Physics NCERT Exemplar Solutions Class 12th Chapter Fourteen

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Explanation- E=hc/ $\lambda$ = $\frac{6.6*{10}^{-34}*3*{10}^8}{6000*{10}^{-10}*1.6*{10}^{-19}}$ = 2.06eV

The incident radiation which is detected by the photodiode having energy should be greater

Than the band-gap. So, it is only valid for diode D2.Then, diode D2 will detect this radiation.

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Explanation- In CE transistor amplifier, the power gain is very high.

In this circuit, the extra power required for amplified output is obtained from DC source.

Thus, the circuit used does not violet the law of conservation.

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Explanation- as we know total voltage amplification = $\frac{outputsignalvoltage}{inputsignalvoltge}$

Av_x= 10, Av_y = 20, Av_z = 30

$?$ v_i = 1mV= 10^-3V

So according to formula

 = (Av_x ) (Av_y ) (Av_z)

$?$ v_o =  (Av_x ) (Av_y ) (Av_z) $?$ v_i

  = 10^{-3 $*20*30*10$} =6V

So in first case output cannot be greater than 6V but in 2^nd case it will work

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Explanation- diode only works in forward biased. So output waveform is

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Explanation- We cannot measure the potential barrier across a p-n junction by a voltmeter because the resistance of voltmeter is very high as compared to the junction resistance.

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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.

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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.

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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

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