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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}.

3.21
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(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}

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