The 741 opAmp DC and Small Signal Analysis Jeremy Andrus For Engineering 332 May 15, 2002 Prof.
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The 741 opAmp DC and Small Signal Analysis Jeremy Andrus For Engineering 332 May 15, 2002 Prof. Ribeiro Overview: Five Parts of the 741 Biasing Currents Input Stage Second Stage Output Stage Short Circuit Protection Overview: 741 Schematic HI Q12 Q13 b Q13 a Q14 Q15 Q9 Q8 Q19 R6 27k Q18 Vo Vi n+ R10 40k Vi nQ1 R7 27k Q2 Q21 R5 39k Q20 Cc Q3 Q4 Q23 30p Q7 Q16 Q17 Q10 Q6 Q5 R8 100 R3 50k R4 5k R1 1k R2 1k Q22 Q24 R11 50k LO Q11 R9 50k Biasing Current Sources Q12 Q9 Q8 R5 39k Q11 Q10 R4 5k Generates the reference bias current through R5 Biasing Current Sources: DC Analysis The opAmp reference current is given by: Iref VCC VEB12 VBE11 VEE R5 For Vcc=Vee=15V and VBE11=VBE12=0.7V, we have IREF=0.73mA Input Stage Vi n+ Vi nQ1 Q2 Q3 Q4 Q7 Q6 Q5 R3 50k R1 1k R2 1k The differential pair, Q1 and Q2 provide the main input Transistors Q5-Q7 provide an active load for the input Input Stage: DC Analysis - 1 Assuming that Q10 and Q11 are matched, we can write the equation from the Widlar current source: IREF VT ln I C10 IC10 R4 Using trial and error, we can solve for IC10, and we get: IC10=19A Input Stage: DC Analysis -2 From symmetry we see that IC1=IC2=I, and if the npn is large, then IE3=IE4=I Analysis continues: Input Stage: DC Analysis -3 Analysis of the active load: Second (Intermediate) Stage Q13 b Q13 a Cc 30p Q16 Q17 R9 50k R8 100 Transistor Q16 acts as an emitter-follower giving this stage a high input resistance Capacitor Cc provides frequency compensation using the Miller compensation technique Second Stage: DC Analysis Neglecting the base current of Q23, IC17 is equal to the current supplied by Q13b IC13b=0.75IREF where P >> 1 Thus: IC13b=550uA=IC17 Then we can also write: VBE17 IC16 IE16 IC17 VT ln I S IB17 618mV IE17 R8 VBE17 R9 16.2A Output Stage Q14 Q15 Q19 R6 27k Q18 Vo R10 40k R7 27k Q21 Q20 Q23 Provides the opAmp with a low output resistance Class AB output stage provides fairly high current load capabilities without hindering power dissipation in the IC Output Stage: DC Analysis Q13a delivers a current of 0.25IREF, so we can say: IC23=IE23=0.25IREF=180A Assuming VBE18 = 0.6V, then IR10=15A, IE18=180-15=165A and IC18=IE18=165A IC19=IE19=IB18+IR10=15.8A Short Circuit Protection Q24 Q22 R11 50k These transistors are normally off They only conduct in the event that a large current is drawn from the output terminal (i.e. a short circuit) DC Analysis Summary DC Collector Currents of the 741 (mA) Q1 9.5 Q8 19 Q13b Q2 9.5 Q9 19 Q14 Q3 9.5 Q10 19 Q15 Q4 9.5 Q11 730 Q16 Q5 9.5 Q12 730 Q17 Q6 9.5 Q13a 180 Q18 Q7 10.5 550 154 0 16.2 550 165 Q19 Q20 Q21 Q22 Q23 Q24 15.8 154 0 0 180 0 741 opAmp Simulation: Schematic HI Rfeedbac k2 Vo Rfeedbac k1 Q12 Q13b Q13a Vin1k Q14 VOFF = 0V VAMPL = 1mV FREQ = 1k Q15 Q9 Q8 20k Vin Q19 R6 27k Vin+ Q18 Vo Vin+ R10 40k VinQ1 0 R7 27k Q2 Q21 R5 39k Q20 Cc Q3 Q4 Q23 Inverting Amplifier with a gain of 20 30p Q7 Q16 Q17 Q6 Q5 R8 100 R3 50k R1 1k 15Vdc Vcc R2 1k Q22 Q24 Vee R11 50k -15Vdc 0 LO R4 5k LO Q10 HI Q11 R9 50k 741 opAmp Simulation: Input 1.0mV 0.5mV 1mV Amplitude 0V -0.5mV -1.0mV 0s 1.0ms 2.0ms V(Vin:+) Time 3.0ms 4.0ms 741 opAmp Simulation: Output 20mV 20mV Amplitude 0V Inverted output -20mV 0s 1.0ms V(Vo) - 28.234mV 2.0ms Time 3.0ms 4.0ms Conclusions The 741 is a versatile opAmp that can be used in a multitude of different ways When you break it down into the different components, it’s operation is actually understandable and comprehendible