# Electronic Circuits Homework 5 > 劉子雍．108502523 > 資訊工程學系三年級 A 班 ## 1. <!-- TODO --> ### (a) Reverse bias. $I = 0$ $V_R = 8 - 5\ (V) = 3\ (V)$ ==**ANS**==: $V_R = 3\ (V)$ ### (b) Forward bias. $V_F \approx 0.7\ (V)$ ==**ANS**==: $V_F \approx 0.7\ (V)$ ### \(c\) Forward bias. $V_F \approx 0.7\ (V)$ ==**ANS**==: $V_F \approx 0.7\ (V)$ ### (d) Forward bias. $V_F \approx 0.7\ (V)$ ==**ANS**==: $V_F \approx 0.7\ (V)$ ## 2. <!-- TODO --> ### (a) ==**ANS**==: Center-tapped full-wave rectifier ### (b) $V_{p(sec)} = V_{p(pri)} \cdot \frac{1}{4} = (V_{rms(pri)} \cdot \sqrt{2}) \cdot \frac{1}{4} = (80 \cdot \sqrt{2}) \cdot \frac{1}{4}\ (V) \approx 28.28\ (V)$ ==**ANS**==: $28.28\ (V)$ ### \(c\) $V_{p(sec.h)} = \frac{V_{p(sec)}}{2} \approx \frac{28.28}{2}\ (V) = 14.14\ (V)$ ==**ANS**==: $14.14\ (V)$ ### (d) $V_{out}: v_{out}(t) = \left\{\begin{array}{l} V_{p(sec.h)}|\sin{(2\pi ft)}| - V_B,\ V_{p(sec.h)}|\sin{(2\pi ft)}| > V_B \\ 0,\ \text{otherwise} \end{array}\right. (V)$ $V_{p(out)} = \max{\{V_{out}\}} = V_{p(sec.h)} - V_B \approx 14.14 - 0.7\ (V) = 13.44\ (V)$ ==**ANS**==: - $x$: time in $s$, $y(x)$: voltage in $V$ - For $f = 1\ (Hz)$ - 🔵 $V_{out}: y(x) = \left\{\begin{array}{l} 14.14 \cdot |\sin{(2\pi fx)}| - V_B,\ 14.14 \cdot |\sin{(2\pi fx)}| > 0.7 \\ 0,\ \text{otherwise} \end{array}\right. (V)$ ![](https://i.imgur.com/ElmYsiO.png) ### (e) $I_{p(out)} = \frac{V_{p(out)}}{R_L} \approx \frac{13.44}{1}\ (mA) = 13.44\ (mA)$ ==**ANS**==: $13.44\ (mA)$ ### (f) $PIV = 2 \cdot V_{p(sec.h)} - V_B \approx 2 \cdot 14.14 - 0.7\ (V) = 27.58\ (V)$ ==**ANS**==: $27.58\ (V)$ ## 3. <!-- TODO --> $V_{rms(pri)} = 120\ (V)$ $V_{p(R_L)} = \frac{V_p(pri)}{2} - V_B = \frac{V_{rms(pri)} \cdot \sqrt{2}}{2} - V_B = \frac{120 \cdot \sqrt{2}}{2} - 0.7\ (V) \approx 84.15\ (V)$ ==**ANS**==: $84.15\ (V)$ ## 4. <!-- TODO --> $V_{p(in)} = V_{rms(in)} \cdot \sqrt{2} = 120 \cdot \sqrt{2}\ (V) \approx 169.70\ (V)$ $V_{in}: v_{in}(t) = V_{p(in)}\sin{(2\pi ft)} \approx 169.70 \cdot \sin{(2\pi ft)}\ (V)$ $V_{sec}: v_{sec}(t) = V_{p(sec)}\sin{(2\pi ft)} \approx \frac{169.70}{3} \cdot \sin{(2\pi ft)}\ (V) \approx 56.57 \cdot \sin{(2\pi ft)}\ (V)$ $V_A: v_A(t) = \left\{\begin{array}{l} v_{sec}(t) - V_B,\ v_{sec}(t) > 2 \cdot V_B \\ - V_B,\ v_{sec}(t) < -2 \cdot V_B \\ 0,\ \text{otherwise} \end{array}\right. (V) \\ \qquad\qquad \approx \left\{\begin{array}{l} 56.57 \cdot \sin{(2\pi ft)} - V_B,\ 56.57 \cdot \sin{(2\pi ft)} > 2 \cdot V_B \\ - V_B,\ v_{sec}(t) < -2 \cdot V_B \\ 0,\ \text{otherwise} \end{array}\right. (V)$ $V'_{B}: v'_{B}(t) = \left\{\begin{array}{l} |v_{sec}(t)| - 2 \cdot V_B,\ |v_{sec}(t)| > 2 \cdot V_B \\ 0,\ \text{otherwise} \end{array}\right. (V) \\ \qquad\qquad = \left\{\begin{array}{l} 56.57 \cdot |\sin{(2\pi ft)}| - 2 \cdot V_B,\ 56.57 \cdot |\sin{(2\pi ft)}| > 2 \cdot V_B \\ 0,\ \text{otherwise} \end{array}\right. (V)$ ==**ANS**==: - The capacitive effects are not calculated precisely. - $x$: time in $s$, $y(x)$: voltage in $V$ - For $f = 60\ (Hz)$ - 🔵 $V_{in}: y(x) = 169.70 \cdot \sin{(2\pi ft)}\ (V)$ - 🟢 $V_A: y(x) = \left\{\begin{array}{l} 56.57 \cdot \sin{(2\pi ft)} - 0.7,\ 56.57 \cdot \sin{(2\pi ft)} > 1.4 \\ - 0.7,\ v_{sec}(t) < -1.4 \\ 0,\ \text{otherwise} \end{array}\right. (V)$ - 🔴 $V'_B: y(x) = \left\{\begin{array}{l} 56.57 \cdot |\sin{(2\pi fx)}| - 1.4,\ 56.57 \cdot |\sin{(2\pi fx)}| > 1.4 \\ 0,\ \text{otherwise} \end{array}\right. (V)$ - 🟠 The capacitive effects (not precisely). - Actual $V_B$: 🔴 $V'_B$ with 🟠 the capacitive effects. ![](https://i.imgur.com/PYzaqVr.png) ## 5. <!-- TODO --> $V_B = 2\ (V)$ $V_{BE} = 0.7\ (V)$ $V_E - V_B - V{BE} = 2 - 0.7\ (V) = 1.3\ (V)$ $I_E = \frac{V_E}{R_E} = \frac{1.3}{1.0}\ (mA) = 1.3\ (mA)$ $I_C = \alpha_{DC} I_E = 0.98 \cdot 1.3\ (mA) \approx 1.27\ (mA)$ $I_B = \frac{I_C}{\beta_{DC}} \approx \frac{1.27}{49}\ (mA) \approx 25.92\ (\mu A)$ ==**ANS**==: $I_B \approx 25.92\ (\mu A),\ I_E = 1.3\ (mA),\ I_C \approx 1.27\ (mA)$ ## 6. <!-- TODO --> $V_B = \left(\frac{R_2}{R_1 + R_2}\right) V_{CC} = \left(\frac{10}{22 + 10}\right) \cdot 12\ (V) = 3.75\ (V)$ $V_E = V_B - V_{BE} = 3.75 - 0.7\ (V) = 3.05\ (V)$ $I_E = \frac{V_E}{R_E} = \frac{3.05}{680}\ (A) \approx 4.48\ (mA)$ $I_C = \left(\frac{\beta_{DC}}{\beta_{DC} + 1}\right) I_E \approx \left(\frac{100}{101}\right) 4.48\ (mA) \approx 4.44\ (mA)$ $V_C = V_{CC} - I_C R_C \approx 12 - 4.44 \cdot 1.2\ (V) \approx 6.67\ (V)$ ==**ANS**==: $V_B = 3.75\ (V),\ V_E = 3.05\ (V),\ I_E \approx 4.48\ (mA),\ I_C \approx 4.44\ (mA),\ V_C \approx 6.67\ (V)$ ## 7. ### (a) $V_B = \left(\frac{R_2}{R_1 + R_2}\right) V_{CC} = \left(\frac{12}{47 + 12}\right) \cdot 18\ (V) \approx 3.66\ (V)$ =**ANS**==: $V_B \approx 3.66\ (V)$ ### (b) $V_E = V_B - V_{BE} \approx 3.66 - 0.7\ (V) = 2.96\ (V)$ ==**ANS**==: $V_E \approx 2.96\ (V)$ ### \(c\) $I_E = \frac{V_E}{R_E} \approx \frac{2.96}{1}\ (mA) = 2.96\ (mA)$ ==**ANS**==: $I_E \approx 2.96\ (mA)$ ### (d) $I_C = \left(\frac{\beta_{DC}}{\beta_{DC} + 1}\right) I_E \approx \left(\frac{75}{75 + 1}\right) 2.96\ (mA) \approx 2.92\ (mA)$ ==**ANS**==: $I_C \approx 2.92\ (mA)$ ### (e) $V_C = V_{CC} - I_C R_C \approx 18 - 2.92 \cdot 3.3\ (V) \approx 8.36\ (V)$ ==**ANS**==: $V_C \approx 8.36\ (V)$ ### (f) $V_{CE} = V_C - V_E \approx 8.36 - 2.96\ (V) = 5.40\ (V)$ ==**ANS**==: $V_{CE} \approx 5.40\ (V)$ ## 8. <!-- TODO --> $V_{p(out)} = V_{p(in)} = \frac{V_CC}{2} = 10\ (V)$ $I_{p(out)} = \frac{V_{p(out)}}{R_L} = \frac{10}{16}\ (A) = 625\ (mA)$ ==**ANS**==: $V_{p(out)} = 10\ (V),\ I_{p(out)} = 625\ (mA)$ ## 9. <!--TODO --> Saturation: $V_{CE} = 0\ (V)$ $I_{C(sat)} = \frac{V_{CC}}{R_C} = \frac{5}{10}\ (mA) = 500\ (\mu A)$ $I_B > \frac{I_{C(sat)}}{\beta_{DC}} = \frac{500}{150}\ (\mu A) \approx 3.33\ (\mu A)$ $V_B = I_B R_B + V_{BE} > 3.33 \cdot 0.1 + 0.7 \approx 1.03\ (V)$ ==**ANS**==: $I_{C(sat)} = 500\ (\mu A),\ I_B > 3.33\ (\mu A),\ V_B > 1.03\ (V)$ ## 10. ### (a) ==**ANS**==: Single-ended mode. ### (b) ==**ANS**==: Differential mode. ### \(c\) ==**ANS**==: Common mode. ## 11. <!-- TODO --> ### (a) ==**ANS**==: $V_{out} = V_{in}$ ### (b) ==**ANS**==: $V_{out} = \left(\frac{R_i + R_f}{R_i}\right) V_{in}$ ### \(c\) ==**ANS**==: $V_{out} = \left(-\frac{R_f}{R_i}\right) V_{in}$ ## 12. ### (a) ==**ANS**==: $A_v = 1$ ### (b) $A_v = -\frac{R_f}{R_i} = -\frac{100}{100} = -1$ ==**ANS**==: $A_v = -1$ ### \(c\) $A_v = \frac{R_i + R_f}{R_i} = \frac{47 + 1000}{47} \approx 22.28$ ==**ANS**==: $A_v \approx 22.28$ ### (d) $A_v = -\frac{R_f}{R_i} = -\frac{330}{33} = -10$ ==**ANS**==: $A_v = -10$ ## 13. ### (a) $V_{+} - V_{-} = 0 - 1\ (V) = -1\ (V) < 0\ (V) \Rightarrow$ maximum negative ==**ANS**==: maximum negative ### (b) $V_{+} - V_{-} = 2 - 0\ (V) = 2\ (V) > 0\ (V) \Rightarrow$ maximum positive ==**ANS**==: maximum positive ### \(c\) $V_{+} - V_{-} = 5 - 7\ (V) = -2\ (V) < 0\ (V) \Rightarrow$ maximum negative ==**ANS**==: maximum negative ## 14. <!-- TODO --> ### (a) $V_{-}: v_{in}(t) = \sin{(2\pi f t)}$ $V_{out}: \left\{\begin{array}{l} V_{out(max)},\ V_{+} - V_{-} > 0 \\ -V_{out(max)},\ \text{otherwise} \end{array}\right. = \left\{\begin{array}{l} 8,\ -\sin{(2\pi f t)} > 0 \\ -8,\ \text{otherwise} \end{array}\right. (V)$ ==**ANS**==: - $x$: time in $s$, $y(x)$: voltage in $V$ - For $f = 1\ (Hz)$ - 🔵 $V_{in}: y(x) = \sin{(2\pi f x)}\ (V)$ - 🔴 $V_{out}: y(x) = \left\{\begin{array}{l} 8,\ -\sin{(2\pi f t)} > 0 \\ -8,\ \text{otherwise} \end{array}\right. (V)$ ![](https://i.imgur.com/mbUysbr.png) ### (b) $V_{+}: v_{in}(t) = 2\sin{(2\pi f t)}$ $V_{out}: \left\{\begin{array}{l} V_{out(max)},\ V_{+} - V_{-} > 0 \\ -V_{out(max)},\ \text{otherwise} \end{array}\right. = \left\{\begin{array}{l} 10,\ 2\sin{(2\pi f t)} - 1 > 0 \\ -10,\ \text{otherwise} \end{array}\right. (V)$ ==**ANS**==: - $x$: time in $s$, $y(x)$: voltage in $V$ - For $f = 1\ (Hz)$ - 🔵 $V_{in}: y(x) = 2\sin{(2\pi f x)}\ (V)$ - 🔴 $V_{out}: y(x) = \left\{\begin{array}{l} 10,\ 2\sin{(2\pi f t)} - 1 > 0 \\ -10,\ \text{otherwise} \end{array}\right. (V)$ ![](https://i.imgur.com/zaD7Q24.png) ## 15. ### (a) $V_{out} = -R_f\left(\frac{1}{R_1} + \frac{1.5}{R_2}\right)\ (V) = -10\left(\frac{1}{10} + \frac{1.5}{10}\right)\ (V) = -2.5\ (V)$ ==**ANS**==: $V_{out} = -2.5\ (V)$ ### (b) $V_{out} = -R_f\left(\frac{-0.1}{R_1} + \frac{1}{R_2} + \frac{0.5}{R_3}\right)\ (V) = -22\left(\frac{-0.1}{10} + \frac{1}{10} + \frac{0.5}{10}\right)\ (V) = -3.08\ (V)$ ==**ANS**==: $V_{out} = -3.08\ (V)$ ## 16. <!-- TODO --> $\frac{\Delta V_{out}}{\Delta t} = -\frac{V_{in}}{R_i C} = -\frac{5}{56 \cdot 0.022}\ (kV/s) \approx -4.06\ (kV/s)$ ==**ANS**==: $-4.06\ (kV/s)$ ## 17. $V_{out}: v_{out}(t) = -\left(\frac{\Delta V_{in}(t)}{\Delta t}\right)R_f C = \left\{\begin{array}{l} -\left(\frac{5}{5}\right)10 \cdot 0.001,\ 2 \mid floor(t/5\ (\mu s)) \\ -\left(\frac{-5}{5}\right)10 \cdot 0.001,\ \text{otherwise} \end{array}\right. (kV) \\ \qquad\qquad\quad = \left\{\begin{array}{l} -10,\ 2 \mid floor(t/5\ (\mu s)) \\ 10,\ \text{otherwise} \end{array}\right. (V)$ ==**ANS**==: - $x$: time in $s$, $y(x)$: voltage in $V$ - 🔵 $V_{in}$ - 🔴 $V_{out}: y(x) = \left\{\begin{array}{l} -10,\ 2 \mid floor(x/5\ (\mu s)) \\ 10,\ \text{otherwise} \end{array}\right. (V)$ ![](https://i.imgur.com/i1vcZK2.png)