Class Note for ECE 304 with Professor Brews at UA
Class Note for ECE 304 with Professor Brews at UA
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This 3 page Class Notes was uploaded by an elite notetaker on Friday February 6, 2015. The Class Notes belongs to a course at University of Arizona taught by a professor in Fall. Since its upload, it has received 25 views.
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Date Created: 02/06/15
ECE 304 Feedback Triple Schematic WP P P P P 9000v R1 R2 R3 5010uA 5010uA PARAMETERS V1 R C1 RCz RC3 OUT3 Q4 lV CC 664mV 054v 6597V QN RF640 o oun OUT2 OUT3E v OUT c BY 10 F u R E1 10 6000uA cc 5960V RE3 10 CH K3 FB Q1 oz Q3 10F RC11354k 10F Q N Q N Q N RL R c2 5k 3631mV 4040m A Rc3 600 VSIN 0 I OUT To 4 R8 R L 100k 0 05v 39 4088uA 6000uA t vcc 9v 1kHz 1 FB m Ro R o939123k VAC 1015mA 105mV RF 331mV 3632mA 1 0k 1 A DOTEMODEL PARAMS 5 1V 2 quot 39 gt c BY 500uA B F 100 3 CBY Ls WA RE1 RE3 E VAF 1E12V 0 model QN NPN BfBF lslS VafVAF FIGURE 1 Feedback triple set up to approximate the circuit in 8amp8 Example 82 Figure 1 shows an amplifier using a feedback triple The collector currents of Q1 and Q3 are chosen to match 8amp8 Example 82 The current of Q1 is adjusted using Rm which differs from 8amp8 Rm 9 kg The current of Q3 is adjusted with R0 The collector current of Q2 cannot be made to fit their value of 1 mA The DC collectorvoltage of Q3 is set using the DC feedback loop at the right of Figure 1 It has been set to obtain Ice 4 mA to agree with 8amp8 A better idea from a design viewpoint would be to center this voltage for the best output swing We ask what kind of feedback is in use here Output side From the figure it looks like we are expecting to use this amplifier as a voltage amplifier it has a voltage driver and the output is the voltage across the load RL However the output voltage at node VOUT is outside the feedback loop so feedback is NOT being used to make this amplifier look like a low impedance Thevenin source The smallsignal impedance looking into node OUT3 is RC3 independent of the feedback loop Instead the output side of the feedback network is in series with the emitter of Q3 appropriate for a current output That is feedback is serving to make the amplifier behave as a better Norton source for the current in the emitter of Q3 CURRENT FEEDBACK FOR A VOLTAGE AMPLIFIER What benefit can this current feedback have for the voltage output The answer is that it makes the voltage gain less sensitive to variations in the transistor properties due to variations in temperature or manufacturing variations in parameters like VAF and 3 Using current feedback in a voltage amplifier may seem bizarre but it has the effect of making the voltage gain insensitive to variations in the main amplifier parameters The reason it works is that the current in the emitter of Q3 is made insensitive to the amplifier parameters and the output voltage isjust BQ3BQ31 153Q3 RLchs 80 making 153 insensitive means the output voltage is insensitive too provided 3Q3 gtgt 1 Input side On the input side the feedback voltage across R51 is in series with the input voltage uBEQ1 to the amplifier suggesting that the input hookup is series which is appropriate for a voltage input J R Brews Page 1 412005 Checks Let s check these observations using PSPICE Figure 2 shows that indeed the gain behavior as a function of RL is exactly determined by the output resistance RC3 unaffected by feedback 60 50 4O 3O 20 1O 0 1EO1 1E02 1EO3 1EO4 1E05 RL 9 El Ratio PSpice FIGURE 2 Comparison of voltage gain from PSPICE with gain based upon divider ratio RLRLRC3 P P P R1 R2 R3 PARAMETERS R C1 RC2 RC3 6 1054v RF 640 oung OUT2 OUT3 v OUT c BY 10 F quot R E1 10 6000uA 1549m OOOmA cc 2 51 1205 Q1 Q2 10quot 100k RCz 5k39 QN QN QN E E 363 1m 4040mA E Rc3 600 39 0 IOUT 39 0 R L 100k 5 4088u 1 R L 6000uA vcc9v m l 39 m A R O9123k quot39 101 5m RF 3631m 3 632mA 100k DOTVMODEL PARAMS BF 100 I is 10fA R E R E3 VAF 1E12v 0 model QN NPN BfBF lslSVafVAF FIGURE 3 Circuit of Figure 1 modified to place the load resistor in the emitter lead of Q3 Using the circuit of Figure 3 we check how the current gain depends on RL inserted in the emitter lead of Q3 122K808m 10 100 10K 10K 100K 10M u 1IiOUT FIGURE 4 Transconductance gain at emitter of Q3 the sudden drop is due to cutoff of Q3 at large R J R Brews Page 2 412005 Figure 4 shows that the transG gain is independent of RL inserted in the emitter lead of Q3 until RL becomes large enough that Q3 goes into cutoff1 That is feedback IS effective in keeping the emitter current fixed independent of the load so the amplifier is a good Norton driver for the load in this position Summary lfthe output variable is sensed by the feedback loop feedback can be used to make the amplifier driver a more ideal Norton or Thevenin source The gain of this driver compared to the original source is a reliable value determined by the feedback network lfthe output variable is not sensed by the feedback loop the amplifier driver is not converted to a more ideal Norton orThevenin source Nonetheless feedback still can be useful to control sensitivity to amplifier variations like temperature or manufacturing variations 1 As feedback keeps the current in RL constant as RL increases of course the voltage drop across RL increases as RL increases The increased drop across RL means that the emitter of Q3 rises in voltage until VBE gt 0 and Q3 cuts off J R Brews Page 3 412005
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