Figure 9.3.1
For the amplifier in Fig. 9.3.1:
(a) Assuming $Q_{3}$ and $Q_{4}$ have very high $\beta$ values, determine the value of $R$ that will provide a dc current of $0.25 \mathrm{~mA}$ for each of $Q_{1}$ and $Q_{2}$. (b) Assuming $Q_{1}$ and $Q_{2}$ have very high $\beta$ values, determine the value of $R_{C}$ that results in a dc voltage of $+3 \mathrm{~V}$ at the collectors of $Q_{1}$ and $Q_{2}$.
(c) What is the input common-mode range for this differential amplifier?
(d) Determine the value of the differential voltage gain $v_{O} / v_{i d}$.
(e) Determine the value of the input differential resistance, $R_{i d}$. Assume $\beta_{1}=\beta_{2}=100$.
(f) If it is required to raise the value of $R_{i d}$ by factor of 5 by including a resistance $R_{E}$ in the emitter of each of $Q_{1}$ and $Q_{2}$, what value of $R_{E}$ is required? What is the resulting value of the differential gain?
(g) If the transistors have an Early voltage $V_{A}=50 \mathrm{~V}$, calculate the worst-case value of the common-mode gain resulting from the resistances $R_{C}$ having finite tolerances of $\pm 1 \%$. Hence, find the value of the CMRR of the original amplifier (i.e., the one without the resistances $R_{E}$ included).
(h) With the two input terminals grounded, calculate the value of the worst-case differential dc voltage at the output resulting from the resistances $R_{C}$ having finite tolerances of $\pm 1 \%$. Hence, find the input offset voltage of the original amplifier (i.e., the one without $R_{E}$ 's).

Question

Figure 9.3.1
For the amplifier in Fig. 9.3.1:
(a) Assuming $Q_{3}$ and $Q_{4}$ have very high $\beta$ values, determine the value of $R$ that will provide a dc current of $0.25 \mathrm{~mA}$ for each of $Q_{1}$ and $Q_{2}$. (b) Assuming $Q_{1}$ and $Q_{2}$ have very high $\beta$ values, determine the value of $R_{C}$ that results in a dc voltage of $+3 \mathrm{~V}$ at the collectors of $Q_{1}$ and $Q_{2}$.
(c) What is the input common-mode range for this differential amplifier?
(d) Determine the value of the differential voltage gain $v_{O} / v_{i d}$.
(e) Determine the value of the input differential resistance, $R_{i d}$. Assume $\beta_{1}=\beta_{2}=100$.
(f) If it is required to raise the value of $R_{i d}$ by factor of 5 by including a resistance $R_{E}$ in the emitter of each of $Q_{1}$ and $Q_{2}$, what value of $R_{E}$ is required? What is the resulting value of the differential gain?
(g) If the transistors have an Early voltage $V_{A}=50 \mathrm{~V}$, calculate the worst-case value of the common-mode gain resulting from the resistances $R_{C}$ having finite tolerances of $\pm 1 \%$. Hence, find the value of the CMRR of the original amplifier (i.e., the one without the resistances $R_{E}$ included). 
(h) With the two input terminals grounded, calculate the value of the worst-case differential dc voltage at the output resulting from the resistances $R_{C}$ having finite tolerances of $\pm 1 \%$. Hence, find the input offset voltage of the original amplifier (i.e., the one without $R_{E}$ 's).

Quizplus · Accepted Answer

The Answer is $Figure 9.3.1 (a) To obtain a 0.25-\mathrm{mA} current in each of Q_{1} and Q_{2} , the current in Q_{3} must be 0.5 \mathrm{~mA} . For Q_{3} and Q_{4} having high \beta values, the current through R and Q_{4} must be 0.5 \mathrm{~mA} , thus R=\frac{10-0.7}{0.5}=18.6 \mathrm{k} \Omega (b) R_{C}=\frac{5-3}{0.25}=8 \mathrm{k} \Omega (c) \begin{aligned} v_{I C M \max } & =V_{C 1,2}+0.4 \ & =3.4 \mathrm{~V} \ v_{I C M \min } & =-5+0.3+0.7=-4 \mathrm{~V} \end{aligned} Thus, -4 \mathrm{~V} \leq v_{I C M} \leq+3.4 \mathrm{~V} (d) \frac{v_{o}}{v_{i d}}=g_{m 1,2} R_{C} where g_{m 1,2}=\frac{I_{C 1,2}}{V_{T}}=\frac{0.25 \mathrm{~mA}}{0.025 \mathrm{~V}}=10 \mathrm{~mA} / \mathrm{V} Thus, \frac{v_{o}}{v_{i d}}=10 imes 8=80 \mathrm{~V} / \mathrm{V} (e) \begin{aligned} R_{i d} & =2 r_{\pi}=2 \frac{\beta}{g_{m}} \ & =\frac{2 imes 100}{10}=20 \mathrm{k} \Omega \end{aligned} (f) To raise R_{i d} by a factor of 5 , we include a resistance R_{E} in each of the emitters of Q_{1} and Q_{2} with R_{E} given by \begin{aligned} R_{E} & =4 r_{e} \ & =4 \frac{V_{T}}{I_{E}} \ & =4 \frac{25 \mathrm{mV}}{0.25 \mathrm{~mA}} \ & =400 \Omega \end{aligned} This will reduce the differential gain by the same factor of 5 , thus \frac{v_{o}}{v_{i d}}=\frac{80}{5}=16 \mathrm{~V} / \mathrm{V} (g) \left|A_{c m} ight|=\left(\frac{R_{C}}{2 R_{E E}} ight)\left(\frac{ riangle R_{C}}{R_{C}} ight) where R_{E E}=\left.r_{o} ight|_{Q_{3}}=\frac{V_{A}}{I_{C 3}}=\frac{50}{0.5}=100 \mathrm{k} \Omega \frac{ riangle R_{C}}{R_{C}}=0.02 Thus, \begin{aligned} \left|A_{c m} ight| & =\frac{8}{2 imes 100} imes 0.02=8 imes 10^{-4} \mathrm{~V} / \mathrm{V} \ \mathrm{CMRR} & =\frac{80}{8 imes 10^{-4}}=10^{5} \end{aligned} or 100 \mathrm{~dB} . (h) \begin{aligned} \left|V_{O} ight| & =0.25\left(R_{C}+0.01 R_{C} ight)-0.25\left(R_{C}-0.01 R_{C} ight) \ & =0.25 imes 0.02 imes 8=0.04 \mathrm{~V} \ & =40 \mathrm{mV} \ V_{O S} & =\frac{40}{80}=0.5 \mathrm{mV} \end{aligned}$

Figure 9.3.1
(a) To obtain a

0.25-\mathrm{mA}

current in each of

Q_{1}

and

Q_{2}

, the current in

Q_{3}

must be

0.5 \mathrm{~mA}

. For

Q_{3}

and

Q_{4}

having high

\beta

values, the current through

R

and

Q_{4}

must be

0.5 \mathrm{~mA}

, thus

R=\frac{10-0.7}{0.5}=18.6 \mathrm{k} \Omega

(b)

R_{C}=\frac{5-3}{0.25}=8 \mathrm{k} \Omega

(c)

\begin{aligned}v_{I C M \max } & =V_{C 1,2}+0.4 \& =3.4 \mathrm{~V} \v_{I C M \min } & =-5+0.3+0.7=-4 \mathrm{~V}\end{aligned}

Thus,

-4 \mathrm{~V} \leq v_{I C M} \leq+3.4 \mathrm{~V}

(d)

\frac{v_{o}}{v_{i d}}=g_{m 1,2} R_{C}

where

g_{m 1,2}=\frac{I_{C 1,2}}{V_{T}}=\frac{0.25 \mathrm{~mA}}{0.025 \mathrm{~V}}=10 \mathrm{~mA} / \mathrm{V}

Thus,

\frac{v_{o}}{v_{i d}}=10 imes 8=80 \mathrm{~V} / \mathrm{V}

(e)

\begin{aligned}R_{i d} & =2 r_{\pi}=2 \frac{\beta}{g_{m}} \& =\frac{2 imes 100}{10}=20 \mathrm{k} \Omega\end{aligned}

(f) To raise

R_{i d}

by a factor of 5 , we include a resistance

R_{E}

in each of the emitters of

Q_{1}

and

Q_{2}

with

R_{E}

given by

\begin{aligned}R_{E} & =4 r_{e} \& =4 \frac{V_{T}}{I_{E}} \& =4 \frac{25 \mathrm{mV}}{0.25 \mathrm{~mA}} \& =400 \Omega\end{aligned}

This will reduce the differential gain by the same factor of 5 , thus

\frac{v_{o}}{v_{i d}}=\frac{80}{5}=16 \mathrm{~V} / \mathrm{V}

(g)

\left|A_{c m} ight|=\left(\frac{R_{C}}{2 R_{E E}} ight)\left(\frac{ riangle R_{C}}{R_{C}} ight)

where

R_{E E}=\left.r_{o} ight|_{Q_{3}}=\frac{V_{A}}{I_{C 3}}=\frac{50}{0.5}=100 \mathrm{k} \Omega

\frac{ riangle R_{C}}{R_{C}}=0.02

Thus,

\begin{aligned}\left|A_{c m} ight| & =\frac{8}{2 imes 100} imes 0.02=8 imes 10^{-4} \mathrm{~V} / \mathrm{V} \\mathrm{CMRR} & =\frac{80}{8 imes 10^{-4}}=10^{5}\end{aligned}

or

100 \mathrm{~dB}

.
(h)

\begin{aligned}\left|V_{O} ight| & =0.25\left(R_{C}+0.01 R_{C} ight)-0.25\left(R_{C}-0.01 R_{C} ight) \& =0.25 imes 0.02 imes 8=0.04 \mathrm{~V} \& =40 \mathrm{mV} \V_{O S} & =\frac{40}{80}=0.5 \mathrm{mV}\end{aligned}

Figure 931 for the Amplifier in Fig Q3Q_{3}Q3​ And Q4Q_{4}Q4​ Have Very High

Figure 931
for the Amplifier in Fig $Q_{3}$ And $Q_{4}$ Have Very High