Tag: semiconductor electronics: materials, devices and simple circuits

Questions Related to semiconductor electronics: materials, devices and simple circuits

For a common emitter circuit if $I _C$ / $I _E$ =0.98 then current gain for common emitter circuit will be

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

  1. 4.9

  2. 98

  3. 49

  4. 9.8

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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

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

  1. 0.20 mA

  2. 0.36 mA

  3. 0.29 mA

  4. 0.45 mA

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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

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

  1. $50$

  2. $75$

  3. $100$

  4. $25$

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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

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
$\dfrac {\delta I _{E}}{\delta I _{C}} = \dfrac {\delta I _{B}}{\delta I _{C}} + 1$
$\dfrac {1}{\alpha} = \dfrac {1}{\beta} + 1$
$\Rightarrow \dfrac {1}{\beta} = \dfrac {1}{\alpha} - 1 = \dfrac {1 - \alpha}{\alpha}$
$\therefore \beta = \dfrac {\alpha}{1 - \alpha}$.