NEET Practice Questions (MCQs) with Answers & Solutions

Integrated circuits are miniature electronic circuit produced within a single crystal of a semiconductors such as silicon. They contain a million or so transistors and resistors or capacitors. They are widely used in memory circuits, micro computers, pocket calculators and electronic watches on account of their low cost and bulk, reliability into specific regions of the semiconductor crystals.

  Voltage gain = $\beta$ $\times$impedance gain

               50 = $\beta$ $\times$$\frac{200}{100}$

                   $\Rightarrow$$\beta$= 25

  Also, power gain = ${\beta }^2$ $\times$ impedance gain

                           = (25)² $\times$$\frac{200}{100}$=1250

A transistor is mostly used in the active region of operation i.e., emitter-base junction is forward biased and the collector-base junction is reverse biased.

The base region must be very thin and lightly doped.

Key Idea    Only signals having wavelength less than threshold wavelength will be detected.

 $$\begin{gathered} \mathrm{Energy} \mathrm{E}=\mathrm{hv}=\mathrm{h}\frac{\mathrm{c}}{\mathrm{\lambda }} \\ \Rightarrow \mathrm{\lambda }=\frac{\mathrm{hc}}{\mathrm{E}} \end{gathered}$$

Substituting the value of h, c and E in the above equation 

                  $\mathrm{\lambda }=\frac{6.6\times {10}^{-34}\times 3\times {10}^8}{2.5\times 1.6\times {10}^{-19}}=5000 \stackrel{\circ }{\mathrm{A}}$

   As  $4000 \stackrel{\circ }{\mathrm{A}}<5000 \stackrel{\circ }{\mathrm{A}}$

 Signal of wavelength $4000 \stackrel{\circ }{\mathrm{A}}$ can be detected by the photodiode.

Neighbour distance of a body centred cubic cell $d=\frac{\sqrt{3}}{2}a, where a is the lattice parameter.$

$\Rightarrow 3.7=\frac{\sqrt{3}a}{2}$

or               $a=\frac{2\times 3.7}{\sqrt{3}}=4.3\stackrel{0}{A}$

Atomic radius is the distance between two nearest neighbours. For fcc crystal

Atomic radius = $\frac{1}{2\sqrt{2}}\times lattice parameter$

or               $r=\frac{a}{2\sqrt{2}}$

or             $r=\frac{3.6}{2\sqrt{2}} \stackrel{0}{A}=1.27 \stackrel{0}{A}$

(b) In forward biasing the P-end is connected to the positive terminal of the battery.

(a) In forward biased PN-junction, external voltage decreases the potential barrier, so current is maximum. While in reversed biased PN-junction, external voltage increases the potential barrier, so the current is very small

(b) The cut in voltage for silicon diode = 0.6 V

(b) Filter circuits are used to get smooth dc $\mathrm{\pi }$ -filter is the best filter.