Tag: electronic devices

Quizzes (9) Questions (113)

Questions Related to electronic devices

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

The current gain $\beta$ of a transistor is $50$. The input resistance of the transistor, when used in the common emitter configuration, is $1$ $k\Omega$. The peak value of the collector a.c. current for an alternating peak input voltage $0.01$ V is?

  1. $100$ $\mu A$
  2. $250$ $\mu A$
  3. $500$ $\mu A$
  4. $800$ $\mu A$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

voltage gain is ${ A } _{ v }=\beta \frac { { R } _{ c } }{ { R } _{ B } } $

${ I } _{ c }=\frac { { A } _{ v }{ V } _{ m } }{ { R } _{ b } } =\frac { \beta { V } _{ m } }{ { R } _{ m } } =\frac { 50\times 0.01 }{ { 10 }^{ 3 } } $
$5\times { 10 }^{ -4 }A=500\mu \alpha$

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

A transistor is used as a common emitter amplifier with a load resistance $2 K ohm $. The input resistance is $150 Ohm $. Base current is charged by $20 \mu A$ which results in a change in collector current by $1.5 mA$. The voltage gain of the amplifier is

  1. $900$
  2. $1000$
  3. $1100$
  4. $1200$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

We know that voltage gain, $\dfrac{V _o}{V _i}=\beta\dfrac{R _L}{R _I}$


Where, $\beta=\dfrac{I _C}{I _B}=\dfrac{1.5mA}{20\mu A}=75$

Hence, voltage gain=$75\times\dfrac{2000}{150}=1000$

Hence, answer is option-(B).