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PHXII14:SEMICONDUCTOR ELECTRONICS- MATERIALS- DEVICES AND SIMPLE CIRCUITS

365303 The diode used in the circuit shown in the figure has a constant voltage drop of 0.5 $V$ at all currents and a maximum power rating of 100 milliwatts. What should be the value of the resistor $R$ , connected in series with the diode, for obtaining maximum current?
supporting img

1

1.5 Ω

2

5 Ω

3

6.67 Ω

4

200 Ω

Explanation:

Current in diode,
$I = \frac{P}{V} = \frac{100 m W}{0.5 V} = 0.2 A$
Voltage drop across resistance,
$R = (1.5 - 0.5) V = 1 V$
$∴$ Resistance $R = \frac{Voltage drop}{Current} = \frac{1 V}{0.2 A} = 5 Ω .$
So correct option is (2)

PHXII14:SEMICONDUCTOR ELECTRONICS- MATERIALS- DEVICES AND SIMPLE CIRCUITS

365304 A diode is connected to 220 $V$ $d c$ in series with a resistor as shown in figure. The current through the resistor is
supporting img

1

5 A

2

20 A

3

50 A

4

10 A

Explanation:

$i = \frac{V}{r} = \frac{220 V}{11 Ω} = 20 A$

PHXII14:SEMICONDUCTOR ELECTRONICS- MATERIALS- DEVICES AND SIMPLE CIRCUITS

365305 Assertion :
In a semiconductor diode, the reverse biased current is due to drift of free electrons and holes.
Reason :
The drift of electrons and holes is due to thermal excitations.

1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.

2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.

3 Assertion is correct but Reason is incorrect.

4 Assertion is incorrect but reason is correct.

PHXII14:SEMICONDUCTOR ELECTRONICS- MATERIALS- DEVICES AND SIMPLE CIRCUITS

365306 What is the current through an ideal $p - n$ junction diode shown in figure below ?
supporting img

1 Zero

2

10 m A

3

20 m A

4

50 m A

Explanation:

Applying KVL, $1 + 100 i - 3 = 0$
$i = \frac{2}{100} \times \frac{1000}{1000}$
$= 20 \times 10^{- 3} A$
$= 20 m A$

PHXII14:SEMICONDUCTOR ELECTRONICS- MATERIALS- DEVICES AND SIMPLE CIRCUITS

365303 The diode used in the circuit shown in the figure has a constant voltage drop of 0.5 $V$ at all currents and a maximum power rating of 100 milliwatts. What should be the value of the resistor $R$ , connected in series with the diode, for obtaining maximum current?
supporting img

1

1.5 Ω

2

5 Ω

3

6.67 Ω

4

200 Ω

Explanation:

Current in diode,
$I = \frac{P}{V} = \frac{100 m W}{0.5 V} = 0.2 A$
Voltage drop across resistance,
$R = (1.5 - 0.5) V = 1 V$
$∴$ Resistance $R = \frac{Voltage drop}{Current} = \frac{1 V}{0.2 A} = 5 Ω .$
So correct option is (2)

PHXII14:SEMICONDUCTOR ELECTRONICS- MATERIALS- DEVICES AND SIMPLE CIRCUITS

365304 A diode is connected to 220 $V$ $d c$ in series with a resistor as shown in figure. The current through the resistor is
supporting img

1

5 A

2

20 A

3

50 A

4

10 A

Explanation:

$i = \frac{V}{r} = \frac{220 V}{11 Ω} = 20 A$

PHXII14:SEMICONDUCTOR ELECTRONICS- MATERIALS- DEVICES AND SIMPLE CIRCUITS

365305 Assertion :
In a semiconductor diode, the reverse biased current is due to drift of free electrons and holes.
Reason :
The drift of electrons and holes is due to thermal excitations.

1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.

2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.

3 Assertion is correct but Reason is incorrect.

4 Assertion is incorrect but reason is correct.

PHXII14:SEMICONDUCTOR ELECTRONICS- MATERIALS- DEVICES AND SIMPLE CIRCUITS

365306 What is the current through an ideal $p - n$ junction diode shown in figure below ?
supporting img

1 Zero

2

10 m A

3

20 m A

4

50 m A

Explanation:

Applying KVL, $1 + 100 i - 3 = 0$
$i = \frac{2}{100} \times \frac{1000}{1000}$
$= 20 \times 10^{- 3} A$
$= 20 m A$

PHXII14:SEMICONDUCTOR ELECTRONICS- MATERIALS- DEVICES AND SIMPLE CIRCUITS

365303 The diode used in the circuit shown in the figure has a constant voltage drop of 0.5 $V$ at all currents and a maximum power rating of 100 milliwatts. What should be the value of the resistor $R$ , connected in series with the diode, for obtaining maximum current?
supporting img

1

1.5 Ω

2

5 Ω

3

6.67 Ω

4

200 Ω

Explanation:

Current in diode,
$I = \frac{P}{V} = \frac{100 m W}{0.5 V} = 0.2 A$
Voltage drop across resistance,
$R = (1.5 - 0.5) V = 1 V$
$∴$ Resistance $R = \frac{Voltage drop}{Current} = \frac{1 V}{0.2 A} = 5 Ω .$
So correct option is (2)

PHXII14:SEMICONDUCTOR ELECTRONICS- MATERIALS- DEVICES AND SIMPLE CIRCUITS

365304 A diode is connected to 220 $V$ $d c$ in series with a resistor as shown in figure. The current through the resistor is
supporting img

1

5 A

2

20 A

3

50 A

4

10 A

Explanation:

$i = \frac{V}{r} = \frac{220 V}{11 Ω} = 20 A$

PHXII14:SEMICONDUCTOR ELECTRONICS- MATERIALS- DEVICES AND SIMPLE CIRCUITS

365305 Assertion :
In a semiconductor diode, the reverse biased current is due to drift of free electrons and holes.
Reason :
The drift of electrons and holes is due to thermal excitations.

1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.

2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.

3 Assertion is correct but Reason is incorrect.

4 Assertion is incorrect but reason is correct.

PHXII14:SEMICONDUCTOR ELECTRONICS- MATERIALS- DEVICES AND SIMPLE CIRCUITS

365306 What is the current through an ideal $p - n$ junction diode shown in figure below ?
supporting img

1 Zero

2

10 m A

3

20 m A

4

50 m A

Explanation:

Applying KVL, $1 + 100 i - 3 = 0$
$i = \frac{2}{100} \times \frac{1000}{1000}$
$= 20 \times 10^{- 3} A$
$= 20 m A$

PHXII14:SEMICONDUCTOR ELECTRONICS- MATERIALS- DEVICES AND SIMPLE CIRCUITS

365303 The diode used in the circuit shown in the figure has a constant voltage drop of 0.5 $V$ at all currents and a maximum power rating of 100 milliwatts. What should be the value of the resistor $R$ , connected in series with the diode, for obtaining maximum current?
supporting img

1

1.5 Ω

2

5 Ω

3

6.67 Ω

4

200 Ω

Explanation:

Current in diode,
$I = \frac{P}{V} = \frac{100 m W}{0.5 V} = 0.2 A$
Voltage drop across resistance,
$R = (1.5 - 0.5) V = 1 V$
$∴$ Resistance $R = \frac{Voltage drop}{Current} = \frac{1 V}{0.2 A} = 5 Ω .$
So correct option is (2)

PHXII14:SEMICONDUCTOR ELECTRONICS- MATERIALS- DEVICES AND SIMPLE CIRCUITS

365304 A diode is connected to 220 $V$ $d c$ in series with a resistor as shown in figure. The current through the resistor is
supporting img

1

5 A

2

20 A

3

50 A

4

10 A

Explanation:

$i = \frac{V}{r} = \frac{220 V}{11 Ω} = 20 A$

PHXII14:SEMICONDUCTOR ELECTRONICS- MATERIALS- DEVICES AND SIMPLE CIRCUITS

365305 Assertion :
In a semiconductor diode, the reverse biased current is due to drift of free electrons and holes.
Reason :
The drift of electrons and holes is due to thermal excitations.

1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.

2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.

3 Assertion is correct but Reason is incorrect.

4 Assertion is incorrect but reason is correct.

PHXII14:SEMICONDUCTOR ELECTRONICS- MATERIALS- DEVICES AND SIMPLE CIRCUITS

365306 What is the current through an ideal $p - n$ junction diode shown in figure below ?
supporting img

1 Zero

2

10 m A

3

20 m A

4

50 m A

Explanation:

Applying KVL, $1 + 100 i - 3 = 0$
$i = \frac{2}{100} \times \frac{1000}{1000}$
$= 20 \times 10^{- 3} A$
$= 20 m A$

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