Semiconductor Electronics: Materials, Devices and Simple Circuits Test 19

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Semiconductor Electronics: Materials, Devices and Simple Circuits Test 19

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Weekly Quiz Competition

Question 1
1 / -0

Which of the following devices can be used as a square law device for modulation.

A
p-n-p transistor in active region

B
p-n junction diode near threshold voltage.

C
Superconductor

D
Inductor

Solution

$$p-n$$ junction diode near threshold voltage is used as a square law device for modulation.
Question 2
1 / -0

What is the order of magnitude of the resistance of a dry human body?

A
$$10\Omega$$

B
$$10^4 \Omega$$

C
$$10$$ M$$\Omega$$

D
$$10\mu\Omega$$

Solution

It is known that the resistance of a dry human body is $$10 k \Omega = 10^{4} \Omega$$.
Question 3
1 / -0

Zener breakdown takes place if:

A
Doped impurity is low

B
Doped impurity is high

C
Less impurity in N-part

D
Less impurity in P-type

Solution

Zener breakdown occurs due to heavily doped diodes.In this process electrons crosses the barrier from
the valence band of the p-type to the conduction band of the lightly filled n-type material.
Question 4
1 / -0

On increasing the temperature of a semi-conductor material

A
Density of charge-carrier as well as their mobilities both increases

B
Density of charge-carriers increases, but their mobilities decreases

C
Density of charge-carries decreases, but their mobilities increases

D
Both density of charge-carriers as well as their mobilities decreases

Solution

On increasing the temperature of a semi-conductor material, the density of charge increases which apparently increases the hinderance and thus decreases their mobility.
Question 5
1 / -0

If the series resistance decreases in an unloaded zener regulator, the zener current

A
Decreases

B
Stays the same

C
Increases

D
Equals the voltage divided by the resistance

Solution

current increases as the resistance decreases, as zener current is inversely proportional to the series resistance
Question 6
1 / -0

For detecting the light_______________.

A
The photodiode has to be forward biased

B
The photodiode has to be reversed biased

C
The LED has to connected in forward bias mode

D
The LED has to be connected in reverse bias mode

Solution

Photo diode is used to detect light.
In reverse biased condition, the width of depletion region increases as the the applied reverse bias voltage increases across the diode. So, by applying a larger voltage, more of the incident photons are converted to electric current thereby increasing the efficiency.
In forward biased condition, the width of depletion region decreases so only a small portion of the incident photons get converted to electric current and hence the efficiency is less.
Hence photo diode in reverse biased condition detects light.
Therefore the correct option is (B).
Question 7
1 / -0

Resistor is coloured as green - orange - yellow - gold,its resistance will be:

A
$$53 \times 10^4 \pm 10\%$$

B
$$54 \times 10^4 \pm 10\%$$

C
$$42 \times 10^4 \pm 10\%$$

D
$$53 \times 10^4 \pm 5\%$$

Solution

REF.Image.
$$ R = ab\times 10^{c}\pm d $$% tolerance
a = 5 (For Green)
b = 3 (For Orange)
c = 4 (For Yellow)
d = % (For Gold)
$$ R = 53\times 10^{4}\pm 5$$% $$ \Omega $$
Question 8
1 / -0

When a p-n junction diode is connected in forward bias its barrier potential

A
decreases and less current flows in the circuit

B
decreases and more current flows in the circuit

C
increases and more current flows in the circuit

D
decreases and no current flows in the circuit

Solution

When a p-n junction diode is connected in forward bias its barrier potential decreases and less current flows in the circuit.
so the option is A.
Question 9
1 / -0

The breakdown in a reverse biased p-n junction diode is more likely to occur due to.

A
large velocity of the minority charge if the doping concentration is small.

B
large velocity of the minority charge carriers if the doping concentration is small.

C
strong electric field in a depletion region if the doping concentration is small.

D
strong electric filed in the deplention region if the doping conentration is large.

Solution

When it comes to he breakdown in a reverse biased PN junction diode, it will probably happen only because of the accumulation of the higher charge at the biased region and large velocity of the minority charge if the doping concentration is small. This is the main cause the breakdown
Question 10
1 / -0

A resistance is shown in the figure. Its value and tolerance are given respectively by:

A
27 $$

\mathrm{K} \Omega

$$, 20%

B
270 $$

\mathrm{K} \Omega

$$, 5%

C
270 $$

\mathrm{K} \Omega

$$, 10%

D
27 $$

\mathrm{K} \Omega

$$, 10%

Solution

$$ Silver \Rightarrow tolerance = 10 %

Red = 2

violet = 7

orange = 3 $$ \Rightarrow value = 27 \times 10 ^{3}$$ $$ \Omega 10$$ % tolerance

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Semiconductor Electronics: Materials, Devices and Simple Circuits Test 19

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Semiconductor Electronics: Materials, Devices and Simple Circuits Test 19

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SHARING IS CARING

Question 1

p-n-p transistor in active region

p-n junction diode near threshold voltage.

Superconductor

Inductor

Solution

Question 2

$$10\Omega$$

$$10^4 \Omega$$

$$10$$ M$$\Omega$$

$$10\mu\Omega$$

Solution

Question 3

Doped impurity is low

Doped impurity is high

Less impurity in N-part

Less impurity in P-type

Solution

Question 4

Density of charge-carrier as well as their mobilities both increases

Density of charge-carriers increases, but their mobilities decreases

Density of charge-carries decreases, but their mobilities increases

Both density of charge-carriers as well as their mobilities decreases

Solution

Question 5

Decreases

Stays the same

Increases

Equals the voltage divided by the resistance

Solution

Question 6

The photodiode has to be forward biased

The photodiode has to be reversed biased

The LED has to connected in forward bias mode

The LED has to be connected in reverse bias mode

Solution

Question 7

$$53 \times 10^4 \pm 10\%$$

$$54 \times 10^4 \pm 10\%$$

$$42 \times 10^4 \pm 10\%$$

$$53 \times 10^4 \pm 5\%$$

Solution

Question 8

decreases and less current flows in the circuit

decreases and more current flows in the circuit

increases and more current flows in the circuit

decreases and no current flows in the circuit

Solution

Question 9

large velocity of the minority charge if the doping concentration is small.

large velocity of the minority charge carriers if the doping concentration is small.

strong electric field in a depletion region if the doping concentration is small.

strong electric filed in the deplention region if the doping conentration is large.

Solution

Question 10

27 $$ \mathrm{K} \Omega $$, 20%

270 $$ \mathrm{K} \Omega $$, 5%

270 $$ \mathrm{K} \Omega $$, 10%

27 $$ \mathrm{K} \Omega $$, 10%

Solution

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27 $$

\mathrm{K} \Omega

$$, 20%

270 $$

\mathrm{K} \Omega

$$, 5%

270 $$

\mathrm{K} \Omega

$$, 10%

27 $$

\mathrm{K} \Omega

$$, 10%