Semiconductor Electronics

This is a  Short Answer Type Questions as classified in NCERT Exemplar

Explanation- A material is a conductor if in its energy band diagram, there is no energy gap between conduction band and valence band. For insulator, the energy gap is large and for

semiconductor the energy gap is moderate.

The energy gap for Sn is 0 eV, for C is 5.4 eV, for Si is 1.1 eV and for Ge is 0.7 eV, related to their atomic size. Therefore Sn is a conductor, C is an insulator and Ge and Si are semiconductors.

This is a  Short Answer Type Questions as classified in NCERT Exemplar

Explanation- The size of the dopant atom should be such that their presence in the pure semiconductor does not distort the semiconductor but easily contribute the charge carriers on forming covalent bonds with Si or Ge atoms, which are provided by group XIII or group XV elements.

This is a Long Answer Type Questions as classified in NCERT Exemplar

Explanation- I_E=I_C+I_B and I_c= $\beta I_B$

I_cRc+V_CE+I_ER_E=V_CC

RI_B+V_BE+I_ER_E=V_cc

I_E=I_c= $\beta I_B$

From above equation

(R+ $\beta R$ _E)I_B=V_CC-V_BE

I_B= $\frac{V_{CC}-V_{BE}}{R+\beta R_E}=\frac{12-0.5}{80+1.2*100}=\frac{11.5}{200}$

( $R_C+R_E$ )= $\frac{V_{CE}-V_{BE}}{I_C}$

( $R_C+R_E$ )=1.56

Rc=1.56-1=0.56kohm

This is a Long Answer Type Questions as classified in NCERT Exemplar

Explanation-I_c= I_E

R_c= 7.8Kohm

I_c (R_c+R_E)+V_CE= 12

(R_E+R_C) $*1*{10}^{-3}+3=12$

R_E+R_C= 9 $*{10}^3=9kohm$

R_E= 9-7.3= 1.2kohm

V_E= I_{E $*$} R_E

 = 1 $*{10}^{-3}*1.2*{10}^3=1.2V$

Voltage V_B=V_E+V_BE= 1.2+0.5= 1.7V

I= $\frac{1.7}{20*{10}^3}=0.085mA$

Resistance R_B= $\frac{12-1.7}{\frac{I_C}{?}+0.085}=\frac{10.3}{0.01+0.085}=108kohm$

This is a Long Answer Type Questions as classified in NCERT Exemplar

Y= A'B+A.B'=Y₁+Y₂

Y₁= A.B and Y₂= A.B'

Y₁ can be obtained as output of AND GATE I for which one Input is of A through NOT GATE and

another input is of B. Y2 can be obtained as output of AND GATE II for which one input is of A

and other input is of B through NOT gate.

Now Y₂ can be obtained as output from or gate, where, Y₁ and Y₂ are input of or gate.

Thus, the given table can be obtained from the logic circuit given below

This is a Long Answer Type Questions as classified in NCERT Exemplar

whem As is used then it will create n type semiconductor

N_e=N_D= $\frac{1}{{10}^6}*5*{10}^{28}=5*{10}^{22}/m^3$

Number of minority carriers n_h= $\frac{{ni}^2}{ne}=\frac{{(1.5*{10}^{16})}^2}{5*{10}^{22}}$ = 0.45 $*{10}^{10}/m^3$

But when B is implanted in Si crystal then p type semiconductor is formed

N_h=N_A= $\frac{200}{{10}^6}*\left(5*{10}^{28}\right)=1*{10}^{25}/m^3$

Minority carriers created in p type is

N_e= $\frac{{ni}^2}{ne}=\frac{({1.5*{10}^{16})}^2}{1*{10}^{25}}=2.25*{10}^{27}/m^3$

Minority charge carriers holes in p type would contribute more in reverse saturatiom current.

This is a Long Answer Type Questions as classified in NCERT Exemplar

This is a Long Answer Type Questions as classified in NCERT Exemplar