Light Emitting Diode (L.E.D), Photodiode, Solar Cell Questions in English

(B) The correct answer is $B$.
$A$ solar cell is a photovoltaic device that converts light energy directly into electrical energy.
Unlike a photodiode or a varactor diode,which require an external bias voltage to function in their specific modes (reverse bias),a solar cell operates in the fourth quadrant of the $I-V$ characteristic curve without any external bias voltage.
It generates its own electromotive force when exposed to light.

(D) The correct answer is $D$. An $LED$ (Light Emitting Diode) is a heavily doped $p-n$ junction diode,not a lightly doped one. When it is forward biased,the recombination of electrons and holes releases energy in the form of photons (light). Options $A$,$B$,and $C$ are correct statements regarding semiconductor diodes.

(A) The energy gap of the $LED$ is given as $E_g = 2.4 \ eV$.
To convert this energy into Joules,we multiply by $1.6 \times 10^{-19} \ J/eV$:
$E = 2.4 \times 1.6 \times 10^{-19} \ J = 3.84 \times 10^{-19} \ J$.
The momentum $p$ of a photon is related to its energy $E$ by the formula $p = \frac{E}{c}$,where $c$ is the speed of light $(3 \times 10^8 \ m/s)$.
Substituting the values:
$p = \frac{3.84 \times 10^{-19}}{3 \times 10^8} \ kg \ ms^{-1}$.
$p = 1.28 \times 10^{-27} \ kg \ ms^{-1}$.
Therefore,the correct option is $A$.

(C) The energy of a photon corresponding to a wavelength $\lambda$ is given by $E = \frac{1240}{\lambda (nm)} eV$.
For $\lambda = 600 nm$,the energy of the incident photon is $E = \frac{1240}{600} \approx 2.07 eV$.
$A$ photodiode can detect light only if the energy of the incident photon is greater than or equal to the band gap energy $(E_{g})$ of the semiconductor material $(E \ge E_{g})$.
Comparing the photon energy $(2.07 eV)$ with the band gaps:
For $D_{1}$: $E_{g} = 2.5 eV$. Since $2.07 eV < 2.5 eV$,$D_{1}$ cannot detect this light.
For $D_{2}$: $E_{g} = 2 eV$. Since $2.07 eV > 2 eV$,$D_{2}$ can detect this light.
For $D_{3}$: $E_{g} = 3 eV$. Since $2.07 eV < 3 eV$,$D_{3}$ cannot detect this light.
Therefore,only $D_{2}$ will be able to detect the light of wavelength $600 nm$.

(C) The energy gap $E_g$ of a semiconductor is related to the wavelength $\lambda$ of the emitted light by the formula: $E_g = \frac{hc}{\lambda}$.
Given $E_g = 1.9 \ eV$. Converting this to Joules: $E_g = 1.9 \times 1.6 \times 10^{-19} \ J$.
Using the values $h = 6.6 \times 10^{-34} \ J \cdot s$ and $c = 3 \times 10^8 \ m/s$:
$\lambda = \frac{hc}{E_g} = \frac{6.6 \times 10^{-34} \times 3 \times 10^8}{1.9 \times 1.6 \times 10^{-19}} \ m$.
$\lambda = \frac{19.8 \times 10^{-26}}{3.04 \times 10^{-19}} \ m \approx 6.513 \times 10^{-7} \ m$.
Rounding to the nearest given option,the wavelength is $6.5 \times 10^{-7} \ m$.

(C) The energy of a photon is given by the formula $E = \frac{hc}{\lambda}$.
Given the band gap energy $E_g = 3 \ eV$.
We know that $hc \approx 1240 \ eV \cdot nm$.
Substituting the values,we get $\lambda = \frac{hc}{E_g} = \frac{1240 \ eV \cdot nm}{3 \ eV}$.
$\lambda \approx 413.33 \ nm$.
Therefore,the wavelength of light it can detect is nearly $413 \ nm$.

(C) Visible light corresponds to the wavelength range of approximately $380 \, nm$ to $750 \, nm$.
Using the relation $E = \frac{hc}{\lambda}$, where $h = 6.63 \times 10^{-34} \, J \cdot s$, $c = 3 \times 10^8 \, m/s$, and $1 \, eV = 1.6 \times 10^{-19} \, J$.
For the longest wavelength of visible light $(\lambda = 750 \, nm)$, the energy is $E = \frac{1240 \, eV \cdot nm}{750 \, nm} \approx 1.65 \, eV$.
In practice, to emit visible light efficiently, the semiconductor material used for the fabrication of visible LEDs must have a band gap of at least $1.8 \, eV$.

(C) Infrared $LED$'s are typically fabricated using Gallium Arsenide $(GaAs)$. While Gallium Arsenide Phosphide $(GaAsP)$ is commonly used for visible red light LEDs,pure Gallium Arsenide is the standard material for infrared emission due to its direct bandgap energy,which corresponds to the infrared spectrum.

(B) $photodiode$ is a semiconductor device that converts light into an electrical current. It is specifically designed to operate under reverse bias conditions and is widely used to detect optical signals.

(D) In a photodiode,the photocurrent is directly proportional to the intensity of the incident light. When photons with energy greater than the bandgap energy strike the depletion region,they create electron-hole pairs. The number of such pairs generated is proportional to the number of incident photons,which is defined by the intensity of the light.

(B) photodiode is a semiconductor device that converts light into electrical current.
When light with energy greater than the bandgap energy of the semiconductor falls on the $p-n$ junction,electron-hole pairs are generated.
The number of electron-hole pairs generated is directly proportional to the number of incident photons.
Since the intensity of light is defined as the number of photons incident per unit area per unit time,the generated photocurrent (and the resulting $emf$ in open-circuit conditions) is directly proportional to the intensity of the incident light.

(C) The visible region of the electromagnetic spectrum corresponds to wavelengths approximately between $400 \ nm$ and $700 \ nm$. The energy $E$ of a photon is given by $E = \frac{hc}{\lambda}$.
For $\lambda = 700 \ nm = 700 \times 10^{-9} \ m$:
$E_{\min} = \frac{6.6 \times 10^{-34} \times 3 \times 10^8}{700 \times 10^{-9} \times 1.6 \times 10^{-19}} \ eV \approx 1.77 \ eV$.
For $\lambda = 400 \ nm = 400 \times 10^{-9} \ m$:
$E_{\max} = \frac{6.6 \times 10^{-34} \times 3 \times 10^8}{400 \times 10^{-9} \times 1.6 \times 10^{-19}} \ eV \approx 3.1 \ eV$.
Thus,the energy band gap required for an $LED$ to emit visible light is in the range of $1.7 \ eV$ to $3.1 \ eV$. The correct option is $(c)$.

(D) The energy of a photon is given by $E = \frac{hc}{\lambda}$.
For a $p-n$ junction to detect a signal,the energy of the incident photon must be greater than or equal to the band gap energy $(E_g = 2.8 \ eV)$.
If $E < E_g$,the photon cannot excite an electron from the valence band to the conduction band,and thus cannot be detected.
Using $h = 6 \times 10^{-34} \ J \cdot s$ and $c = 3 \times 10^8 \ m/s$,we have $hc = 18 \times 10^{-26} \ J \cdot m$.
Converting $E_g$ to Joules: $2.8 \ eV = 2.8 \times 1.6 \times 10^{-19} \ J = 4.48 \times 10^{-19} \ J$.
The threshold wavelength $\lambda_0$ is given by $\lambda_0 = \frac{hc}{E_g} = \frac{18 \times 10^{-26}}{4.48 \times 10^{-19}} \approx 4.017 \times 10^{-7} \ m = 401.7 \ \text{nm}$.
Any wavelength $\lambda > \lambda_0$ will have energy $E < E_g$ and will not be detected.
Among the given options,$600 \ nm$ is the only wavelength greater than $401.7 \ nm$.

(B) The energy of the emitted photon is given by the relation $E = \frac{hc}{\lambda}$.
Given the energy gap $E = 1.9 \ eV$.
We know that $hc \approx 1240 \ eV \cdot nm$.
Therefore,the wavelength $\lambda = \frac{hc}{E} = \frac{1240 \ eV \cdot nm}{1.9 \ eV} \approx 652.6 \ nm$.
The wavelength range for red light is approximately $620 \ nm$ to $750 \ nm$.
Since $652.6 \ nm$ falls within this range,the color of the light emitted by the $LED$ is red.

(D) An $LED$ (Light Emitting Diode) is a $p-n$ junction diode that emits light when forward biased.
Key characteristics of LEDs include:
$1$. They operate at low voltages compared to incandescent bulbs.
$2$. They do not require any warm-up time.
$3$. They have very fast on-off switching capabilities,making them ideal for high-speed communication and display technologies.
$4$. The bandwidth of light emitted is typically narrow,not $4000 \ Å - 7000 \ Å$ (which covers the entire visible spectrum).
Therefore,the correct statement is that they have fast on-off switching.

(C) In a photodiode,the reverse saturation current is primarily due to the flow of minority charge carriers. When light is incident on the photodiode,it generates electron-hole pairs,which increases the number of minority charge carriers. Because the initial number of minority carriers is very small,even a small amount of incident light causes a significant fractional change in their concentration. This leads to a noticeable change in the reverse current,making the photodiode highly sensitive to light. In contrast,in forward bias,the current is dominated by majority carriers,which are already present in large numbers,making the fractional change due to incident light negligible.

(A) The energy band gap of $Si$ is approximately $1.14 \ eV$ and for $GaAs$ it is $1.42 \ eV$.
Solar cells are designed to absorb photons from the solar spectrum. The solar radiation spectrum has a peak intensity around $1.5 \ eV$ to $2.0 \ eV$.
For a material to be an efficient solar cell,its band gap should be close to the energy range of the maximum solar irradiance (typically $1.0 \ eV$ to $1.8 \ eV$).
Since the band gaps of $Si$ $(1.14 \ eV)$ and $GaAs$ $(1.42 \ eV)$ fall within this optimal range,they are preferred materials.
The reason $(R)$ states that the band gaps are 'much below' the energy level of maximum solar irradiance,which is incorrect because they are actually very close to the optimal range.
Therefore,the assertion $(A)$ is correct,but the reason $(R)$ is incorrect.

(A) Given: Maximum power of $LED$ $P_{max} = 2 \text{ mW}$,Input voltage $V_{in} = 5 \text{ V}$,Resistance $R = 1 \text{ k}\Omega$,and current through $R$ is $I_R = 0.5 \text{ mA}$.
$1$. The voltage across the $LED$ $(V_{LED})$ is the same as the voltage across the resistor $R$ because they are in parallel. Thus,$V_{LED} = I_R \times R = 0.5 \text{ mA} \times 1 \text{ k}\Omega = 0.5 \text{ V}$.
$2$. The maximum current allowed through the $LED$ is $I_{LED} = \frac{P_{max}}{V_{LED}} = \frac{2 \text{ mW}}{0.5 \text{ V}} = 4 \text{ mA}$.
$3$. The total current flowing through the series resistor $R_S$ is $I_S = I_{LED} + I_R = 4 \text{ mA} + 0.5 \text{ mA} = 4.5 \text{ mA}$.
$4$. The voltage across $R_S$ is $V_{R_S} = V_{in} - V_{LED} = 5 \text{ V} - 0.5 \text{ V} = 4.5 \text{ V}$.
$5$. The minimum resistance $R_S$ is $R_S = \frac{V_{R_S}}{I_S} = \frac{4.5 \text{ V}}{4.5 \text{ mA}} = 1 \text{ k}\Omega$. Since $1 \text{ k}\Omega$ is not among the options,and the question asks for the minimum value to ensure safety,we re-evaluate the circuit. If the diode is considered ideal with a forward voltage drop of $0.5 \text{ V}$,the calculation holds. Given the options,there might be a typo in the provided options or the circuit parameters. Based on standard competitive exam patterns for this specific problem,$1 \text{ k}\Omega$ is the calculated value.