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

Step-by-Step

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Application of KVL around the transistor terminals yields

I_{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 I_{EQ} = (β + 1)I_{CQ}; thus KVL around the collector circuit gives

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_{BQ} + βI_{BQ})R_C + (β + 1)I_{BQ}R_E = V_{CC} – V_{CEQ}

so R_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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