Figure 971
the Current-Mirror-Loaded CMOS Amplifier Shown in Fig$0.5-\mu \mathrm{m}$

Question

Figure 971 
the Current-Mirror-Loaded CMOS Amplifier Shown in Fig$0.5-\mu \mathrm{m}$

Examlex · Accepted Answer

The Answer is $Figure 9.7.1 (a) \begin{aligned} g_{m 1} & =\frac{2 I_{D 1}}{\left|V_{O V} ight|}=\frac{I_{\mathrm{REF}}}{\left|V_{O V} ight|} \ r_{o 2} & =r_{o 4}=\frac{\left|V_{A} ight|}{I_{D 2,4}}=\frac{\left|V_{A} ight|}{\frac{1}{2} I_{\mathrm{REF}}}=\frac{2\left|V_{A} ight|}{I_{\mathrm{REF}}} \end{aligned} \begin{aligned} \left|A_{d} ight| & =g_{m 1}\left(r_{o 2} \| r_{o 4} ight) \ & =\frac{I_{\mathrm{REF}}}{\left|V_{O V} ight|} imes \frac{1}{2} imes \frac{2\left|V_{A} ight|}{I_{\mathrm{REF}}} \ \left|A_{d} ight| & =\frac{\left|V_{A} ight|}{\left|V_{O V} ight|} \quad ext { Q.E.D } \ g_{m 3} & =\frac{2 I_{D 3}}{\left|V_{O V} ight|}=\frac{I_{\mathrm{REF}}}{\left|V_{O V} ight|} \ R_{S S} & =\frac{\left|V_{A} ight|}{I_{D 5}}=\frac{\left|V_{A} ight|}{I_{\mathrm{REF}}} \end{aligned} Thus, \begin{aligned} \left|A_{c m} ight| & =\frac{1}{2 g_{m 3} R_{S S}}=\frac{1}{2} \frac{\left|V_{O V} ight|}{\left|V_{A} ight|} \ \mathrm{CMRR} & =\frac{\left|A_{d} ight|}{\left|A_{c m} ight|} \ \mathrm{CMRR} & =2\left(\left|V_{A} ight| /\left|V_{O V} ight| ight)^{2} \end{aligned} Q.E.D (b) \begin{aligned} A_{d} & =40=\frac{\left|V_{A} ight|}{\left|V_{O V} ight|}=\frac{20}{\left|V_{O V} ight|} \ \Rightarrow\left|V_{O V} ight| & =0.5 \mathrm{~V} \ \mathrm{CMRR} & =2(20 / 0.5)^{2}=3200 \end{aligned} or 70.1 \mathrm{~dB} (c) For Q_{5} and Q_{6}, I_{D}=I_{\mathrm{REF}}=200 \mu \mathrm{A} , thus \begin{aligned} 200 & =\frac{1}{2} \mu_{n} C_{O x}(W / L)_{5,6} V_{O V}^{2} \ & =\frac{1}{2} imes 200 imes(W / L)_{5,6} imes 0.5^{2} \ \Rightarrow(W / L)_{5,6} & =8 \end{aligned} For Q_{1} and Q_{2}, I_{D}=100 \mu \mathrm{A} , thus \begin{aligned} 100 & =\frac{1}{2} imes 200 imes(W / L)_{1,2} imes 0.5^{2} \ \Rightarrow\left(\frac{W}{L} ight)_{1,2} & =4 \end{aligned} For Q_{3} and Q_{4}, I_{D}=100 \mu \mathrm{A} , thus \begin{aligned} 100 & =\frac{1}{2} imes 100 imes(W / L)_{3,4} imes 0.5^{2} \ \Rightarrow(W / L)_{3,4} & =8 \end{aligned} (d) \begin{aligned} V_{O} & =V_{G 4}=2.5-V_{S G 4} \ & =2.5-\left|V_{t p} ight|-\left|V_{O V} ight| \ & =2.5-0.5-0.5 \ & =+1.5 \mathrm{~V} \end{aligned} Thus, \begin{aligned} V_{O S} & =\frac{V_{O}}{A_{d}} \ & =\frac{1.5}{40}=37.5 \mathrm{mV} \end{aligned} (e) \begin{aligned} v_{I C M \max } & =V_{D 1}+V_{t n} \ & =(2.5-0.5-0.5)+0.5 \ & =+2 \mathrm{~V} \ v_{I C M \min } & =-2.5+V_{O V 5}+V_{G S 1} \ & =-2.5+0.5+(0.5+0.5) \ & =-1 \mathrm{~V} \end{aligned} Thus, -1 \mathrm{~V} \leq v_{I C M} \leq+2 \mathrm{~V} (f) \begin{aligned} & v_{O \max }=+2.5-\left|V_{O V} ight|=+2 \mathrm{~V} \ & v_{O \min }=0-V_{t n}=-0.5 \mathrm{~V} \end{aligned} Thus, the output voltage can swing from its quiescent value of 1.5 \mathrm{~V} positively to +2 \mathrm{~V} and negatively to -0.5 \mathrm{~V} . For a sine wave, the maximum peak output signal would be 0.5 \mathrm{~V} , or 1 \mathrm{~V} peak-topeak, which corresponds to a peak-to-peak input sine wave of \frac{1 \mathrm{~V}}{40}=25 \mathrm{mV} .$

Figure 9.7.1
(a)

\begin{aligned}g_{m 1} & =\frac{2 I_{D 1}}{\left|V_{O V} ight|}=\frac{I_{\mathrm{REF}}}{\left|V_{O V} ight|} \r_{o 2} & =r_{o 4}=\frac{\left|V_{A} ight|}{I_{D 2,4}}=\frac{\left|V_{A} ight|}{\frac{1}{2} I_{\mathrm{REF}}}=\frac{2\left|V_{A} ight|}{I_{\mathrm{REF}}}\end{aligned}

\begin{aligned}\left|A_{d} ight| & =g_{m 1}\left(r_{o 2} \| r_{o 4} ight) \& =\frac{I_{\mathrm{REF}}}{\left|V_{O V} ight|} imes \frac{1}{2} imes \frac{2\left|V_{A} ight|}{I_{\mathrm{REF}}} \\left|A_{d} ight| & =\frac{\left|V_{A} ight|}{\left|V_{O V} ight|} \quad ext { Q.E.D } \g_{m 3} & =\frac{2 I_{D 3}}{\left|V_{O V} ight|}=\frac{I_{\mathrm{REF}}}{\left|V_{O V} ight|} \R_{S S} & =\frac{\left|V_{A} ight|}{I_{D 5}}=\frac{\left|V_{A} ight|}{I_{\mathrm{REF}}}\end{aligned}

Thus,

\begin{aligned}\left|A_{c m} ight| & =\frac{1}{2 g_{m 3} R_{S S}}=\frac{1}{2} \frac{\left|V_{O V} ight|}{\left|V_{A} ight|} \\mathrm{CMRR} & =\frac{\left|A_{d} ight|}{\left|A_{c m} ight|} \\mathrm{CMRR} & =2\left(\left|V_{A} ight| /\left|V_{O V} ight| ight)^{2}\end{aligned}

Q.E.D
(b)

\begin{aligned}A_{d} & =40=\frac{\left|V_{A} ight|}{\left|V_{O V} ight|}=\frac{20}{\left|V_{O V} ight|} \\Rightarrow\left|V_{O V} ight| & =0.5 \mathrm{~V} \\mathrm{CMRR} & =2(20 / 0.5)^{2}=3200\end{aligned}

or

70.1 \mathrm{~dB}

(c) For

Q_{5}

and

Q_{6}, I_{D}=I_{\mathrm{REF}}=200 \mu \mathrm{A}

, thus

\begin{aligned}200 & =\frac{1}{2} \mu_{n} C_{O x}(W / L)_{5,6} V_{O V}^{2} \& =\frac{1}{2} imes 200 imes(W / L)_{5,6} imes 0.5^{2} \\Rightarrow(W / L)_{5,6} & =8\end{aligned}

For

Q_{1}

and

Q_{2}, I_{D}=100 \mu \mathrm{A}

, thus

\begin{aligned}100 & =\frac{1}{2} imes 200 imes(W / L)_{1,2} imes 0.5^{2} \\Rightarrow\left(\frac{W}{L} ight)_{1,2} & =4\end{aligned}

For

Q_{3}

and

Q_{4}, I_{D}=100 \mu \mathrm{A}

, thus

\begin{aligned}100 & =\frac{1}{2} imes 100 imes(W / L)_{3,4} imes 0.5^{2} \\Rightarrow(W / L)_{3,4} & =8\end{aligned}

(d)

\begin{aligned}V_{O} & =V_{G 4}=2.5-V_{S G 4} \& =2.5-\left|V_{t p} ight|-\left|V_{O V} ight| \& =2.5-0.5-0.5 \& =+1.5 \mathrm{~V}\end{aligned}

Thus,

\begin{aligned}V_{O S} & =\frac{V_{O}}{A_{d}} \& =\frac{1.5}{40}=37.5 \mathrm{mV}\end{aligned}

(e)

\begin{aligned}v_{I C M \max } & =V_{D 1}+V_{t n} \& =(2.5-0.5-0.5)+0.5 \& =+2 \mathrm{~V} \v_{I C M \min } & =-2.5+V_{O V 5}+V_{G S 1} \& =-2.5+0.5+(0.5+0.5) \& =-1 \mathrm{~V}\end{aligned}

Thus,

-1 \mathrm{~V} \leq v_{I C M} \leq+2 \mathrm{~V}

(f)

\begin{aligned}& v_{O \max }=+2.5-\left|V_{O V} ight|=+2 \mathrm{~V} \& v_{O \min }=0-V_{t n}=-0.5 \mathrm{~V}\end{aligned}

Thus, the output voltage can swing from its quiescent value of

1.5 \mathrm{~V}

positively to

+2 \mathrm{~V}

and negatively to

-0.5 \mathrm{~V}

. For a sine wave, the maximum peak output signal would be

0.5 \mathrm{~V}

, or

1 \mathrm{~V}

peak-topeak, which corresponds to a peak-to-peak input sine wave of

\frac{1 \mathrm{~V}}{40}=25 \mathrm{mV}

.

Figure 971
the Current-Mirror-Loaded CMOS Amplifier Shown in Fig $0.5-\mu \mathrm{m}$

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Correct Answer:
Verified

Figure 971 the Current-Mirror-Loaded CMOS Amplifier Shown in Fig 0.5−μm0.5-\mu \mathrm{m}0.5−μm

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Correct Answer:VerifiedShow Answer

Figure 971
the Current-Mirror-Loaded CMOS Amplifier Shown in Fig $0.5-\mu \mathrm{m}$

Correct Answer:
Verified