Chap7part3 - Ch 7 Frequency Response Part 3 1 ECES 352...

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Unformatted text preview: Ch. 7 Frequency Response Part 3 1 ECES 352 Winter 2007 Common-Base (CB) Amplifier * DC biasing ● Calculate I C , I B , V CE ● Determine related small signal equivalent circuit parameters h Transconductance g m h Input resistance r π * Midband gain analysis * Low frequency analysis ● Gray-Searle (Short Circuit) Technique h Determine pole frequencies ω PL1 , ω PL2 , ... ω PLn ● Determine zero frequencies ω ZL1 , ω ZL2 , ... ω ZLn * High frequency analysis ● Gray-Searle (Open Circuit) Technique Input at emitter, output at collector. Ch. 7 Frequency Response Part 3 2 ECES 352 Winter 2007 CB Amplifier - DC Analysis (Same as CE Amplifier) * GIVEN: Transistor parameters: ● Current gain β = 200 ● Base resistance r x = 65 Ω ● Base-emitter voltage V BE,active = 0.7 V ● Resistors: R 1 =10K, R 2 =2.5K, R C =1.2K, R E =0.33K * Form Thevenin equivalent for base; given V CC = 12.5V ● R Th = R B = R 1 ||R 2 = 10K||2.5K = 2K ● V Th = V BB = V CC R 2 / [R 1 +R 2 ] = 2.5V ● KVL base loop h I B = [V Th-V BE,active ] / [R Th +(β +1)R E ] h I B = 26 μA * DC collector current I C = β I B I C = 200(26 μ A) = 5.27 mA * Transconductance g m = I C / V T ; V T = k B T/q = 26 m R 1 = 10K R 2 = 2.5K R C = 1.2K R E = 0.33K Ch. 7 Frequency Response Part 3 3 ECES 352 Winter 2007 * Construct small signal ac equivalent circuit ( set DC supply to ground ) * Substitute small signal equivalent circuit ( hybrid-pi model ) for transistor * Neglect all capacitances ● Coupling and emitter bypass capacitors become shorts at midband frequencies (~ 10 5 rad/s) h Why? Impedances are negligibly small, e.g. few ohms because C C1 , C C2 , C E ~ few μF (10-6 F) ● Transistor capacitances become open circuits at midband frequencies h Why? Impedances are very large, e.g. ~ 10’s M Ω because C π , C μ ~ pF (10-12 F) * Calculate small signal voltage gain A Vo = V o /V s CB Amplifier - Midband Gain Analysis Ω = = = 10 ) 1 )( / 10 ( 1 1 5 F s rad C Z C μ ϖ Ω = = = 7 5 10 ) 1 )( / 10 ( 1 1 pF s rad C Z C ϖ High and Low Frequency AC Equivalent Circuit Ch. 7 Frequency Response Part 3 4 ECES 352 Winter 2007 CB Amplifier - Midband Gain Analysis ( 29 ( 29 ( 29 ( 29 [ ] [ ] [ ] [ ] ( 29 ( 29 ( 29 ( 29 dB dB A V V A K K K K K K K r R R r R V V K K K r r I I r V V K K V mA R R g V R R V g V V V V V V V V V V A Vo Vo e E s e E s e x e C L m C L m o s e e o s o Vo 14 ) 20 . log( 20 / 20 . 001 . 94 . 218 001 . 0050 . 5 0050 . 0051 . 33 . 5 0051 . 33 . 94 . 065 . 97 . 97 . ) ( 218 9 2 . 1 / 206- = = =-- = = = + = + =- = +- = +- =- =- =- =- = = = π π π π π π π π π π K R K R K R K R K R K R S E C L 5 5 . 2 10 33 . 2 . 1 9 2 1 = = = = = = Equivalent resistance r e K K K r r r g r r r g r r r r r g r V V I V r r r g V r g V I r V V g I E node at KCL I V r x m x m x m e e e e m m e m e e e e 0051 ....
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Chap7part3 - Ch 7 Frequency Response Part 3 1 ECES 352...

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