Physics

Semiconductors and Diodes

461 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 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

The current gain for a transistor working as common base amplifier is $0.96$. lf the emitter current is $7.2 mA$, the base current will be :

  1. $0.42 mA$
  2. $0.49 mA$
  3. $0.29 mA$
  4. $0.35 mA$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Here, $\displaystyle \alpha =0.96,{ I } _{ e }=7.2mA$
$\displaystyle \frac { { I } _{ c } }{ { I } _{ e } } =0.96$
$\displaystyle { I } _{ c }={ I } _{ e }\times 0.96=7.2\times 0.96=6.912mA$
Now,
$\displaystyle { I } _{ b }={ I } _{ c }-{ I } _{ c }=.2-6.912=0.29mA$

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

For a common emitter circuit if $I _C$ / $I _E$ =0.98 then current gain for common emitter circuit will be

  1. 4.9

  2. 98

  3. 49

  4. 9.8
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Given that $\frac{I _{c}}{I _{e}}=0.98$

$\Rightarrow I _{c}=0.98I _{e}$
So we get $I _{b}=I _{e}-I _{c}=0.02I _{e}$
Therefore the current gain is $\frac{I _{c}}{I _{b}}=\frac{0.98}{0.02}=49$
Therefore the correct option is $C$.

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

In a p-n-p transistor, working as a common base amplifier, current gain is 0.96 and emitter current is 7.2mA. The base current is

  1. 0.20 mA

  2. 0.36 mA

  3. 0.29 mA

  4. 0.45 mA
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Current gain $=\dfrac{I _{C}}{I _{B}}$
$I _{C}$ = collector current
$I _{B}$ = base current
So, $I _{B}=\dfrac{I _{C}}{current\, gain}= \dfrac{7.2\, mA}{0.96}=7.5\, mA$
As $I _{E}=I _{B}+I _{C}$
So, $I _{B}=I _{E}-I _{C}$
$=(7.5-7.2)\, mA=0.3\, mA$
$\approx 0.29\,mA$

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

A transistor is operated in common emitter configuration at $V _c=2V$ such that a change in the base current from $100\mu A$ to $300\mu A$ produces a change in the collector current from 10mA to 20mA. The current gain is

  1. $50$
  2. $75$
  3. $100$
  4. $25$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The base current changes from $100\mu A$ to $300 \mu A$ and produces a change in the collector current from $10mA$ to $20mA$

The change in collector current is $10mA$
The change in base current is $200\mu A$
The current gain is $\frac{10000}{200}=50$
Therefore option $C$ is correct.

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 } } $
$5\times { 10 }^{ -4 }A=500\mu \alpha$