begin{array}{l} V_{TN}=0.5V\\ K{\text{'}}_n=80\mu A/V^2\\ \lambda =0.02V^{-1}\\ $ {(W/L)}_1=12$$ {(W/L)}_2=6$$ {(W/L)}_3=3$\end{array}$

  $V_{DS2}=2\text{V}$

Calculation:

  

According to the circuit the reference current will be,

  $I_{REF}=\frac{K{\text{'}}_n}{2}{\left(\frac{W}{L}\right)}_1{(V_{GS1}-V_{TN1})}^2(1+\lambda V_{DS1})$

Substitute given values in above expression,

  $\begin{array}{}$ I_{REF}=\frac{80\times {10}^{-6}}{2}\left(12\right){(V_{GS1}-0.5)}^2(1+(0.02)V_{DS1})$$ I_{REF}=480\times {10}^{-6}{(V_{GS1}-0.5)}^2\left[1+(0.02)V_{DS1}\right]\left(1\right)$\end{array}$

The gate- to- source voltage $V_{GS1}$ will be,

  $V_{GS1}=\frac{\sqrt{\frac{{\left(W/L\right)}_3}{{\left(W/L\right)}_1}}}{1+\sqrt{\frac{{\left(W/L\right)}_3}{{\left(W/L\right)}_1}}}(V^{+}-V^{-})+\frac{1-\sqrt{\frac{{\left(W/L\right)}_3}{{\left(W/L\right)}_1}}}{1+\sqrt{\frac{{\left(W/L\right)}_3}{{\left(W/L\right)}_1}}}{V}_{TN}$

Substitute given values in above expression,

  $V_{GS1}=\frac{\sqrt{\frac{3}{12}}}{1+\sqrt{\frac{3}{12}}}(3-(-3)+\frac{1-\sqrt{\frac{3}{12}}}{1+\sqrt{\frac{3}{12}}}(0.5)$

  $\begin{array}{l}{V}_{GS1}=\frac{0.5}{1.5}(6)+\frac{0.5}{1.5}(0.5)\\ V_{GS1}=2+0.167\\ V_{GS1}=2.167V\end{array}$

As $V_{DS1}=V_{GS1}=2.167V$ so substitute the values in equation (1),

  $\begin{array}{l}$ I_{REF}=480\times {10}^{-6}{(2.167-0.5)}^2\left[1+(0.02)(2.167)\right]$I_{REF}=1.39mA\end{array}$

Now calculate load current $I_O$ for $V_{DS2}=2V$ ,

  $I_O=\frac{{(W/L)}_2}{{(W/L)}_1}\left(\frac{1+\lambda V_{DS2}}{1+\lambda V_{DS1}}\right)I_{REF}$

Substitute given values in above expression

  $\begin{array}{l}{I}_O=\frac{6}{12}\left(\frac{1+(0.02)(2)}{1+(0.02)(2.167)}\right)(1.39\times {10}^3)\\ I_O=0.6936mA\end{array}$

To determine

The load current $I_O$ for given value of $V_{DS2}$ .

Answer to Problem 10.6TYU

  $I_O=0.72mA$

Explanation of Solution

Given Information:

  $\begin{array}{l}{V}^{+}=3V\\ V^{-}=-3V\end{array}$

  $\begin{array}{l} V_{TN}=0.5V\\ K{\text{'}}_n=80\mu A/V^2\\ \lambda =0.02V^{-1}\\ $ {(W/L)}_1=12$$ {(W/L)}_2=6$$ {(W/L)}_3=3$\end{array}$

  $V_{DS2}=4V$

Calculation:

Consider the given circuit.

  

According to the circuit the reference current will be,

  $I_{REF}=\frac{K{\text{'}}_n}{2}{\left(\frac{W}{L}\right)}_1{(V_{GS1}-V_{TN1})}^2(1+\lambda V_{DS1})$

Substitute given values in above expression,

  $\begin{array}{}$ I_{REF}=\frac{80\times {10}^{-6}}{2}\left(12\right){(V_{GS1}-0.5)}^2(1+(0.02)V_{DS1})$$ I_{REF}=480\times {10}^{-6}{(V_{GS1}-0.5)}^2\left[1+(0.02)V_{DS1}\right]\left(1\right)$\end{array}$

The gate- to- source voltage $V_{GS1}$ will be,

  $V_{GS1}=\frac{\sqrt{\frac{{\left(W/L\right)}_3}{{\left(W/L\right)}_1}}}{1+\sqrt{\frac{{\left(W/L\right)}_3}{{\left(W/L\right)}_1}}}(V^{+}-V^{-})+\frac{1-\sqrt{\frac{{\left(W/L\right)}_3}{{\left(W/L\right)}_1}}}{1+\sqrt{\frac{{\left(W/L\right)}_3}{{\left(W/L\right)}_1}}}{V}_{TN}$

Substitute given values in above expression,

  $V_{GS1}=\frac{\sqrt{\frac{3}{12}}}{1+\sqrt{\frac{3}{12}}}(3-(-3)+\frac{1-\sqrt{\frac{3}{12}}}{1+\sqrt{\frac{3}{12}}}(0.5)$

  $\begin{array}{l}{V}_{GS1}=\frac{0.5}{1.5}(6)+\frac{0.5}{1.5}(0.5)\\ V_{GS1}=2+0.167\\ V_{GS1}=2.167V\end{array}$

As $V_{DS1}=V_{GS1}=2.167V$ so substitute the values in equation (1),

  $\begin{array}{l}$ I_{REF}=480\times {10}^{-6}{(2.167-0.5)}^2\left[1+(0.02)(2.167)\right]$I_{REF}=1.39mA\end{array}$

Now calculate load current $I_O$ for $V_{DS2}=4V$ ,

  $I_O=\frac{{(W/L)}_2}{{(W/L)}_1}\left(\frac{1+\lambda V_{DS2}}{1+\lambda V_{DS1}}\right)I_{REF}$

Substitute given values in above expression

  $\begin{array}{l}{I}_O=\frac{6}{12}\left(\frac{1+(0.02)(4)}{1+(0.02)(2.167)}\right)(1.39\times {10}^3)\\ I_O=0.72mA\end{array}$