Tag: electronic devices

Questions Related to electronic devices

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.