Semiconductors and Electronic Devices MCQ Questions & Answers in Modern Physics | Physics

The current will flow through $${R_L}$$ when the diode is forward biased.

The most important characteristic of a vacuum diode is the plate characteristic which gives the relation between plate voltage and plate current for a given cathode temperature.
It may be noted from the plate characteristics that all the curves are coincident at low plate voltage where the negative space charge is most effective in limiting plate current. This low plate voltage region [region $$oa$$ in figure] is known as space charge limited region. In this region, the plate current increases as the plate voltage is increased, because more positive plate attracts electrons from the space charge at a greater rate. In the space charge limited region the plate current is given by the relation $$i = k{V^{\frac{3}{2}}}$$

i.e. $$i \propto {V^{\frac{3}{2}}}$$
where, $$k$$ is constant.

Since in forward biasing, negative of the battery is connected to $$n$$-type side, and the positive terminal of battery is connected to $$p$$-type side hence an electric field is created in direction opposite to the $$E$$ in the potential barrier region. Thus, depletion layer of $$p$$ - $$n$$ junction diode decreases.

$$V-I$$  characteristics of a solar cell is shown. Where, A represents open circuit voltage (i.e $$I = 0,V = emf$$   ) and $$B$$ shows short circuit voltage (i.e. $$I = I,V = 0$$  ).

In pure semiconductor electron-hole pair $$ = 7 \times {10^{15}}/{m^3}$$
$$\eqalign{ & {n_{{\text{initial}}}} = {n_h} + {n_e} = 14 \times {10^{15}}\,{\text{after doping donor Impurity}} \cr & {N_D} = \frac{{5 \times {{10}^{28}}}}{{{{10}^7}}} = 5 \times {10^{21}}\,\,{\text{and}}\,\,{n_e} = \frac{{{N_D}}}{2} \cr & = 2.5 \times {10^{21}} \cr & {\text{So,}}\,\,{n_{{\text{final}}}} = {n_h} + {n_e} \cr & \Rightarrow {n_{{\text{final}}}} \approx {n_e} \approx 2.5 \times {10^{21}}\,\,\left( {\because {n_e} > > {n_h}} \right) \cr & {\text{Factor}} = \frac{{{n_{{\text{final}}}} - {n_{{\text{initial}}}}}}{{{n_{{\text{initial}}}}}} \cr & = \frac{{2.5 \times {{10}^{21}} - 14 \times {{10}^{15}}}}{{14 \times {{10}^{15}}}} \approx \frac{{2.5 \times {{10}^{21}}}}{{14 \times {{10}^{15}}}} = 1.8 \times {10^5} \cr} $$

Peak value of rectified output voltage = peak value of input voltage - barrier voltage
$$ = 2 - 0.7 = 1.3\,V.$$

The current will flow through $${R_L}$$ when the diode is forward biased.

The output $$DC$$  component of half wave rectifier is given by
$${V_{{\text{output}}}} = \frac{{{\text{peak}}\,{\text{voltage}}}}{\pi } = \frac{{10}}{\pi }V$$

$$X = \overline{\overline {AB}} = A \cdot B\left( {{\text{i}}{\text{.e}}{\text{.}}\,{\text{AND}}\,{\text{gate}}} \right)$$
If the output of NAND gate is connected to the input of NOT gate (made from NAND gate by joining two inputs) from the given figure then we get back an AND gate.

AC input is applied across $$A$$ and $$C$$ and output is taken across $$BD.$$

When positive cycle is fed to $$AC,{D_1}$$   and $${D_4}$$ conduct, when negative cycle is fed to $$AC,{D_3}$$   and $${D_2}$$ conduct in the same direction. Output across $$BD$$  is thus full wave rectified.