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

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. $620\ V$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

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

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

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

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

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
Reveal answer Fill a bubble to check yourself
B Correct answer
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  $200.$

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 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
Reveal answer Fill a bubble to check yourself
B Correct answer
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

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