Exam 1: MOS Field-Effect Transistors Mosfets

Correct Answer:

Verified

(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,
$$v_{O}=-0.927 \mathrm{~V}$$

Correct Answer:

Verified

(a)


Figure 5.1.1(a)

Correct Answer:

Verified

(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
$$\begin{aligned}R_{D} & =\frac{5-V_{D}}{I_{D}}=\frac{5-(-1)}{0.25} \\& =24 \mathrm{k} \Omega\end{aligned}$$