# Microelectronic Circuit Design

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CHAPTER 9 9.1 Since V REF = − 1.25 V , and v I = − 1.6 V , Q 1 is off and Q 2 is conducting. v C 1 = 0 V and v C 2 = − α F I EE R C ≅ − I EE R C = − 2 mA ( ) 350 ( ) = − 0.700 V 9.2 I C 2 I C 1 = exp V BE V T ⇒ ∆ V BE = 0.025ln 0.995 α F I EE 0.005 α F I EE = 0.132 V (a) v I = V REF + ∆ V BE = − 1.25 + 0.132 = − 1.12 V v I = V REF + ∆ V BE = − 1.25 0.132 = − 1.38 V (b) v I = V REF + ∆ V BE = − 2.00 + 0.132 = − 1.87 V v I = V REF + ∆ V BE = − 2.00 0.132 = − 2.13 V 9.3 Since V REF = − 2 V , and v I = − 1.6 V , Q 2 is off and Q 1 is conducting. v C 2 = 0 V and v C 1 = − α F I EE R C ≅ − I EE R C = − 2.5 mA ( ) 700 ( ) = − 1.75 V Note that Q 1 is beginning to enter the saturation region of operation, but V BC = +0.15 V is not really enough to turn on the collector-base diode. (See Problems 9.5 or 5.61.) 9.4 v I = V REF + 0.3 V Q 1 on; Q 2 off. I C 1 = α F I EE I EE = 0.3 mA | I C 2 = 0 v C 1 = 0 I C 1 R 1 + R C ( ) = − 0.3 mA 3.33 k Ω+ 2 k ( ) = − 1.60 V v C 2 = 0 I C 1 R 1 = − 0.3 mA 3.33 k ( ) = − 0.999 V 9-1

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