Tag: the bipolar junction transistor

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Questions Related to the bipolar junction transistor

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 
$A _r = \dfrac{\Delta I _C \times R _L}{\Delta I _b \times R _i}$
$= \dfrac{10^{-3} \times 5 \times 10^3}{15 \times 10^{-6} \times 330} \approx 1010$
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
$1900\times 0.95 = 1805$.