133
154
196
105
Given Vi = 10 V, Rb = 400 kΩ = 400 × 10³ Ω, Rc = 3 kΩ = 3 × 10³ Ω, Vbe = 0, Vce = 0, VCC = 10 V
As Vi − Vbe = RbIb ∴ 10 − 0 = (400 × 10³)Ib
Ib = 10/400 × 10³ = 25 × 10⁻⁶ A = 25 μA
and VCC − Vce = IcRc
Ic = 10/3 × 10³ = 3.33 × 10⁻³ = 3.33 mA
β = Ic/Ib = 3.33 × 10⁻³ / 25 × 10⁻⁶ = 133.