Question 3.SP.21: The Si Darlington transistor pair of Fig. 3-21 has negligibl......
The Si Darlington transistor pair of Fig. 3-21 has negligible leakage current, and β_1 = β_2 = 60. Let R_1 = R_2 = 1 MΩ, R_E = 500 Ω, and V_{CC} = 12 \text{V}. Find (a) I_{EQ2}, (b) V_{CEQ2}, and (c) I_{CQ1}.
(a) A Thévenin equivalent for the circuit to the left of terminals a, b has
V_{Th} = \frac{R_2}{R_1 + R_2} V_{CC} = \frac{1 × 10^6}{1 × 10^6 + 1 × 10^6}12 = 6 \text{V}
and R_{Th} = \frac{R_1R_2}{R_1 + R_2} = \frac{(1 × 10^6)(1 × 10^6)}{1 × 10^6 + 1 × 10^6} = 500 kΩ
With the Thévenin circuit in place, KVL gives
V_{Th} = I_{BQ1}R_{Th} + V_{BEQ1} + V_{BEQ2} + I_{EQ2}R_E (1)
Realizing that
I_{EQ2} = (β_2 + 1) I_{BQ2} = (β_2 + 1)(β_1 + 1)I_{BQ1}
we can substitute for I_{BQ1} in (1) and solve for I_{EQ2}, obtaining
I_{EQ2} = \frac{(β_1 + 1)(β_2 + 1)(V_{Th} – V_{BEQ1} – V_{BEQ2})}{R_{Th} + (β_1 + 1)(β_2 + 1)R_E} = \frac{(60 + 1)(60 + 1)(6 – 0.7 – 0.7)}{500 × 10^3 + (60 + 1)(60 + 1)(500)} = 7.25 \text{mA}
(b) By KVL,
V_{CEQ2} = V_{CC} – I_{EQ2}R_E = 12 – (7.25 × 10^{-3})(500) = 8.375 \text{V}
(c) From (3.1) and (3.2),
α(≡ h_{FB}) ≡ \frac{I_C – I_{CBO}}{I_E} (3.1)
β(≡ h_{FE}) ≡ \frac{α}{1 – α} ≡ \frac{I_C – I_{CEO}}{I_B} (3.2)
I_{CQ1} = \frac{β_1}{β_1 + 1}I_{EQ1} = \frac{β_1}{β_1 + 1}I_{BQ2} = \frac{β_1}{β_1 + 1}\frac{I_{EQ2}}{β_2 + 1} = \frac{60}{60 + 1}\frac{7.25 × 10^{-3}}{60 + 1} = 116.9 μ\text{A}