Physics

Semiconductors and Diodes

457 Questions

Semiconductors and diodes form the foundation of modern electronics, covering topics like intrinsic carrier concentration and Zener breakdown. Questions often explore the characteristics of bipolar junction transistors (BJT) and the properties of doped silicon materials. This topic is crucial for physics and electronics engineering competitive exams.

Zener diode breakdownBJT circuit analysisIntrinsic carrier concentrationDoping in semiconductorsEmitter follower circuit

Semiconductors and Diodes Questions

Multiple choice zener diode special purpose diodes electronic devices semiconductor electronics: materials, devices and simple circuits physics

Which of the following is/are true regarding  breakdown voltage of Zener diode?

  1. If the reverse bias voltage across a p-n junction diode is increased, at a particular voltage the reverse current suddenly decreases to a large value

  2. The holes in the n-side and the conduction electrons in the p-side are accelerated due to the reverse bias voltage.

  3. The voltage at which the rate of creation of hole-electron pairs is increased leading to the increased current is called avalanche breakdown

  4. None of the above
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Consider applying an external voltage to the junction with the the negative terminal connected to the P type material and the positive terminal connected to the N type material. For the P-N junction this would require injecting electrons into the P type material. These electrons would recombine with holes and therefore further deplete the majority charge carriers in the P type semiconductor. 


Also removing electrons from the N type material would obviously cause further depletion the free electrons (the majority charge carriers in the N type semiconductor). This causes a widening of the depletion layer and an increase in the charge at the junction which reinforces the barrier voltage to oppose the applied voltage. The net result is that when the P-N junction is reversed biased it blocks current flow.

Multiple choice zener diode special purpose diodes electronic devices semiconductor electronics: materials, devices and simple circuits physics

Which of the following semi-conducting devices is used as voltage regulator?

  1. Zener diode

  2. LASER diode

  3. Photo diode

  4. Solar cell
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Zener diode is used as a voltage regulator device because the voltage across the zener diode in reverse bias is constant after reaching certain value i.e. breakdown voltage.

Multiple choice zener diode special purpose diodes electronic devices semiconductor electronics: materials, devices and simple circuits physics

What is true about the breakdown voltage in a zener diode?

  1. It decreases when current increases.

  2. It destroys the diode.

  3. It equals the current times the resistance.

  4. It is approximately constant.
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

As the current in the zener diode increases, the voltage across the diode remains constant. This voltage will be equal to the zener breakdown voltage

Multiple choice transistor and its types the bipolar junction transistor electronic devices semiconductor electronics: materials, devices and simple circuits physics

A transistor based radio receiver set $($effective resistance of the order of $18$ $ohms$$)$ operates on a $9V$ dc battery. If this replaced by a dc power supply with rating $9V$, $500V$ then

  1. Receiver will work normally

  2. Receiver will give distorted output

  3. Receiver will get burnt

  4. Power supply will get over heated
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Answer is A.

In this case, the 9 V battery is replaced with a 9 V power supply. Here, there is no change in the power supply to the radio receiver.
Hence, the receiver will work fine as the voltage supply or power supply remains the same.

Multiple choice transistor and its types the bipolar junction transistor electronic devices semiconductor electronics: materials, devices and simple circuits physics

For a transistor amplifier in common emitter configuration for load impedance of 1 k ( $h _{fe}$ = 50 and $h _{oe}$ = $25 \times 10^{-6}$) the current gain is

  1. -48.78

  2. -15.7

  3. -24.8

  4. -5.2
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Given,

Transistor amplifier in $C-E$ configuration.
Load impedance $'R _L'=1K\Omega$
$h _{fe}=50$
$h _{oe}=25\times 10^{-6}$
To find the current gain $=A _i=?$
$A _i=-\cfrac{h _{fe}}{1+h _{oe}.R _{L}}=\cfrac{50}{1+25\times 10^{-6}\times 1\times 10^{3}}$
$=\cfrac{50}{1+25\times10^{-3}}$
$=-48.76$

Multiple choice transistor and its types the bipolar junction transistor electronic devices semiconductor electronics: materials, devices and simple circuits physics

The input resistance of a common emitter amplifier is $330 \Omega$ and the load resistance is $5 k \Omega$. A change of base current is $15 \mu A$ results in the change of collector current by $1 mA$. The voltage gain of amplifier is

  1. $1000$
  2. $10001$
  3. $1010$
  4. $1100$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation
Given: $\Delta I _C = 1 mA = 10^{-3} A$
$\Delta I _b = 15 \mu A = 15 \times 10^{-6}A$
$R _L = 5 k\Omega = 5 \times 10^3 \Omega$
$Ri = 330 \Omega$
The voltage gain of an amplifier 
$A _r = \dfrac{\Delta I _C \times R _L}{\Delta I _b \times R _i}$
$= \dfrac{10^{-3} \times 5 \times 10^3}{15 \times 10^{-6} \times 330} \approx 1010$
Multiple choice transistor and its types the bipolar junction transistor electronic devices semiconductor electronics: materials, devices and simple circuits physics

The relationship between current gain $\alpha$ in Common Base [CB] mode and current gain $\beta$ in Common Emitter [CE] mode is

  1. $\beta = \alpha + 1$
  2. $\beta = \dfrac{\alpha}{1 - \alpha}$
  3. $\beta = \dfrac{\alpha}{1 + \alpha}$
  4. $\beta = 1 - \alpha$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

We define both current gains as $\beta = \dfrac{I _c}{I _b}$ and $\alpha = \dfrac{I _c}{I _e}$

Using $I _e = I _c + I _b$
Dividing both sides by $I _c$, we get $\dfrac{I _e}{I _c} = 1+\dfrac{I _b}{I _c}$
Or $\dfrac{1}{\alpha} = 1+\dfrac{1}{\beta}$
Or $\dfrac{1}{\beta} = \dfrac{1}{\alpha}-1$
$\implies$ $\beta = \dfrac{\alpha}{1-\alpha}$

Multiple choice transistor and its types the bipolar junction transistor electronic devices semiconductor electronics: materials, devices and simple circuits physics

In a common base transistor circuit, $I _C$ is the output current and $I _E$ is the input current. The current gain a pc is 

  1. Equal to one

  2. Greater than one

  3. Less than one

  4. None of these
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

In common base transistor circuit, the current gain $\left ( \alpha _{DC}=\dfrac{I _C}{I _E} \right )$ is less than one.

Multiple choice transistor and its types the bipolar junction transistor electronic devices semiconductor electronics: materials, devices and simple circuits physics

For a transistor in common base, the current gain is $0.95$. If the load resistance is $400\ k\Omega$ and input resistance is $200\Omega$, then the voltage gain and power gain will be

  1. $1900$ and $1800$
  2. $1900$ and $1805$
  3. $5525$ and $3591$
  4. $1805$ and $1900$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Given, $\alpha = 0.95, R _{0} = 400\times 10^{3}\Omega$
$R _{i} = 200\Omega$
As, voltage gain $= \alpha \dfrac {R _{0}}{R _{i}} = 0.95\times \dfrac {400\times 10^{3}}{200} = 1900$
Power gain $=$ Voltage gain $\times$ Current gain
$1900\times 0.95 = 1805$.

Multiple choice transistor and its types the bipolar junction transistor electronic devices semiconductor electronics: materials, devices and simple circuits physics

In a transistor amplifier, the two a.c. current gains $\alpha$ and $\beta$ are defined as $\alpha =\delta I _{C}/ \delta I _{E}$ and $\beta = \delta I _{C}/ \delta I _{B}$.
The relation between $\alpha$ and $\beta$ is

  1. $\beta = \dfrac {1 + \alpha}{\alpha}$
  2. $\beta = \dfrac {1 - \alpha}{\alpha}$
  3. $\beta = \dfrac {\alpha}{1 - \alpha}$
  4. $\beta = \dfrac {\alpha}{1 + \alpha}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

$I _{E} = I _{B} + I _{C}$
Differentiate w.r.t. to $I _{C}$ on both side
$\dfrac {\delta I _{E}}{\delta I _{C}} = \dfrac {\delta I _{B}}{\delta I _{C}} + 1$
$\dfrac {1}{\alpha} = \dfrac {1}{\beta} + 1$
$\Rightarrow \dfrac {1}{\beta} = \dfrac {1}{\alpha} - 1 = \dfrac {1 - \alpha}{\alpha}$
$\therefore \beta = \dfrac {\alpha}{1 - \alpha}$.

Multiple choice transistor and its types the bipolar junction transistor electronic devices semiconductor electronics: materials, devices and simple circuits physics

In CE NPN transistor ${10}^{10}$ electrons enter the emitter in ${10}^{-6}$s when it is connected to battery. About $5$% electrons recombine with holes in the base. The current gain of the transistor is______
$\left( e=1.6\times { 10 }^{ -19 }C \right) $

  1. $0.98$
  2. $19$
  3. $49$
  4. $0.95$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
Given:
The number of electrons entering the emitter is $10^{10}\ electrons$.
The time taken by the electrons is $10^{-6}\ s$.

Current gain common emitter 
$\beta=\cfrac { { I } _{ C } }{ { I } _{ B } } $
The emitter current is given as:
$I _E=\dfrac qt$

$\Rightarrow\dfrac{10^{10}\times1.6\times10^{-19}}{10^{-6}}$

$I _E=1.6\ mA$

Now, $5\%$ of the electrons recombine in the base region. So the base current will be:
$I _B=5\%\times1.6\ mA$
$I _B=0.08\ mA$

We know that the emitter current is the sum of the base current and collector current.
$I _E=I _B+I _C$

So, $I _C=1.6-0.08\ =\ 1.52\ mA$

$\implies\beta=\cfrac { 1.52 }{ 0.08 } $
$\quad =19$
Multiple choice transistor and its types the bipolar junction transistor electronic devices semiconductor electronics: materials, devices and simple circuits physics

In a common emitter configuration with suitable bias, it is given than ${R} _{L}$ is the load resistance and ${R} _{BE}$ is small signal dynamic resistance (input side). Then, voltage gain, current gain and power gain are given, respectively, by:
$\beta$ is current gain, ${ I } _{ B },{ I } _{ C },{ I } _{ E }$ are respectively base, collector and emitter currents:

  1. $\beta \cfrac { { R } _{ L } }{ { R } _{ BE } } ,\cfrac { \Delta { I } _{ E } }{ \Delta { I } _{ B } } ,{ \beta }^{ 2 }\cfrac { { R } _{ L } }{ { R } _{ BE } } $
  2. ${ \beta }^{ 2 }\cfrac { { R } _{ L } }{ { R } _{ BE } } ,\cfrac { \Delta { I } _{ C } }{ \Delta { I } _{ B } } ,\beta \cfrac { { R } _{ L } }{ { R } _{ BE } } $
  3. ${ \beta }^{ 2 }\cfrac { { R } _{ L } }{ { R } _{ BE } } ,\cfrac { \Delta { I } _{ C } }{ \Delta { I } _{ E } } ,{ \beta }^{ 2 }\cfrac { { R } _{ L } }{ { R } _{ BE } } $
  4. $\beta \cfrac { { R } _{ L } }{ { R } _{ BE } } ,\cfrac { \Delta { I } _{ C } }{ \Delta { I } _{ B } } ,{ \beta }^{ 2 }\cfrac { { R } _{ L } }{ { R } _{ BE } } $
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Voltage gain = $\dfrac{V _{CE}}{V _{BE}} =   \beta \dfrac{R _L}{R _{BE}}$

Current gain = $\beta = \dfrac{I _C}{I _B}$
Power gain = $voltage \ gain \times current \ gain = \beta^2 \dfrac{R _L}{R _{BE}}$

Multiple choice transistor and its types the bipolar junction transistor electronic devices semiconductor electronics: materials, devices and simple circuits physics

A common-emitter transistor amplifier has a current gain of 50. If the load resistance is $4k\Omega$ and input resistance is $500\Omega $, then the voltage gain of the amplifier is:

  1. 160

  2. 200

  3. 300

  4. 400
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

In common-emitter transistor amplifier,

$\beta=50$
$R _i=500\Omega$
$R _0=4k\Omega$
$\beta=\dfrac{I _c}{I _B}=50$
Voltage gain of the amplifier,
$A _v=\dfrac{V _0}{V _i}$

$A _v=\dfrac{V _{CE}}{V _{BE}}=\dfrac{I _CR _0}{I _BR _i}$

$A _v=\dfrac{50\times 4\times 1000}{500}=400$
The correct option is D.

Multiple choice transistor and its types the bipolar junction transistor electronic devices semiconductor electronics: materials, devices and simple circuits physics

The current gain $\beta$ of a transistor is $50$. The input resistance of the transistor, when used in the common emitter configuration, is $1$ $k\Omega$. The peak value of the collector a.c. current for an alternating peak input voltage $0.01$ V is?

  1. $100$ $\mu A$
  2. $250$ $\mu A$
  3. $500$ $\mu A$
  4. $800$ $\mu A$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

voltage gain is ${ A } _{ v }=\beta \frac { { R } _{ c } }{ { R } _{ B } } $

${ I } _{ c }=\frac { { A } _{ v }{ V } _{ m } }{ { R } _{ b } } =\frac { \beta { V } _{ m } }{ { R } _{ m } } =\frac { 50\times 0.01 }{ { 10 }^{ 3 } } $
$5\times { 10 }^{ -4 }A=500\mu \alpha$

Multiple choice transistor and its types the bipolar junction transistor electronic devices semiconductor electronics: materials, devices and simple circuits physics

A transistor is used as a common emitter amplifier with a load resistance $2 K ohm $. The input resistance is $150 Ohm $. Base current is charged by $20 \mu A$ which results in a change in collector current by $1.5 mA$. The voltage gain of the amplifier is

  1. $900$
  2. $1000$
  3. $1100$
  4. $1200$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

We know that voltage gain, $\dfrac{V _o}{V _i}=\beta\dfrac{R _L}{R _I}$


Where, $\beta=\dfrac{I _C}{I _B}=\dfrac{1.5mA}{20\mu A}=75$

Hence, voltage gain=$75\times\dfrac{2000}{150}=1000$

Hence, answer is option-(B).