When the base current in a transistor is changed from 30 - A to 80 - A- the collector current is changed from 1-0 mA to 3-5 mA- Find the current gain -

Given δ l_n=80 μ A 30 μ A =50 μ A=50× 10^ 6 A δ l_c=3.5 μ A 1 mA =2.5 mA=2.5× 10^ 3 A β=(δ l_n/δ l_c )V_co (current) =2.5× 10^ 3/50× 10^ 6=250/50=50 So, ...

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Standard XII

Physics

Transistor as an Amplifier

When the base...

Question

When the base current in a transistor is changed from $30 μ A$ to $80 μ A,$ the collector current is changed from 1.0 mA to 3.5 mA. Find the current gain $β .$

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Solution

Given $δ l_{n} = 80 μ A - 30 μ A$

$= 50 μ A = 50 \times 10^{- 6} A$

$δ l_{c} = 3.5 μ A - 1 m A$

$= 2.5 m A = 2.5 \times 10^{- 3} A$

$β = (\frac{δ l_{n}}{δ l_{c}}) V_{c o} (c u r r e n t)$

$= \frac{2.5 \times 10^{- 3}}{50 \times 10^{- 6}} = \frac{250}{50} = 50$

So, Current gain = 50

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When the base current in a transistor is changed from 30 μA to 80 μA, the collector current is changed from 1.0 mA to 3.5 mA. Find the current gain β.

Given δln=80 μA−30 μA =50 μA=50×10−6 A δlc=3.5 μA−1 mA =2.5 mA=2.5×10−3 A β=(δlnδlc)Vco (current) =2.5×10−350×10−6=25050=50 So, Current gain = 50

When the base current in a transistor is changed from $30 μ A$ to $80 μ A,$ the collector current is changed from 1.0 mA to 3.5 mA. Find the current gain $β .$

Given $δ l_{n} = 80 μ A - 30 μ A$

$= 50 μ A = 50 \times 10^{- 6} A$

$δ l_{c} = 3.5 μ A - 1 m A$

$= 2.5 m A = 2.5 \times 10^{- 3} A$

$β = (\frac{δ l_{n}}{δ l_{c}}) V_{c o} (c u r r e n t)$

$= \frac{2.5 \times 10^{- 3}}{50 \times 10^{- 6}} = \frac{250}{50} = 50$

So, Current gain = 50