Figure 9.2.1
The differential amplifier in Fig. 9.2.1 utilizes current-source loads and is biased with a modified Wilson current mirror. All transistors are operated at the same overdrive voltage $\left|V_{O V}\right|=0.2 \mathrm{~V}$ and have equal Early voltage, $\left|V_{A}\right|=4 \mathrm{~V}$.
(a) Find the differential gain.
(b) Find the CMRR (in $\mathrm{dB}$ ) assuming that the
only mismatch is that between $g_{m 1}$ and $g_{m 2}$ with $\triangle g_{m} / g_{m}=0.01$.
(c) If the Wilson mirror is replaced with a simple current mirror, what does the CMRR become?

Question

Quizplus · Accepted Answer

The Answer is $Figure 9.2.1 (a) A_{d}=g_{m 1}\left(r_{o 1} \| r_{o 3} ight) where \begin{aligned} g_{m 1} & =\frac{2 I_{D}}{\left|V_{O V} ight|} \ r_{o 1} & =r_{o 3}=\frac{\left|V_{A} ight|}{I_{D}} \ A_{d} & =\frac{2 I_{D}}{\left|V_{O V} ight|} imes \frac{1}{2} \frac{\left|V_{A} ight|}{I_{D}} \ & =\frac{\left|V_{A} ight|}{\left|V_{O V} ight|}=\frac{4}{0.2}=20 \mathrm{~V} / \mathrm{V} \end{aligned} (b) \mathrm{CMRR}=\left(2 g_{m 1} R_{S S} ight) /\left(\frac{ riangle g_{m}}{g_{m}} ight) where \begin{aligned} R_{S S} & =\left(g_{m 7} r_{o 7} ight) r_{o 9} \ & =\frac{2 I_{D 7}}{\left|V_{O V} ight|} imes \frac{V_{A}}{I_{D 7}} imes \frac{V_{A}}{I_{D 9}} \ & =\frac{2 V_{A}}{\left|V_{O V} ight|} \frac{V_{A}}{I_{D 9}} \end{aligned} But I_{D 9}=2 imes \frac{I}{2}=I and g_{m 1}=\frac{2 I_{D 1}}{\left|V_{O V} ight|}=\frac{2 imes\left(\frac{I}{2} ight)}{\left|V_{O V} ight|}=\frac{I}{\left|V_{O V} ight|} Thus, \begin{aligned} \mathrm{CMRR} & =\frac{2 I}{\left|V_{O V} ight|} imes \frac{2\left|V_{A} ight|}{\left|V_{O V} ight|} imes \frac{\left|V_{A} ight|}{I} imes \frac{1}{0.01} \ & =400\left|\frac{V_{A}}{V_{O V}} ight|^{2}=400 imes\left(\frac{4}{0.2} ight)^{2} \ & =1.6 imes 10^{5} \end{aligned} or 20 \log \left(1.6 imes 10^{5} ight)=104 \mathrm{~dB} (c) If the Wilson mirror is replaced with a simple current mirror, R_{S S}=r_{o}=\frac{\left|V_{A} ight|}{I} Thus, \begin{aligned} \mathrm{CMRR} & =\frac{2 I}{\left|V_{O V} ight|} imes \frac{\left|V_{A} ight|}{I} imes \frac{1}{0.01} \ & =200\left|\frac{V_{A}}{V_{O V}} ight|=200 imes\left(\frac{4}{0.2} ight) \ & =4000 \end{aligned} or 72 \mathrm{~dB}$

Figure 9.2.1 (a)

A_{d}=g_{m 1}\left(r_{o 1} \| r_{o 3} ight)

where

\begin{aligned}g_{m 1} & =\frac{2 I_{D}}{\left|V_{O V} ight|} \r_{o 1} & =r_{o 3}=\frac{\left|V_{A} ight|}{I_{D}} \A_{d} & =\frac{2 I_{D}}{\left|V_{O V} ight|} imes \frac{1}{2} \frac{\left|V_{A} ight|}{I_{D}} \& =\frac{\left|V_{A} ight|}{\left|V_{O V} ight|}=\frac{4}{0.2}=20 \mathrm{~V} / \mathrm{V}\end{aligned}

(b)

\mathrm{CMRR}=\left(2 g_{m 1} R_{S S} ight) /\left(\frac{ riangle g_{m}}{g_{m}} ight)

where

\begin{aligned}R_{S S} & =\left(g_{m 7} r_{o 7} ight) r_{o 9} \& =\frac{2 I_{D 7}}{\left|V_{O V} ight|} imes \frac{V_{A}}{I_{D 7}} imes \frac{V_{A}}{I_{D 9}} \& =\frac{2 V_{A}}{\left|V_{O V} ight|} \frac{V_{A}}{I_{D 9}}\end{aligned}

But

I_{D 9}=2 imes \frac{I}{2}=I

and

g_{m 1}=\frac{2 I_{D 1}}{\left|V_{O V} ight|}=\frac{2 imes\left(\frac{I}{2} ight)}{\left|V_{O V} ight|}=\frac{I}{\left|V_{O V} ight|}

Thus,

\begin{aligned}\mathrm{CMRR} & =\frac{2 I}{\left|V_{O V} ight|} imes \frac{2\left|V_{A} ight|}{\left|V_{O V} ight|} imes \frac{\left|V_{A} ight|}{I} imes \frac{1}{0.01} \& =400\left|\frac{V_{A}}{V_{O V}} ight|^{2}=400 imes\left(\frac{4}{0.2} ight)^{2} \& =1.6 imes 10^{5}\end{aligned}

or

20 \log \left(1.6 imes 10^{5} ight)=104 \mathrm{~dB}

(c) If the Wilson mirror is replaced with a simple current mirror,

R_{S S}=r_{o}=\frac{\left|V_{A} ight|}{I}

Thus,

\begin{aligned}\mathrm{CMRR} & =\frac{2 I}{\left|V_{O V} ight|} imes \frac{\left|V_{A} ight|}{I} imes \frac{1}{0.01} \& =200\left|\frac{V_{A}}{V_{O V}} ight|=200 imes\left(\frac{4}{0.2} ight) \& =4000\end{aligned}

or

72 \mathrm{~dB}

Figure 921 the Differential Amplifier in Fig ∣VOV∣=0.2 V\left|V_{O V}\right|=0.2 \mathrm{~V}∣VOV​∣=0.2 V And Have Equal Early Voltage

Figure 921
the Differential Amplifier in Fig $\left|V_{O V}\right|=0.2 \mathrm{~V}$ And Have Equal Early Voltage