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Semiconductor Electronics Material Devices and Simple Circuits

150950 In a Zener diode, the reverse bias voltage is $3 V$ and the width of the depletion region is $300 \AA$ the electric field intensity will be $V / cm$

1

10^{4}

2

10^{6}

3

10^{8}

4

10^{- 2}

Explanation:

B original image
Given,
$Δ V = 3 Volt$
Width of depletion region $(d) = 300 \AA$
We know that,
Electric field intensity
$E = \frac{Δ V}{d} = \frac{3}{300 \times 10^{- 10} m}$
$E = 1 \times 10^{8} \frac{V}{m}$
$E = 10^{6} V / cm$

GUJCET 2014

Semiconductor Electronics Material Devices and Simple Circuits

150951 Two amplifiers are connected one after the other in series (cascaded). The first amplifier has a voltage gain of 10 and second has a voltage gain of 20 . If the input signal is $0.01$ volt, what is the output ac signal?

1

1.5 V

2

2 V

3

2.5 V

4

3.0 V

Explanation:

B Given that,
Voltage gain, $A_{1} = 10 V$
Voltage gain, $A_{2} = 20 V$
Net voltage gain, $A_{V} = A_{1} \times A_{2}$
$A_{V} = 10 \times 20$
$= 200 Volt$
We know that,
$V_{0} = A_{V} \times V_{in}$
$Here, V_{in} = .01 V$
$V_{0} = 200 \times 0.01$
$V_{0} = 2 V$ Here,

COMEDK 2014

Semiconductor Electronics Material Devices and Simple Circuits

150956 A Zener diode, having breakdown voltage equal to $15 V$ is used in a voltage regulator circuit shown in figure. The current through the diode is
original image

1

10 mA

2

15 mA

3

20 mA

4

5 mA

Explanation:

D original image
By ohm's Law,
$V = I \cdot R$
So, current through $1 k Ω$ resistance -
$I_{1 k Ω} = \frac{V}{R} = \frac{15}{1000}$
$I_{1 k Ω} = .015 A$
$= 15 mA$
Now, current through $250 Ω$ resistance -
$I_{250 Ω} = \frac{20 - 15}{250} = \frac{5}{250}$
$I_{250 Ω} = 20 mA$
$I_{250 Ω} = I_{Z} + I_{1 k Ω}$
$I_{Z} = I_{250} - I_{1 k Ω}$
$I_{Z} = (20 - 15)$
$I_{Z} = 5 mA$

VITEEE-2014

Semiconductor Electronics Material Devices and Simple Circuits

150960 In the circuit shown below if the current through the $500 Ω$ resistor is $25 mA$ and that through the load $R_{L}$ is $5 mA$ , then the maximum current through the Zener diode is
original image

1

5 mA

2

20 mA

3

30 mA

4

125 mA

Explanation:

B Given that,
$R = 500 Ω$
$I_{R} = 25 mA$
$I_{L} = 5 mA$
$I_{Z} = ?$
$I_{Z} = I_{R} - I_{L}$
$I_{Z} = 25 - 5$
$I_{Z} = 20 mA$

J and K CET-2016

Semiconductor Electronics Material Devices and Simple Circuits

150950 In a Zener diode, the reverse bias voltage is $3 V$ and the width of the depletion region is $300 \AA$ the electric field intensity will be $V / cm$

1

10^{4}

2

10^{6}

3

10^{8}

4

10^{- 2}

Explanation:

B original image
Given,
$Δ V = 3 Volt$
Width of depletion region $(d) = 300 \AA$
We know that,
Electric field intensity
$E = \frac{Δ V}{d} = \frac{3}{300 \times 10^{- 10} m}$
$E = 1 \times 10^{8} \frac{V}{m}$
$E = 10^{6} V / cm$

GUJCET 2014

Semiconductor Electronics Material Devices and Simple Circuits

150951 Two amplifiers are connected one after the other in series (cascaded). The first amplifier has a voltage gain of 10 and second has a voltage gain of 20 . If the input signal is $0.01$ volt, what is the output ac signal?

1

1.5 V

2

2 V

3

2.5 V

4

3.0 V

Explanation:

B Given that,
Voltage gain, $A_{1} = 10 V$
Voltage gain, $A_{2} = 20 V$
Net voltage gain, $A_{V} = A_{1} \times A_{2}$
$A_{V} = 10 \times 20$
$= 200 Volt$
We know that,
$V_{0} = A_{V} \times V_{in}$
$Here, V_{in} = .01 V$
$V_{0} = 200 \times 0.01$
$V_{0} = 2 V$ Here,

COMEDK 2014

Semiconductor Electronics Material Devices and Simple Circuits

150956 A Zener diode, having breakdown voltage equal to $15 V$ is used in a voltage regulator circuit shown in figure. The current through the diode is
original image

1

10 mA

2

15 mA

3

20 mA

4

5 mA

Explanation:

D original image
By ohm's Law,
$V = I \cdot R$
So, current through $1 k Ω$ resistance -
$I_{1 k Ω} = \frac{V}{R} = \frac{15}{1000}$
$I_{1 k Ω} = .015 A$
$= 15 mA$
Now, current through $250 Ω$ resistance -
$I_{250 Ω} = \frac{20 - 15}{250} = \frac{5}{250}$
$I_{250 Ω} = 20 mA$
$I_{250 Ω} = I_{Z} + I_{1 k Ω}$
$I_{Z} = I_{250} - I_{1 k Ω}$
$I_{Z} = (20 - 15)$
$I_{Z} = 5 mA$

VITEEE-2014

Semiconductor Electronics Material Devices and Simple Circuits

150960 In the circuit shown below if the current through the $500 Ω$ resistor is $25 mA$ and that through the load $R_{L}$ is $5 mA$ , then the maximum current through the Zener diode is
original image

1

5 mA

2

20 mA

3

30 mA

4

125 mA

Explanation:

B Given that,
$R = 500 Ω$
$I_{R} = 25 mA$
$I_{L} = 5 mA$
$I_{Z} = ?$
$I_{Z} = I_{R} - I_{L}$
$I_{Z} = 25 - 5$
$I_{Z} = 20 mA$

J and K CET-2016

Semiconductor Electronics Material Devices and Simple Circuits

150950 In a Zener diode, the reverse bias voltage is $3 V$ and the width of the depletion region is $300 \AA$ the electric field intensity will be $V / cm$

1

10^{4}

2

10^{6}

3

10^{8}

4

10^{- 2}

Explanation:

B original image
Given,
$Δ V = 3 Volt$
Width of depletion region $(d) = 300 \AA$
We know that,
Electric field intensity
$E = \frac{Δ V}{d} = \frac{3}{300 \times 10^{- 10} m}$
$E = 1 \times 10^{8} \frac{V}{m}$
$E = 10^{6} V / cm$

GUJCET 2014

Semiconductor Electronics Material Devices and Simple Circuits

150951 Two amplifiers are connected one after the other in series (cascaded). The first amplifier has a voltage gain of 10 and second has a voltage gain of 20 . If the input signal is $0.01$ volt, what is the output ac signal?

1

1.5 V

2

2 V

3

2.5 V

4

3.0 V

Explanation:

B Given that,
Voltage gain, $A_{1} = 10 V$
Voltage gain, $A_{2} = 20 V$
Net voltage gain, $A_{V} = A_{1} \times A_{2}$
$A_{V} = 10 \times 20$
$= 200 Volt$
We know that,
$V_{0} = A_{V} \times V_{in}$
$Here, V_{in} = .01 V$
$V_{0} = 200 \times 0.01$
$V_{0} = 2 V$ Here,

COMEDK 2014

Semiconductor Electronics Material Devices and Simple Circuits

150956 A Zener diode, having breakdown voltage equal to $15 V$ is used in a voltage regulator circuit shown in figure. The current through the diode is
original image

1

10 mA

2

15 mA

3

20 mA

4

5 mA

Explanation:

D original image
By ohm's Law,
$V = I \cdot R$
So, current through $1 k Ω$ resistance -
$I_{1 k Ω} = \frac{V}{R} = \frac{15}{1000}$
$I_{1 k Ω} = .015 A$
$= 15 mA$
Now, current through $250 Ω$ resistance -
$I_{250 Ω} = \frac{20 - 15}{250} = \frac{5}{250}$
$I_{250 Ω} = 20 mA$
$I_{250 Ω} = I_{Z} + I_{1 k Ω}$
$I_{Z} = I_{250} - I_{1 k Ω}$
$I_{Z} = (20 - 15)$
$I_{Z} = 5 mA$

VITEEE-2014

Semiconductor Electronics Material Devices and Simple Circuits

150960 In the circuit shown below if the current through the $500 Ω$ resistor is $25 mA$ and that through the load $R_{L}$ is $5 mA$ , then the maximum current through the Zener diode is
original image

1

5 mA

2

20 mA

3

30 mA

4

125 mA

Explanation:

B Given that,
$R = 500 Ω$
$I_{R} = 25 mA$
$I_{L} = 5 mA$
$I_{Z} = ?$
$I_{Z} = I_{R} - I_{L}$
$I_{Z} = 25 - 5$
$I_{Z} = 20 mA$

J and K CET-2016

Semiconductor Electronics Material Devices and Simple Circuits

150950 In a Zener diode, the reverse bias voltage is $3 V$ and the width of the depletion region is $300 \AA$ the electric field intensity will be $V / cm$

1

10^{4}

2

10^{6}

3

10^{8}

4

10^{- 2}

Explanation:

B original image
Given,
$Δ V = 3 Volt$
Width of depletion region $(d) = 300 \AA$
We know that,
Electric field intensity
$E = \frac{Δ V}{d} = \frac{3}{300 \times 10^{- 10} m}$
$E = 1 \times 10^{8} \frac{V}{m}$
$E = 10^{6} V / cm$

GUJCET 2014

Semiconductor Electronics Material Devices and Simple Circuits

150951 Two amplifiers are connected one after the other in series (cascaded). The first amplifier has a voltage gain of 10 and second has a voltage gain of 20 . If the input signal is $0.01$ volt, what is the output ac signal?

1

1.5 V

2

2 V

3

2.5 V

4

3.0 V

Explanation:

B Given that,
Voltage gain, $A_{1} = 10 V$
Voltage gain, $A_{2} = 20 V$
Net voltage gain, $A_{V} = A_{1} \times A_{2}$
$A_{V} = 10 \times 20$
$= 200 Volt$
We know that,
$V_{0} = A_{V} \times V_{in}$
$Here, V_{in} = .01 V$
$V_{0} = 200 \times 0.01$
$V_{0} = 2 V$ Here,

COMEDK 2014

Semiconductor Electronics Material Devices and Simple Circuits

150956 A Zener diode, having breakdown voltage equal to $15 V$ is used in a voltage regulator circuit shown in figure. The current through the diode is
original image

1

10 mA

2

15 mA

3

20 mA

4

5 mA

Explanation:

D original image
By ohm's Law,
$V = I \cdot R$
So, current through $1 k Ω$ resistance -
$I_{1 k Ω} = \frac{V}{R} = \frac{15}{1000}$
$I_{1 k Ω} = .015 A$
$= 15 mA$
Now, current through $250 Ω$ resistance -
$I_{250 Ω} = \frac{20 - 15}{250} = \frac{5}{250}$
$I_{250 Ω} = 20 mA$
$I_{250 Ω} = I_{Z} + I_{1 k Ω}$
$I_{Z} = I_{250} - I_{1 k Ω}$
$I_{Z} = (20 - 15)$
$I_{Z} = 5 mA$

VITEEE-2014

Semiconductor Electronics Material Devices and Simple Circuits

150960 In the circuit shown below if the current through the $500 Ω$ resistor is $25 mA$ and that through the load $R_{L}$ is $5 mA$ , then the maximum current through the Zener diode is
original image

1

5 mA

2

20 mA

3

30 mA

4

125 mA

Explanation:

B Given that,
$R = 500 Ω$
$I_{R} = 25 mA$
$I_{L} = 5 mA$
$I_{Z} = ?$
$I_{Z} = I_{R} - I_{L}$
$I_{Z} = 25 - 5$
$I_{Z} = 20 mA$

J and K CET-2016

Question Bank Series

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