Q5Analog Electronics

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(B) For fig. 7 Calculate the quiescent points (ICQ and VCEQ) and also determine VCECut-off and IC Saturation Given - \beta = 100 VCC = 16 V R1 = 47K \Omega R2 = 12K \Omega RC = 2200 \Omega RE = 1800 \Omega

Answer 1

Given: $β = 100$ , $V_{C C} = 16 V$ , $R_{1} = 47 k Ω$ , $R_{2} = 12 k Ω$ , $R_{C} = 2200Ω$ , $R_{E} = 1800Ω$

V_{T H} = V_{C C} \times \frac{R _{2}}{R _{1} + R _{2}} = 16 \times \frac{12 k}{47 k + 12 k} = 3.25 V

R_{T H} = R_{1} ∥ R_{2} = \frac{47 k \times 12 k}{59 k} = 9.56 k Ω

I_{C Q} = \frac{V _{T H} - V _{B E}}{R _{E} + R _{T H} / β} = \frac{3.25 - 0.7}{1800 + 95.6} = \frac{2.55}{1895.6} \approx 1.345 mA

V_{C E Q} = V_{C C} - I_{C Q} (R_{C} + R_{E}) = 16 - 1.345 \times 1 0^{- 3} \times 4000 \approx 10.62 V

V_{C E (c u t o f f)} = V_{C C} = 16 V

I_{C (s a t)} = \frac{V _{C C}}{R _{C} + R _{E}} = \frac{16}{4000} = 4 mA

Q5Analog Electronics

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