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Question: The current gain \(\left( \beta \right)\) of a transistor in a common base mode is \(40\) . To chang...

The current gain (β)\left( \beta \right) of a transistor in a common base mode is 4040 . To change the collector current by 160mA160\,mA , the necessary change in the base current is (at constant VCE{V_{CE}} )
A) 0.25μ0.25\,\mu
B) 4μA4\,\mu A
C) 4mA4\,mA
D) 40mA40\,mA

Explanation

Solution

Use the formula of the current gain and rearrange it to obtain the value of the change in the base current. Substitute the values of the current gain and the change in the collector current of the transistor, in the rearranged formula to know the answer.

Useful formula:
The formula of the current gain is given by
β=ΔIcΔIb\beta = \dfrac{{\Delta {I_c}}}{{\Delta {I_b}}}
Where β\beta is the current gain of the transistor, ΔIc\Delta {I_c} is the change in the collector current and the ΔIb\Delta {I_b} is the change in the base current.

Complete step by step solution:
It is given that the
Current gain of the transistor in the common base mode, β=40\beta = 40
The change in the collector current, ΔIc=160mA\Delta {I_c} = 160\,mA
By using the formula of the current gain,
β=ΔIcΔIb\beta = \dfrac{{\Delta {I_c}}}{{\Delta {I_b}}}
By rearranging the above formula, in order to obtain the value of the change in the base current,
ΔIb=ΔIcβ\Delta {I_b} = \dfrac{{\Delta {I_c}}}{\beta }
Substituting the values of the change in the collector current and the change in the collector current given in the question in the above equation,
ΔIb=160mA40\Delta {I_b} = \dfrac{{160\,mA}}{{40}}
By performing the basic arithmetic division, the result obtained is
ΔIb=4mA\Delta {I_b} = 4\,mA
Hence the change in the base current is obtained as 4mA4\,mA .

Thus the option (C) is correct.

Note: The current gain of the transistor in the common base mode shows that the small change in the base current can cause the greater changes in the collector current of the transistor. The value of the current can be maximum up to 200200in case of the standard transistor.