741 Internal Circuits
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Transcript of 741 Internal Circuits
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The 741 opAmp
DC and Small Signal AnalysisMADAN SHARMA
For EEC-501
AUG 14, 2012
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Overview: Five Parts of the 741
Biasing Currents
Input Stage
Second Stage Output Stage
Short Circuit Protection
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Overview: 741 Schematic
Q2
Q19
Vo
R1150k
Q21
Q12
R10
40k
Q10R950k
Q5
HI
R8100
Cc
30p
Q6
R539k
Q13a
Q9
R21k
Q1
R350k
R7
27k
Q16
Q14
Q20
Q8
Q13b
Q23Q3
Vin+
Q22Q24
Q15
R11k
Q7
Q17
Q18
LO
Q4
R627k
Vin-
R45k
Q11
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Biasing Current Sources
Generates the reference bias currentthrough R5
Q10
Q9
Q12
R5
39k
R4
5k
Q11
Q8
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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
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Input Stage
The differential pair, Q1 and Q2 provide the main inputand hih input resistance,Q3,Q4 PNP transistor withcommon base provide high voltage gain.
Transistors Q5-Q7 provide an active load for the input
Q6
Q3
Q1
Q7
Q5
R11k
Vin-Vin+
Q4
R350k
R21k
Q2
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Input Stage:
DC Analysis - 1Assuming that Q10 and Q11 are
matched, we can write the equation from
the Widlar current source:VT ln
IREF
IC10
IC10 R4
Using trial and error, we can solve for
IC10, and we get: IC10=19A
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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:
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Input Stage:DC Analysis -3
Analysis of the active load:
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Second (Intermediate) Stage
Transistor Q16 acts as an emitter-follower givingthis stage a high input resistance
Capacitor Cc provides frequency compensationusing the Miller compensation technique
Q13b
R950k
Q13a
Q16
Q17
Cc
30p
R8100
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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 VT ln
IC17
IS
618mV
IC16 IE16 IB17
IE17 R8 VBE17
R9
16.2A
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Output Stage
Provides the opAmp with a low output resistance
Class AB output stage provides fairly high current loadcapabilities without hindering power dissipation in the IC
Q15
Q18
Q23
Q21
Q14
Vo
Q19
R7
27k
R6
27k
Q20
R10
40k
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Output Stage:DC Analysis
Q13a delivers a current of 0.25IREF, so wecan 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
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Short Circuit Protection
These transistors are normally off
They only conduct in the event that a largecurrent is drawn from the output terminal (i.e. ashort circuit)
R1150k
Q24
Q22
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DC Analysis Summary
Q1 9.5 Q8 19 Q13b 550 Q19 15.8
Q2 9.5 Q9 19 Q14 154 Q20 154
Q3 9.5 Q10 19 Q15 0 Q21 0
Q4 9.5 Q11 730 Q16 16.2 Q22 0
Q5 9.5 Q12 730 Q17 550 Q23 180
Q6 9.5 Q13a 180 Q18 165 Q24 0
Q7 10.5
DC Collector Currents of the 741 (mA)
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741 opAmp Simulation: Schematic
Q2
0
Q19
Vo
R1150k
Q21
Q12
R10
40k
Q10
R9
50k
Inverting Amplifier with a gain of 20
Rf eedback1
1k
Rf eedback2
20k
Q5
HI
Vo
LO
R8100
Cc
30p
Q6
HI
R539k
Q13a
Q9
R21k
Vcc15Vdc
Vee-15Vdc
Vin
FREQ = 1kVAMPL = 1mVVOFF = 0V
Q1
R350k
R7
27k
Q16
Q14
Q20
Vin-
Q8
Q13b
Q23Q3
Vin+
Q22Q24
Q15
R11k
Q7
Q17
Q18
LO
Q4
R627k
Vin+
Vin-
R45k
0
Q11
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741 opAmp Simulation: Input
Time
0s 1.0ms 2.0ms 3.0ms 4.0ms
V(Vin:+)
-1.0mV
-0.5mV
0V
0.5mV
1.0mV
1mV Amplitude
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741 opAmp Simulation: Output
Time
0s 1.0ms 2.0ms 3.0ms 4.0ms
V(Vo) - 28.234mV
-20mV
0V
20mV
Inverted output
20mV Amplitude
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Conclusions
The 741 is a versatile opAmp that can beused in a multitude of different ways
When you break it down into the differentcomponents, its operation is actuallyunderstandable and comprehendible