Semiconductor Electronics: Materials, Devices and Simple Circuits Test 48

Result

Semiconductor Electronics: Materials, Devices and Simple Circuits Test 48

Score
-
out of -
Rank
-
out of -

TIME Taken - -

SHARING IS CARING

If our Website helped you a little, then kindly spread our voice using Social Networks. Spread our word to your readers, friends, teachers, students & all those close ones who deserve to know what you know now.

Weekly Quiz Competition

Question 1
1 / -0

The material used in solar cells contains

A
$$C_s$$

B
$$Si$$

C
$$Sn$$

D
None of the above

Solution

The first-generation cells are known as the conventional, water based or traditional cells which are made up of crystalline silicon. And the materials like monocrystalline silicon and polysilicon are used in commercial predominant technology. So, silicon is used as the material in solar cells.
Question 2
1 / -0

When a diode is heavily doped the

A
Zener voltage will be low

B
Avalanche voltage will be high

C
Depletion region will be thin

D
Leakage current will be low

Solution

The value of reverse voltage at which this occurs is controlled by the amount ot doping of the diode. A heavily doped diode has a low Zener breakdown voltage, while a lightly doped diode has a high Zener breakdown voltage. At voltages above approximately 8V, the predominant mechanism is the avalanche breakdown.
Question 3
1 / -0

The number of particles crossing unit area perpendicular to X-axis in unit time is given by $$ \dfrac { D\left( n_ 2-n_ 1 \right) }{ \left( x_ 2-x_ 1 \right) } $$ where $$ n_1 and n_2 $$ are number of particle per unit volume respectively at and $$ x_2 $$ The dimension of the diffusion constant D are

A
$$ \left[ LT^ { -1} \right] $$

B
$$ \left[ L^2 T^ { -1} \right] $$

C
[LT]

D
$$ [L^{-1} T ] $$

Solution

$$\begin{array}{l} \left[ n \right] ={ \left[ L \right] ^{ -2 } }\times { \left[ T \right] ^{ -1 } } \\ \left[ { { n_{ 2 } }-{ n_{ 1 } } } \right] ={ \left[ L \right] ^{ -3 } } \\ \left[ { { x_{ 2 } }-{ x_{ 1 } } } \right] =\left[ L \right] \\ \left[ n \right] =\left[ d \right] \times \left[ { { n_{ 2 } }-{ n_{ 1 } } } \right] /\left[ { { x_{ 2 } }-{ x_{ 1 } } } \right] \\ put\, \, the\, \, above\, \, values\, \, in:- \\ { \left[ L \right] ^{ -2 } }\times { \left[ T \right] ^{ -1 } }=\left[ d \right] \times { \left[ L \right] ^{ -3 } }/\left[ L \right] \\ \left[ d \right] ={ \left[ L \right] ^{ -2 } }\times { \left[ T \right] ^{ -1 } }\times \left[ L \right] /{ \left[ L \right] ^{ -3 } } \\ \left[ d \right] ={ \left[ L \right] ^{ 2+1+3 } }\times { \left[ T \right] ^{ -1 } } \\ \left[ d \right] ={ \left[ L \right] ^{ 2 } }\times { \left[ T \right] ^{ -1 } }=\left[ { { L^{ 2 } }{ T^{ -1 } } } \right] \end{array}$$
Hence,
option $$(B)$$ is correct answer.
Question 4
1 / -0

A light emitting diode (LED) has a voltage drop of 2 V across it and passes a current of 10 mA. When it operates with a 6 V battery through a limiting resistor R, the value of R is

A
200 $$\Omega$$

B
400 $$\Omega$$

C
40 $$k\Omega$$

D
4 $$k\Omega$$

Solution

Light emitting diode is a forward biased P-N junction which emits light.

The voltage across it = 2 v

Battery voltage = 6 v

Hence total voltage in the circuit = V = 6 - 2 = 4 v

since a LED has 0 resistance in the forward biased region, total resistance in the circuit

= 0 + r = r

current I = 10 mA = 10/1000 A = 0.01 A

we know that,

R = V/I

=> r = 4/0.01 = 400Ω

Hence the value of the limiting resistor is 400Ω
Question 5
1 / -0

The mobility of hole in a semiconductor depend on

A
Electric field

B
Potential difference

C
current

D
Mass

Solution

Semiconductor mobility depends on the impurity concentrations including donor and acceptor concentrations, defect concentration, temperature, and electron and hole concentrations. Mobility describes the relation between drift velocity of electrons or holes and an applied electric field in a solid.
Question 6
1 / -0

If potential $$V=100\pm0.5$$Volt and current $$I=10\pm0.2$$ amp are given to us, then what will be the value of resistance

A
$$10\pm0.7$$ohm

B
$$5\pm2$$ohm

C
$$0.1\pm0.2$$ohm

D
none of these

Solution
Question 7
1 / -0

For detecting intensity of light we use

A
Photodiode in forward bias

B
Photodiode in reverse bias

C
LED in forward bias

D
LED in reverse bias

Solution
Question 8
1 / -0

In PN-junction diode the reverse saturation current is $$10 ^ { - 5 }$$ amp at $$27 ^ { \circ } C$$. The forward current for a voltage of 0.2 volt is

A
$$2037.6 \times 10 ^ { - 3 } \mathrm { amp }$$

B
$$203.76 \times 10 ^ { - 3 } \mathrm { amp }$$

C
$$20.376 \times 10 ^ { - 3 }amp$$

D
$$2.0376 \times 10 ^ { 3 }amp$$

Solution

The forward current $$\large i=\;{i_s}\left( {{e^{ev/kT}} - 1} \right)$$
$$={10^{ - 5}}\left[ {{e^{\cfrac{{1.6 \times {{10}^{ - 19}} \times 0.2}}{{1.4 \times {{10}^{ - 23}} \times 300}} - 1}}} \right]$$
$$ = {10^{ - 5}}\left[ {2038.6 - 1} \right] = 20.376 \times {10^{ - 3}}A$$
Hence,
option $$(C)$$ is correct answer.
Question 9
1 / -0

Two identical p-n junction may be connected in series with a battery in three ways as shown in the, adjoining figure. The potential drop across the p-n junctions are equal in

A
First and second circuits

B
Second and third circuits

C
Third and third circuits

D
All of these.

Solution

In case (i) and (ii) P is connected to positive terminal of supply and N is connected to negative terminal of supply hence diode is forward biased it means diode is in conduction
N is connected to positive terminal of supply and P is connected to negative terminal of supply hence diode is Reverse biased hence no conduction is seen
Question 10
1 / -0

The intrinsic charge carrier density in germanium crystal at $$ 300 K is 2.5 \times 10^{13} / cm^3 $$. density in an n-type germanium crystal at 300 K be $$ 5 \times 10^{16} /cm^3 $$,the hole density in this n-type crystal at 300 K would be

A
$$ 2.5 \times 10^{13} / cm^3 $$

B
$$ 5 \times 10^6 / cm^3 $$

C
$$ 1.25 \times 10^{10} / cm^3 $$

D
$$ 0.2 \times 10^4 / cm^3 $$

Solution

$$\large \begin{array}{l} nP=n{ i^{ 2 } }\left[ { By\, \, Mass\, \, action\, \, law } \right] \\ n=\frac { { 2.5\times 2.5\times { { 10 }^{ 26 } } } }{ { 5\times { { 10 }^{ 16 } } } } \\ n=1.25\times { 10^{ 10 } }/c{ m^{ 3 } } \\ Hence, \\ option\, \, C\, \, is\, correct\, \, answer. \end{array}$$

User

Question Analysis

Correct -
Wrong -
Skipped -

My Perfomance

Score
-
out of -
Rank
-
out of -

Re-Attempt Weekly Quiz Competition

Semiconductor Electronics: Materials, Devices and Simple Circuits Test 48

Result

Semiconductor Electronics: Materials, Devices and Simple Circuits Test 48

Score

-

Rank

-

SHARING IS CARING

Question 1

$$C_s$$

$$Si$$

$$Sn$$

None of the above

Solution

Question 2

Zener voltage will be low

Avalanche voltage will be high

Depletion region will be thin

Leakage current will be low

Solution

Question 3

$$ \left[ LT^ { -1} \right] $$

$$ \left[ L^2 T^ { -1} \right] $$

[LT]

$$ [L^{-1} T ] $$

Solution

Question 4

200 $$\Omega$$

400 $$\Omega$$

40 $$k\Omega$$

4 $$k\Omega$$

Solution

Question 5

Electric field

Potential difference

current

Mass

Solution

Question 6

$$10\pm0.7$$ohm

$$5\pm2$$ohm

$$0.1\pm0.2$$ohm

none of these

Solution

Question 7

Photodiode in forward bias

Photodiode in reverse bias

LED in forward bias

LED in reverse bias

Solution

Question 8

$$2037.6 \times 10 ^ { - 3 } \mathrm { amp }$$

$$203.76 \times 10 ^ { - 3 } \mathrm { amp }$$

$$20.376 \times 10 ^ { - 3 }amp$$

$$2.0376 \times 10 ^ { 3 }amp$$

Solution

Question 9

First and second circuits

Second and third circuits

Third and third circuits

All of these.

Solution

Question 10

$$ 2.5 \times 10^{13} / cm^3 $$

$$ 5 \times 10^6 / cm^3 $$

$$ 1.25 \times 10^{10} / cm^3 $$

$$ 0.2 \times 10^4 / cm^3 $$

Solution

User

Question Analysis

My Perfomance

Score

-

Rank

-

Results Announcement on: coming Monday

Stay Tuned: We’ll update you on your result via email or SMS.