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Questions Related to physics

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

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

  1. $\beta = \alpha + 1$

  2. $\beta = \dfrac{\alpha}{1 - \alpha}$

  3. $\beta = \dfrac{\alpha}{1 + \alpha}$

  4. $\beta = 1 - \alpha$

Show answer
Correct Option: B
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 $\dfrac{1}{\beta} = \dfrac{1}{\alpha}-1$
$\implies$ $\beta = \dfrac{\alpha}{1-\alpha}$

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 

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

  1. Equal to one

  2. Greater than one

  3. Less than one

  4. None of these

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Correct Option: C
Explanation:

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

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

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

  1. $1900$ and $1800$

  2. $1900$ and $1805$

  3. $5525$ and $3591$

  4. $1805$ and $1900$

Show answer
Correct Option: B
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
$1900\times 0.95 = 1805$.

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

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

  1. $\beta = \dfrac {1 + \alpha}{\alpha}$

  2. $\beta = \dfrac {1 - \alpha}{\alpha}$

  3. $\beta = \dfrac {\alpha}{1 - \alpha}$

  4. $\beta = \dfrac {\alpha}{1 + \alpha}$

Show answer
Correct Option: C
Explanation:

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

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) $

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

  1. $0.98$

  2. $19$

  3. $49$

  4. $0.95$

Show answer
Correct Option: B
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 } $
$\quad =19$

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:

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

  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 } } $

Show answer
Correct Option: D
Explanation:

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

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

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:

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

  1. 160

  2. 200

  3. 300

  4. 400

Show answer
Correct Option: D
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.

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?

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

  1. $100$ $\mu A$

  2. $250$ $\mu A$

  3. $500$ $\mu A$

  4. $800$ $\mu A$

Show answer
Correct Option: C
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 } } $
$5\times { 10 }^{ -4 }A=500\mu \alpha$

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

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

  1. $900$

  2. $1000$

  3. $1100$

  4. $1200$

Show answer
Correct Option: B
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).

A square loop of side 12 cm and resistance 0.60$\Omega$ is placed vertically in the east-west plane. A uniform magnetic field of 0.1 T is setup across the plane in north-east direction. The magnetic field is decreased to zero in 0.6 s at a steady rate. The magnitude of current during this time interval is

physics magnetic fields and electromagnetism magnetic flux density magnetic flux electromagnetic induction

  1. $1.42 \times 10^{-3} A$

  2. $2.67 \times 10^{-3} A$

  3. $3.41\times 10^{-3} A$

  4. $4.21 \times 10^{-3} A$

Show answer
Correct Option: B
Explanation:

Here, Area $A=l^2=(12cm)^2=1.4\times 10^{-2} m^2$
$R=0.60\omega, B _1=0.10 T,\theta=45^0$
$B _2=0,dt=0.6$ s
Initial flux,
$\phi _1=B _1Acos\theta$
      $=0.10\times1.4\times10^{-2}\times cos 45^0$
      $=9.8\times 10^{-4}$

final flux, $\phi _2$=0
Induced emf,$E=\dfrac{| d\phi |}{dt}=\dfrac{|\phi _2-\phi _1|}{dt}$
                      $E=\dfrac{|9.8\times 10^{-4}|}{0.6}s\\,\,\,\,\,=1.6\times10^{-3}V$
Current, $I=\dfrac{E}{R}=\dfrac{1.6\times\times 10^{-3}}{0.6}=2.67 \times 10^{-3}$