In the figure- given that VBB supply can vary from 0 to 5-0 V- VCC-5 V- -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-

At saturation, V_CE=0 nbsp; Considering output loop, V_CE=V_CC I_CR_C ⇒ I_C=V_CCR_C=5 × 10^ 3 A Current gain, β_dc =I_CI_B ⇒ I_B=5 × 10^ 3200=25 μA nbsp ...

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Byju's Answer

Standard XII

Physics

Inductor

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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20 μ A and 2.8 V

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25 μ A and 3.5 V

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25 μ A and 2.8 V

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Solution

The correct option is C $25 μ A and 3.5 V$
$At saturation, V_{C E} = 0$
Considering output loop,

$V_{C E} = V_{C C} - I_{C} R_{C}$
$\Rightarrow I_{C} = \frac{V_{C C}}{R_{C}} = 5 \times 10^{- 3} A$

$Current gain, β_{d c} = \frac{I_{C}}{I_{B}}$
$\Rightarrow I_{B} = \frac{5 \times 10^{- 3}}{200} = 25 μ A$

At input loop,

$V_{B B} = I_{B} R_{B} + V_{B E}$
$\Rightarrow V_{B B} = (25 μ A) (100 k Ω) + 1 V$
$\Rightarrow V_{B B} = 3.5 V$

Hence, option $(A)$ is correct.

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In the figure, given that VBB supply can vary from 0 to 5.0 V, VCC=5 V,βdc=200, RB=100 kΩ,RC=1 kΩ 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 C 25 μA and 3.5 VAt saturation, VCE=0 Considering output loop, VCE=VCC−ICRC ⇒IC=VCCRC=5×10−3 A Current gain, βdc=ICIB ⇒IB=5×10−3200=25 μA At input loop, VBB=IBRB+VBE ⇒VBB=(25 μA)(100 kΩ)+1 V ⇒VBB=3.5 V Hence, option (A) is correct.

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: