Semiconductor Electronics: Materials, Devices and Simple Circuits Test 64

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Semiconductor Electronics: Materials, Devices and Simple Circuits Test 64

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Question 1
1 / -0

Two junction diodes one of germanium $$(Ge)$$ and other of silicon $$(Si)$$ are connected as shown in figure to a battery of emf $$12\ v$$ and a load resistance $$10\ k\ \Omega$$. The germanium diode conducts at $$0.3\ V$$ and silicon diode at $$0.7\ V$$. When a current flows in the circuit, then the potential of terminal $$Y$$ will be:

A
$$12\ V$$

B
$$11\ V$$

C
$$11.3\ V$$

D
$$11.7\ V$$

Solution

Ge conducts at $$0.3\ V$$ and $$Si$$ Conducts at $$0.7\ V$$. Both $$Ge$$ and $$Si$$ are connected in parallel. When current begins to flow, then the potential difference remains at $$0.3\ V$$, so no current flows through Si-diode.
$$\therefore$$ Potential difference across resistive load $$= 12 - 0.3$$
$$= 11.7\ V$$
$$\therefore$$ Potential of $$Y = 11.7\ V$$.
Question 2
1 / -0

With the help of the arrangement shown in figure which of the following gates could be realised?

A
OR

B
NAND

C
AND

D
NOR

Solution
Question 3
1 / -0

Mobilities of electrons and holes in a sample of intrinsic germanium at room temperature are $$0.36m^2V^{-1}s^{-1}$$ and $$0.17m^2V^{-1}s^{-1}$$. The electron and hole densities are each equal to $$2.5\times 10^{19}m^3$$. The electrical conductivity of germanium is

A
$$4.24Sm^{-1}$$

B
$$2.12Sm^{-1}$$

C
$$1.09Sm^{-1}$$

D
$$0.47Sm^{-1}$$

Solution

Given in question,
Mobilities of electrons$$\mu_e=0.36\times m^2V^{-1}s^{-1}$$
Mobilities o holes$$\mu_h=0.17\times m^2V^{-1}s^{-1}$$
densities of electron$$=$$ densities of holes$$=2.5\times10^{19}m^{-3}$$
As we know, conductivity,
$$\sigma = \dfrac{1}{p}=e({\mu}_e n_e+{\mu}_n n_n)$$
$$=1.6\times 10^{-19}[0.36\times2.5\times10^{19}+0.17\times2.5\times 10^{19})]$$
$$=2.12 Sm^{-1}$$
So the electrical conductivies of germanium is$$2.12Sm^{-1}$$
Question 4
1 / -0

At absolute zero of temperature, the electrical conductivity of a pure semiconductor is

A
Zero

B
About $$10^3 (\Omega m)^{-1}$$

C
About $$10^7 (\Omega m)^{-1}$$

D
Infinity

Solution
Question 5
1 / -0

In a diode detector, output circuit consist of $$R=1M\Omega \quad and\quad C=1pF$$. Calculate the carrier frequency it can detect.

A
$$3 \times 10^ 6Hz$$

B
$$1 \times 10^ 6Hz$$

C
$$30 \times 10^ 6Hz$$

D
$$4 \times 10^ 6Hz$$

Solution
Question 6
1 / -0

If the lattice constant of this semiconductor is decreased, then which of the following is correct?

A
All $$E_{c}, E_{g}, E_{v}$$ increase

B
$$ E_{c}$$ and $$ E_{v}$$ increases but $$ E_{g}$$ decrease

C
$$ E_{c}$$ and $$ E_{v}$$ decreases but $$ E_{g}$$ increase

D
All $$E_{c}, E_{g}, E_{v}$$ decrease

Solution
Question 7
1 / -0

The arrangement of $$NAND$$ gates shown below effectively work as

A
$$AND$$ gate

B
$$OR$$ gate

C
$$NAND$$ gate

D
$$NOR$$ gate

Solution
Question 8
1 / -0

A semiconductor X is made by doping a germanium crystal with arsenic $$(Z=33)$$. A second semiconductor Y is made by doping germanium with indium$$(Z=49)$$. The two are joined end to end and connected to a battery as shown. Which of the following statements is correct?

A
X is P-type, Y is N-type and the junction is forward biased

B
X is N-type, Y is P-type and the junction is forward biased

C
X is P-type, Y is N-type and the junction is reverse biased

D
X is N-type, Y is P-type and the junction is reverse biased

Solution

X is N-type, Y is P-type and the junction is reverse biased
Question 9
1 / -0

The electrical conductivity of a semiconductor increases when electromagnetic radiation of wavelength shorter than 2480nm is incident on it. The band gap(in eV) for the semiconductor is

A
0.9

B
0.7

C
0.5

D
0.1

Solution
Question 10
1 / -0

There photo dlodes $$D_{1}, D_{2}$$ and $$D_{3}$$ are made of semiconductor having band gap $$2.5\ eV, 2\ eV$$ and $$3\ eV$$ respectively. Which one will be able to detect light of wavelength $$6000\ A^o$$?

A
$$D_{1}$$

B
$$D_{2}$$

C
$$D_{3}$$

D
$$D_{1}$$ and $$D_{2}$$ both

Solution

Given that,
Band gap of $${{D}_{1}}=2.5\,eV$$
Band gap of $${{D}_{2}}=2\,eV$$
Band gap of $${{D}_{3}}=3\,eV$$
Wave length $$\lambda =6000\overset{\circ }{\mathop{A}}\,$$
Now, the wave length for $$2.5\ eV$$
$$ {{\lambda }_{1}}=\dfrac{hc}{E} $$
$$ {{\lambda }_{1}}=\dfrac{6.6\times {{10}^{-34}}\times 3\times {{10}^{8}}}{2.5\times 1.6\times {{10}^{-19}}} $$
$$ {{\lambda }_{1}}=4.95\times {{10}^{-7}}\,m $$
Now, for $$2\ eV$$
$$ {{\lambda }_{2}}=\dfrac{hc}{E} $$
$$ {{\lambda }_{2}}=\dfrac{6.6\times {{10}^{-34}}\times 3\times {{10}^{8}}}{2\times 1.6\times {{10}^{-19}}} $$
$$ {{\lambda }_{2}}=6.2\times {{10}^{-7}}\,m $$
Now, for $$3\ eV$$
$$ {{\lambda }_{3}}=\dfrac{hc}{E} $$
$$ {{\lambda }_{3}}=\dfrac{6.6\times {{10}^{-34}}\times 3\times {{10}^{8}}}{3\times 1.6\times {{10}^{-19}}} $$
$$ {{\lambda }_{3}}=4.13\times {{10}^{-7}}\,m $$
For detection of optical signal the wave length of incident energy radiation must be greater.
So, only $${{D}_{2}}$$can detect the radiation