A transistor is connected in the CE configuration. The voltage drop across the load resistance (Rc) 3 k Ω is 6 V. Find the base current. The current gain of the transistor is 0.97. Answer (2.06 ｘ10⁻³ A) 



Given:

Load resistance = `\R_C` = 3 kΩ = 3ｘ10³ Ω

Voltage drop at load = `\V_C` =  6 V

Current gain = β =  0.97


To Find:

Base Current = `\I_B` =  ?



Solution:


Using the Current gain β formula


β = `\frac {I_C}{I_B}` 

or 

`\I_B` = `\frac {I_C}{β}` -----------(1)

where `\I_C` is unknown and can be calculated by Using Ohm's Law

 `\I_C`  = `\frac {V_C}{R}`

putting values

`\I_C`  = `\frac {6 V}{3ｘ10³ Ω}`

`\I_C`  = 2ｘ10⁻³ A 


Now putting the corresponding values in equation (1)

`\I_B` = `\frac {2ｘ10⁻³ A }{0.97}`

`\I_B` = 2.06ｘ10³ A 

or

`\I_B` = 2.06 mA -------------Ans.





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