Question

Consider the circuit shown in Figure 1. Transistors Q1 and Q2 are identical and can be
describe with the following parameters: I_S = 5 ×10⁻¹⁶ A, β = 100 and V_A = 5 V.

+ 1kΩ Vcc= 2.5 V Vina Q1 Vout Q2 + -1.5 V 31k92

1. when V_in = 1,5V I_CQ1=I_CQ2.

2. Write an equation for currents I_C1 and I_C2 in terms of the voltages V_in(VBQ1) and V_BQ2. For it
you can omit the Early effect and assume I_E ≈ I_C. Solve the equations to find so
Analyze the value of Vin that simulated in the previous point. how are these equations obtained?

Answer 1

. Solution Vcc=2.5v Пқ-— 91 Vo 82 Vin o R Ici Icz 1R2 Analyzing gi Base voltage Vin. Vin - 017 = Tert 1 k-2 Iez) So,

As, IBI Ich IB2 Ier Icz B 3 So, Vin - 0.7 B (I81 + 132) ... Analyzing 92: VB92 -0.7 - B. ( 1B1 t. I82) So, from and ② say that, then condition for operating in Active regions for both 91 and 92 we can VBy2 = Vin. it, Vine > VBg2 then, IBG2 as BE funetion Of 92 will be reverse biased, Vin-017) MA and Iel = 0. so, Ica 3 is, Vir < VBg2, thew, Ici - 0 in similar logic and Ica- VB42 -0.7 mA 97) B 1.5 V, for, IMA Nin VBL 2 1.5 -0.7 2 Il2 = 004 mA <Ans> And, Tela IB2 IBI