Question 1

Figure 5.6.1
The transistors in the circuit of Fig. 5.6.1 have $k_{n}=k_{p}=2 \mathrm{~mA} / \mathrm{V}^{2}$ and $V_{t n}=-V_{t p}=0.4 \mathrm{~V}$. Find $v_{O}$ for each of the following cases:
(a) $v_{I}=0 \mathrm{~V}$
(b) $v_{I}=+1 \mathrm{~V}$
(c) $v_{I}=-1 \mathrm{~V}$
(d) $v_{I}=+2 \mathrm{~V}$
(e) $v_{I}=-2 \mathrm{~V}$

Accepted Answer

(a)

Figure 5.6.2
From Fig. 5.6.2, we see that when $v_{I}=0 \mathrm{~V}$, both transistors are cut off and
$$v_{O}=0 \mathrm{~V}$$
(b)

With $v_{I}=+1 \mathrm{~V}, Q_{P}$ cannot conduct and can be eliminated, reducing the circuit to that shown in Fig. 5.6.3. Since $v_{G D N}=0 \mathrm{~V}, Q_{N}$ will be operating in saturation with
$$i_{D}=\frac{1}{2} k_{n}\left(v_{G S}-V_{t n}\right)^{2}$$
But
$$i_{D}=\frac{v_{O}}{10 \mathrm{k} \Omega}=0.1 v_{O}, \quad \mathrm{~mA}$$
and
$$v_{G S}=v_{I}-v_{O}=1-v_{O}$$
Thus,
$$\begin{aligned}
0.1
 v_{O} & =\frac{1}{2} \times 2\left(1-v_{O}-0.4\right)^{2} \
& =\left(0.6-v_{O}\right)^{2} \
& =0.36-1.2 v_{O}+v_{O}^{2}
\end{aligned}$$
which can be rearranged in the form
$$v_{O}^{2}-1.3 v_{O}+0.36=0$$
This quadratic has two solutions: $0.775 \mathrm{~V}$, which is physically meaningless since $v_{G S}$ becomes $1-$ $0.775=0.225 \mathrm{~V}$, which is less than $V_{t n}$; and $0.4 \mathrm{~V}$, which is possible. Thus,
$$v_{O}=0.4 \mathrm{~V}$$
(c)

Figure 5.6.4
With $v_{I}=-1 \mathrm{~V}, Q_{N}$ cannot conduct and can be eliminated, reducing the circuit to that shown in Fig. 5.6.4. Since $v_{G D P}=0 \mathrm{~V}, Q_{P}$ will be operating in saturation. We recognize this situation to be the complement of that in (b) above. Thus, we do not need to do the analysis again and we can simply write
$$v_{O}=-0.4 \mathrm{~V}$$
(d)

With $v_{I}=+2 \mathrm{~V}, Q_{P}$ cannot conduct and can be removed, thus reducing the circuit to that shown in Fig. 5.6.5. Since $v_{G D N}=1 \mathrm{~V}$, which is greater than $V_{t n}, Q_{N}$ will be operating in the triode region. Thus,
$$i_{D}=k_{n}\left[\left(v_{G S}-V_{t n}\right) v_{D S}-\frac{1}{2} v_{D S}^{2}\right]$$
Here,
$$\begin{aligned}
i_{D} & =\frac{v_{O}}{10 \mathrm{k} \Omega}=0.1 v_{O}, \quad \mathrm{~mA} \
v_{G S} & =2-v_{O} \
v_{D S} & =1-v_{O}
\end{aligned}$$
Thus,
$$\begin{aligned}
0.1
 v_{O} & =2\left[\left(1.6-v_{O}\right)\left(1-v_{O}\right)-\frac{1}{2}\left(1-v_{O}\right)^{2}\right] \
& =2\left[1.6-2.6 v_{O}+v_{O}^{2}-\frac{1}{2}+v_{O}-\frac{1}{2} v_{O}^{2}\right] \
& =v_{O}^{2}-3.2 v_{O}+2.2 \
& \Rightarrow v_{O}^{2}-3.3 v_{O}+2.2=0
\end{aligned}$$
which has the following solution
$$v_{O}=\frac{+3.3 \pm \sqrt{3.3^{2}-8.8}}{2}=0.927 \mathrm{~V} \text { or } 2.37 \mathrm{~V}$$
The second solution is physically meaningless as $v_{O}$ exceeds $v_{D}$. Thus,
$$v_{O}=0.927 \mathrm{~V}$$
(e)

Figure 5.6.6
With $v_{I}=-2 \mathrm{~V}, Q_{N}$ cannot conduct and the circuit reduces to that shown in Fig. 5.6.6. We recognize this situation to be the complement of that in (d) above. Thus,
$$v_{O}=-0.927 \mathrm{~V}$$

Question 2

Figure 5.1.1
(a) For the circuit shown in Fig. 5.1.1(a), show (neglecting $\lambda$ ) that
$$V=V_{t}+\sqrt{\frac{2 I}{k_{n}^{\prime}(W / L)}}$$
(b) The MOSFETs in the circuit of Fig. 5.1.1(b) [6 points] have $V_{t}=0.4 \mathrm{~V}, k_{n}^{\prime}=0.4 \mathrm{~mA} / \mathrm{V}^{2}$, and $\lambda=0$. Find $(W / L)_{1},(W / L)_{2}$, and $(W / L)_{3}$ to obtain the reference voltages shown.

Accepted Answer

(a)

Figure 5.1.1(a)

Question 3

Figure 5.4.1
The MOSFET in the circuit of Fig. 5.4.1 has $V_{t}=$ $1 \mathrm{~V}$ and $k_{n}=2 \mathrm{~mA} / \mathrm{V}^{2}$, and the Early effect can be neglected.
(a) Find the values of $R_{S}$ and $R_{D}$ that result in the MOSFET operating with an overdrive voltage of $0.5 \mathrm{~V}$ and a drain voltage of $1.5 \mathrm{~V}$. What is the resulting $I_{D}$ value?
(b) If $R_{L}$ is reduced from $15 \mathrm{k} \Omega$ to $10 \mathrm{k} \Omega$, what does $V_{D}$ become? 
(c) If $R_{L}$ is disconnected, what does $V_{D}$ become? 
(d) With $R_{L}$ disconnected, what is the largest $R_{D}$ that can be used while the MOSFET is remaining in saturation?

Accepted Answer

(b) Refer to Figure 5.4 .3 below. 
  
We assume that with $R_{L}$ reduced to $10 \mathrm{k} \Omega$, the MOSFET remains in saturation. Thus, $I_{D}$ remains unchanged at 0.25 $\mathrm{mA}$. Figure 5.4.3 shows the circuit before and after Thévenin theorem is applied to simplify the drain circuit. Now,
$$V_{D}=2.5-0.25 \times 5=1.25 \mathrm{~V}$$
Thus, $V_{D}>V_{G}$, and saturation-mode operation is confirmed. The drain voltage becomes
$$V_{D}=1.25 \mathrm{~V}$$
(c)

Figure 5.4.4
If $R_{L}$ is disconnected, the circuit becomes as in Fig. 5.4.4. Assuming that the transistor remains in saturation, then $I_{D}=0.25 \mathrm{~mA}$ and
$$V_{D}=5-0.25 \times 10=2.5 \mathrm{~V}$$
which is greater than $V_{G}$, thus confirming saturation-mode operation. The drain voltage is now
$$V_{D}=+2.5 \mathrm{~V}$$

(d)
  
Figure 5.4.5
With $R_{L}$ disconnected and the MOSFET operating at the edge of saturation, $V_{D G}=-V_{t}$, thus
$$V_{D}=-1 \mathrm{~V}$$
and the current remains unchanged,
$$I_{D}=0.25 \mathrm{~mA}$$
From the circuit in Fig. 5.4.5, we can write
$$\begin{aligned}
R_{D} & =\frac{5-V_{D}}{I_{D}}=\frac{5-(-1)}{0.25} \
& =24 \mathrm{k} \Omega
\end{aligned}$$

Question 4

Figure 5.5.1
The MOSFETs in the circuits of Fig. 5.5.1 have $k=1 \mathrm{~mA} / \mathrm{V}^{2},\left|V_{t}\right|=1 \mathrm{~V}$, and $\lambda=0$.
(a) For the circuit in Fig. 5.5.1(a), find the values of $R_{D}$ and $R_{S}$ that result in the MOSFET operating at the edge of the saturation region with $I_{D}=0.1 \mathrm{~mA}$.
(b) For the circuit in Fig. 5.5.1(b), find $I_{D}$ and $R_{D}$.
(c) For the circuit in Fig. 5.5.1(c), find $I_{D N}, I_{D P}$, and $V_{O}$. Also, find the drain-to-source incremental resistance of each of $Q_{N}$ and $Q_{P}$.

Accepted Answer

The answer of

Figure 5.5.1
The MOSFETs in the circuits...

Question 5

Figure 5.7.1
The MOSFETs in the circuit of Fig. 5.7.1 have $\mu_{n} C_{o x}=400 \mu \mathrm{A} / \mathrm{V}^{2}, V_{t}=0.4 \mathrm{~V}, \lambda=0, L=$ $0.4 \mu \mathrm{m}, W_{1}=2 \mu \mathrm{m}$, and $W_{2}=W_{3}=10 \mu \mathrm{m}$. Find $R$ to obtain a reference current $I_{\mathrm{REF}}$ of $40 \mu \mathrm{A}$. Also, find the values of $V_{1}, I_{2}, V_{2}$, and $V_{3}$.

Accepted Answer

The answer of  
Figure 5.7.1
The MOSFETs in the circuit...

Question 6

Figure 5.3.1
The MOSFET in Fig. 5.3.1 has $V_{t}=0.5 \mathrm{~V}, k_{n}^{\prime}=$ $0.4 \mathrm{~mA} / \mathrm{V}^{2}, V_{A}=10 \mathrm{~V}$, and $W / L=10$. Find the value of $R_{D}$ that results in
i. $V_{D}=0.1 \mathrm{~V}$
ii. $V_{D}=1.5 \mathrm{~V}$
In each case, find $I_{D}$ and the incremental drain-tosource resistance of the MOSFET.

Accepted Answer

The answer of

Figure 5.3.1
The MOSFET in Fig. 5.3.1...

Question 7

(a)

(b)
Figure 5.2.1
The MOSFETs in the circuits of Fig. 5.2.1 have $V_{t}=0.4 \mathrm{~V}, k_{n}^{\prime}=0.4 \mathrm{~mA} / \mathrm{V}^{2}, \lambda=0$, and $L=$ $0.4 \mu \mathrm{m}$.
(a) For the circuit in Fig. 5.2.1(a), find the values of $W$ and $R_{D}$ to operate the MOSFET at $I_{D}=$ $0.2 \mathrm{~mA}$ and $V_{D}=0.6 \mathrm{~V}$.
(b) For the circuit in Fig. 5.2.1(b), find $W_{1}, W_{2}$, and $W_{3}$ to obtain $V_{1}=0.5 \mathrm{~V}$ and $V_{2}=1.1 \mathrm{~V}$.

Accepted Answer

The answer of  
(a)

(b)
Figure 5.2.1
The MOSFETs in...

Exam 1: MOS Field-Effect Transistors Mosfets