Figure 561
the Transistors in the Circuit of Fig$k_{n}=k_{p}=2 \mathrm{~mA} / \mathrm{V}^{2}$

Question

Figure 561 
the Transistors in the Circuit of Fig$k_{n}=k_{p}=2 \mathrm{~mA} / \mathrm{V}^{2}$

Examlex · Accepted Answer

The Answer is (a)
$(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} ight)^{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} imes 2\left(1-v_{O}-0.4 ight)^{2} \ & =\left(0.6-v_{O} ight)^{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} ight) v_{D S}-\frac{1}{2} v_{D S}^{2} ight] 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} ight)\left(1-v_{O} ight)-\frac{1}{2}\left(1-v_{O} ight)^{2} ight] \ & =2\left[1.6-2.6 v_{O}+v_{O}^{2}-\frac{1}{2}+v_{O}-\frac{1}{2} v_{O}^{2} ight] \ & =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} ext { 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}$

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)
$(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} ight)^{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} imes 2\left(1-v_{O}-0.4 ight)^{2} \ & =\left(0.6-v_{O} ight)^{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} ight) v_{D S}-\frac{1}{2} v_{D S}^{2} ight] 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} ight)\left(1-v_{O} ight)-\frac{1}{2}\left(1-v_{O} ight)^{2} ight] \ & =2\left[1.6-2.6 v_{O}+v_{O}^{2}-\frac{1}{2}+v_{O}-\frac{1}{2} v_{O}^{2} ight] \ & =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} ext { 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}$

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} ight)^{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} imes 2\left(1-v_{O}-0.4 ight)^{2} \& =\left(0.6-v_{O} ight)^{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)
$(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} ight)^{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} imes 2\left(1-v_{O}-0.4 ight)^{2} \ & =\left(0.6-v_{O} ight)^{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} ight) v_{D S}-\frac{1}{2} v_{D S}^{2} ight] 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} ight)\left(1-v_{O} ight)-\frac{1}{2}\left(1-v_{O} ight)^{2} ight] \ & =2\left[1.6-2.6 v_{O}+v_{O}^{2}-\frac{1}{2}+v_{O}-\frac{1}{2} v_{O}^{2} ight] \ & =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} ext { 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}$

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)
$(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} ight)^{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} imes 2\left(1-v_{O}-0.4 ight)^{2} \ & =\left(0.6-v_{O} ight)^{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} ight) v_{D S}-\frac{1}{2} v_{D S}^{2} ight] 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} ight)\left(1-v_{O} ight)-\frac{1}{2}\left(1-v_{O} ight)^{2} ight] \ & =2\left[1.6-2.6 v_{O}+v_{O}^{2}-\frac{1}{2}+v_{O}-\frac{1}{2} v_{O}^{2} ight] \ & =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} ext { 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}$

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} ight) v_{D S}-\frac{1}{2} v_{D S}^{2} ight]

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} ight)\left(1-v_{O} ight)-\frac{1}{2}\left(1-v_{O} ight)^{2} ight] \& =2\left[1.6-2.6 v_{O}+v_{O}^{2}-\frac{1}{2}+v_{O}-\frac{1}{2} v_{O}^{2} ight] \& =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} ext { 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)
$(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} ight)^{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} imes 2\left(1-v_{O}-0.4 ight)^{2} \ & =\left(0.6-v_{O} ight)^{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} ight) v_{D S}-\frac{1}{2} v_{D S}^{2} ight] 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} ight)\left(1-v_{O} ight)-\frac{1}{2}\left(1-v_{O} ight)^{2} ight] \ & =2\left[1.6-2.6 v_{O}+v_{O}^{2}-\frac{1}{2}+v_{O}-\frac{1}{2} v_{O}^{2} ight] \ & =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} ext { 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}$

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}

Figure 561
the Transistors in the Circuit of Fig $k_{n}=k_{p}=2 \mathrm{~mA} / \mathrm{V}^{2}$

Essay

Correct Answer:
Verified

Figure 561 the Transistors in the Circuit of Fig kn=kp=2 mA/V2k_{n}=k_{p}=2 \mathrm{~mA} / \mathrm{V}^{2}kn​=kp​=2 mA/V2

Essay

Correct Answer:VerifiedShow Answer

Figure 561
the Transistors in the Circuit of Fig $k_{n}=k_{p}=2 \mathrm{~mA} / \mathrm{V}^{2}$

Correct Answer:
Verified