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# System prepended metadata

title: Electronic Circuits Homework 5

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# 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)

