Consider the circuit shown in the figure below-Transistor has - - 50- VBE - 0-7V- VCEsat-0-2V- The maximum value of RB upto which BJT remains in saturaion is k-

From the circuit diagram, V_CC V_CEsat/R_C = I_C_sat ⇒ nbsp;I_C_sat = 12 0.2/4=2.95mA ⇒ nbsp;I_B_min = I_C_sat/β= 2.95/50 = 0.059mA For BJT, to operate in ...

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Standard X

Physics

Combination of Resistors

Consider the ...

Question

Consider the circuit shown in the figure below:

Transistor has $β = 50, V_{B E} = 0.7 V, V_{C E s a t} = 0.2 V .$ The maximum value of $R_{B}$ upto which BJT remains in saturaion is _______ $k Ω$ .
72.88

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Solution

The correct option is A 72.88
From the circuit diagram, $\frac{V_{C C} - V_{C E s a t}}{R_{C}} = I_{C_{s a t}}$
$\Rightarrow I_{C_{s a t}} = \frac{12 - 0.2}{4} = 2.95 m A$
$\Rightarrow I_{B_{m i n}} = \frac{I_{C_{s a t}}}{β} = \frac{2.95}{50} = 0.059 m A$
For BJT, to operate in saturation region;
$I_{B} \geq I_{B m i n}$
$\frac{5 - 0.7}{R_{B}} \geq 0.059 m A$
$R_{B} \leq 72.88 k Ω$
$R_{B m a x} = 72.88 k Ω$

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Consider the circuit shown in the figure below: Transistor has β=50,VBE=0.7V,VCEsat=0.2V. The maximum value of RB upto which BJT remains in saturaion is _______kΩ.72.88

The correct option is A 72.88From the circuit diagram, VCC−VCEsatRC=ICsat ⇒ICsat=12−0.24=2.95mA ⇒IBmin=ICsatβ=2.9550=0.059mA For BJT, to operate in saturation region; IB≥IBmin 5−0.7RB≥0.059mA RB≤72.88kΩ RBmax=72.88kΩ

Consider the circuit shown in the figure below:

Transistor has $β = 50, V_{B E} = 0.7 V, V_{C E s a t} = 0.2 V .$ The maximum value of $R_{B}$ upto which BJT remains in saturaion is _______ $k Ω$ .
72.88