Exam 8: Operational Amplifier Circuits, Filters, Oscillators and CMOS Digital Logic Circuits

Correct Answer:

Verified

(a)


Figure 14.1.2
Figure 14.1.2 shows the $s$ -plane locations of the poles. Here,
$$\begin{aligned}\omega_{p} & \equiv \text { frequency of the passband edge }=10^{4} \mathrm{rad} / \mathrm{s} \\\epsilon & \equiv \text { passband ripple factor }=1\end{aligned}$$
The pair of complex-conjugate poles $p_{1}$ and $p_{1}^{*}$ have
$$\omega_{01}=\omega_{p}=10^{4} \mathrm{rad} / \mathrm{s}$$
and
$$\begin{aligned}\frac{\omega_{01}}{2 Q_{1}} & =\omega_{01} \cos 72^{\circ} \\\Rightarrow Q_{1} & =\frac{1}{2 \cos 72^{\circ}}=1.618\end{aligned}$$
The pair of complex-conjugate poles $p_{2}$ and $p_{2}^{*}$ have
$$\omega_{02}=\omega_{p}=10^{4} \mathrm{rad} / \mathrm{s}$$
and
$$Q_{2}=\frac{1}{2 \cos 36^{\circ}}=0.618$$
The real pole $p_{3}$ has a frequency
$$\omega_{03}=\omega_{p}=10^{4} \mathrm{rad} / \mathrm{s}$$
Thus, the transfer function of the fifth-order lowpass filter is
$T(s)=\frac{10^{20}}{\left(s+10^{4}\right)\left(s^{2}+s \frac{10^{4}}{1.618}+10^{8}\right)\left(s^{2}+s \frac{10^{4}}{0.618}+10^{8}\right)}$
where the numerator was found by enforcing the condition $T(0)=1$

(b) Refer to Figure 14.1.3


Figure 14.1.3
Each of the pair of complex-conjugate poles will be realized utilizing the low-pass circuit shown in Fig. 14.1.3. Here,
$$\begin{aligned}C & =10 \mathrm{nF} \\\omega_{0} & =10^{4}=\frac{1}{C R} \\R & =\frac{1}{\omega_{0} C}=\frac{1}{10^{4} \times 10 \times 10^{-9}} \\& =10 \mathrm{k} \Omega\end{aligned}$$
For the circuit realizing $\left(p_{1}, p_{1}^{*}\right)$ ,
$$Q R=1.618 \times 10=16.18 \mathrm{k} \Omega$$
For the circuit realizing $\left(p_{2}, p_{2}^{*}\right)$ ,
$$Q R=0.618 \times 10=6.18 \mathrm{k} \Omega$$
Each circuit has a unity de gain.


Figure 14.1.4
Figure 14.1.4 shows the circuit for realizing the real-axis pole, $p_{3}$ . Here
$$C=10 \mathrm{nF} \quad \text { and } \quad R=\frac{1}{\omega_{0} C}=10 \mathrm{k} \Omega$$
The feedback resistance is selected equal to $R$ so as to obtain a unity de gain.
Placing the three sections in cascade provides the realization of the fifth-order Butterworth lowpass filter. The resulting circuit is shown in Fig. 14.1.5


(c)
$$\begin{aligned}|T| & =\frac{1}{\sqrt{1+\left(\omega / \omega_{p}\right)^{2 N}}} \\A & =10 \log \left[1+\left(\omega / \omega_{p}\right)^{2 N}\right], \mathrm{dB}\end{aligned}$$
where
$$\omega_{p}=10^{4} \mathrm{rad} / \mathrm{s} \quad \text { and } \quad N=5$$
At $\omega=\omega_{S}=2 \times 10^{4} \mathrm{rad} / \mathrm{s}$ ,
$$\begin{aligned}A & =10 \log \left(1+2^{10}\right) \\& =30.1 \mathrm{~dB}\end{aligned}$$


Figure 14.1.6

Correct Answer:

Verified

(a) For matching,
$$\begin{aligned}k_{n} & =k_{p} \\\mu_{n} C_{o x}\left(\frac{W}{L}\right)_{N} & =\mu_{p} C_{o x}\left(\frac{W}{L}\right)_{P} \\\mu_{n} \times \frac{0.27}{0.18} & =\frac{1}{3} \mu_{n} \times \frac{W_{p}}{0.18} \\\Rightarrow W_{p} & =3 \times 0.27=0.81 \mu \mathrm{m}\end{aligned}$$

(b) For $v_{I}=1.8 \mathrm{~V}, Q_{N}$ will be operating in the triode region:
$$i_{D N}=k_{n}\left[\left(v_{G S}-V_{t n}\right) v_{D S}-\frac{1}{2} v_{D S}^{2}\right]$$
For $v_{O}=0.1 \mathrm{~V}$ ,
$$\begin{aligned}i_{D N} & =0.4 \times \frac{0.27}{0.18}\left[(1.8-0.4) \times 0.1-\frac{1}{2} \times 0.1^{2}\right] \\& =0.081 \mathrm{~mA} \\& =81 \mu \mathrm{A}\end{aligned}$$

Correct Answer:

Verified

Figure 16.2.1
(a) For the inverter switching to occur at $0.9 \mathrm{~V}$ , which is $V_{D D} / 2, Q_{N}$ and $Q_{P}$ must be matched:
$$\begin{aligned}k_{n} & =k_{p} \\\mu_{n} C_{O x}\left(\frac{W}{L}\right)_{N} & =\mu_{p} C_{O x}\left(\frac{W}{L}\right)_{P}\end{aligned}$$
Thus,
$$\frac{(W / P)_{P}}{(W / L)_{N}}=\frac{\mu_{n}}{\mu_{p}}=4$$
Since $L_{N}=L_{P}=0.18 \mu \mathrm{m}$ and
$$\begin{aligned}\left(\frac{W}{L}\right)_{N} & =1.5 \\\Rightarrow W_{N} & =1.5 \times 0.18=0.27 \mu \mathrm{m} \\W_{P} & =4 W_{N}=4 \times 0.27=1.08 \mu \mathrm{m}\end{aligned}$$
we obtain
$$\left(\frac{W}{L}\right)_{P}=\frac{1.08 \mu \mathrm{m}}{0.18 \mu \mathrm{m}}$$
(b)
$$\begin{aligned}V_{O H} & =V_{D D}=1.8 \mathrm{~V} \\V_{O L} & =0 \mathrm{~V} \\V_{I L} & =\frac{1}{8}\left(3 V_{D D}+2 V_{t}\right) \\& =\frac{1}{8}(3 \times 1.8+2 \times 0.4) \\& =0.775 \mathrm{~V} \\N M_{L} & =V_{I L}-V_{O L}=0.775-0=0.775 \mathrm{~V} \\N M_{H} & =N M_{L}=0.775 \mathrm{~V}\end{aligned}$$
(c) At the switching point, $Q_{N}$ and $Q_{P}$ operate in saturation:
$$\begin{aligned}I_{D} & =\frac{1}{2} \mu_{n} C_{o x}\left(\frac{W}{L}\right)_{N}\left(\frac{V_{D D}}{2}-V_{t}\right)^{2} \\& =\frac{1}{2} \times 0.4 \times 1.5(0.9-0.4)^{2} \\& =0.075 \mathrm{~mA}=75 \mu \mathrm{A}\end{aligned}$$
(d) For $v_{I}=1.8 \mathrm{~V}, Q_{N}$ is operating in the triode region:
$$i_{D N}=k_{n}\left[\left(v_{G S}-V_{t n}\right) v_{D S}-\frac{1}{2} v_{D S}^{2}\right]$$
For $v_{G S}=1.8 \mathrm{~V}$ and $v_{D S}=0.4 \mathrm{~V}$ ,
$$\begin{aligned}i_{D N} & =0.4 \times 1.5\left[(1.8-0.4) 0.4-\frac{1}{2} \times 0.4^{2}\right] \\& =0.288 \mathrm{~mA}\end{aligned}$$