The base current in a transistor circuit changes from \[45\mu A\] to \[140\mu A\]. Accordingly, the collector current changes from $0.2 mA$ to $4.00 mA$. The gain in current is:
A) 9.5
B) 1
C) 40
D) 20
Answer
283.8k+ views
Hint: For a transistor in common base configuration the current gain is defined as the ratio of change in collector current and the change in emitter current when the voltage between base and collector is kept constant.
Formulae used:
Current gain of a transistor circuit is given by:
\[{g_I} = \dfrac{{\Delta {I_C}}}{{\Delta {I_B}}}\]..............(1)
Where,
\[{g_I}\]denotes the current gain of the transistor,
\[\Delta {I_C}\]denotes change in collector current of the transistor,
\[\Delta {I_B}\] denotes change in base current of the transistor.
Complete step by step solution:
Here, base current changes from \[45\mu A\] to \[140\mu A\].
Here, collector current changes from 0.2 mA to 4 mA.
To find: Gain in current for the transistor.
From the given information get the change in base current as:
\[
\Delta {I_B} = 140\mu A - 45\mu A = 95\mu A \\
\therefore \Delta {I_B} = 9.5 \times {10^{ - 5}}A \\
\]....................(2)
Similarly, get the change in collector current as:
\[
\Delta {I_C} = 4mA - 0.2mA = 3.8mA \\
\therefore \Delta {I_C} = 3.8 \times {10^{ - 3}}A \\
\]...................(3)
Substitute the obtained value of \[\Delta {I_B}\]and \[\Delta {I_C}\]from eq.(2) and eq.(3) respectively into the eq.(1) to obtain current gain \[{g_I}\]as:
\[
{g_I} = \dfrac{{3.8 \times {{10}^{ - 3}}A}}{{9.5 \times {{10}^{ - 5}}A}} = \dfrac{{3800}}{{95}} \\
\therefore {g_I} = 40 \\
\]
Gain in current for the transistor is got to be (C), 40.
Additional information:
Current gain of a transistor is completely an intrinsic property of the transistor itself. This quantity is not very much precise in general and varies quite abruptly depending on its production. When the transistor is used as a current gaining device then its quality is measured by this current gain value.
Note: A student might make a mistake while using the current gain formula. Just remember how actually the transistor works. At the base of the transistor a small amount of current is injected. This allows the load to draw a large current through the collector. This ratio of change in collector current for the change in base current gives the current gain of the transistor.
Formulae used:
Current gain of a transistor circuit is given by:
\[{g_I} = \dfrac{{\Delta {I_C}}}{{\Delta {I_B}}}\]..............(1)
Where,
\[{g_I}\]denotes the current gain of the transistor,
\[\Delta {I_C}\]denotes change in collector current of the transistor,
\[\Delta {I_B}\] denotes change in base current of the transistor.
Complete step by step solution:
Here, base current changes from \[45\mu A\] to \[140\mu A\].
Here, collector current changes from 0.2 mA to 4 mA.
To find: Gain in current for the transistor.
From the given information get the change in base current as:
\[
\Delta {I_B} = 140\mu A - 45\mu A = 95\mu A \\
\therefore \Delta {I_B} = 9.5 \times {10^{ - 5}}A \\
\]....................(2)
Similarly, get the change in collector current as:
\[
\Delta {I_C} = 4mA - 0.2mA = 3.8mA \\
\therefore \Delta {I_C} = 3.8 \times {10^{ - 3}}A \\
\]...................(3)
Substitute the obtained value of \[\Delta {I_B}\]and \[\Delta {I_C}\]from eq.(2) and eq.(3) respectively into the eq.(1) to obtain current gain \[{g_I}\]as:
\[
{g_I} = \dfrac{{3.8 \times {{10}^{ - 3}}A}}{{9.5 \times {{10}^{ - 5}}A}} = \dfrac{{3800}}{{95}} \\
\therefore {g_I} = 40 \\
\]
Gain in current for the transistor is got to be (C), 40.
Additional information:
Current gain of a transistor is completely an intrinsic property of the transistor itself. This quantity is not very much precise in general and varies quite abruptly depending on its production. When the transistor is used as a current gaining device then its quality is measured by this current gain value.
Note: A student might make a mistake while using the current gain formula. Just remember how actually the transistor works. At the base of the transistor a small amount of current is injected. This allows the load to draw a large current through the collector. This ratio of change in collector current for the change in base current gives the current gain of the transistor.
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