1
TS EAMCET 2020 (Online) 10th September Evening Shift
MCQ (Single Correct Answer)
+1
-0

A Zener diode is connected to battery and a load resistance as shown below

TS EAMCET 2020 (Online) 10th September Evening Shift Physics - Semiconductor Devices and Logic Gates Question 9 English

The currents $I, I_Z$ and $I_L$ respectively are

A

$10 \mathrm{~mA}, 5 \mathrm{~mA}, 5 \mathrm{~mA}$

B

$15 \mathrm{~mA}, 7.5 \mathrm{~mA}, 7.5 \mathrm{~mA}$

C

$12.5 \mathrm{~mA}, 5 \mathrm{~mA}, 7.5 \mathrm{~mA}$

D

$12.5 \mathrm{~mA}, 7.5 \mathrm{~mA}, 5 \mathrm{~mA}$

2
TS EAMCET 2020 (Online) 10th September Evening Shift
MCQ (Single Correct Answer)
+1
-0

A semiconductor is doped with phosphorous atoms as impurity. The impurity levels created in the semiconductor are close to the

A

top of the valence band

B

bottom of the conduction band

C

bottom of the valence band

D

top of the conduction band

3
TS EAMCET 2020 (Online) 10th September Morning Shift
MCQ (Single Correct Answer)
+1
-0

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

TS EAMCET 2020 (Online) 10th September Morning Shift Physics - Semiconductor Devices and Logic Gates Question 11 English

A

TS EAMCET 2020 (Online) 10th September Morning Shift Physics - Semiconductor Devices and Logic Gates Question 11 English Option 1

B

TS EAMCET 2020 (Online) 10th September Morning Shift Physics - Semiconductor Devices and Logic Gates Question 11 English Option 2

C

TS EAMCET 2020 (Online) 10th September Morning Shift Physics - Semiconductor Devices and Logic Gates Question 11 English Option 3

D

TS EAMCET 2020 (Online) 10th September Morning Shift Physics - Semiconductor Devices and Logic Gates Question 11 English Option 4

4
TS EAMCET 2020 (Online) 10th September Morning Shift
MCQ (Single Correct Answer)
+1
-0

The Boolean expression of the circuit given in figure is

TS EAMCET 2020 (Online) 10th September Morning Shift Physics - Semiconductor Devices and Logic Gates Question 12 English

A

$Y=A+\bar{B}$

B

$Y=\overline{A+B}$

C

$Y=\bar{A}+B$

D

$Y=A+B$

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