# Chap7part4 - Emitter-Follower(EF Amplifier DC biasing C B...

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Ch. 7 Frequency Response Part 4 1 ECES 352 Winter 2007 Emitter-Follower (EF) 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 h Determine pole frequencies PH1 PH2 PHn High and Low Frequency AC Equivalent Circuit

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Ch. 7 Frequency Response Part 4 2 ECES 352 Winter 2007 EF Amplifier - DC Analysis (Nearly the 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 = 0 K R E = 0.33K Note: Only difference here from CE case is V CE is larger C
Ch. 7 Frequency Response Part 4 3 ECES 352 Winter 2007 EF Amplifier - Midband Gain Analysis ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 998 . 0 905 . 1 9 . 1 65 065 . 0 2 005 . 0 65 065 . 0 2 999 . 0 65 065 . 0 65 015 . 0 66 1 1 1 1 66 97 . 0 201 33 . 0 9 1 = = + + + = + + + = = + = + = = + = + = + = = = + = + = = = K K K K K K K K K R r R R R r R V V K K K R r R V V V V V V V V V K K K r r g R R V V g r V R R V V V V V V V V V V V V A i x B S i x B s b i x i b i o o i m E L m E L o s b b i i o s o Vo π = = = = = = = 5 005 . 0 5 . 2 10 33 . 0 0 9 2 1 K R K R K R K R K R K R S E C L Equivalent input resistance R i ( 29 K K V V r r V V V I V R o o i i 65 66 1 97 . 0 1 = + = + = + = = ( 29 ( 29 ( 29 ( 29 ( 29 dB dB A A Vo Vo 1 . 0 987 . 0 log 20 ) ( 987 . 0 998 . 0 999 . 0 015 . 0 66 - = = = = V i + _ R i V b + _ I π K V mA g r V mA mV mA V I g m T C m 97 . 0 / 206 200 / 206 26 27 . 5 = = = = = = β DC analysis is nearly the same! I B , I C and g m are all the same. Only V CE is different since R C =0. V O NOTE: Voltage gain is only ~1! This is a characteristic of the EF amplifier! Cannot get voltage gain >1 for this amplifier!

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Ch. 7 Frequency Response Part 4 4 ECES 352 Winter 2007 Analysis of Low Frequency Poles Gray-Searle (Short Circuit) Technique * Draw low frequency AC circuit Substitute AC equivalent circuit for transistor (hybrid-pi for bipolar transistor) Include coupling capacitors C C1 , C C2 Ignore (remove) all transistor capacitances C π , C μ * Turn off signal source, i.e. set V s = 0 Keep source resistance R S in circuit (do not remove) * Consider the circuit one capacitor C x at a time Replace all other capacitors with short circuits Solve remaining circuit for equivalent resistance R x seen by the selected capacitor Calculate pole frequency using Repeat process for each capacitor finding equivalent resistance seen and corresponding pole frequency * Calculate the final low 3 dB frequency using x x Px C R 1 = ϖ = + = = x x Pn P P Px LP C R 1 ...
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Chap7part4 - Emitter-Follower(EF Amplifier DC biasing C B...

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