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Semiconductor Electronics Material Devices and Simple Circuits

150705 Which of the following depicts the output of the full wave rectifier with capacitor filter for the following $AC$ input?
original image

1

2

3

4

Explanation:

C The output of full wave rectifier with capacitor filter for the given $AC$ input is -
original image

TS- EAMCET-10.09.2020

Semiconductor Electronics Material Devices and Simple Circuits

150712 In the diode-based rectifier circuit given below, if $V_{S} = V_{m} \sin ω t$ and the diode is ideal, then the average value of $V_{L}$ is
original image

1

\frac{R_{L}}{(R_{L} + R_{S})} \frac{V_{m}}{π}

2

R_{L} V_{m} \sin ω t

3

\frac{R_{L}}{(R_{L} + R_{S})} V_{m}

4

\frac{R_{L}}{(R_{L} + R_{S})} V_{m} \sin ω t

Explanation:

B Given that,
$V_{s} = V_{m} \sin ω t$
original image
Which is $A C$ voltage and $V_{m}$ is the maximum value of voltage in the given circuit diode will only conduct current in forward biased so output across $R_{L}$ will be the average voltage,
$V_{av} = \frac{V_{m}}{π}$
And, Average current,
$I_{av} = \frac{V_{av}}{R}$

Here, Total Resistance, $R = R_{S} + R_{L}$
$I_{av} = \frac{V_{m}}{(R_{S} + R_{L}) π}$
Now, voltage across $R_{L}$ will be -
$V_{L} = I_{a v} R_{L}$
$V_{L} = \frac{V_{m}}{(R_{S} + R_{L}) π} \times R_{L}$
$V_{L} = (\frac{R_{L}}{(R_{L} + R_{S})}) \frac{V_{m}}{π}$

TS- EAMCET-07.05.2018

Semiconductor Electronics Material Devices and Simple Circuits

150713 What is the current through the circuit and the potential difference across the diode shown in figure. The drift current for the diode is $30 μ A$ .
original image

1

30 μ A, 5.99 V

2

30 μ A, 5 V

3

20 μ A, 6 V

4

20 μ A, 5.99 V

Explanation:

B Given that,
$I_{D} = 30 μ A = 30 \times 10^{- 6} A$
Since both the resistor and diode are connected in series so the amount of current in both of them will be remains the same which is $30 μ A$ .
original image
Then potential drop across the resistor will be equal to,
$V = I R$
$V = 30 \times 10^{- 6} \times 20 = 600 \times 10^{- 6} V$
$V = 0.6 mV$
Then potential drop across the diode is,
$V_{D} = [6 - 0.6 \times 10^{- 3}]$
$V_{D} = [6 - 0.0006]$
$V_{D} = 5.99 V$ Hence, $I_{D} = 30 μ A$ and $V_{D} = 5.99 V$

Manipal UGET -2020

Semiconductor Electronics Material Devices and Simple Circuits

150716 Assuming that the junction diode is ideal, the current in the arrangement shown in
original image

1

20 mA

2

10 mA

3

40 mA

4

30 mA

Explanation:

B original image
Given,
$R = 100 Ω$
We know that,
$V = I R$
$I = \frac{V}{R} = \frac{3 - 1}{100}$
$I = \frac{2}{100} = \frac{1}{50}$
$I = 20 mA$

MHT-CET 2019

Semiconductor Electronics Material Devices and Simple Circuits

150705 Which of the following depicts the output of the full wave rectifier with capacitor filter for the following $AC$ input?
original image

1

2

3

4

Explanation:

C The output of full wave rectifier with capacitor filter for the given $AC$ input is -
original image

TS- EAMCET-10.09.2020

Semiconductor Electronics Material Devices and Simple Circuits

150712 In the diode-based rectifier circuit given below, if $V_{S} = V_{m} \sin ω t$ and the diode is ideal, then the average value of $V_{L}$ is
original image

1

\frac{R_{L}}{(R_{L} + R_{S})} \frac{V_{m}}{π}

2

R_{L} V_{m} \sin ω t

3

\frac{R_{L}}{(R_{L} + R_{S})} V_{m}

4

\frac{R_{L}}{(R_{L} + R_{S})} V_{m} \sin ω t

Explanation:

B Given that,
$V_{s} = V_{m} \sin ω t$
original image
Which is $A C$ voltage and $V_{m}$ is the maximum value of voltage in the given circuit diode will only conduct current in forward biased so output across $R_{L}$ will be the average voltage,
$V_{av} = \frac{V_{m}}{π}$
And, Average current,
$I_{av} = \frac{V_{av}}{R}$

Here, Total Resistance, $R = R_{S} + R_{L}$
$I_{av} = \frac{V_{m}}{(R_{S} + R_{L}) π}$
Now, voltage across $R_{L}$ will be -
$V_{L} = I_{a v} R_{L}$
$V_{L} = \frac{V_{m}}{(R_{S} + R_{L}) π} \times R_{L}$
$V_{L} = (\frac{R_{L}}{(R_{L} + R_{S})}) \frac{V_{m}}{π}$

TS- EAMCET-07.05.2018

Semiconductor Electronics Material Devices and Simple Circuits

150713 What is the current through the circuit and the potential difference across the diode shown in figure. The drift current for the diode is $30 μ A$ .
original image

1

30 μ A, 5.99 V

2

30 μ A, 5 V

3

20 μ A, 6 V

4

20 μ A, 5.99 V

Explanation:

B Given that,
$I_{D} = 30 μ A = 30 \times 10^{- 6} A$
Since both the resistor and diode are connected in series so the amount of current in both of them will be remains the same which is $30 μ A$ .
original image
Then potential drop across the resistor will be equal to,
$V = I R$
$V = 30 \times 10^{- 6} \times 20 = 600 \times 10^{- 6} V$
$V = 0.6 mV$
Then potential drop across the diode is,
$V_{D} = [6 - 0.6 \times 10^{- 3}]$
$V_{D} = [6 - 0.0006]$
$V_{D} = 5.99 V$ Hence, $I_{D} = 30 μ A$ and $V_{D} = 5.99 V$

Manipal UGET -2020

Semiconductor Electronics Material Devices and Simple Circuits

150716 Assuming that the junction diode is ideal, the current in the arrangement shown in
original image

1

20 mA

2

10 mA

3

40 mA

4

30 mA

Explanation:

B original image
Given,
$R = 100 Ω$
We know that,
$V = I R$
$I = \frac{V}{R} = \frac{3 - 1}{100}$
$I = \frac{2}{100} = \frac{1}{50}$
$I = 20 mA$

MHT-CET 2019

Semiconductor Electronics Material Devices and Simple Circuits

150705 Which of the following depicts the output of the full wave rectifier with capacitor filter for the following $AC$ input?
original image

1

2

3

4

Explanation:

C The output of full wave rectifier with capacitor filter for the given $AC$ input is -
original image

TS- EAMCET-10.09.2020

Semiconductor Electronics Material Devices and Simple Circuits

150712 In the diode-based rectifier circuit given below, if $V_{S} = V_{m} \sin ω t$ and the diode is ideal, then the average value of $V_{L}$ is
original image

1

\frac{R_{L}}{(R_{L} + R_{S})} \frac{V_{m}}{π}

2

R_{L} V_{m} \sin ω t

3

\frac{R_{L}}{(R_{L} + R_{S})} V_{m}

4

\frac{R_{L}}{(R_{L} + R_{S})} V_{m} \sin ω t

Explanation:

B Given that,
$V_{s} = V_{m} \sin ω t$
original image
Which is $A C$ voltage and $V_{m}$ is the maximum value of voltage in the given circuit diode will only conduct current in forward biased so output across $R_{L}$ will be the average voltage,
$V_{av} = \frac{V_{m}}{π}$
And, Average current,
$I_{av} = \frac{V_{av}}{R}$

Here, Total Resistance, $R = R_{S} + R_{L}$
$I_{av} = \frac{V_{m}}{(R_{S} + R_{L}) π}$
Now, voltage across $R_{L}$ will be -
$V_{L} = I_{a v} R_{L}$
$V_{L} = \frac{V_{m}}{(R_{S} + R_{L}) π} \times R_{L}$
$V_{L} = (\frac{R_{L}}{(R_{L} + R_{S})}) \frac{V_{m}}{π}$

TS- EAMCET-07.05.2018

Semiconductor Electronics Material Devices and Simple Circuits

150713 What is the current through the circuit and the potential difference across the diode shown in figure. The drift current for the diode is $30 μ A$ .
original image

1

30 μ A, 5.99 V

2

30 μ A, 5 V

3

20 μ A, 6 V

4

20 μ A, 5.99 V

Explanation:

B Given that,
$I_{D} = 30 μ A = 30 \times 10^{- 6} A$
Since both the resistor and diode are connected in series so the amount of current in both of them will be remains the same which is $30 μ A$ .
original image
Then potential drop across the resistor will be equal to,
$V = I R$
$V = 30 \times 10^{- 6} \times 20 = 600 \times 10^{- 6} V$
$V = 0.6 mV$
Then potential drop across the diode is,
$V_{D} = [6 - 0.6 \times 10^{- 3}]$
$V_{D} = [6 - 0.0006]$
$V_{D} = 5.99 V$ Hence, $I_{D} = 30 μ A$ and $V_{D} = 5.99 V$

Manipal UGET -2020

Semiconductor Electronics Material Devices and Simple Circuits

150716 Assuming that the junction diode is ideal, the current in the arrangement shown in
original image

1

20 mA

2

10 mA

3

40 mA

4

30 mA

Explanation:

B original image
Given,
$R = 100 Ω$
We know that,
$V = I R$
$I = \frac{V}{R} = \frac{3 - 1}{100}$
$I = \frac{2}{100} = \frac{1}{50}$
$I = 20 mA$

MHT-CET 2019

Semiconductor Electronics Material Devices and Simple Circuits

150705 Which of the following depicts the output of the full wave rectifier with capacitor filter for the following $AC$ input?
original image

1

2

3

4

Explanation:

C The output of full wave rectifier with capacitor filter for the given $AC$ input is -
original image

TS- EAMCET-10.09.2020

Semiconductor Electronics Material Devices and Simple Circuits

150712 In the diode-based rectifier circuit given below, if $V_{S} = V_{m} \sin ω t$ and the diode is ideal, then the average value of $V_{L}$ is
original image

1

\frac{R_{L}}{(R_{L} + R_{S})} \frac{V_{m}}{π}

2

R_{L} V_{m} \sin ω t

3

\frac{R_{L}}{(R_{L} + R_{S})} V_{m}

4

\frac{R_{L}}{(R_{L} + R_{S})} V_{m} \sin ω t

Explanation:

B Given that,
$V_{s} = V_{m} \sin ω t$
original image
Which is $A C$ voltage and $V_{m}$ is the maximum value of voltage in the given circuit diode will only conduct current in forward biased so output across $R_{L}$ will be the average voltage,
$V_{av} = \frac{V_{m}}{π}$
And, Average current,
$I_{av} = \frac{V_{av}}{R}$

Here, Total Resistance, $R = R_{S} + R_{L}$
$I_{av} = \frac{V_{m}}{(R_{S} + R_{L}) π}$
Now, voltage across $R_{L}$ will be -
$V_{L} = I_{a v} R_{L}$
$V_{L} = \frac{V_{m}}{(R_{S} + R_{L}) π} \times R_{L}$
$V_{L} = (\frac{R_{L}}{(R_{L} + R_{S})}) \frac{V_{m}}{π}$

TS- EAMCET-07.05.2018

Semiconductor Electronics Material Devices and Simple Circuits

150713 What is the current through the circuit and the potential difference across the diode shown in figure. The drift current for the diode is $30 μ A$ .
original image

1

30 μ A, 5.99 V

2

30 μ A, 5 V

3

20 μ A, 6 V

4

20 μ A, 5.99 V

Explanation:

B Given that,
$I_{D} = 30 μ A = 30 \times 10^{- 6} A$
Since both the resistor and diode are connected in series so the amount of current in both of them will be remains the same which is $30 μ A$ .
original image
Then potential drop across the resistor will be equal to,
$V = I R$
$V = 30 \times 10^{- 6} \times 20 = 600 \times 10^{- 6} V$
$V = 0.6 mV$
Then potential drop across the diode is,
$V_{D} = [6 - 0.6 \times 10^{- 3}]$
$V_{D} = [6 - 0.0006]$
$V_{D} = 5.99 V$ Hence, $I_{D} = 30 μ A$ and $V_{D} = 5.99 V$

Manipal UGET -2020

Semiconductor Electronics Material Devices and Simple Circuits

150716 Assuming that the junction diode is ideal, the current in the arrangement shown in
original image

1

20 mA

2

10 mA

3

40 mA

4

30 mA

Explanation:

B original image
Given,
$R = 100 Ω$
We know that,
$V = I R$
$I = \frac{V}{R} = \frac{3 - 1}{100}$
$I = \frac{2}{100} = \frac{1}{50}$
$I = 20 mA$

MHT-CET 2019