In a common emitter amplifier, the phase difference between input signal and output signal is:

1. zero             

2. π               

3. π2               

4. π4

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In an NPN transistor amplifier, if 98% of electrons emitted from the emitter reach the collector, then the value of collector current for the base current of 40 μA will be:

1. 1.96 mA             

2. 1.92 mA             

3. 1.96 A             

4. 1.92 A

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The energy band diagrams for semiconductor samples of silicon are as shown. We can assert that:

 

1. Sample \(X\) is undoped while samples \(Y\) and \(Z\) have been doped with a third group and a fifth group impurity respectively.
2. Sample \(X\) is undoped while both samples \(Y\) and \(Z\) have been doped with a fifth group impurity.
3. Sample \(X\) has been doped with equal amounts of third and fifth group impurities while samples \(Y\) and \(Z\) are undoped.
4. Sample \(X\) is undoped while samples \(Y\) and \(Z\) have been doped with a fifth group and a third group impurity respectively.
Subtopic:  Energy Band theory |
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If the reverse bias in a junction diode is changed from \(5~\text V\) to \(15~\text V\) then the value of current changes from \(38~\mu \text{A}\) to \(88~\mu \text{A}.\) The resistance of the junction diode will be:
1. \(4\times10^{5}\) 
2. \(3\times10^{5}\)
3. \(2\times10^{5}\)
4. \(10^{6}\)

Subtopic:  PN junction |
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In a common emitter circuit, if VCC is changed by 0.2 V, collector current changes by 4 x 10–3 mA. The output resistance will be:

1. 10 kΩ 

2. 30 kΩ 

3. 50 kΩ 

4. 70 kΩ

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An LED is constructed from a \(\mathrm{p\text{-}n}\) junction diode using \(\mathrm{GaAsP}.\) The energy gap is \(1.9~\text{eV}.\) The wavelength of the light emitted will be equal to:
1. \(10.4 \times 10^{-26}~ \text{m}\)
2. \(654~ \text{nm}\)
3. \(654~ \text{m}\)
4. \(654\times 10^{-11}~\text{m}\)

Subtopic:  Applications of PN junction |
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Level 1: 80%+
NEET - 2019
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In a given circuit as shown the two input waveforms \(A\) and \(B\) are applied simultaneously. The resultant waveform \(Y\) is:
     

1. 2.
3. 4.
Subtopic:  Logic gates |
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In the circuit shown in the figure, the maximum output voltage \(V_0\) is:

     

1. \(0\) V
2. \(5\) V
3. \(10\) V
4. \(\frac{5}{\sqrt{2}}~\text{V}\)

Subtopic:  Rectifier |
Level 3: 35%-60%
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In the circuit given below, if \(V(t)\) is the sinusoidal voltage source, then the voltage drop \(V_{AB}(t)\) across the resistance \(R\):

1. is half-wave rectified.
2. is full-wave rectified.
3. has the same peak value in the positive and negative half-cycles.
4. has different peak values during the positive and negative half-cycles.
Subtopic:  Rectifier |
Level 3: 35%-60%
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A \(2\) V battery is connected across the points \(A\) and \(B\) as shown in the figure given below. Assuming that the resistance of each diode is zero in forward bias and infinity in reverse bias, the current supplied by the battery when its positive terminal is connected to \(A\) is:

1. \(0.2\) A 2. \(0.4\) A
3. zero 4. \(0.1\) A
Subtopic:  PN junction |
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Level 1: 80%+
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