Tag: electronic devices

Questions Related to electronic devices

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

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

  1. -48.78

  2. -15.7

  3. -24.8

  4. -5.2

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Correct Option: A
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 $=A _i=?$
$A _i=-\cfrac{h _{fe}}{1+h _{oe}.R _{L}}=\cfrac{50}{1+25\times 10^{-6}\times 1\times 10^{3}}$
$=\cfrac{50}{1+25\times10^{-3}}$
$=-48.76$

The voltage gain of an amplifier with 9 negative feedback is 10. The voltage gain without feedback will be

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

  1. 90

  2. 100

  3. 10

  4. 1.25

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Correct Option: B
Explanation:

Let the input voltage be $v _{i}$ and the output voltage be $v _{0}$

We have $v _{0}=10v _{i}$ and $(v _{i}-0.09v _{0})A=0.01v _{0}$ , where $A$ is voltage gain without feedback
$\Rightarrow A=\frac{v _{0}}{0.01v _{0}}=100$
Therefore option $B$ is correct

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

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

  1. $1000$

  2. $10001$

  3. $1010$

  4. $1100$

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Correct Option: C
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 
$A _r = \dfrac{\Delta I _C \times R _L}{\Delta I _b \times R _i}$
$= \dfrac{10^{-3} \times 5 \times 10^3}{15 \times 10^{-6} \times 330} \approx 1010$

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. $\beta = 1 - \alpha$

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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 $1900$

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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

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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.