In the circuit shown in the figure, the input voltage Vi is 20 V , VBE = 0 and VCE = 0 .The values of IB , IC and β are given by

Question

In the circuit shown in the figure, the input voltage Vi is 20 V , VBE = 0 and VCE = 0 .The values of IB , IC and β are given by (A) IB = 40 μ A, IC = 10 mA , β = 250 (B) IB = 25 μ A, IC = 5 mA , β = 200 (C) IB = 20 μ A, IC = 5 mA , β = 250 (D) IB = 40 μ A, IC = 5 mA , β = 125

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/34ed3d3a9dd25de4b58a55bb066108c9-.png" /> VBE = 0 VCE = 0 Vb = 0 Ic = ((20 -0)/4 ×103) Ic = 5×10-3 = 5 mA Vi = VBE + IB RB Vi = 0 + IB RB 20 = IB ×500 ×103 IB = (20/500 ×103) = 40 μ A β = (Ic/Ib) = (25 ×10-3/40 ×10-6 ) = 125

Q. In the circuit shown in the figure, the input voltage $V_{i}$ is $20 V, V_{BE} = 0$ and $V_{CE} = 0$ .The values of $I_{B}, I_{C}$ and $β$ are given by

$I_{B} = 40 μ A, I_{C} = 10 m A, β = 250$

$I_{B} = 25 μ A, I_{C} = 5 m A, β = 200$

$I_{B} = 20 μ A, I_{C} = 5 m A, β = 250$

$I_{B} = 40 μ A, I_{C} = 5 m A, β = 125$

Q. In the circuit shown in the figure, the input voltage Vi​ is 20V,VBE​=0 and VCE​=0 .The values of IB​,IC​ and β are given by

IB​=40μA,IC​=10mA,β=250

IB​=25μA,IC​=5mA,β=200

IB​=20μA,IC​=5mA,β=250

IB​=40μA,IC​=5mA,β=125

VBE​=0 VCE​=0 Vb​=0 Ic​=4×103(20−0)​ Ic​=5×10−3=5mA Vi​=VBE​+IB​RB​ Vi​=0+IB​RB​ 20=IB​×500×103 IB​=500×10320​=40μA β=Ib​Ic​​=40×10−625×10−3​=125

Q. In the circuit shown in the figure, the input voltage $V_{i}$ is $20 V, V_{BE} = 0$ and $V_{CE} = 0$ .The values of $I_{B}, I_{C}$ and $β$ are given by

$I_{B} = 40 μ A, I_{C} = 10 m A, β = 250$

$I_{B} = 25 μ A, I_{C} = 5 m A, β = 200$

$I_{B} = 20 μ A, I_{C} = 5 m A, β = 250$

$I_{B} = 40 μ A, I_{C} = 5 m A, β = 125$