Chapter 5 Solved Problems

(a) $$\begin{array}{rcll}L& = & 1.5L_{\text{min}}=1.5\times 0.13=0.195 \mu \text{m}& \\ C_{ox}& =& {\displaystyle \frac{{𝜖}_{ox}}{t_{ox}}}& \\ \text{ where}& & & \\ {𝜖}_{ox}& =& \mathrm{3.9 }{𝜖}_0=3.9\times 8.854\times 10^{-12}& \\ & =& 3.45\times 10^{-11}\text{ F/m}& \\ C_{ox}& =& {\displaystyle \frac{3.45\times 10^{-11}\text{ F/m}}{2.7\times 10^{-9}\text{ m}}}& \\ & =& 1.28\times 10^{-2}{\text{ F/m}}^2& \\ & =& 1.28\times 10^{-2}\times 10^{15}\times 10^{-12}\text{ fF∕μ}{\text{m}}^2& \\ & =& 12.8\text{ fF∕μ}{\text{m}}^2& \\ k_n^{\prime }& =& {\mu }_n{C}_{ox}& \\ & =& 400 ({\text{cm}}^2\mathrm{∕}\text{V·s}\mathrm{)}\times 12.8\mathrm{ (}\text{fF}\mathrm{∕}\mu {\text{m}}^2\mathrm{)}& \\ & =& 400\times 10^8\mathrm{ (}\mu {\text{m}}^2\mathrm{∕}\text{V·s}\mathrm{)}\times 12.8\times 10^{-15}\mathrm{ (}\text{F}\mathrm{∕}\mu {\text{m}}^2\mathrm{)}& \\ & =& 512\times 10^{-6}\mathrm{(}\text{F/V·s}\mathrm{)}=512\times 10^{-6}\mathrm{ (}{\text{A/V}}^2\mathrm{)}& \\ & =& 512 \mu {\text{A/V}}^2& \\ k_n& =& k_n^{\prime }\mathrm{ (}W\mathrm{∕}L\mathrm{)}& \\ & =& 512\times {\displaystyle \frac{1.3}{0.195}}=3413 \mu {\text{A/V}}^2& \\ k_n& =& 3.413{\text{ mA/V}}^2& \end{array}$$
(b) When the MOSFET operates in saturation, we have

$$I_D={\displaystyle \frac{1}{2}}{k}_n{V}_{OV}^2$$

Thus, $$\begin{array}{rcll}100& = & {\displaystyle \frac{1}{2}}\times 3413\times V_{OV}^2& \\ \Rightarrow V_{OV}& =& 0.24\text{ V}& \end{array}$$ To operate in saturation, $V_{DS}$ must at least be equal to $V_{OV}$, thus

$$V_{DS\text{min}}=0.24\text{ V}$$

The gate-to-source voltage is

$$V_{GS}=V_{tn}+V_{OV}=0.4+0.24=0.64\text{ V}$$

(c) When $v_{DS}$ is small,

$$i_D\simeq k_n{V}_{OV}{v}_{DS}$$

and

$$r_{DS}\equiv {\displaystyle \frac{v_{DS}}{i_D}}=1\mathrm{∕}{k}_n{V}_{OV}$$

Thus, for $r_{DS}=2\text{ k}\Omega$, $$\begin{array}{rcll}2\times 10^3& = & {\displaystyle \frac{1}{3.413\times 10^{-3}{V}_{OV}}}& \\ \Rightarrow V_{OV}& = & 0.15\text{ V}& \end{array}$$ and, correspondingly,

$$V_{GS}=0.4+0.15=0.55\text{ V}$$

If $V_{GS}$ is doubled, we obtain

$$V_{GS}=2\times 0.55=1.1\text{ V}$$

and

$$V_{OV}=1.1-0.4=0.7\text{ V}$$

Thus, correspondingly, $r_{DS}$ becomes $$\begin{array}{rcll}{r}_{DS}& = & {\displaystyle \frac{1}{k_n{V}_{OV}}}={\displaystyle \frac{1}{3.413\times 10^{-3}\times 0.7}}& \\ & = & 418.6 \Omega & \end{array}$$ As $V_{GS}$ is reduced, $r_{DS}$ increases, becoming infinite when the channel disappears, which occurs as $V_{OV}$ reaches zero or, correspondingly,

$$V_{GS}=V_{tn}=0.4\text{ V}$$

(a) When the transistor operates in saturation, we obtain

$$I_D={\displaystyle \frac{1}{2}}{\mu }_n{C}_{ox}\left({\displaystyle \frac{W}{L}}\right)V_{OV}^2$$

where $$\begin{array}{rcll}{\mu }_n{C}_{ox}& = & 45\mathrm{0 }\left({\text{cm}}^2\mathrm{∕}\text{V·s}\right)\times 1\mathrm{2.8 }(\text{fF/}\mu {\text{m}}^2)& \\ & =& 450\times 10^8\left(\mu {\text{m}}^2\mathrm{∕}\text{V·s}\right)\times 12.8\times 10^{-15}(\text{F/}\mu {\text{m}}^2)& \\ & =& 576\times 10^{-6}\left(\text{F/V·s}\right)& \\ & =& 576 \mu {\text{A/V}}^2& \end{array}$$ To obtain $I_D=100 \mu \text{A}$, $V_{OV}$ can be found from $$\begin{array}{rcll}100& = & {\displaystyle \frac{1}{2}}\times 576\times {\displaystyle \frac{2 \mu \text{m}}{0.2 \mu \text{m}}}\times V_{OV}^2& \\ \Rightarrow V_{OV}& =& 0.186\text{ V}& \end{array}$$ Correspondingly,

$$V_{GS}=V_{tn}+V_{OV}=0.4+0.186=0.586\text{ V}$$

At the edge of saturation,

$$V_{DS}=V_{DS\text{min}}=V_{OV}=0.186\text{ V}$$

(b) If $V_{DS}$ is lowered below $V_{DS\text{min}}$, the transistor operates in the triode region, thus $$\begin{array}{rcll}{i}_D& = & {\mu }_n{C}_{ox}\left({\displaystyle \frac{W}{L}}\right)\left[\left(V_{GS}-V_{tn}\right)v_{DS}-{\displaystyle \frac{1}{2}}{v}_{DS}^2\right]& \\ & =& 576\times {\displaystyle \frac{2}{0.2}}\left(0.186v_{DS}-0.5v_{DS}^2\right)& \\ & & & \end{array}$$ For $$\begin{array}{lll}{v}_{DS}& = & \mathrm{0.5 }{V}_{DS\text{min}}=0.5\times 0.186=0.093\text{ V}\end{array}$$ we obtain $$\begin{array}{r}\\ i_D& = & 576\times 10\left(0.186\times 0.093-0.5\times 0.093^2\right)& \\ & =& 74.7 \mu \text{A}& \end{array}$$ For $$\begin{array}{rcl}{v}_{DS}& = & \mathrm{0.1 }{V}_{DS\text{min}}=0.1\times 0.186=0.0186\text{ V}\end{array}$$ we get $$\begin{array}{rcll}{i}_D& = & 576\times 10 (0.186\times 0.0186-0.5\times 0.0186^2)& \\ & =& 18.9 \mu \text{A}& \end{array}$$ (c) For $V_{GS}=0.6$ V (i.e., $V_{OV}=0.2$ V) and $V_{DS}=0.3$ V, the MOSFET will be operating in saturation with $$\begin{array}{rcll}{I}_D& = & {\displaystyle \frac{1}{2}}{\mathrm{ \mu }}_n{C}_{ox}\left({\displaystyle \frac{W}{L}}\right)V_{OV}^2& \\ & =& {\displaystyle \frac{1}{2}}\times 576\times {\displaystyle \frac{2}{0.2}}\times 0.2^2& \\ & =& 115.2 \mu \text{A}& \end{array}$$ Now, if $v_{GS}$ is increased by a 10-mV increment, then

$$v_{GS}=0.6+0.010=0.610\text{ V}$$

and the current becomes $$\begin{array}{rcll}{i}_D& = & {\displaystyle \frac{1}{2}}\times 576\times {\displaystyle \frac{2}{0.2}}\times {(0.610-0.4)}^2& \\ & & & \\ & =& 127\mu \text{A}& \end{array}$$ Thus, $i_D$ increases by an increment

$$\mathrm{△}{i}_D=127-115.2=1\mathrm{1.8 }\mu \text{A}$$

If $v_{GS}$ is decreased by 10 mV, we obtain

$$v_{GS}=0.6-0.010=0.590\text{ V}$$

and the current becomes $$\begin{array}{rcll}{i}_D& = & {\displaystyle \frac{1}{2}}\times 576\times {\displaystyle \frac{2}{0.2}}\times {(0.590-0.4)}^2& \\ & & & \\ & =& 104\mu \text{A}& \end{array}$$ Thus, the change in $i_D$ is

$$\mathrm{△}{i}_D=104-115.2=-11.2\mu \text{A}$$

Observe that the incremental changes in $i_D$ are almost equal, indicating that the operation is almost linear. Linearity improves if the incremental changes in $v_{GS}$ are made smaller. (For instance, try $\mathrm{△}{v}_{GS}=\pm 5$ mV.)

Figure 5.3.1

Refer to Fig. 5.3.1 and observe that since $V_{DS}=V_{GS}=V_t+V_{OV}$, we have

$$V_{DS}>V_{OV}$$

and thus the MOSFET is operating in the saturation region. Thus, ignoring channel-length modulation, we can write

$$I_D={\displaystyle \frac{1}{2}}{k}_n^{\prime }{\displaystyle \frac{W}{L}{\left(V_{GS}-V_t\right)}^2}$$

Substituting the given data, we obtain

$$40={\displaystyle \frac{1}{2}}\times 400\times \left({\displaystyle \frac{W}{L}}\right){\left(0.6-V_t\right)}^2$$

(1)

and

$$90={\displaystyle \frac{1}{2}}\times 400\times \left({\displaystyle \frac{W}{L}}\right){\left(0.7-V_t\right)}^2$$

(2)

Dividing Eq. (2) by Eq. (1), we obtain

$$\begin{array}{rcll}{\displaystyle \frac{9}{4}}& = & {\displaystyle \frac{{\left(0.7-V_t\right)}^2}{{\left(0.6-V_t\right)}^2}}& \\ \Rightarrow {\displaystyle \frac{3}{2}}& =& {\displaystyle \frac{0.7-V_t}{0.6-V_t}}& \end{array}$$

which results in

$$V_t=0.4\text{ V}$$

Substituting for $V_t$ into Eq. (1) gives

$$\begin{array}{rcll}40& = & 200\times \left({\displaystyle \frac{W}{L}}\right)\times 0.04& \\ \Rightarrow {\displaystyle \frac{W}{L}}& =& 5& \end{array}$$

Operation with $V_{DS}=V_{GS}=V_t+V_{OV}$ means $V_{DS}>V_{OV}$ and thus the MOSFET is in the saturation region. Thus, neglecting channel-length modulation, we can write for $I_D$,

$$\begin{array}{rcll}{I}_D& = & {\displaystyle \frac{1}{2}}{k}_n{\left(V_{GS}-V_t\right)}^2& \\ & =& {\displaystyle \frac{1}{2}}\times 4\times {\left(0.6-0.35\right)}^2& \\ & =& 0.125\text{ mA}& \end{array}$$

The voltage $V_{DS}$ can be reduced to a value equal to $V_{OV}$ while the MOSFET remains in the saturation region, that is,

$$V_{DS\text{min}}=0.6-0.35=0.25\text{ V}$$

A transistor having twice the value of $W$ will have twice the value of $k_n$ and thus the current will be twice as large, that is,

$$I_D=2\times 0.125=0.25\text{ mA}$$

The linear resistance $r_{DS}$ is given by

$$r_{DS}={\displaystyle \frac{1}{k_n\left(V_{GS}-V_t\right)}}$$

With $V_t=0.35$ V and with $V_{GS}$ varying over the range 0.5 V to 1 V, $r_{DS}$ will vary over the range

$$r_{DS}={\displaystyle \frac{1}{0.15k_n}}\text{ to}{\displaystyle \frac{1}{0.65k_n}}$$

For the first device with $k_n=4$ mA/V, $r_{DS}$ will vary over the range

$$r_{DS}={\displaystyle \frac{1}{0.15\times 4}}=1.67\text{ k}\Omega$$

to

$$r_{DS}={\displaystyle \frac{1}{0.65\times 4}}=0.38\text{ k}\Omega$$

The wider device has $k_n=8$ mA/V and thus its $r_{DS}$ will vary over the range

$$r_{DS}=0.833 \text{k}\mathrm{\Omega } \text{to}0.192 \text{k}\mathrm{\Omega }$$

(a)

$$\begin{array}{rcll}{V}_A& = & V_A^{\prime }L=5\times 0.26=1.3\text{ V}& \\ \lambda & =& {\displaystyle \frac{1}{V_A}}={\displaystyle \frac{1}{1.3}}=0.77{\text{ V}}^{-1}& \end{array}$$

(b) Since $V_{DS}=0.65$ V is greater than $V_{OV}$, the NMOS transistor is operating in saturation. Thus, $$\begin{array}{rcll}{I}_D& = & {\displaystyle \frac{1}{2}}{k}_n^{\prime }\left({\displaystyle \frac{W}{L}}\right)V_{OV}^2\left(1+\lambda V_{DS}\right)& \\ & =& {\displaystyle \frac{1}{2}}\times 500\times {\displaystyle \frac{2.6}{0.26}}\times 0.2^2\times \left(1+0.77\times 0.65\right)& \\ & =& 150\mu \text{A}& \end{array}$$

(c)

$$r_o={\displaystyle \frac{V_A}{I_D^{\text{'}}}}$$

where $I_D^{\text{'}}$ is the drain current without taking channel-length modulation into account, thus

$$\begin{array}{rcll}{I}_D& = & {\displaystyle \frac{1}{2}}{k}_n^{\prime }\left({\displaystyle \frac{W}{L}}\right)V_{OV}^2& \\ & =& {\displaystyle \frac{1}{2}}\times 500\times {\displaystyle \frac{2.6}{0.26}}\times 0.2^2& \\ & =& 100\mu \text{A}& \end{array}$$

Hence,

$$r_o={\displaystyle \frac{1.3\text{ V}}{100 \mu \text{A}}}={\displaystyle \frac{1.3\text{ V}}{0.1\text{ mA}}}=13\text{ k}\Omega$$

(d) If $V_{DS}$ is increased to 1.3 V, $I_D$ becomes

$$\begin{array}{rcll}{I}_D& = & {\displaystyle \frac{1}{2}}\times 500\times {\displaystyle \frac{2.6}{0.26}}\times 0.2^2\left(1+0.77\times 1.3\right)& \\ & =& 200 \mu \text{A}& \end{array}$$

That is, $I_D$ increases by 50 $\mu \text{A}$. Alternatively, we can use $r_o$ to determine the increase in $I_D$ as

$$\begin{array}{rcll}\mathrm{△}{I}_D& = & {\displaystyle \frac{\mathrm{△}{V}_{DS}}{r_o}}& \\ & =& {\displaystyle \frac{0.65\text{ V}}{13\text{ k}\Omega }}=0.05\text{ mA}=50\mu \text{A}& \end{array}$$

which is identical to the result obtained directly.

Figure 5.6.1

(a) For the transistor to conduct, $v_G$ must be lower than $v_S$ by at least $\mid V_{tp}\mid$, that is, by 0.5 V. Thus the transistor conducts for $v_G\le 1.8-0.5$, or $v_G\le 1.3$ V.
(b) For the transistor to operate in the triode region, the drain voltage must be higher than the gate voltage by at least $\mid V_{tp}\mid$ volts, thus

$$v_D\ge v_G+0.5\text{ V}$$

(c) For the transistor to operate in the saturation region, the drain voltage cannot exceed the gate voltage by more than $\mid V_{tp}\mid$, that is,

$$v_D\le v_G+0.5\text{ V}$$

(d) When the transistor is operating in saturation, we obtain

$$I_D={\displaystyle \frac{1}{2}}{k}_p^{\prime }\left({\displaystyle \frac{W}{L}}\right) {\left|V_{OV}\right|}^2$$

Substituting the given values, we obtain

$$\begin{array}{rcll}20& = & {\displaystyle \frac{1}{2}}\times 100\times 10{\left|V_{OV}\right|}^2& \\ \Rightarrow \mathrm{\mid }{V}_{OV}\mid & =& 0.2\text{ V}& \end{array}$$

which is obtained when

$$\begin{array}{rcll}{v}_G& = & V_{DD}-V_{SG}& \\ & =& 1.8-\left(\left|V_{tp}\right|+\left|V_{OV}\right|\right)& \\ & =& 1.8-(0.5+0.2)=1.1\text{ V}& \end{array}$$

For this value of $v_G$, the range that $v_D$ is allowed to have while the transistor remains in saturation is

$$v_D\le v_G+\left|V_{tp}\right|$$

that is,

$$v_D\le 1.6\text{ V}$$

(e)

$$r_o = {\displaystyle \frac{\mid V_A\mid }{I_D^{\prime }}} = {\displaystyle \frac{1}{\left|\lambda \right|I_D^{\prime }}}$$

where $I_D^{\prime }$ is the value of $I_D$ without channel-length modulation taken into account, that is,

$$\begin{array}{rcll}{I}_D^{\prime }& = & {\displaystyle \frac{1}{2}}{k}_p^{\prime }\left({\displaystyle \frac{W}{L}}\right) {\left|V_{OV}\right|}^2& \\ & =& {\displaystyle \frac{1}{2}}\times 100\times 10\times 0.2^2=20 \mu \text{A}& \end{array}$$

Thus,

$$r_o={\displaystyle \frac{1}{0.2\times 20}}=0.25\text{ M}\Omega$$

(f)

$$I_D={\displaystyle \frac{1}{2}}{k}_p^{\prime }\left({\displaystyle \frac{W}{L}}\right)V_{OV}^2\left(1+\mid \lambda \mid V_{SD}\right)$$

At $V_D=1$ V, we have $V_{SD}=1.8-1=0.8$ V, and $I_D={\displaystyle \frac{1}{2}}\times 100\times 10\times 0.2^2\left(1+0.2\times 0.8\right)=23.2 \mu \text{A}$.

At $V_D=0$ V, we get $V_{SD}=1.8-0=1.8$ V, and $I_D={\displaystyle \frac{1}{2}}\times 100\times 10\times 0.2^2\left(1+0.2\times 1.8\right)=27.2 \mu \text{A}$

Thus, for

$$\mathrm{△}{V}_{SD}=1.8-0.8=1\text{ V},$$

the current changes by

$$\mathrm{△}{I}_D=27.2-23.2=4 \mu \text{A}$$

indicating that the output resistance $r_o$ is

$$r_o={\displaystyle \frac{\mathrm{△}{V}_D}{\mathrm{△}{I}_D}}={\displaystyle \frac{1\text{ V}}{4 \mu \text{A}}}=0.25\text{ M}\Omega$$

which is the same value found in (e).

Figure 5.7.1

(a) Refer to the circuit in Fig 5.7.1. For $V_D=+0.5$ V, the transistor is operating in saturation since $V_{\mathrm{D }}\mathrm{>}{\mathrm{ V}}_G$. Thus,

$$I_D={\displaystyle \frac{1}{2}}{k}_n^{\prime }\left({\displaystyle \frac{W}{L}}\right)V_{OV}^2$$

where we have utilized the given information that $\lambda =0$. To obtain $I_D=180 \mu \text{A}$, the required $V_{OV}$ can be found from

$$\begin{array}{rcll}180& = & {\displaystyle \frac{1}{2}}\times 400\times 10V_{OV}^2& \\ \Rightarrow V_{OV}& =& 0.3\text{ V}& \end{array}$$

The value of $V_{GS}$ can be found as

$$V_{GS}=V_{tn}+V_{OV}=0.5+0.3=0.8\text{ V}$$

from which $V_S$ can be determined as

$$V_S=V_G-V_{GS}=0-0.8=-0.8\text{ V}$$

The required value of $R_S$ can now be found from

$$\begin{array}{rcll}{R}_S& = & {\displaystyle \frac{V_S - (-V_{SS})}{I_D}}& \\ & =& {\displaystyle \frac{-0.8 - (-1)}{180 \mu \text{A}}}={\displaystyle \frac{0.2\text{ V}}{0.18\text{ mA}}}=1.11\text{ k}\Omega & \end{array}$$

Finally, the value of $R_D$ can be found from

$$\begin{array}{rcll}{R}_D& = & {\displaystyle \frac{V_{DD} - V_D}{I_D}}& \\ & =& {\displaystyle \frac{1 - 0.5}{0.18\text{ mA}}}=2.78\text{ k}\Omega & \end{array}$$

Figure 5.7.3

Figure 5.7.3 shows the designed circuit with the component values and the values of current and voltages.
(b) If $R_S$ is replaced by a constant-current source $I$, as shown in Fig. 5.7.2, the value of $I$ must be equal to the desired value of $I_D$, that is, $180 \mu \text{A}$ or $0.18\text{ mA}$.

Figure 5.7.2

(c) Refer to Fig. 5.7.1.
As $R_D$ is increased, $V_D$ decreases as

$$\begin{array}{rcll}{V}_D& = & 1-I_D{R}_D& \\ & =& 1-0.18R_D& \end{array}$$

Eventually, $V_D$ falls below $V_G$ by $V_{tn}$ at which point the transistor leaves the saturation region and enters the triode region. This occurs at

$$V_D=V_G-V_{tn}=0-0.5=-0.5\text{ V}$$

The corresponding value of $R_D$ can be found from

$$\begin{array}{rcll}-0.5& = & 1-0.18\times R_D& \\ \Rightarrow R_D& =& 8.33\text{ k}\Omega & \end{array}$$

Figure 5.8.1

(a) With $V_D=0$ V, the transistor will be operating in the saturation region since $V_D=V_G$. Thus,

$$I_D={\displaystyle \frac{1}{2}}{k}_p^{\prime }\left({\displaystyle \frac{W}{L}}\right) {\left|V_{OV}\right|}^2$$

where we have taken into account that $\lambda =0$ as stated. To obtain $I_D=0.1\text{ mA}=100 \mu \text{A}$, the required value of $\mid V_{OV}\mid$ can be found as follows:

$$\begin{array}{rcll}100& = & {\displaystyle \frac{1}{2}}\times 100\times 20{\left|V_{OV}\right|}^2& \\ \Rightarrow \mathrm{\mid }{V}_{OV}\mathrm{\mid }& =& 0.316\text{ V}& \end{array}$$

The value of $V_{SG}$ can now be found as

$$V_{SG}=\mathrm{\mid }{V}_{tp}\mathrm{\mid }+\mathrm{\mid }{V}_{OV}\mathrm{\mid }=0.5+0.316=0.816\text{ V}$$

Thus,

$$V_S=V_{SG}=0.816\text{ V}$$

The required value of $R_S$ can be determined from

$$\begin{array}{rcll}{R}_S& = & {\displaystyle \frac{V_{DD} - V_S}{I_D}}& \\ & =& {\displaystyle \frac{1 - 0.816}{0.1}}=1.84\text{ k}\Omega & \end{array}$$

Finally, the required value of $R_D$ can be found from

$$\begin{array}{rcll}{R}_D& = & {\displaystyle \frac{V_D - \left(-V_{SS}\right)}{I_D}}& \\ & =& {\displaystyle \frac{0 - \left(-1\right)}{0.1}}=10\text{ k}\Omega & \end{array}$$

The designed circuit with component values and current and voltage values is shown in Figure 5.8.2. The reader can check the calculations directly on the circuit diagram.

Figure 5.8.2

(b) Refer to Figure 5.8.1. The transistor remains in saturation as long as $V_D$ does not increase above $V_G$ by more than $\mid V_{tp}\mid$. Since $V_G=0$ and $\mathrm{\mid }{V}_{tp}\mathrm{\mid }=0.5$ V, the maximum allowable value of $V_D$ is

$$V_{D\text{max}}=+0.5\text{ V}$$

To obtain this value of $V_D$, $R_D$ must be increased to

$$R_D={\displaystyle \frac{V_{D\text{ max}}-\left(-1\right)}{0.1\text{ mA}}}={\displaystyle \frac{0.5+1}{0.1}}=15\text{ k}\Omega$$

5.9

Figure 5.9.1

The current $I$ through the voltage divider $R_{G1}$− $R_{G2}$ can be found as

$$\begin{array}{rcll}I& = & {\displaystyle \frac{V_{DD}}{R_{G1}+R_{G2}}}& \\ & =& {\displaystyle \frac{5\text{ V}}{3\text{ M}\Omega +2\text{ M}\Omega }}={\displaystyle \frac{5\text{ V}}{5\text{ M}\Omega }}=1 \mu \text{A}& \end{array}$$

The voltage $V_G$ at the gate can now be found as

$$V_G=I{R}_{G2}=1 \mu \text{A}\times 2\text{ M}\Omega =2\text{ V}$$

The voltage $V_S$ is given by

$$\begin{array}{rcll}{V}_S& = & V_G-V_{GS}=V_G-\left(V_t+V_{OV}\right)& \\ & =& 2-\left(0.5+V_{OV}\right)& \end{array}$$

$$V_S=1.5-V_{OV}$$

(1)

But $V_S$ can be expressed in terms of $I_D$ as

$$V_S=I_D{R}_S=I_D\times 6.5=6.5I_D$$

Thus,

$$6.5I_D=1.5-V_{OV}$$

(2)

We do not know whether the transistor is operating in the saturation region or in the triode region. Therefore, we must make an assumption about the region of operation, complete the analysis, and then use the results obtained to check the validity of our assumption. If our assumption proves valid, our work is done. Otherwise, we must redo the analysis assuming the other mode of operation. Since the $i$- $v$ relationships that describe the saturation-region operation are simpler than those that apply in the triode region, we normally assume operation in the saturation region, unless of course there is an indication of triode-mode operation.

Assuming that the transistor in the circuit of Figure 5.9.1 is operating in saturation, we can write

$$I_D={\displaystyle \frac{1}{2}}{k}_n{V}_{OV}^2={\displaystyle \frac{1}{2}}\times 10V_{OV}^2$$

$I_D=5V_{OV}^2$

(3)

Substituting for $I_D$ from Eq. (3) into Eq. (2) gives

$$6.5\times 5V_{OV}^2=1.5-V_{OV}$$

which can be rearranged into the form

$$32.5V_{OV}^2+V_{OV}-1.5=0$$

Solving this quadratic equation yields

$$V_{OV}=0.2\text{ V}\text{ or}-0.23\text{ V}$$

Obviously, the negative value is physically meaningless and can be discarded. Thus,

$$V_{OV}=0.2\text{ V}$$

and

$$I_D=5V_{OV}^2=5\times 0.2^2=0.2\text{ mA}$$

We are now ready to check the validity of our assumption of saturation mode operation. Referring to the circuit in Figure 5.9.1, we can find the voltage $V_D$ as follows:

$$\begin{array}{rcll}{V}_D& = & V_{DD}-I_D{R}_D& \\ & =& 5-0.2\times 12.5=2.5\text{ V}& \end{array}$$

which is greater than $V_G\left(2\text{ V}\right)$ confirming that the transistor is operating in saturation, as assumed. Figure 5.9.2 shows the circuit together with the values of all node voltages and branch currents. The reader is encouraged to check their results by doing a few calculations directly on the circuit.

Figure 5.9.2

5.10

From Fig. 5.10.1 in the problem statement we observe that transistor $Q_1$ together with its associated resistors is an identical circuit to that analyzed in the solution to Problem 5.9 (see Fig. 5.9.1). Since the gate terminal of $Q_2$ draws zero current, transistor $Q_2$ together with its associated resistances do not change the currents and voltages in $Q_1$ and its associated resistances. Thus, we need to only concern ourselves with the analysis of the part of the circuit shown in Figure 5.10.2, where $V_{GS}$ is found from

$$V_{G2}=V_{D1}=2.5\text{ V}$$

Figure 5.10.2

Since we do not know whether $Q_2$ is operating in saturation or in the triode region, we shall assume saturation-mode operation and, of course, we will have to check the validity of this assumption. We can now write

$\begin{array}{cll}{I}_{D2}& = & {\displaystyle \frac{1}{2}}{k}_p{\left|V_{OV2}\right|}^2\\ & = & {\displaystyle \frac{1}{2}}\times 12.5{\left|V_{OV2}\right|}^2\end{array}$

Thus,

$$I_{D2}=6.25\mathrm{\mid }{V}_{OV2}{\mathrm{\mid }}^2,\text{ mA}$$

(1)

Another relationship between $I_{D2}$ and $\mid V_{OV2}\mid$ can be obtained as follows:

$$\begin{array}{rcll}{V}_{S2}& = & V_{G2}+V_{SG2}& \\ & =& V_{G2}+\left(\mathrm{\mid }{V}_{tp}\mathrm{\mid }+\mathrm{\mid }{V}_{OV2}\mathrm{\mid }\right)& \\ & =& 2.5+0.5+\mathrm{\mid }{V}_{OV2}\mathrm{\mid }& \\ & =& 3+\mathrm{\mid }{V}_{OV2}\mathrm{\mid }& \end{array}$$

But

$$\begin{array}{cll}{I}_{D2}& = & {\displaystyle \frac{V_{DD}-V_{S2}}{R_{S2}}}\\ & = & {\displaystyle \frac{5-(3+\left|V_{OV2}\right|)}{7.2}}\\ I_{D2}& = & {\displaystyle \frac{2-\left|V_{OV2}\right|}{7.2}}, \mathrm{m}\mathrm{A}\end{array}$$

(2)

Equating $I_{D2}$ from Eqs. (1) and (2) results in

$$6.25\mathrm{\mid }{V}_{OV2}{\mathrm{\mid }}^2={\displaystyle \frac{2-\mathrm{\mid }{V}_{OV2}\mathrm{\mid }}{7.2}}$$

which can be rearranged into the form

$$45\mathrm{\mid }{V}_{OV2}{\mathrm{\mid }}^2+\mathrm{\mid }{V}_{OV2}\mathrm{\mid }-2=0$$

This quadratic equation can be solved to obtain

$$\mid V_{OV2}\mathrm{\mid }=0.2\text{ V}\text{ or}-0.22\text{ V}$$

Obviously, the negative solution is physically meaningless, thus

$$\mathrm{\mid }{V}_{OV2}\mathrm{\mid }=0.2\text{ V}$$

and

$$I_{D2}=6.25\times 0.2^2=0.25\text{ mA}$$

We are now ready to check the validity of our assumption of saturation-mode operation. We can do this by finding $V_{D2}$:

$$V_{D2}=I_{D2}{R}_{D2}=0.25\times 8=2\text{ V}$$

which is lower than the voltage at the gate ( V _G2 = 2.5 V), confirming saturation-mode operation. The voltage V _S2 can be found as

$$V_{S2}=V_{DD}-I_{D2}{R}_{S2}=5-0.25\times 7.2=+3.2\text{ V}$$

Figure 5.10.3

Finally, Fig. 5.10.3 shows the complete circuit with all currents and voltages. The values associated with $Q_1$ are those obtained in the solution for Problem 5.9. The reader is urged to make a few quick checks on the results displayed in Fig. 5.10.3.

5.11

Figure 5.11.1

Refer to Fig. 5.11.1. We assume that the transistor is operating in the saturation mode, thus

$$I_D={\displaystyle \frac{1}{2}}{k}_n{V}_{OV}^2$$

where we have taken account of the stated value $\lambda =0$. To obtain $I_D=0.5$ mA, the required value of $V_{OV}$ can be found from

$$\begin{array}{rcll}0.5& = & {\displaystyle \frac{1}{2}}\times 4V_{OV}^2& \\ \Rightarrow V_{OV}& =& 0.5\text{ V}& \end{array}$$

Now, since

$$V_{DS}=V_D-V_S=5-2=3\text{ V}$$

is greater than $V_{OV}$, the MOSFET is operating in saturation, as assumed. The required value of $R_S$ can be determined as follows:

$$R_S={\displaystyle \frac{V_S}{I_D}}={\displaystyle \frac{2}{0.5}}=4\text{ k}\Omega$$

and the required value of $R_D$ can be determined as follows:

$$R_D={\displaystyle \frac{V_{DD}-V_D}{I_D}}={\displaystyle \frac{10-5}{0.5}}=10\text{ k}\Omega$$

The voltage at the gate $V_G$ is found as

$$V_G=V_{GS}+V_S$$

where

$$V_{GS}=V_t+V_{OV}=0.5+0.5=1\text{ V}$$

Thus,

$$V_G=1+2=3\text{ V}$$

The resistance $R_{G1}$ can be found as follows:

$$R_{G1}={\displaystyle \frac{V_{DD}-V_G}{I_{}}}={\displaystyle \frac{10-3}{1\mu \text{A}}}=7\text{ M}\Omega$$

and, finally, $R_{G2}$ can be found as

$$R_{G2}={\displaystyle \frac{V_G}{I_{}}}={\displaystyle \frac{3\text{ V}}{1\mu \text{A}}}=3\text{ M}\Omega$$