# Sol3 - 120 k Ω 240 k Ω V CC = 15 V 150 k Ω 100 k Ω...

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Unformatted text preview: 120 k Ω 240 k Ω V CC = + 15 V 150 k Ω 100 k Ω ECE 255 ELECTRONIC ANALYSIS AND DESIGN Fall 2009 Homework 3 Solutions Problem 1 (Text 5.86) Find the Q-point for the circuit shown below if R 1 = 120 k Ω , R 2 = 240 k Ω , R E = 100 k Ω , R C = 150 k Ω , β F = 100, V BE = 0.7 V, and the positive power supply voltage is 15 V. TH 120 k V 15V 5.0 V (120 k 240 k) Ω = × = + Ω and TH R 120 k 240 k 80 k = Ω = Ω ( 29 ( 29 ( 29 TH BE C TH E V V 100 5.0 0.7 .430 I 42.24 A R ( 1) R 80 k 101 100 k 10.18k β-- = = = = μ + β + + ( 29 ( 29( 29 CE 101 V 15 42.24 A 150 k 42.24 A 100 k 4.40 V 100 =- μ Ω - μ Ω = Problem 2 (Text 5.87) (a) Design a four-resistor bias network for an npn transistor to give I C = 1 mA, V CE = 5 V, and V E = 2 V if V CC = 12 V and β F = 100. Assume V = 0.7 V. Since V E = 2 V and |I E | = I C + I B ⇒ E 2 V R 1.98k 1mA .01mA = = Ω + Since V Rc = (12 – 5 – 2) V = 5 V ⇒ C 5V R 5.0 k 1mA = = Ω Set I R1 = 20 I B = 0.2 mA, V B = (2.0 + 0.7) V ⇒ ( 29 2 12 2.7 V R 46.5k 0.2 mA- = = Ω Then I R2 = 19 I B and V B = 2.7 V ⇒ 1 2.7 V R 14.2 k 0.19 mA = = Ω (b) Replace the exact values determined in part (a) with the nearest 5% values found in the resistor table and find the resulting Q-point....
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Sol3 - 120 k Ω 240 k Ω V CC = 15 V 150 k Ω 100 k Ω...

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