Semiconductor Electronics: Materials, Devices and

Here diode-D₁ is forced biased and diode-D₂ is reversed biased, so

$R_{AB}=\left(20//20\right)+15=25\Omega$

$E_9=\frac{hc}{\lambda }=\frac{1242}{\lambda \left(nm\right)}=\frac{1242}{400}=3.105$

 $i_{\left(zener-max\right)}=25mA$

20 – i_max R – 8 = 0

i_max R = 12

At minimum zener current $\left(\mu A\right):-$

$20-i_{min}R-i_{min}{R}_L=0$

$\frac{R}{R_L}=\frac{12}{8}=\frac{3}{2}$

${\mathcal{l}}_{min}R=12$

$i_{min}{R}_L=8$

At maxm zener current –

$20-i_{max}R-8=0$

$i_L=O\left\{as{i}_{z}ma{x}^m=25mA\right\}$

i_maxR = 12v

25 * 103 R = 12

$R=\frac{12*10^3}{25}=12*40=480\Omega$

g_eff = g – $\left(\frac{{\rho }_w}{{\rho }_b}\right)g$

$T=2\pi \sqrt[]{\frac{\mathcal{l}}{g}}$

$T\text{'}=2\pi \sqrt[]{\frac{\mathcal{l}}{g_{eff}}}$

$\Rightarrow T\text{'}=\sqrt[]{\frac{5}{4}}T$

= $\sqrt[]{\frac{5}{4}}*10$

= $5\sqrt[]{5}sec$

So, x = 5

$I_L=\frac{V}{R_L}=\frac{5}{1k\Omega }$

$l_L=5*10^{-3}A=5mA$

I_b = 10 µA

I_C = 1.5 mA

R_L = 50 kW or (Rc)

Base – emitter voltage = 10 mv

$R_B=\frac{V_B}{I_B}=\frac{10*10^{-3}}{10*10^{-6}}=10^3\Omega$

$A_v=\left(\frac{\Delta l_C}{\Delta l_B}\right)*\left(\frac{R_C}{R_B}\right)$

= $\frac{1.5*10^{-3}}{10*10^{-6}}*\frac{5*10^3}{10^3}$  

A_v = 750

According to Work energy theorem, we can write

$K_f-K_i=W_{ElectricForce}\Rightarrow \frac{1}{2}m{v}^2-\frac{1}{2}m{v}_0^2=-eV\Rightarrow v^2=v_0^2-\frac{2eV}{m}$

$\Rightarrow v^2={\left(6.0*10^5\right)}^2-\frac{2*1.6*10^{-19}}{9*10^{-31}}=\frac{324-128}{9}*10^{10}=\frac{196}{9}*10^{10}\Rightarrow V=\frac{14}{3}*10^{5}m/s$

Given

V_L = V_C = 2V_R, f = 50Hz

$L=\frac{1}{k\pi }mH$

since, V_L = V_C.

then

$V_{net}=\sqrt[]{{\left(V_L-V_C\right)}^2+{\left(V_R\right)}^2}$

V_R = V_Net = 220V

I = $\frac{V_{Net}}{Z}=\frac{220}{\sqrt[]{{\left(X_L-X_C\right)}^2+12^2}}$

$I=44A$

$V_L=I{x}_L=2v_R$

= I _xL = 440

x_L = 10

WL = 10

2πfL = 10

$L=\frac{1}{\frac{1}{100}\pi }k=\frac{1}{100}=0.01\approx 0$

 A p-type semiconductor is electrically neutral despite having more holes, because the number of positively charged holes is exactly balanced/equal by the number of negatively charged acceptor ions introduced during doping. so practically untill any volatage is applied the semiconductor remains chargeless in other words doesn't produce any current even after doping.

 As per the NCERT Textbooks information"Although the number of holes is more than the number of electrons in a p-type semiconductor, the material as a whole is electrically neutral because the charge of holes is balanced by the negatively charged acceptor ions.”