Semiconductor Electronics: Materials, Devices and

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

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

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.”

As per NCERT Textbboks"If forward current is too large, it can produce large heating and damage the junction. So, a resistor is used in series with the diode to limit the current in the circuit.”

It means when a semiconductor diode is connected to a source under high current it causes excess heat due to more electrical energy. This excess heat is responsible to damage the junction in semiconductor diode permanently. Student can check out NCERT Solutions for Semicondutor Electronics of class 12 physics.  

14.14 A and B are the inputs of the given circuit. The output of the first NOR gate is $\stackrel{?}{A}$ +$\stackrel{?}{B}$ . It can be observed from the following figure that the inputs of the second NOR gate become the output of the first one.

Hence, the output of the combination is given as:

Y = $\stackrel{?}{\stackrel{?}{A+B}+\stackrel{?}{A+B}}$  =  $\stackrel{?}{\stackrel{?}{A}.\stackrel{?}{B}}$  + $\stackrel{?}{\stackrel{?}{A}.\stackrel{?}{B}}$    = = $\stackrel{?}{\stackrel{?}{A}.\stackrel{?}{B}}$    = $\stackrel{?}{A}$  + $\stackrel{?}{B}$  = $\stackrel{?}{A+B}$

The truth table for this operation is given as:

This is the truth table of an or gate. Hence, this circuit functions as an or gate.

This is the truth table of an OR gate. Hence, this circuit functions as an OR gate.

14.13 The output of the left NAND gate will be $\stackrel{?}{A.B}$ , as shown in the following figure:

Hence, the output of the combination of two NAND gates is given as:

Y = ( $\stackrel{?}{A.B}$ ).( $\stackrel{?}{A.B}$ ) = $\stackrel{?}{A.B}$ + $\stackrel{?}{A.B}$  = AB

Hence the circuit functions as an AND gate.

$\stackrel{?}{A}$ is the output of the upper left of the NAND gate and $\stackrel{?}{B}$  is the output of the lower half of the NAND gate, as shown in the following figure.

Hence, the output of the combination of the NAND gates will be given as:

Y = $\stackrel{?}{A}$ . $\stackrel{?}{B}$  = $\stackrel{?}{A}$ + $\stackrel{?}{B}$  = A + B

Hence, this circuit functions as an OR gate.