Semiconductor Electronics

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,

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

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

( $R_C+R_E$ )=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

Explanation- a) In V- graph of condition (i), a reverse characteristics is shown in fig. (c). Here A is connected to n - side of p-n junction and B is connected to p -side of p-n junction I with a resistance in series.

(b) In V

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

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

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

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

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

Minority carriers created in p type is

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

Mi

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

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

Explanation- When the input voltage is equal to or less than 5 V, diode will be revers biased. It will offer high resistance in comparison to resistance ( ) R in series. Now, diode appears in open circuit. The input waveform is

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

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

Explanation- voltage across R_B= 10V

Resistance R_B= 400kohm

V_BE= 0, V_CE= 0, R_c=3kohm

I_B= voltage /R_B= 10/400 $\times$  10³= 25 $\times$ 10^-6A

Voltage across R_c= 10V

I_c=voltage/R_c= 10/3 $\times$ 10³= 3.33 $\times$ 10^-3

$\beta =\frac{I_c}{I_B}=\frac{3.33\times {10}^{-3}}{25\times {10}^{-6}}=133$