Q: Suppose a pure Si crystal has \$5\times 10^{28}\$ atoms \$m^{-3}\$. it is doped by 1 ppm concentration of pentavalent As. Calculate the number of electrons and holes. Given that \$n_{1}=1.5\times 10^{16}m^{-3}\$.

1 atom of Si doped out of \$10^{6}\$ atoms (1 ppm of As)\$ \Rightarrow \$ In \$ 5 \times 10^{28}\$ atoms net doped =\$\frac{5\times10^{28}}{10^{6}} = 5\times 10^{22} \$ atoms1 As atom creates \$1 e^{\circleddash }\$ excenso, no. of excen electron \$ = 5 \times 10^{22} = n_{e}\$also, \$ n_{e}n_{n}=n_{1}^{2}\$\$ \Rightarrow n_{n}=\frac{n_{1}^{2}}{n_{e}} = \frac{(1.5\times10^{16})^{2}}{5\times 10^{22}} = 4.5\times 10^{9}\$\$ \therefore \$ No. of holes formed \$ = 4.5 \times 10^{9}.\$as \$ n_{e} > n_{n} \rightarrow \$ It is a N - type semiconductor formed.

Q: What is the depletion region in pn junction?

Depletion region or depletion layer is a region in a P-N junction diode where no mobile charge carriers are present. Depletion layer acts like a barrier that opposes the flow of electrons from n-side and holes from p-side

Q: Assuming that the two diodes \$D_1\$ and \$D_2\$ used in the electric circuit shown in the figure are ideal, find out the value of the current flowing through \$1\Omega\$ resistor.

\$D_2\$ is in reverse bias so no current through it.Now \$I = \dfrac{\varepsilon}{R + r} = \dfrac{6}{2 + 1} = 2A\$

Q: Determine the current through zener diode for the circuit shown in figure is: (Given: zener diode break down voltage \$V_z=5.6V\$)

For zener brak down potential difference across \$800\Omega \$ resistor will be \$5.6V\$4\$V_z=5.6V\$\$i_2=\dfrac{V_z}{800}=\dfrac{5.6}{800}=7mA\$\$\Delta V\$ across \$200\Omega =9-5.6=3.4V\$\$i_1=\dfrac{3.4}{200}=17mA\$\$i_1=1_2+i_z\$\$i_z=17mA-7mA=10mA\$

Question 1

Explain the working of P-N junction diode in forward and reverse biased mode.

Accepted Answer

A P-N junction diode is said to be forward biased when the positive terminal of a cell or battery is connected to the p-side of the junction and the negative terminal to the $n$ side.When diode is forward-biased the depletion region narrows and consequently, the potential barrier is lowered. This causes the majority charge carriers of each region to cross into the other region. The electrons travel from the n-side to the p-side and go to the positive terminal of the battery. The holes that travel from the p-side to the n-side combine with the electrons injected into the n-region from the negative terminal of the battery. This way the diode conducts when forward-biased.Reverse-biased pn junction diodeA pn-junction diode is said to be reverse biased when the positive terminal of a cell or battery is connected to the n-side of the junction and the negative terminal to the p-side.When reverse biased, the depletion region widens and the potential barrier is increased. The polarity of the battery extracts the majority charge carriers of each region. The holes in the p-region from the electrons injected into the p-region from the negative terminal of the battery. The electrons in the n-region go to the positive terminal of the battery. This way, the majority charge carrier concentration in each region decreases against the equilibrium values and the reverse biased junction diode has a high resistance. Thus, the diffusion current across the junction becomes zero. Thus, the diode does not conduct when reverse biased and is said to be in a quiescent or non-conducting state i.e., it acts as an open switch.

Question 2

Suppose a pure Si crystal has $5\times 10^{28}$ atoms $m^{-3}$. it is doped by 1 ppm concentration of pentavalent As. Calculate the number of electrons and  holes. Given that $n_{1}=1.5\times 10^{16}m^{-3}$.

Accepted Answer

1 atom of Si doped out of $10^{6}$ atoms (1 ppm of As)$ \Rightarrow $ In $ 5 \times 10^{28}$ atoms net doped =$\frac{5\times10^{28}}{10^{6}} = 5\times 10^{22} $ atoms1 As atom creates $1 e^{\circleddash }$ excenso, no. of excen electron $ = 5 \times 10^{22}  = n_{e}$also, $ n_{e}n_{n}=n_{1}^{2}$$ \Rightarrow n_{n}=\frac{n_{1}^{2}}{n_{e}} = \frac{(1.5\times10^{16})^{2}}{5\times 10^{22}} = 4.5\times 10^{9}$$ \therefore $ No. of holes formed $ = 4.5 \times 10^{9}.$as $ n_{e} > n_{n} \rightarrow $ It is a N - type semiconductor formed.

Question 3

Why is a photodiode is used in reverse bias?

Accepted Answer

When a photo- diode is reverse biased, the width of depletion layer increases as compared to forward biased and a small reverse current (dark current) flows through the diode. Now, when the light is incident on the junction, electron-hole pairs are generated in depletion layer in a big amount (due to broad depletion layer) and these charge carriers can easily cross the barrier, hence contribute to current across the diode. We can say that in reverse bias, diode changes the incident light to current, more significantly due to broad depletion layer i.e. photocurrent is significant in reverse bias as compared to the forward bias current.

Question 4

What is the depletion region in pn junction?

Accepted Answer

Depletion region or depletion layer is a region in&#10;a P-N junction diode where no mobile charge carriers are present. Depletion&#10;layer acts like a barrier that opposes the flow of electrons from n-side&#10;and holes from p-side

Question 5

A p-n photodiode is made of a material with a band gap of $2.0   eV$. The minimum frequency of the radiation that can be absorbed by the material is nearly :

Accepted Answer

p-n photodiode is a semiconductor diode that produces a significant current when illuminated with sufficient frequency of radiation. For absorbing minimum frequency of radiation the condition is given below,Energy of radiation =  band gap energyi.e., $hv=2.0eV$or  $v=\dfrac { 2.0\times 1.6\times { 10 }^{ -19 } }{ 6.6\times { 10 }^{ -34 } } \approx 5\times { 10 }^{ 14 }Hz$

Question 6

Write down 5 difference between N-type and P-type semiconductors.

Accepted Answer

Difference between the P and N type semiconductors Sl No P-type Semi conductor N-type semi conductor 1P-type semiconductor is formed due to the dopping of III group elements i.e. Boron, Aluminium, Thallium.N-type semi conductor is formed due to dopping of Nitrogen, Phosporus, Arsenic, Antimony, Bismoth. 2These are also known as Trivalent semi conductors.These are also known pentavalent semiconductor. 3P-type semiconductors is positive type semiconductor it means it deficiency of 1 electron is required.N-type semiconductor is negative type semi-conductor it means excess of 1 electron is required. 4In P-type semiconductor majority charge carries are holes and minority charge carries are electrons.In N-type semiconductor majority charge carries are electrons and minority charge carries are hole. 5A hole indicates a missing electron. In this no. of holes is more than the no. of electrons.In N-type semiconductor the no. of holes is less than the no. of free electron.

Question 7

Assuming that the two diodes $D_1$ and $D_2$ used in the electric circuit shown in the figure are ideal, find out the value of the current flowing through $1\Omega$ resistor.

Accepted Answer

$D_2$ is in reverse bias so no current through it.Now $I = \dfrac{\varepsilon}{R + r} = \dfrac{6}{2 + 1} = 2A$

Question 8

Determine the current through zener diode for the circuit shown in figure is: (Given: zener diode break down voltage $V_z=5.6V$)

Accepted Answer

For zener brak down potential difference across $800\Omega $  resistor will be $5.6V$4$V_z=5.6V$$i_2=\dfrac{V_z}{800}=\dfrac{5.6}{800}=7mA$$\Delta V$ across $200\Omega =9-5.6=3.4V$$i_1=\dfrac{3.4}{200}=17mA$$i_1=1_2+i_z$$i_z=17mA-7mA=10mA$

Question 9

In a p-n junction diode, the current I can be expressed as $I = I_0\,exp\, \left( \dfrac{eV}{2K_BT} - 1 \right)$, where $I_0$ is called the reverse saturation current, $V$ is the voltage across the diode and is positive for forward bias and negative for reverse bias, and $I$ is the current through the diode, $k_B$ is the Boltzmann constant $(8.6 \times 10^{-5} eV/K)$ and $T$ is the absolute temperature. If for a given diode $I_o = 5 \times 10^{-12}$$ A$ and $T = 300\ K$, then(a) What will be the forward current at a forward voltage of $0.6\ V$?(b) What will be the increase in the current if the voltage across the diode is increased to $0.7\ V$?(c) What is the dynamic resistance?(d) What will be the current if reverse bias voltage changes from $1\ V$ to $2\ V$?

Accepted Answer

(a)In a p-n junction diode, the expression for current is $I=I_0exp[\dfrac{eV}{2k_BT}-1]$where $I_0$ is the reverse saturation current = $5\times 10^{-12}A$$T=$Absolute temperature = $300\ K$$V$ is the voltage across diode.For forward voltage $V = 0.6\ V$ $I=5\times 10^{-12}\times exp[\dfrac{1.6\times 10^{-19}\times 0.6}{1.376\times 10^{-23}\times 300}-1] = 0.0256\ A$&#10;&#9;&#10;&#9;&#10;&#9;&#10;&#10;&#10;(b)&#10;&#10;In&#10;a p-n junction diode, the expression for current is&#10;$I=I_0exp[\dfrac{eV}{2k_BT}-1]$&#10;&#10;where&#10;$I_0$ is the reverse saturation current = $5\times 10^{-12}A$&#10;&#10;$T=$Absolute&#10;temperature=$300\ K$&#10;&#10;$V$ is the voltage across diode.&#10;&#10;For&#10;forward voltage, $V = 0.7\ V$ &#10;&#10;$I=5\times&#10;10^{-12}\times exp[\dfrac{1.6\times 10^{-19}\times 0.7}{1.376\times&#10;10^{-23}\times 300}-1]=1.257\ A$&#10;&#10;Hence, increase in current = $1.257\ A-0.0256\ A=1.23\ A$&#10;&#10;(c)&#10;&#10;In&#10;a p-n junction diode, the expression for current is&#10;$I=I_0exp[\dfrac{eV}{2k_BT}-1]$&#10;&#10;where&#10;$I_0$ is the reverse saturation current = $5\times 10^{-12}A$&#10;&#10;$T=$Absolute&#10;temperature=$300\ K$&#10;&#10;$V$ is the voltage across diode.&#10;&#10;For&#10;forward voltage $V = 0.6\ V$ &#10;&#10;$I=5\times&#10;10^{-12}\times exp[\dfrac{1.6\times 10^{-19}\times 0.6}{1.376\times&#10;10^{-23}\times 300}-1]=0.0256\ A$&#10;&#10;For&#10;forward voltage $V = 0.7\ V$ &#10;&#10;$I=5\times&#10;10^{-12}\times exp[\dfrac{1.6\times 10^{-19}\times 0.7}{1.376\times&#10;10^{-23}\times 300}-1]=1.257\ A$&#10;&#10;Hence&#10;increase in current=$1.257\ A-0.0256\ A = 1.23\ A$&#10;&#10;Dynamic&#10;resistance = Change in voltage /Change in current&#10;&#10;$=\dfrac{0.7-0.6}{1.23}=0.081\Omega$&#10;&#10;(d)&#10;&#10;If&#10;reverse bias voltage changes from $1\ V$ to $2\ V$, the current will be&#10;almost constant, because dynamic resistance in the reverse bias is&#10;infinite.

Question 10

When a p-n junction diode is reverse biased

Accepted Answer

When a p-n junction is reversed biased, the negative terminal of the battery attracts the free holes in the p-type towards itself whereas the positive terminal attracts the free electrons in n-type towards itself. Hence, the electrons and the holes move away from the junction which results in increasing of the depletion width as well as the potential barrier.