Practice Start Test

Semiconductor

JEE Physics · 70 questions · Page 4 of 7 · Click an option or "Show Solution" to reveal answer

Q31
An NPN transistor is used in common emitter configuration as an amplifier with 1 kΩ\Omega load resistance. Signal voltage of 10 mV is applied across the base-emitter. This produces a 3 mA change in the collector current and 15μA change in the base current of the amplifier. The input resistance and voltage gain are :
A 0.67 kW, 200
B 0.33 kW, 1.5
C 0.67 kW, 300
D 0.33 kW, 300
Correct Answer
Option C
Solution

Input current = 15 × 10–6 Output current = 3 × 10–3 Resistance out put = 1000 Vinput = 10 × 10–3 Now Vinput = rinput × iinput 10 × 10–3 = rinput × 15 × 10–6 rinput =

20003=0.67 KΩ{{2000} \over 3} = 0.67\,K\Omega

Voltage gain =

VoutputVinput=1000×3×10−310×10−3=300{{{V_{output}}} \over {{V_{input}}}} = {{1000 \times 3 \times {{10}^{ - 3}}} \over {10 \times {{10}^{ - 3}}}} = 300
Q32
When a diode is forward biased, it has a voltage drop of 0.5 V. The safe limit of current through the diode is 10 mA. If a battery of emf 1.5 V is used in the circuit, the value of minimum resistance to be connected in series with the diode so that the current does not exceed the safe limit is
A 50 Ω\Omega
B 200 Ω\Omega
C 300 Ω\Omega
D 100 Ω\Omega
Correct Answer
Option D
Solution

1.5 – 0.5 – R ×\times 10–2 = 0 ⇒\Rightarrow R = 100

Ω\Omega
Q33
If a semiconductor photodiode can detect a photon with a maximum wavelength of 400 nm, then its band gap energy is : Planck’s constant h = 6.63 × \times 10–34 J.s. Speed of light c = 3 × \times 108 m/s
A 1.5 eV
B 2.0 eV
C 3.1 eV
D 1.1 eV
Correct Answer
Option C
Solution
E=hcλE = {{hc} \over \lambda }

=

6.63×10−34×3×108400×10−9{{6.63 \times {{10}^{ - 34}} \times 3 \times {{10}^8}} \over {400 \times {{10}^{ - 9}}}}

=

1240400{{1240} \over {400}}

eV = 3.1 eV

Q34
Which of the following will NOT be observed when a multimeter (operating in resistance measuring mode) probes connected across a component, are just reversed?
A Multimeter shows NO deflection in both cases i.e. before and after reversing the probes if the chosen component is metal wire.
B Multimeter shows a deflection, accompanied by a splash of light out of connected component in one direction and NO deflection on reversing the probes if the chosen component is LED.
C Multimeter shows an equal deflection in both cases i.e. before and after reversing the probes if the chosen component is resistor.
D Multimeter shows NO deflection in both cases i.e. before and after reversing the probes if the chosen component is capacitor.
Correct Answer
Option D
Solution

(1) Multimeter shows deflection when it connects with capacitor.

(2) If we assume that LED has negligible resistance then multimeter shows no deflection for the forward bias but when it connects in reverse direction, it break down occurs so splash of light out.

(3) The resistance of metal wire may be taken zero, so no deflection in multimeter.

(4) No matter, how we connect the resistance across multimeter.

It shows same deflection.

Q35
With increasing biasing voltage of a photodiode, the photocurrent magnitude :
A Increases initially and after attaining certain value, it decreases
B Increases linearly
C Increases initially and saturates finally
D Remains constant
Correct Answer
Option C
Solution

In photodiode, photocurrent increases with increasing biasing voltage and then becomes saturated.

Q36
If an emitter current is changed by 4 mA, the collector current changes by 3.5 mA. The value of β\beta will be :
A 0.875
B 0.5
C 3.5
D 7
Correct Answer
Option D
Solution

Given, emitter current, IE = 4 mA Collector current, IC = 3.5 mA Current gain in common base amplifier,

α=ICIE\alpha = {{{I_C}} \over {{I_E}}}
⇒α=3.54=78\Rightarrow \alpha = {{3.5} \over 4} = {7 \over 8}

Also, current gain in common emitter amplifier,

β=α1−α\beta = {\alpha \over {1 - \alpha }}
⇒β=7/81−7/8\Rightarrow \beta = {{7/8} \over {1 - 7/8}}
β=7\beta = 7
Q37
Given below are two statements : Statement I : PN junction diodes can be used to function as transistor, simply by connecting two diodes, back to back, which acts as the base terminal. Statement II : In the study of transistor, the amplification factor β\beta indicates ratio of the collector current to the base current. In the light of the above statements, choose the correct answer from the options given below.
A Both Statement I and Statement II are false
B Statement I is false but Statement II is true
C Both Statement I and Statement II are true
D Statement I is true but Statement II is false
Correct Answer
Option B
Solution

S-1 : Statement 1 is false because in case of two discrete back to back connected diodes, there are four doped regions instead of three and there is nothing that resembles a thin base region between an emitter and a collector.

S-2 : Statement-2 is true, as we know that, amplification factor (β\beta) is the ratio of collector current to base current.

β=ICIB\beta = {{{I_C}} \over {{I_B}}}
Q38
Zener breakdown occurs in a p −- n junction having p and n both :
A heavily doped and have wide depletion layer.
B lightly doped and have narrow depletion layer.
C heavily doped and have narrow depletion layer.
D lightly doped and have wide depletion layer.
Correct Answer
Option C
Solution

The zener breakdown occurs in the heavily doped p-n junction diode.

Heavily doped p-n junction diodes have narrow depletion region.

Q39
For extrinsic semiconductors; when doping level is increased;
A Fermi-level of both p-type and n-type semiconductors will go upward for T > TFK and downward for T < TFK, where TF is Fermi temperature.
B Fermi-level of p-type semiconductor will go upward and Fermi-level of n-type semiconductors will go downward
C Fermi-level of p and n-type semiconductors will not be affected.
D Fermi-level of p-type semiconductors will go downward and Fermi-level of n-type semiconductor will go upward.
Correct Answer
Option D
Solution

In n-type semiconductor pentavalent impurity is added.

Each pentavalent impurity donates a free electron.

So the Fermi-level of n-type semiconductor will go upward.

And In p-type semiconductor trivalent impurity is added.

Each trivalent impurity creates a hole in the valence band.

So the Fermi-level of p-type semiconductor will go downward.

Q40
LED is constructed from Ga-As-P semiconducting material. The energy gap of this LED is 1.9 eV. Calculate the wavelength of light emitted and its colour. [h = 6.63 ×\times 10−-34 Js and c = 3 ×\times 108 ms−-1]
A 654 nm and orange colour
B 654 nm and red colour
C 1046 nm and red colour
D 1046 nm and blue colour
Correct Answer
Option B
Solution

We know that

E=hcλE = {{hc} \over \lambda }
λ=hcE⇒1240(in eV)E(in eV)\lambda = {{hc} \over E} \Rightarrow {{1240(in\,eV)} \over {E(in\,eV)}}
λ=12401.9\lambda = {{1240} \over {1.9}}

= 652.63 nm ≈\approx 654 nm Wavelength of red light is 620 nm to 750 nm

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