Semiconductor Devices Test

Result

Semiconductor Devices Test - 105

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Weekly Quiz Competition

Question 1
4 / -1

The combination of gates shown below produces

A
AND gate

B
XOR gate

C
NOR gate

D
NAND gate

Solution
Question 2
4 / -1

The figure shows two NAND gates followed by a NOR gate. The system is equivalent to the following logic gate

A
OR

B
AND

C
NAND

D
None of the above

Solution
Question 3
4 / -1

The voltage gain of the following amplifier is

A
10

B
100

C
1000

D
9.9

Solution
Question 4
4 / -1

Amplification factor of a triode is 10. When the plate potential is 200 volt and grid potential is – 4 volt, then the plate current of 4 mA is observed. If plate potential is changed to 160 volt and grid potential is kept at – 7 volt, then the plate current will be

A
1.69 mA

B
3.95 mA

C
2.87 mA

D
7.02 mA

Solution
Question 5
4 / -1

On applying a potential of –1 volt at the grid of a triode, the following relation between plate voltage V_p (volt) and plate current I_p (in mA) is found I_p = 0.125 V_p – 7.5. If on applying –3 volt potential at grid and 300 V potential at plate, the plate current is found to be 5 mA, then amplification factor of the triode is

A
100

B
50

C
30

D
20

Solution
Question 6
4 / -1

The temperature (T) dependence on resistivity (ρ) of asemiconductor is represented by

A

B

C

D
Question 7
4 / -1

In a forward biased PN–junction diode, the potentialbarrier in the depletion region is of the form

A

B

C

D
Question 8
4 / -1

The current between charge density and distance near P–N junction will be

A

B

C

D

Solution
Question 9
4 / -1

The resistance of a germanium junction diode whose V – I is shown in figure is (V_k = 0.3V )

A
5 k Ω

B
0.2 kΩ

C
2.3 kΩ

D

Solution
Question 10
4 / -1

The output in the circuit of figure is taken across a capacitor. It is as shown in figure

A

B

C

D

Solution