Question 3.SP.15: Find the value of the emitter resistor RE that, when added t......

Find the value of the emitter resistor RER_E that, when added to the Si transistor circuit of Fig. 3-17, would bias for operation about VCEQ=5 VV_{CEQ} = 5  \text{V}. Let ICEO=0, β=80, RF=220 kΩ, RC=2 kΩI_{CEO} = 0,  β = 80,  R_F = 220  kΩ,  R_C = 2  kΩ, and VCC=12 VV_{CC} = 12  \text{V}.

3.17
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Application of KVL around the transistor terminals yields
IBQ=VCEQ – VBEQRF=5 – 0.7220 × 103=19.545 μAI_{BQ} = \frac{V_{CEQ}  –  V_{BEQ}}{R_F} = \frac{5  –  0.7}{220  ×  10^{3}} = 19.545  μ\text{A}

Since leakage current is zero, (3.1) and (3.2) give IEQ=(β+1)ICQI_{EQ} = (β + 1)I_{CQ}; thus KVL around the collector circuit gives
α(hFB)IC – ICBOIEα(≡ h_{FB}) ≡ \frac{I_C  –  I_{CBO}}{I_E}          (3.1)
β(hFE)α1 – αIC – ICEOIBβ(≡ h_{FE}) ≡ \frac{α}{1  –  α} ≡ \frac{I_C  –  I_{CEO}}{I_B}          (3.2)
(IBQ+βIBQ)RC+(β+1)IBQRE=VCC – VCEQ(I_{BQ} + βI_{BQ})R_C + (β + 1)I_{BQ}R_E = V_{CC}  –  V_{CEQ}

so      RE=VCC – VCEQ – (β+1)IBQRC(β+1)IBQ=12 – 5 – (80 + 1)(19.545 × 106)(2 × 103)(80 + 1)(19.545 × 106)=2.42 kΩR_E = \frac{V_{CC}  –  V_{CEQ}  –  (β + 1)I_{BQ}R_C}{(β + 1)I_{BQ}} = \frac{12  –  5  –  (80  +  1)(19.545  ×  10^{-6})(2  ×  10^{3})}{(80  +  1)(19.545  ×  10^{-6})} = 2.42  kΩ

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