741 internal circuits

7/29/2019 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

741 internal circuits

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