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Electrical Engineering BA (B), Analog Electronics, ET064G 7.5 Credits ET065G 6 Credits. Exam question. An amplifier have the following open-loop gain: Draw a detailed asymptotic bode-diagram for the open-loop gain. Both the amplitude and phase should be clearly visualized. (2p)
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Electrical Engineering BA (B), Analog Electronics,ET064G 7.5 CreditsET065G 6 Credits Muhammad Amir Yousaf
Examquestion • An amplifier have the following open-loop gain: • Draw a detailed asymptotic bode-diagram for the open-loop gain. Both the amplitude and phase should be clearly visualized. (2p) • What is the gain-bandwidth product for the amplifier? (1p) • What is the lowest amplification the amplifier can be used (AVmin), without any stabilization network assuming 45˚ phase margin, (ΦM). (2p) • Stabilize the amplifier with a dominant pole compensation so that it can be useful as a 20 dB amplifier with ΦM>45˚. Muhammad Amir Yousaf
100db 80 60 40 20 Muhammad Amir Yousaf 10 100 100K 10M 1K 10K 1M
0 -45 -90 -135 • -180 • -225 • -270 Muhammad Amir Yousaf 10 100 100K 10M 1K 10K 1M
0 100db -45 -90 80 -135 60 A feedback factor of -80db or Acl of 80 dbcan make this circuitstable. • -180 40 • -225 20 • -270 Muhammad Amir Yousaf 10 100 100K 10M 1K 10K 1M 1M
0 A feedback factor of -80db or Acl of 80 dbcan make this circuitstable. 100db -45 -90 80 (Acl)db -135 60 • -180 40 (AolB)db • -225 20 • -270 Muhammad Amir Yousaf 10 100 100K 10M 1K 10K 1M 1M
0 b) Highestbandwidth 10K withoutstabilization 100db -45 -90 80 (Acl)db -135 60 • -180 40 • -225 20 • -270 Muhammad Amir Yousaf 10 100 100K 10M 1K 10K 1M 1M
0 A feedback factor of -80db or Acl of 80 dbcan make this circuitstable. 100db -45 -90 80 -135 60 • -180 40 • -225 20 • -270 Muhammad Amir Yousaf 10 100 100K 10M 1K 10K 1M 1M
LeadCompensation Muhammad Amir Yousaf
LeadCompensation Muhammad Amir Yousaf
0 A Zero is placed at 10KHz to eliminate the pole at this frequency Aol 100db -45 -90 80 PhaseResponse -135 60 • -180 40 GainResponse ’LeadCompensation Circuit’ • -225 20 • -270 Muhammad Amir Yousaf 10 100 100K 10M 1K 10K 1M 1M
0 A Zero gives +45 degreephaseshift at 10KHz Aol 100db -45 -90 80 PhaseResponse -135 60 • -180 40 • 90 PhaseResponse ’LeadCompensation Circuit’ • 45 • -225 20 0 • -270 Muhammad Amir Yousaf 10 100 100K 10M 1K 10K 1M 1M
0 Adding Aol and Leadcompensationcircuit’sGain. Aol 100db -45 -90 80 Resultant Aol after compensation -135 60 • -180 40 GainResponse ’LeadCompensation Circuit’ • -225 20 • -270 Muhammad Amir Yousaf 10 100 100K 10M 1K 10K 1M 1M
0 A Zero gives +45 degreephaseshift at 10KHz -45 AmplifierPhaseResponse -90 -135 Resultant phaseresponse • -180 • 90 PhaseResponse ’LeadCompensation Circuit’ • 45 • -225 0 • -270 Muhammad Amir Yousaf 10 100 100K 10M 1K 10K 1M 1M
0 Aol and phaseresponse after compensation Aol 100db -45 -90 80 After compensation, we get -135degree phase shift at 1MHz -135 60 • -180 40 Gain at 1MHz is 40dB. This means that the feedback factor must be atleast -40dB to get Loop Gain(AolB) = 0db a@ 1MHz • -225 20 So amplifier is stable for Acl >= 40dB • -270 Muhammad Amir Yousaf 10 100 100K 10M 1K 10K 1M 1M
But our requirement was to make it stable for 20dB amplification. • Still not achieved…… Muhammad Amir Yousaf
LeadCompensation Here is the solution. This factorcan be adjusted for lowergain in Leadcompensationciruits Muhammad Amir Yousaf
100db 0 Amplifiershould be stabilized with 20dB gain. 80 -45 Aol 60 -90 40 -135 • -180 20 CompensationcircuitGainResponse is adjusted by selecting proper values of R1,R2 0 • -225 -20 • -270 Muhammad Amir Yousaf 10 100 100K 10M 1K 10K 1M 1M
100db 0 Resultant GianResponse 80 -45 Aol 60 -90 40 -135 • -180 20 CompensationcircuitGainResponse is adjusted by selecting proper values of R1,R2 0 • -225 -20 • -270 Muhammad Amir Yousaf 10 100 100K 10M 1K 10K 1M 1M
100db 0 Resultant GianResponse 80 -45 Aol 60 -90 40 -135 Nowwehave 20 dB at 1MHz @ -135 degreephaseshift • -180 20 Feedback factorshould be -20 and Acl = 20dB…. Achieved 0 • -225 -20 • -270 Muhammad Amir Yousaf 10 100 100K 10M 1K 10K 1M 1M