e.m.f of the source , E = 1.5 
Voltage drop across the diode , V_d = 0.5 V
Maximum power rating of the diode 
I = (P / Vd)
I = 0.2A
Potential drop across resistance R
V = E - V_d
= 1 V
R = V/1 = 1/0.2 = 5 \(\Omega \)

Given, wavelength of light,
A = 6000\(\overset{o}{A}\) = 6000 x 10^-10 m
\(\therefore\) Energy of the light photon,
\(E=\frac{h c}{\lambda}=\frac{6.6 \times 10^{-34} \times 3 \times 10^{8}}{6000 \times 10^{-10} \times 1.6 \times 10^{-19}} \mathrm{eV}=2.06 \mathrm{eV}\)
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.

(i) It is because reverse break down voltage of LED is very low i.e., nearly 5 V.
(ii) Because solar cell can work with any light whose photon energy is more than the band gap energy.
(ill) The heavy doping of p and n sides of p-n junction, makes the deplection region very thin, hence for a small reverse bias voltage, electric field is very high.

Principle A solar cell works on the principle of photo voltaic effect according to which when light photons of energy greater than energy band gap of a semiconductor are incident on p-n junction of that semiconductor, electron-hole pairs are generated which give rise to an emf. Thus, working principle of a solar cell is same as that of a photodiode. However, no bias is applied in a solar cell and the junction area is kept much larger so that more solar radiation may be incident.

Generation of emf: Three basic processes are involved in the generation of emf by a solar cell when solar radiations are incident on it. These are
(a) the generation of electron-hole pairs close to the junction due to incidence of light with photo energy hv ≥ b.
(b) the separation of electrons and holes due to the electric field of the depletion region. So, electrons are swept to n-side and holes to p-side.
(c) the electrons reaching the 11-side are collected by the front contact and holes reaching p -side are collected by the back contact. Thus, p-side becomes positive and n-side become negative giving rise 1Q a photovoltage. When an external load RL is connected as shown in figure, a photocurrent IL begins to flow through the load.

Two processes that takes place during the formation of p-n junction are diffusion and drift of charge carriers. In an n-type semiconductor, the concentration of electrons is more than that of holes. Similarly, in a p-type semiconductor, the concentration of holes is more than that of electrons. Formation of depletion region during formation of p-n junction and due to the concentration gradient across p and n-sides, holes diffuse from p-side to n-side
(p ⟶ n) and electrons diffuse from n-side to p-side (n ⟶ p). The diffused charge carriers combine with their counterparts in the immediate vicinity of the junction and neutralise.

This sets up potential difference across the junction and an internal electric field E; directed from n-side to p-side. The equilibrium is established when the field E; becomes strong enough to stop further diffusion of the majority charge carriers (however, it helps the minority charge carriers to drift across the junction). The region on either side of the junction which becomes depleted (free) from the mobile charge carriers is called depletion region or depletion layer. The potential difference developed across the depletion region is called the potential barrier.

\(\sigma = \ n_{ e } \ \mu _{ e }=10^{ 17 }\times 1.6\times 10^{ -19 }\times 3800\)
\(60.8\ S\ cm^{ -1 }\)
\(\\ \frac { \sigma _{ d } }{ \sigma _{ i } } =\frac { n_{ d } }{ n_{ i } } =\frac { 10^{ 17 } }{ 2.5\times 10^{ 13 } } =4\times 10^{ 3 }:1\)

Here, \(E=53meV=53\times 10^{ -3 } \ eV\)
\(=53\times 10^{ -3 }\times 1.6\times 10^{ -19 }J\)
\(=53\times 1.6\times 10^{ -22 }J\)
Max. Wavelength, \(\lambda _{ max }=\frac { hc }{ E } \)
\(=\frac { \left( 6.63\times 10^{ -34 } \right) \times (3\times 10^{ 8 }) }{ 53\times 1.6\times 10^{ -22 } } =2.35\times 10^{ -5 }m\)

\({ V }_{ c }={ V }_{ s }+{ V }_{ F }={ V }_{ s }+\frac { { V }_{ 0 } }{ 100 } \)
or
\({ V }_{ c }=0.2+\frac { -10 }{ 100 } \)
\(=0.2-0.1=0.1\)
\({ A }_{ 0 }=\frac { { V }_{ 0 } }{ { V }_{ c } } =\frac { -10 }{ 0.1 } =\frac { -10 }{ 0.1 } =-100\)
\({ A }_{ 0 }=\frac { -10 }{ 0.2 } =-50\)

(a), (b), (d)
Since ripple factor with capacitor filter is:
\(\gamma =\frac { 1 }{ \sqrt [ 4 ]{ 3R_{ L } } { C }v } ,\)
So \(\gamma \propto \frac { 1 }{ R_{ L } } ,\ \gamma \propto \frac { 1 }{ C } ,\gamma \propto \frac { 1 }{ v } \)
So to reduce \(\gamma \) , Increase \(R_{ L }\) , C and V.

(a), (c) Knowledge based question

Voltage gain of the first amplifier, V₁ = 10
Voltage gain of the second amplifier, V₂ = 20
Input signal voltage, V_i = 0.01 V
Output AC signal voltage = V_o
The total voltage gain of a two-stage cascaded amplifier is given by the product of voltage gains of both the stages, i.e.,
V = V₁ x V₂
= 10 x 20 = 200
We have the relation:
\(V=\frac{V_{0}}{V_{1}}\)
V₀ = V x V_i
= 200 x 0.01 = 2 V
Therefore, the output AC signal of the given amplifier is 2 V.