0

The bipolar junction transistor - class-XII

Description: the bipolar junction transistor
Number of Questions: 36
Created by:
Tags: physics semiconductor electronics: materials, devices and simple circuits electronic devices
Attempted 0/36 Correct 0 Score 0

The input resistance of a silicon transistor is $1 k\Omega $.If base current is changed by $100 \mu A$ it causes the change in collector current by $2 mA$. This transistor is used as a CE amplifier with a load resistance of $5 k\Omega$ What is the ac voltage gain of amplifier?

  1. $10$

  2. 44100$

  3. $500$

  4. $200$


Correct Option: D

In a silicon transistor the base current is changed by $25\mu A$. This results in a change of $0.04\ V$ in the base to emitter voltage and a change of $4\ mA$ in the collector current. The trans-conductance of the transistor (in $\Omega^{-1}$) is

  1. $0.1$

  2. $1$

  3. $1000$

  4. $0.01$


Correct Option: C

For a common emitter configuration, If $\alpha $and $\beta $ have thier usual meanings. The incorrect relationship between $\alpha $and $\beta $ is:

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

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

  3. $\alpha =\dfrac { { \beta }^{ 2 } }{ 1+{ \beta }^{ 2 } } $

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


Correct Option: B,D

For a common emitter connection the values of constant collector current and base current are $5\ mA$ and $50$$\mu A$ respectively. The current gain will be

  1. $10$

  2. $20$

  3. $40$

  4. $100$


Correct Option: D
Explanation:

Given that:
$I _c = 5\ mA$
$I _b = 50\ \mu A = 5 \times 10^{-2}\ mA$
The current gain is
$\beta = \dfrac{I _c}{I _b}$
$\beta = \dfrac{5}{5 \times 10^{-2}} = 100$

In a silicon transistor, the base  current is changed by 20 change of 0.02V in base to emitter voltage and a change of 2mA in the collector current.
(a) Find the input  resistance ${ \beta  } _{ ac }$ 
(b) If this transistor is used as an-amplifier. Find the voltage gain of the-amplifier with the load resistance 5k amplifier.

  1. a)100

    b)-500

  2. a)1000

    b)-500

  3. a)2000

    b)-5000

  4. a)10

    b)-5


Correct Option: A
Explanation:

(a) $n=\frac { { \triangle v } _{ BE } }{ { \triangle | } _{ B } } =\frac { 0.02 }{ 20\times { 10 }^{ -6 } } =1k\Omega $
${ \beta  } _{ ac }=\frac { { \triangle | } _{ c } }{ { \triangle | } _{ B } } =\frac { 2\times { 10 }^{ -3 } }{ 20\times { 10 }^{ -6 } } =100$
(b) Voltage gain,${ A } _{ v }=-{ g } _{ m }{ R } _{ L }=-\left( 0.1 \right) \left( 5\times { 10 }^{ 3 } \right) =-500$. $\left( since\quad { g } _{ m }={ \beta  } _{ ac }/n \right) $

The ac current gain of a transistor is $120$. What is the change in the collector current in the transistor whose base current changes by $100\ \mu A$?

  1. $6\ mA$

  2. $12\ mA$

  3. $3\ mA$

  4. $24\ mA$


Correct Option: B
Explanation:

Given that,

Current gain$\beta =120\,A$

Base current $\Delta {{I} _{b}}=100\,\mu A$


We know that,

  $ {{\beta } _{ac}}=\dfrac{\Delta {{I} _{C}}}{\Delta {{I} _{B}}} $

 $ \Delta {{I} _{C}}=120\times 100\times {{10}^{-6}} $

 $ \Delta {{I} _{C}}=12\times {{10}^{-3}}\,A $

 $ \Delta {{I} _{C}}=12\,mA $

Hence, the collector current in the transistor is $12\ mA$

 

Current gain of a transistor is $0.99$. If it is used in common emitted mode to form a transistor amplifier with input resistance $1\ k\Omega$ and load resistance $10\ k\Omega$, then the voltage gain of the amplifier is

  1. $99$

  2. $0.99$

  3. $9.9$

  4. $990$


Correct Option: C

For a CE transistor amplifier, the audio signal voltage across the collector resistance of $2k\Omega $ is 2 V.Suppose the current amplification factor of the transistor is 100. The value of ${R _B}$ in series with ${V _{BB}}$ the supply of 2V, if the DC base current has to be 10 times the signal current is 

  1. $4k\Omega $

  2. $14k\Omega $

  3. $28k\Omega $

  4. $54k\Omega $


Correct Option: C

When a collector to emitter voltage is constant in a transistor, the collector current changes by $8.2mA$ when the emitter current changes by $8.3mA$. The value of forward current ratio $B$ is

  1. $82$

  2. $83$

  3. $8.2$

  4. $8.3$


Correct Option: A

In the CB mode of a transistor when the collector voltage is changed by $0.5V$. The collector current changes by $0.05mA$. The output resistance will be

  1. $10k\Omega$

  2. $20k\Omega$

  3. $5k\Omega$

  4. $2.5k\Omega$


Correct Option: A

Consider an n-p-n transistor amplifier in common-emitter configuration. The current gain of the transistor is $100$. if the collector current changes by $1mA$. What will be the change in emitter current.

  1. $1.1mA$

  2. $1.01mA$

  3. $0.01mA$

  4. $10mA$


Correct Option: B

Power gain for $N - P - N$ transistor is $10^6$, input resistance $100 \Omega$ and output resistance $1000 \,\Omega$. find out current gain.

  1. $100$

  2. $150$

  3. $200$

  4. $50$


Correct Option: A
Explanation:

Power gain = $(Current \, gain)^2 \left(\dfrac{R _{out}}{R _{in}} \right)$
$10^6 = \beta^2 \times \dfrac{10^4}{100}$
$\beta = 100$

A transistor, when connected in common emitter mode, has a:

  1. high input resistance and a low output resistance

  2. medium input resistance and high output resistance

  3. very low input resistance and a low output resistance

  4. high input resistance and a high output resistance


Correct Option: C

What is the voltage gain in a common emitter amplifier, where input resistance is $3.2$ and load resistance $24$$\Omega, \beta$$ = 0.6$?

  1. $8.4$

  2. $4.8$

  3. $2.4$

  4. $480$


Correct Option: B
Explanation:

Voltage gain, $\displaystyle A _v = \beta \dfrac{R _L}{R _i} = 0.6 \times \dfrac{24}{3} = 4.8$

In a common emitter transistor amplifier $\beta$ $= 60$, $R _0$ $= 5000$$\Omega$ and internal resistance of a transistor is $500$$\Omega$. The voltage amplification of amplifier will be

  1. $500$

  2. $460$

  3. $600$

  4. $560$


Correct Option: C
Explanation:

Voltage amplification, $\displaystyle A _v = \beta \dfrac{R _0}{R _i} = 60 \times \dfrac{5000}{500} = 600$

In a triode, $g$$ _m$ $= 2$ $\times$$ 10$$^{-3}$ $\Omega^{-1}$, $\mu$ $= 42$, resistance of load, $R = 50k\Omega$. The voltage amplification obtained from this triode will be

  1. $30.42$

  2. $29.57$

  3. $28.18$

  4. $27.15$


Correct Option: B
Explanation:

Voltage gain, $\displaystyle A _v = \beta \dfrac{R _L}{R _i} = 0.6 \times \dfrac{24}{3} = 4.8$

The grid voltage of any triode valve is changed from -1 volt to -3 volt and the mutual conductance is 3$\times$ 10$^{-4}$ mho. The change in plate circuit current will be

  1. 0.8 mA

  2. 0.6 mA

  3. 0.4 mA

  4. 1 mA


Correct Option: B
Explanation:

$\displaystyle g _m = \frac{\Delta I _p}{\Delta V _g}$

or $\Delta I _p = g _m \times \Delta V _g = 3 \times 10^{-4} \times [-3-(-1)]$
$= - 0.6 \times 10^{-3} A$ = shortage of $0.6 \times 10^{-3}A$

In common emitter amplifier, the current gain is $62$.The collector resistance and input resistance are $5\ k$$\Omega$ an $500$ $\Omega$ respectively. If the input voltage is $0.01\ V$, the output voltage is

  1. $0.62\ V$

  2. $6.2\ V$

  3. $62\ V$

  4. $620\ V$


Correct Option: B
Explanation:

$\displaystyle \dfrac{V _0}{V _m} = \dfrac{R _0}{R _{in}} \times \beta = \dfrac{5 \times 10^3 \times 62}{500} = 10 \times 62 = 620$

$V _0 = 620 \times V _{in} = 620 \times 0.01 = 6.2\ V$

$\therefore V _0$ = $6.2\ V$

The current gain $\beta$-may be defined as

  1. the ratio of change in collector current to the change in emitter current for a constant collector voltage in a common base arrangement

  2. the ratio of change in collector current to the change in base current at constant collector voltage in a common emitter circuit

  3. the ratio of change in emitter current to the change in base current for constant emitter voltage in common emitter circuit

  4. the ratio of change in base current to the change in collector current in constant collector voltage in common emitter circuit


Correct Option: B
Explanation:

By definition, current gain  $\beta$ is defined as the ratio of change in collector current to change in base current for a constant collector voltage in a common emitter circuit.

$\beta =  \dfrac{\Delta i _C}{\Delta i _B}$   or simply    $\beta = \dfrac{I _C}{I _B}$
Generally,  $\beta$ has value between  $20$ and  $200.$

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


Correct Option: A
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.

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. $0.35 mA$


Correct Option: C
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,
$\displaystyle { I } _{ b }={ I } _{ c }-{ I } _{ c }=.2-6.912=0.29mA$

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


Correct Option: C
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 $C$.

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


Correct Option: C
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, $I _{B}=I _{E}-I _{C}$
$=(7.5-7.2)\, mA=0.3\, mA$
$\approx 0.29\,mA$

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. $25$


Correct Option: C
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.

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


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

  1. 90

  2. 100

  3. 10

  4. 1.25


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

  1. $1000$

  2. $10001$

  3. $1010$

  4. $1100$


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

  1. $\beta = \alpha + 1$

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

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

  4. $\beta = 1 - \alpha$


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 

  1. Equal to one

  2. Greater than one

  3. Less than one

  4. None of these


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

  1. $1900$ and $1800$

  2. $1900$ and $1805$

  3. $5525$ and $3591$

  4. $1805$ and $1900$


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

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

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

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

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


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) $

  1. $0.98$

  2. $19$

  3. $49$

  4. $0.95$


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:

  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 } } $


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:

  1. 160

  2. 200

  3. 300

  4. 400


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?

  1. $100$ $\mu A$

  2. $250$ $\mu A$

  3. $500$ $\mu A$

  4. $800$ $\mu A$


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

  1. $900$

  2. $1000$

  3. $1100$

  4. $1200$


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).

- Hide questions