Tag: semiconductor electronics: materials, devices and simple circuits

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

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

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

  1. $100$

  2. $150$

  3. $200$

  4. $50$

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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:

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

  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

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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$?

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

  1. $8.4$

  2. $4.8$

  3. $2.4$

  4. $480$

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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

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

  1. $500$

  2. $460$

  3. $600$

  4. $560$

Show answer
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

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

  1. $30.42$

  2. $29.57$

  3. $28.18$

  4. $27.15$

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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

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

  1. 0.8 mA

  2. 0.6 mA

  3. 0.4 mA

  4. 1 mA

Show answer
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

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

  1. $0.62\ V$

  2. $6.2\ V$

  3. $62\ V$

  4. $620\ V$

Show answer
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

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

  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

Show answer
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

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

  1. Receiver will work normally

  2. Receiver will give distorted output

  3. Receiver will get burnt

  4. Power supply will get over heated

Show answer
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 :

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

  1. $0.42 mA$

  2. $0.49 mA$

  3. $0.29 mA$

  4. $0.35 mA$

Show answer
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,
$\displaystyle { I } _{ b }={ I } _{ c }-{ I } _{ c }=.2-6.912=0.29mA$