Fig.P.18.2 shows a transistor that operates a relay as the switch S is closed. The relay is energized by a current of 10 mA. Calculate the value R, which will just make the relay operate. The current gain β of the transistor is 200. When the transistor conducts, its `\V_{BC}` can be assumed to be 0.6 V. (Ans: 168 kΩ)

Given:

Collector Current = `\I_C` =  10 mA = 10 ï½˜10⁻³ A

Current gain = Î² = 200

Voltage at Base Current = `\V_{B_C}` =  0.6 V

Voltage at Collector Current = `\V_{C_C}` =  9 V (given in the Figure)


To Find:

Resistance at Base = `\R_B` =  ?



Solution:


We know that

`\V_{C_C}` = `\I_B` `\R_B`  +  `\V_{B_C}`

or 

`\R_B` = `\frac {V_{C_C} - V_{B_C} }{I_B}` -----------(1)

where `\I_B` is unknown and can be calculated from the gain formula

β = `\frac {I_C}{I_B}`

or

 `\I_B` = `\frac {I_C}{β}`

putting values

`\I_B` = `\frac {10 ｘ10⁻³ A}{200}`

`\I_B`  = 0.05ｘ10⁻³ A 


Now putting the corresponding values in equation (1)

`\R_B` = `\frac {9 V - 0.6 V }{0.05ｘ10⁻³ A}`

`\R_B` = `\frac {8.4 V }{0.05ｘ10⁻³ A}`

`\R_B` = 168ｘ10³ Ω 

or

`\R_B` = 168 KΩ -------------Ans.



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