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Emitter-Follower (EF) Amplifier. DC biasing Calculate I C , I B , V CE Determine related small signal equivalent circuit parameters Transconductance g m Input resistance r π Midband gain analysis Low frequency analysis Gray-Searle (Short Circuit) Technique
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Emitter-Follower (EF) Amplifier • DC biasing • Calculate IC, IB, VCE • Determine related small signal equivalent circuit parameters • Transconductance gm • Input resistance rπ • Midband gain analysis • Low frequency analysis • Gray-Searle (Short Circuit) Technique • Determine pole frequencies ωPL1, ωPL2, ... ωPLn • Determine zero frequencies ωZL1, ωZL2, ... ωZLn • High frequency analysis • Gray-Searle (Open Circuit) Technique • Determine pole frequencies ωPH1, ωPH2, ... ωPHn • Determine zero frequencies ωZH1, ωZH2, ... ωZHn High and Low Frequency AC Equivalent Circuit Ch. 7 Frequency Response Part 4
EF Amplifier - DC Analysis (Nearly the Same as CE Amplifier) • GIVEN: Transistor parameters: • Current gain β= 200 • Base resistance rx = 65 Ω • Base-emitter voltage VBE,active = 0.7 V • Resistors: R1=10K, R2=2.5K, RC=1.2K, RE=0.33K • Form Thevenin equivalent for base; given VCC= 12.5V • RTh = RB = R1||R2 = 10K||2.5K = 2K • VTh = VBB = VCC R2 / [R1+R2] = 2.5V • KVL base loop • IB = [VTh-VBE,active] / [RTh+(β +1)RE] • IB = 26 μA • DC collector current IC = βIB IC = 200(26 μ A) = 5.27 mA • Transconductance gm= IC / VT ;VT = kBT/q = 26 mV gm = 5.27 mA/26 mV = 206 mA/V • Input resistancerπ = β / gm = 200/[206 mA/V]= 0.97 K • Check on transistor region of operation • KVL collector loop • VCE = VCC- (β +1) IBRE = 10.8 V (was 4.4 V for CE amplifier) (okay since not close to zero volts). R1 = 10K R2 = 2.5K RC = 0 K RE = 0.33K Note: Only difference here from CE case is VCE is larger since RC was left out here in EF amplifier. Ch. 7 Frequency Response Part 4
EF Amplifier - Midband Gain Analysis DC analysis is nearly the same! IB , IC and gm are all the same. Only VCE is different since RC=0. Iπ + + Ri Vb Vi VO _ _ Equivalent input resistance Ri NOTE: Voltage gain is only ~1! This is a characteristic of the EF amplifier! Cannot get voltage gain >1 for this amplifier! Ch. 7 Frequency Response Part 4
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 CC1, CC2 • Ignore (remove) all transistor capacitances Cπ , Cμ • Turn off signal source, i.e. set Vs= 0 • Keep source resistance RS in circuit (do not remove) • Consider the circuit one capacitor Cx at a time • Replace all other capacitors with short circuits • Solve remaining circuit for equivalent resistance Rx 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 Ch. 7 Frequency Response Part 4
Emitter Follower - Analysis of Low Frequency PolesGray-Searle (Short Circuit) Technique Input coupling capacitor CC1 = 2 μF IX Iπ Ri Vi Ch. 7 Frequency Response Part 4
Emitter Follower - Analysis of Low Frequency PolesGray-Searle (Short Circuit) Technique • Output coupling capacitor CC2 = 3 μF Iπ Ve • Low 3 dB frequency re Ie IX So dominant low frequency pole is due to CC1 ! Ch. 7 Frequency Response Part 4
Emitter Follower - Low Frequency Zeros • What are the zeros for the EF amplifier? • For CC1 and CC2 , we get zeros at ω = 0 since ZC = 1 / ωC and these capacitors are in the signal line, i.e. ZC at ω = 0 so Vo 0. Ch. 7 Frequency Response Part 4
Emitter Follower - Low Frequency Poles and ZerosMagnitude Bode Plot Ch. 7 Frequency Response Part 4
Emitter Follower - Low Frequency Poles and ZerosPhase Shift Bode Plot Ch. 7 Frequency Response Part 4
Analysis of High Frequency Poles Gray-Searle (Open Circuit) Technique • Draw high frequency AC equivalent circuit • Substitute AC equivalent circuit for transistor (hybrid-pi model for transistor with Cπ, Cμ) • Consider coupling and emitter bypass capacitors CC1 and CC2 as shorts • Turn off signal source, i.e. set Vs = 0 • Keep source resistance RS in circuit • Neglect transistor’s output resistance ro • Consider the circuit one capacitor Cx at a time • Replace all other transistor capacitors with open circuits • Solve remaining circuit for equivalent resistance Rx seen by the selected capacitor • Calculate pole frequency using • Repeat process for each capacitor • Calculate the final high frequency pole using Ch. 7 Frequency Response Part 4
Emitter Follower - Analysis of High Frequency PolesGray-Searle (Open Circuit) Technique Ie • Redrawn High Frequency Equivalent Circuit zπ =1/yπ E Ie Zeq Ch. 7 Frequency Response Part 4
Emitter Follower - Analysis of High Frequency PolesGray-Searle (Open Circuit) Technique ZB’ zπ =1/yπ Replace this with this. Modified Equivalent Circuit ZB’ Looks like a resistor in parallel with a capacitor. Ch. 7 Frequency Response Part 4
Emitter Follower - Analysis of High Frequency PolesGray-Searle (Open Circuit) Technique RxCπ • Pole frequency for Cπ =17 pF Ch. 7 Frequency Response Part 4
Emitter Follower - Analysis of High Frequency PolesGray-Searle (Open Circuit) Technique • Pole frequency for Cμ =1.3 pF Ch. 7 Frequency Response Part 4
Emitter Follower - Analysis of High Frequency PolesGray-Searle (Open Circuit) Technique • Alternative Analysis for Pole Due to Cπ Ix-Iπ Ix Vx Iπ E Ie Ie+gmVπ We get the same result here for the high frequency pole associated with Cπ as we did using the equivalent circuit transformation. Ch. 7 Frequency Response Part 4
Emitter Follower - Analysis of High Frequency PolesGray-Searle (Open Circuit) Technique • Alternative Analysis for Pole Due to Cµ Ix-Iπ Ix Vx Iπ E Iπ+gmVπ We get the same result here for the high frequency pole associated with Cµ as we did using the equivalent circuit transformation. Ch. 7 Frequency Response Part 4
Emitter Follower - High Frequency Zeros • What are the high frequency zeros for the EF amplifier? • Voltage gain can be written as • When Vo/Vπ = 0, we have found a zero. • For Cμ, we get Vo 0 when ω since the node B’ will be shorted to ground and Vπ = 0 . • Similarly, we get a zero from Cπ when yπ + gm = 0 since we showed earlier that • Also, can see this from Ch. 7 Frequency Response Part 4
Emitter Follower - High Frequency Poles and ZerosMagnitude Ch. 7 Frequency Response Part 4
Emitter Follower - High Frequency Poles and ZerosPhase Shift Ch. 7 Frequency Response Part 4
Comparison of EF to CE Amplifier (For RS = 5Ω ) CE EF Midband Gain Low Frequency Poles and Zeros High Frequency Poles and Zeroes Better low frequency response ! Much better high frequency response ! Ch. 7 Frequency Response Part 4
Conclusions • Voltage gain • Can get good voltage gain from CE but NOT from EF amplifier (AV 1). • Low frequency performance better for EF amplifier. • EF amplifier gives much better highfrequency performance! • CE amplifier has dominant pole at 5.0x107 rad/s. • EF amplifier has dominant pole at 1.0x1010 rad/s. • Bandwidth approximately 200 X larger! • Miller Effect multiplication of C by the gain is avoided in EF. • Current gain • For CE amplifier, current gain is high = Ic/Ib • For EF amplifier, current gain is also high Ie/Ib = +1 ! • Frequency dependence of current gain similar to voltage gain. • Input and output impedances are different for the two amplifiers! Ch. 7 Frequency Response Part 4