Question 5.18: Objective: Determine the currents in a two-transistor curren...

The reference current is
I_{1} = \frac{− (V^{−}  +  V_{B E} (on))}{R_{1}} = \frac{−((−5)  +  0.7)}{10} = 0.43  mA
From Equation (5.46), the bias current I_{Q} is
I_{C2} = I_{Q} = \frac{I_{1}}{ \left(1  +  \frac{2}{β} \right)} = \frac{0.43}{ \left( 1  +  \frac{2}{50} \right)} = 0.413  mA
The base currents are then
I_{B1} = I_{B2} = \frac{I_{C2}}{β} = \frac{0.413}{50} ⇒ 8.27  µA
Comment: For relatively large values of current gain β, the bias current I_{Q} is essentially the same as the reference current I_{1}.