Semiconductor Electronics

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Explanation- In elemental semiconductor, the band gap is such that the emission are in infrared region and not in visible region.

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Explanation- 

OR GATE gives the desired output.

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Explanation- as base current I_B = $\frac{V_{BB}-V_{BE}}{R_i}$

As R_i is increased I_B is decreased.

And collector current is I_c = I_B as I_B decreased I_c also decreased and the rading of voltmeter and ammeter also decreased.

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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.