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Physics Question on Semiconductor electronics: materials, devices and simple circuits

In the figure, given that VBBV_{BB} supply can vary from 00 to 5.0V,VCC=5V,βdc=200,RB=100kΩ,RC=lkΩ5.0 \,V, V_{CC} = 5\,V, \beta_{dc} = 200, R_B = 100 \,k\,\Omega, R_C=l\, k\Omega and VBE=1.0VV_{BE}=1.0\, V, The minimum base current and the input voltage at which the transistor will go to saturation, will be, respectively :

A

20μA20\,\mu A and 3.5V3.5\,V

B

25μA25\,\mu A and 3.5V3.5\,V

C

25μA25\,\mu A and 2.5V2.5\,V

D

20μA20\,\mu A and 2.8V2.8\,V

Answer

25μA25\,\mu A and 3.5V3.5\,V

Explanation

Solution

At saturation, VCE_{CE} = 0
VCE=VCCICRCV_{CE} \, = \, V_{CC} \, - \, I_C R_C
IC=VCCRC=5×103A\Rightarrow \, \, I_C \, = \frac{V_{CC}}{R_C} \, = \, 5 \, \times \, 10^{-3} \, A
Given
βdc=ICIB\beta_{dc} \, = \, \frac{I_C}{I_B}
IB=5×103200I_B \, = \, \frac{5 \times \, 10^{-3}}{200}
IB=25μAI_B \, = \, 25 \, \mu A
At input side
VBB=IBRB+VBEV_{BB} \, = \, I_B \, R_B \, + \, V_{BE}
=(25mA)(100kΩ)+1V\, \, \, \, =(25 mA)(100 k \Omega) \, + \, 1V
VBB=3.5VV_{BB} \, \, = \, \, 3.5 \, V