Electronic Circuits Homework 2

# Electronic Circuits Homework 2 > 劉子雍．108502523 > 資訊工程學系三年級 A 班 ## 1. ### (a) $R_{13} = 68\ (k\Omega)$ $R_7 = 33\ (k\Omega)$ $R_{14} = 47\ (k\Omega)$ $R_{16} = 22\ (k\Omega)$ $R_{13} + R_7 + R_{14} + R_{16} = 170\ (k\Omega)$ ==**ANS**==: $170\ (k\Omega)$ ### (b) $R_{12} = 10\ (\Omega)$ $R_8 = 18\ (\Omega)$ $R_6 = 22\ (\Omega)$ $R_{12} + R_8 + R_6 = 50\ (\Omega)$ ==**ANS**==: $50\ (\Omega)$ ## 2. ### (a) $V_2 = 15 - 2 - 3.2 - 1 - 1.5 - 0.5\ (V) = 6.8\ (V)$ ==**ANS**==: $V_2 = 6.8\ (V)$ ![](https://i.imgur.com/JMLvYLl.png) ### (b) $V_S = 8\ (V) \times \frac{R + R + 2R + 3R + 4R}{R} = 8\ (V) \times 11 = 88\ (V)$ ==**ANS**==: $V_S = 88\ (V)$ ![](https://i.imgur.com/TD9FJFp.png) ## 3. ### (a) $V_{A-GND} = 15\ (V)$ $V_{B-GND} = 15\ (V) \times \frac{10 + 3.3\ (k\Omega)}{5.6 + 10 + 3.3\ (k\Omega)} = 15\ (V) \times \frac{13.3}{18.9} \approx 10.56\ (V)$ $V_{C-GND} = 15\ (V) \times \frac{3.3\ (k\Omega)}{5.6 + 10 + 3.3\ (k\Omega)} = 15\ (V) \times \frac{3.3}{18.9} \approx 2.62\ (V)$ ==**ANS**==: $V_{A-GND} = 15\ (V),\ V_{B-GND} \approx 10.56\ (V),\ V_{C-GND} \approx 2.62\ (V)$ ### (b) $V_{OUT_{min}} = 12\ (V) \times \frac{680\ (\Omega)}{470 + 1000 + 680\ (\Omega)} = 12\ (V) \times \frac{680}{2150} \approx 3.80\ (V)$ $V_{OUT_{max}} = 12\ (V) \times \frac{1000 + 680\ (\Omega)}{470 + 1000 + 680\ (\Omega)} = 12\ (V) \times \frac{1680}{2150} \approx 9.38\ (V)$ ==**ANS**==: $V_{OUT_{min}} \approx 3.80\ (V),\ V_{OUT_{max}} \approx 9.38\ (V)$ ## 4. $I = 20\ (mA) = 0.02\ (A)$ $V_2 = 0.02\ (A) \times 100\ (\Omega) = 2\ (V)$ $R_5 = \frac{6.6\ (V)}{0.02\ (A)} = 330\ (\Omega)$ $V_3 = V_4 = \frac{20 - 2 - 6.6}{2}\ (V) = 5.7\ (V)$ $R_3 = R_4 = \frac{5.7\ (V)}{0.02\ (A)} = 285\ (\Omega)$ $P_6 = 112\ (mW) = 0.112\ (W)$ $R_6 = \frac{0.112\ (W)}{(0.02\ (A))^2} = 280\ (\Omega)$ $V_6 = 0.02\ (A) \times 280\ (\Omega) = 5.6\ (V)$ $V_1 = 30 - 2 - 5.7 -5.7 - 6.6 - 5.6 = 4.4\ (V)$ $R_1 = \frac{4.4\ (V)}{0.02\ (A)} = 220\ (\Omega)$ ==**ANS**==: $R_1 = 220\ (\Omega),\ R_3 = 285\ (\Omega),\ R_4 = 285\ (\Omega),\ R_5 = 330\ (\Omega),\ R_6 = 280\ (\Omega) \\ V_1 = 4.4\ (V),\ V_2 = 2\ (V),\ V_3 = 5.7\ (V),\ V_4 = 5.7\ (V),\ V_6 = 5.6\ (V)$ ## 5. $R_{(1)-(4)} = R_{10} + R_1 + R_7 + R_8 = 1 + 2.2 + 0.56 + 0.47\ (k\Omega) = 4.23\ (k\Omega)$ $R_{(2)-(3)} = R_6 + R_2 + R_4 + R_{11} = 3.3 + 4.7 + 5.6 + 10\ (k\Omega) = 23.6\ (k\Omega)$ $R_{(5)-(6)} = R_5 + R_3 + R_9 + R_{12} = 3.9 + 1 + 8.2 + 6.8\ (k\Omega) = 19.9\ (k\Omega)$ ==**ANS**==: $R_{(1)-(4)} = 4.23\ (k\Omega),\ R_{(2)-(3)} = 23.6\ (k\Omega),\ R_{(5)-(6)} = 19.9\ (k\Omega)$ ## 6. ### (a) $R_T = R_1 \mid\mid R_2 = (1/4.7 + 1/2.2)^{-1}\ (k\Omega) \approx 1.50\ (k\Omega)$ ==**ANS**==: $R_T \approx 1.50\ (k\Omega)$ ### (b) $R_T = R_1 \mid\mid R_2 = (1/27 + 1/56)^{-1}\ (\Omega) \approx 18.22\ (\Omega)$ ==**ANS**==: $R_T \approx 18.22\ (\Omega)$ ### $c$ $R_T = R_1 \mid\mid R_2 = (1/1.5 + 1/2.2)^{-1}\ (k\Omega) \approx 0.89\ (k\Omega)$ ==**ANS**==: $R_T \approx 0.89\ (k\Omega)$ ## 7. $R_2 = R_3$ $I_2 = I_3 = \frac{50 - 10 - 25\ (mA)}{2} = 7.5\ (mA)$ ==**ANS**==: $7.5\ (mA)$ for both $R_2$ and $R_3$ separately ![](https://i.imgur.com/RSp0Wpu.png) ## 8. $V_S = 1\ (mA) \times 50\ (\Omega) = 0.05\ (V) = 50\ (mV)$ $R_2 = \frac{50\ (mV)}{2\ (mA)} = 25\ (\Omega)$ $R_3 = \frac{50\ (mV)}{0.5\ (mA)} = 100\ (\Omega)$ $R_4 = \frac{50\ (mV)}{7.5 - 1 - 2 - 0.5\ (mA)} = \frac{50\ (mV)}{4\ (mA)} = 12.5\ (\Omega)$ ==**ANS**==: $R_2 = 25\ (\Omega),\ R_3 = 100\ (\Omega),\ R_4 = 12.5\ (\Omega)$ ## 9. $R_{(1)-(2)} = R_{11} \mid\mid R_3 = (1/1800 + 1/330)^{-1}\ (k\Omega) \approx 278.87\ (k\Omega)$ $R_{(3)-(4)} = R_7 \mid\mid R_8 \mid\mid R_6 \mid\mid R_4 = (1/820 + 1/680 + 1/1000 + 1/270)^{-1} \approx 135.25\ (k\Omega)$ $R_{(5)-(6)} = R_{12} \mid\mid R_9 \mid\mid R_{10} \mid\mid R_2 \mid\mid R_1 \mid\mid R_5 \\ \qquad\quad = (1/100 + 1/390 + 1/1200 + 1/220 + 1/100 + 1/560)^{-1} \approx 33.64\ (k\Omega)$ ==**ANS**==: $R_{(1)-(2)} \approx 278.87\ (k\Omega),\ R_{(3)-(4)} \approx 135.25\ (k\Omega),\ R_{(5)-(6)} \approx 33.64\ (k\Omega)$ ## 10. ### (a) $R_T = R_1 \mid\mid R_2 \mid\mid (R_3 + R_4 \mid\mid R_5) = R_1 \mid\mid R_2 \mid\mid (R_3 + (1/4.7 + 1/4.7)^{-1}\ (k\Omega)) \\ \quad \approx R_1 \mid\mid R_2 \mid\mid (R_3 + 2.4\ (k\Omega)) = R_1 \mid\mid R_2 \mid\mid 5.7\ (k\Omega) = (1/10 + 1/2.7 + 1/5.7)^{-1}\ (k\Omega) \\ \quad \approx 1.5\ (k\Omega)$ ==**ANS**==: $R_T \approx 1.5\ (k\Omega)$ ### (b) $I_T \approx \frac{6\ (V)}{1.5\ (k\Omega)} = 4\ (mA)$ ==**ANS**==: $I_T \approx 4\ (mA)$ ## 11. $R_X = \frac{R_2 R_V}{R_4} = \frac{2.2 \times 5}{1.5}\ (k\Omega) \approx 7.3\ (k\Omega)$ ==**ANS**==: $R_X \approx 7.3\ (k\Omega)$ ## 12.  $R_1 = 1 + 0.005 \times (65-25)\ (k\Omega) = 1.2\ (k\Omega)$ $V_{OUT_{A-B}} = 9 \times (\frac{R_3}{R_1 + R_3} - \frac{R_4}{R_2 + R_4})\ (V) = 9 \times (\frac{1}{2.2} - \frac{1}{2})\ (V) \approx -0.4\ (V)$ ==**ANS**==: $V_{OUT_{A-B}} \approx -0.4\ (V)$ ## 13.  ### (a) $V_{TH} = 2.5 \times \frac{78}{100 + 78 + 47}\ (V) \approx 0.87\ (V)$ $R_{TH} = R_4 + R_2 \mid\mid (R_1 + R_3) = R_4 + R_2 \mid\mid 147\ (\Omega) = R_4 + (1/78 + 1/147)^{-1}\ (\Omega) \\ \qquad \approx R_4 + 51\ (\Omega) = 73\ (\Omega)$ ==**ANS**==: $V_{TH} \approx 0.87\ (V),\ R_{TH} \approx 73\ (\Omega)$ ### (b) $V_{TH} = 3 \times \frac{R_1}{R_1 + R_2}\ (V) = 3 \times \frac{100}{100 + 270}\ (V) \approx 3 \times 0.27\ (V) = 0.81\ (V)$ $R_{TH} = R_1 \mid\mid R_2 = (1/100 + 1/270)^{-1}\ (\Omega) \approx 73\ (\Omega)$ ==**ANS**==: $V_{TH} \approx 0.81\ (V),\ R_{TH} \approx 73\ (\Omega)$ ### $c$ $V_{TH} = 1.5 \times \frac{R_2}{R_1 + R_2}\ (V) = 1.5 \times \frac{56}{100 + 56}\ (V) \approx 1.5 \times 0.36\ (V) = 0.54\ (V)$ $R_{TH} = R_1 \mid\mid R_2 = (1/100 + 1/56)^{-1}\ (k\Omega) \approx 36\ (k\Omega)$ ==**ANS**==: $V_{TH} \approx 0.54\ (V),\ R_{TH} \approx 36\ (k\Omega)$ ## 14.  ### (a) $R_{(S:\ 1V)} = R_1 + R_2 \mid\mid R_3 = 100 + (1/56 + 1/27)^{-1}\ (\Omega) \approx 118\ (\Omega)$ $I_{1(S:\ 1V)} = \frac{1\ (V)}{118\ (\Omega)} \approx 8.47\ (mA)$ $I_{3(S:\ 1V)} = 8.47 \times \frac{1/27}{1/56 + 1/27}\ (mA) \approx 5.71\ (mA)$ $R_{(S:\ 1.5V)} = R_2 + R_1 \mid\mid R_3 = 56 + (1/100 + 1/27)^{-1}\ (\Omega) \approx 77\ (\Omega)$ $I_{2(S:\ 1.5V)} = -\frac{1.5\ (V)}{77\ (\Omega)} \approx -19.48\ (mA)$ $I_{3(S:\ 1.5V)} = 19.48 \times \frac{1/27}{1/100 + 1/27}\ (mA) \approx 15.34\ (mA)$ $I_3 = I_{3(S:\ 1V)} + I_{3(S:\ 1.5V)} \approx 5.71 + 15.34\ (mA) = 21.05\ (mA)$ ==**ANS**==: $I_3 \approx 21.05\ (mA)$ ### (b) $I_{2(S:\ 1V)} = I_{1(S:\ 1V)} - I_{3(S:\ 1V)} \approx 8.47 - 5.71\ (mA) = 2.76\ (mA)$ $I_2' = I_{2(S:\ 1V)} + I_{2(S:\ 1.5V)} \approx 2.76 - 19.48\ (mA) = -16.72\ (mA)$ $I_2 = | I_2' | \approx 16.72\ (mA)$ ==**ANS**==: $I_2 \approx 16.72\ (mA)$ ## 15.  $R_6 \mid\mid (R_2 \mid\mid R_5 + R_1) = R_6 \mid\mid ((1/3.3 + 1/3.3)^{-1} + 1.0\ (k\Omega)) \approx R_6 \mid\mid 2.6\ (k\Omega) \\ \quad = (1/4.7 + 1/2.6)^{-1}\ (k\Omega) \approx 1.7\ (k\Omega)$ $V_{TH} = 50 \times \frac{R_6 \mid\mid (R_2 \mid\mid R_5 + R_1)}{R_3 + R_6 \mid\mid (R_2 \mid\mid R_5 + R_1)}\ (V) \approx 50 \times \frac{1.7}{4.7 + 1.7}\ (V) \approx 13.28\ (V)$ $R_{TH} = R_3 \mid\mid R_6 \mid\mid (R_2 \mid\mid R_5 + R_1) \approx R_3 \mid\mid 1.7\ (k\Omega) = (1/4.7 + 1/1.7)^{-1}\ (k\Omega) \\ \qquad \approx 1.25\ (k\Omega)$ $V_4 = V_{TH} \times \frac{R_4}{R_{TH} + R_4} \approx 13.28 \times \frac{10}{1.25 + 10}\ (V) \approx 11.80\ (V)$ ==**ANS**==: $V_4 \approx 11.80\ (V)$