Calculate the value of emitter current for a transistor with α_dc = 0.98, I_CBO = 5 μA and I_B = 95 μA.

Common Emitter(CE) Configuration:

In CE configuration input is connected between base and emitter while the output is taken between collector and emitter.

I_E = I_B + I_C  

I_C = β I_B + I_CEO 

IC = α I_E + ICBO

I_C = α (I_C + I_B) + ICBO

IC (1 - α ) = α I_B + ICBO

In CE configuration, when I_B = 0 then I_C = I_CEO

Where, α = Current gain

β = Current Amplification Factor

IE, IB, IC = Emitter, Base and Collector current respectively

ICEO = Collector emitter cutoff current

ICBO = Collector base cutoff current

Calculation: 

Given: α = 0.98, ICBO = 5 μA, IB = 95 μA

I_C = 49 x (95 × 10^-6) + 250 × 10^-6 = 4905 × 10^-6 A