Semiconductor Electronics · Question 43

The input signal given to a CE amplifier having a voltage gain of 150 is Vi = 2 (15t + 3 ). The corresponding output signal will be

1. 2 (15t + 5 6 )
2. 300 (15t + 4 3 )
3. 300 (15t + 3 )
4. 75 (15t + 2 3 )

Which logic gate is represented by the following combination of logic gates? [figure]

1. AND
2. NOR
3. OR
4. NAND

If in a p-n junction, a square input signal of 10 V is applied, as shown, [figure] then the output across RL will be

1. [figure]
2. [figure]
3. [figure]
4. [figure]

The given graph represents V-I characteristic for a semiconductor device. [figure] Which of the following statement is correct ?

1. It is V-I characteristic for solar cell where, point A represents open circuit voltage and point B short circuit current.
2. It is for a solar cell and points A and B represent open circuit voltage and current, respectively.
3. It is for a photodiode and points A and B represent open circuit voltage and current, respectively.
4. It is for a LED and points A and B represent open circuit voltage and short circuit current, respectively.

The barrier potential of a p-n junction depends on Which one of the following is correct ?

1. A and B only
2. B only
3. B and C only
4. A, B and C

In a n-type semiconductor, which of the following statements is true?

1. Holes are minority carriers and pentavalent atoms are dopants.
2. Holes are majority carriers and trivalent atoms are dopants.
3. Electrons are majority carriers and trivalent atoms are dopants.
4. Electrons are minority carriers and pentavalent atoms are dopants.

The output (X) of the logic circuit shown in figure will be [figure]

1. X = A . B
2. X = A + B
3. X = A . B
4. X = A . B

In a common emitter (CE) amplifier having a voltage gain G, the transistor used has transconductance 0.03 mho and current gain 25. If the above transistor is replaced with another one with transconductance 0.02 mho and current gain 20, the voltage gain will be

1. 13G
2. 54G
3. 23G
4. 1.5G

The output from a NAND gate is divided into two in parallel and fed to another NAND gate. The resulting gate is a [figure]

1. AND gate
2. NOR gate
3. OR gate
4. NOT gate

In an unbiased p-n junction, holes diffuse from the p-region to n-region because of

1. The attraction of free electrons of n-region.
2. The higher hole concentration in p-region than that in n-region.
3. The higher concentration of electrons in the n-region than that in the p-region.
4. The potential difference across the p-n junction.