In the figure- given that VBB supply can vary from 0 to 5-0 V- VCC - 5V- -dc - 200- RB - 100 k- RC - 1 k- and VBE - 1-0 V- The minimum base current and the input voltage at which the transistor will go to saturation- will be respectively-

The correct option is B 25 μ A and 3.5 V At saturation, V_CE = 0 V_CE = V_CC I_CR_C I_C = V_CCR_C = 5 × 10^ 3 A Given β_dc = I_ ...

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Byju's Answer

Standard XII

Physics

Reverse Bias

In the figure...

Question

In the figure, given that $V_{B B}$ supply can vary from $0$ to $5.0 V, V_{C C} = 5 V, β_{d c} = 200, R_{B} = 100 k Ω, R_{C} = 1 k Ω$ and $V_{B E} = 1.0 V$ . The minimum base current and the input voltage at which the transistor will go to saturation, will be respectively:

20 μ A

and

3.5 V

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25 μ A

and

3.5 V

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25 μ A

and

2.5 V

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20 μ A

and

2.8 V

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Solution

The correct option is B $25 μ A$ and $3.5 V$
At saturation, $V_{C E} = 0$
$V_{C E} = V_{C C} - I_{C} R_{C}$
$I_{C} = \frac{V_{C C}}{R_{C}} = 5 \times 10^{- 3} A$
Given
$β_{d c} = \frac{I_{C}}{I_{B}}$
$I_{B} = \frac{5 \times 10^{- 3}}{200}$
$I_{B} = 25 m u A$
At input side
$V_{B B} = I_{B} R_{B} + V_{B E}$
$= (25 m A) (100 k Ω) + 1 V$
$V_{B B} = 3.5 V$

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In the figure, given that VBB supply can vary from 0 to5.0V,VCC=5V,βdc=200,RB=100kΩ,RC=1kΩ and VBE=1.0V. The minimum base current and the input voltage at which the transistor will go to saturation, will be respectively:

The correct option is B 25μA and 3.5VAt saturation, VCE=0VCE=VCC−ICRCIC=VCCRC=5×10−3AGivenβdc=ICIBIB=5×10−3200IB=25muAAt input sideVBB=IBRB+VBE=(25mA)(100kΩ)+1VVBB=3.5V

In the figure, given that $V_{B B}$ supply can vary from $0$ to $5.0 V, V_{C C} = 5 V, β_{d c} = 200, R_{B} = 100 k Ω, R_{C} = 1 k Ω$ and $V_{B E} = 1.0 V$ . The minimum base current and the input voltage at which the transistor will go to saturation, will be respectively: