數位電子學期末考古
===============
## ch13
> In a series RLC circuit, resonance occurs when
> $$\omega L = \dfrac{1}{\omega C}$$
> $\Rightarrow 2\pi f_r L=\dfrac{1}{2\pi f_r C}$
>
> $\Rightarrow 2\pi f_r \times 0.0001 = \dfrac{1}{2\pi f_r \times 0.0000000022}$
>
> $\Rightarrow(f_r)^2=\dfrac{1}{4\pi^2\times0.00000000000022}$
>
> $\Rightarrow f_r = \sqrt{\dfrac{1}{4\pi^2\times0.00000000000022}}=\dfrac{1}{2\pi10^{-6}}\sqrt{\dfrac{1}{22}}\cong339.319(kHz)$
>
> $f_1=\dfrac{-RC+\sqrt{R^2C^2+4LC}}{2\pi 2LC}=\dfrac{-150\times0.0000000022+\sqrt{22500\times0.00000000000000000484+4\times0.0001\times0.0000000022}}{2\pi2\times0.0001\times0.0000000022}=88.077(kHz)$
>
> $f_2=\dfrac{RC+\sqrt{R^2C^2+4LC}}{2\pi 2LC}=\dfrac{150\times0.0000000022+\sqrt{22500\times0.00000000000000000484+4\times0.0001\times0.0000000022}}{2\pi2\times0.0001\times0.0000000022}=326.809(kHz)$
>
> $BW=f_2-f_1=326.809-88.077=238.732(kHz)$
>
> $\Rightarrow$ band-stop
> [name=張聖岳][time=Thu, Jan 2, 2020 5:56 AM]

> low-pass
> $Z=R+j\omega L=39000+j2\pi f10=39000+62.832fj$

> $I_L=\dfrac{V_S}{j\omega L}=-j\dfrac{1.5}{2\pi fL}=-j\dfrac{1.5}{2\pi\times2000\times0.0008}=-149j(mA)$
>
> $I_R=\dfrac{V_S}{R}=\dfrac{1.5}{12}=125(mA)$
>
> $I_{total}=I_R+I_L=125-149j=\sqrt{125^2+149^2}\angle-39.994^\circ=194.489\angle-39.994^\circ(mA)$
>
> $\tan^{-1}\dfrac{125}{-149}=-39.994^\circ$
>
> [name=張聖岳][time=Thu, Jan 2, 2020 6:21 AM]


> $Y = \dfrac{1}{R}+\dfrac{1}{j\omega L}=\dfrac{1}{560}-j\dfrac{1}{2\pi f\times0.025}=0.001786-0.003783j=0.004183\angle25.255^\circ$
>
> $X = \dfrac{1}{Y}=239.063\angle25.255^\circ$
## ch15

> (a) The extra electrons are not attached to any atom, then they become free electrons. So we can form the n-type semiconductor by add Sb into Si; We can form the n-type semiconductor by add B into Si.
> (b) 4
> [name=張聖岳][time=Wed, Jan 1, 2020 6:04 PM]

> (a) forward-biased
> (b) 0.7(V)
> [name=張聖岳][time=Wed, Jan 1, 2020 7:04 PM]

> (a) reverse-bias
> (b) -3(V)
> [name=張聖岳][time=Wed, Jan 1, 2020 11:59 PM]

> (a) -3(V) (b) 0.7(V)
> [name=張聖岳][time=Wed, Jan 1, 2020 8:08 PM]

## ch16
### 16-1

> $V_B\cong(\dfrac{R_2}{R_1+R_2})V_{CC}=\dfrac{10}{32}\times12=3.75(V)$
>
> $V_E=V_B-0.7=3.05(V)$
>
> $I_E=\dfrac{V_E}{R_E}=\dfrac{3.05}{100}=30.5(mA)$
>
> 
> [name=張聖岳][time=Wed, Jan 1, 2020 6:52 PM]

> $V_B = (\dfrac{R_2}{R_1+R_2})V_{CC}=(\dfrac{10}{22+10})\times30=9.375(V)$
>
> $V_B=V_E+V_{BE}\Rightarrow V_E=V_B-V_{BE}=9.375-0.7=8.675(V)$
>
> $I_E=\dfrac{V_E}{R_E}=\dfrac{8.675}{1000}=8.675(mA)$
>
> $I_C\cong I_E=8.675(mA)$
>
> $\beta_{DC}=\dfrac{I_C}{I_B}\Rightarrow I_B=\dfrac{I_C}{\beta_{DC}}=\dfrac{8.675}{100}=86.75(\mu A)$
>
> $V_C=V_{CC}-I_CR_C=30-1\times8.675=21.325(V)$
>
> [name=張聖岳][time=Wed, Jan 1, 2020 7:52 PM]

> $V_B = (\dfrac{R_2}{R_1+R2})V_{CC} =\dfrac{10}{10+22}\times12=3.75(V)$
>
> $V_B = V_E + V_{BE} \Rightarrow V_E = V_B-V_{BE} = 3.75-0.7 = 3.05(V)$
>
> $I_E = \dfrac{V_E}{R_E} = \dfrac{3.05}{680} = 4.485(mA)$
>
> $I_C \cong I_E = 4.485(mA)$
>
> $V_C = V_{CC} - I_CR_C = 12-4.485\times1200=12-5.382=6.618(V)$
>
> [name=張聖岳][time=Wed, Jan 1, 2020 8:45 PM]

### 16-3

> Supposed that the transistor is saturated
> $\Rightarrow V_{CE}\cong0(V), V_{BE}=0.7(V)$
>
> $I_{C(sat)}\cong\dfrac{V_{CC}}{R_C}=\dfrac{5}{10000}=0.5(mA)$
>
> $\beta_{DC} = \dfrac{I_C}{I_B}\Rightarrow I_B=\dfrac{I_C}{beta_{DC}}=\dfrac{0.5}{150}=3.333(\mu A)$
>
> $V_{R_B} = I_B\times R_B=3.333\times100=333.3(mV)$
>
> $V_{R_B}=V_{IN}-0.7 \Rightarrow V_{IN}=V_{R_B}+0.7=1.0\bar{3}(V)$
>
> [name=張聖岳][time=Wed, Jan 1, 2020 11:49 PM]

> [name=楊淳安][time=Thu, Jan 2, 2020 12:04 AM]

> (a.)
> $V_{CE} = V_{CC} = 10(V)$
>
> (b.)
> $I_{C(sat)} = \dfrac{V_{CC}}{R_C} = \dfrac{10}{1000} = 10(mA)$
>
> (c.)
> $\beta_{DC} = \dfrac{I_C}{I_B} \Rightarrow I_B = \dfrac{I_C}{\beta_{DC}}=\dfrac{10}{200}=50(\mu A)$
>
> (d.)
> $V_{R_B} = V_{IN}-0.7=5-0.7=4.3(V)$
> $R_B = \dfrac{V_{R_B}}{I_B}=\dfrac{4.3}{0.00005}=86(k\Omega)$
>
> [name=張聖岳][time=Thu, Jan 2, 2020 12:19 AM]

> 

## ch17

> $A_{cl} = \dfrac{R_f+R_i}{R_i} \Rightarrow R_f = \dfrac{R_f+R_i}{A_{cl}} = \dfrac{R_f+1000}{50}=\dfrac{1000}{49}=20.408(\Omega)$
>
> $A_{cl}=-\dfrac{R_f}{R_i}\Rightarrow R_f=-{A_{cl}}{R_i}=-(-300)\times10000=3(M\Omega)$
>
> [name=張聖岳][time=Thu, Jan 2, 2020 3:12 AM]

## ch18

### 18-1

> $V_{REF}=(\dfrac{10}{100+10})\times15=1.164(V)$
> 在正弦波圖畫一條水平線(y=1.164),與正弦波相交處為畫垂直線，轉換為Vp=12V的數位訊號。
> [name=張聖岳][time=Thu, Jan 2, 2020 3:22 AM]

> 以水平線(y=0)為基準,與正弦波相交處為畫分隔線，轉換為Vp=8 or 10V的數位訊號。
> [name=張聖岳]

### 18-2

> $V_{OUT}=-\sum_{i=1}^{4}\dfrac{R_f}{R_i}V_i=-(\dfrac{10}{10}\times2+\dfrac{10}{33}\times3+\dfrac{10}{91}\times3+\dfrac{10}{180}\times6)=-(2+0.909+0.330+0.222)=-3.461(V)$
> [name=張聖岳][time=Thu, Jan 2, 2020 3:45 AM]

### 18-3

> 

>