L07 Op amps - Massachusetts Institute of Technologyweb.mit.edu/6.101/www/s2017/handouts/L07.pdf · • Op Amps • 741, 356 • Imperfections • Op‐amp applications 6.101 Spring

• Op Amps• 741, 356• Imperfections• Op‐amp applications

6.101 Spring 2017 Lecture 7 1

Acknowledgements: Ron Roscoe,Neamen, Donald: Microelectronics Circuit Analysis and Design, 3rd Edition

JFET Application Current Source

• Household application: battery charger (car, laptop, mp3 players)

• Differential amplifier current source

• Ramp waveform generator• High Speed DA converter using

capacitors• Simple circuit: 2N5459

Nchannel JFET

2

1

P

GSDSSD V

vIi

6.101 Spring 2017 2

+15V

2N5459

IDSS = current with VGS=0

VP = pinchoff voltage

iD

6.101 Spring 2017 3

Op‐Amps

• Active device: V0 = a(V+-V-); note that it is the difference of the input voltage!

• a=open loop gain ~ 105 – 106

• Most applications use negative feedback.• Comparator: no feedback• Active device requires power. No shown for

simplicity.• Classics op-amps: 741, 357 ~ $0.20; one, two or

four in a package. • Newer op-amps operate at <3.3V (OPA369 1.8V)

V+

VoV-

Lecture 7

Op‐Amp Packaging

6.101 Spring 2017 4

LM324

LF353

Lecture 7

356 JFET Input Op‐amp

6.101 Spring 2017 5Lecture 7

JFET Differential Pair

6.101 Spring 2017 6

Small Signal Model

741 Circuit

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22 transistors, 11 resistors, 1 capacitor, 1 diode

Lecture 7

Differential (Emitter Coupled) Pair

6.101 Spring 2017 8

BJT Diff PairSmall Signal Model

Differential Pair – Common Mode Voltage

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Small Signal Model

Differential Pair – Differential Mode Voltage

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MOSFET Differential Pair

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Small Signal Model

+

6.101 Spring 2017 12

Virtual Node Analysis

V1

V2 Vout = a(V2-V1)

Vx

-

+

a • If a>>1 and a>> β then v1 ~ v2

• Current into input terminals zero by design

• Typical values: a~100,000 & a β >>1

• ok for a=a(s) and β = β(s) as long as a(s) β(s) >> 1

• β is the loop transfer function(not to be confused of β of a BJT)

• a β is the loop gain

outx VVV 1

121 VaVaVV x

21 1 VaVaV x

V1 Vx1

1 a

a1 a

V2 V2

a = gainβ = feedback or loop function

Lecture 7

Op Amps – Virtual Node

• With negative feedback, output will drive the input voltage difference to zero => V+ = V‐

• Input current = 0

6.101 Spring 2017 13

Benefits of FeedbackStabilize gain against device variations, temperature, aging

Reduce distortion by the feedback factor [(1+aβ)]

Input and output impedances adjusted by (1+aβ)

Gain determined by passive components

Disadvantages of Feedback

Loss of gain; need more stages Greater tendency for instability (oscillations)

Lecture 7

221 111 VV

aa

aVV x

741 Op Amp Max Ratings

6.101 Spring 2017 14

Need +Vcc, -Vee for operationcommon mode voltage appears at both inputs

Idiot proof

Lecture 7

741 Electrical Characteristics

6.101 Spring 2017 15

Almost zero

Lecture 7

LF356


not rail to rail

LM6132 – Rail to Rail Output


Rail to Rail Input


Decibel (dB)

i

o

VVdB log20

i

o

PPdB log10

6.101 Spring 2017 19

3 dB point =

log10(2)=.301 100 dB = 100,000 = 105

Lecture 7

80 dB = 10,000 = 104

60 dB = 1,000 = 103

40 dB = 100 = 102

half power point

741 Open Loop Frequency Gain


Non‐Inverting Amplifer

6.101 Spring 2017 21

-

++15

A

2

3

4

76

vin+

vout

_

+

_ -15

Rs

R2

R1

v+

v-

1

2

1

21

;21

1

21

1

1

;

RRAor

RRR

vv

vRR

Rvso

vvbutvRR

Rv

vvthereforecurrentinputZero

v

in

out

outin

out

in

AAA

vv

RRR

vin

out

1

21

1

for finite A

β (not to be confused with βof a BJT)

741 Open Loop Frequency Gain

6.101 Spring 2017 22

Examples at 1 Hz, 1000 Hz, and 10kHz

Voltage gain Av=40dB = 100; R2= 100k, R1= 1k; [101 = 40.1dB!] β=0.01

At 1 Hz, Avol = 100 dB = 1 x 105 = 100,000.

At 1000 Hz, Avol = 60 dB = 103 = 1000.

At 10 kHz, Avol = 42 dB = 1.26 x 102 = 126.

-

++15

A

2

3

4

76

vin+

vout

_

+

_ -15

Rs

R2

R1

v+

v-

dBA

AAv 401001010

01.10110

1 3

5

5

5

dBAAAv 2.399.90

111000

10110

01.10110

1

3

3

3

dBA

AAv 9.348.5526.2

12626.11

12601.1261

1261

β is the loop transfer functionaβ is the loop gain

Lecture 7

741 vs 356 Comparison741 356

Input device BJT JFETInput bias current 0.5uA 0.0001uA

Input resistance 0.3 MΩ 106 MΩSlew rate* 0.5 v/μs 7.5 v/μsGain Bandwidth product 1 Mhz 5 MhzOutput short circuit duration continuous continuous

Identical pin out

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* comparators have >50 v/μs slew rate

Lecture 7

So Why BJT in Op‐amps?

6.101 Spring 2017 24

BJTs have higher transconductance (gain), better consistency in spec between pieces, and in some applications, lower noise than FETs.

Like most JFET op amps, the LF356 has a relatively high offset voltage, and relatively high drifts. BJT op-amps tend to have much lower offset voltage and drifts.

Gain Bandwidth Product = Constant(No free lunch)


Gain: 60dB = 103

Bandwidth = 5x103

Gain Bandwidth product = 5x106

Op‐Amp Imperfections – Real World

• Input offset voltage• Input Current Bias• Input Offset Current• Finite Output Voltage Swing• Finite Current• Finite Gain, gain bandwidth product• Voltage Noise – Johnson Noise• Phase Shifts• Slew Rate


Input Offset Voltage *


* Analog Devices MT-037 Tutorial

741: 6000μv357: 10,000μv

Offset Adjustments


Input Bias Current *


The input offset current, IOS,is the difference between IB– and IB+, or IOS = IB+ − IB–.

* Analog Devices MT-038 Tutorial

741: 200na357: 0.05na

Inverting Amplifier Bias Current Compensation


-

+

+15

A

2

3 4

76

VIN

+VOUT

_

+

_ -15

RIN

RF

IIN IFIB

R1

IB+

VOFF

_

o

OUTIN

Vat offset no for condtion aas thus

:so 0, V want weand Vsignal, input no withbut

1

1BB

1

1 11 :

11I-;11I-

0

0

RRR

RRIR

RRV

RVVI

RVV

IRVIII

FIN

FINB

FINOFF

F

OUTOFFB

IN

OFFIN

BOFF

FBIN

-

+

+15

AVOLVDIFF

-15

RF//RIN

R1

IB

IB

+VOUT

_

1

1

// 0

//

RRRifV

ARIRRIV

INFOUT

VOLBINFBOUT

1

1

// 0

//

RRRifV

ARIRRIV

INFOUT

VOLBINFBOUT

Common Mode Rejection Ratio CMRR


CMRR: ratio of the common-mode gain to differential-mode gain.

Example, if a differential input change of Y volts produces a change of 1 V at the output, and a common-mode change of X volts produces a similar change of 1 V, then the CMRR is X/Y.

CMRR often expressed in dB:

CM

OL

AACMRR log20

Inverting Amplifer – Virtual Ground Analysis


AssumptionsInfinite input impedance:

-

+

+15

A

2

34

7

6

vin

+vout

_

+

_ -15

Rin

Rf

iin if

vin

f

in

out

f

out

in

in

fin

ARR

vv

Rv

Rv

ii

000

0

i i0 0;

A v because v is grounded 0 .

inRfR

VAA

Schmitt Trigger• Schmitt trigger have

different triggers points for rising edge and falling edge.

• Can be used to reduce false triggering

• This is NOT a negative feedback circuit.

Vin Vo

V-

V+

R1

R2


Schmitt Trigger + RC Feedback = Oscillator

• 741 op‐amp.

• R1=10k, R2=4.7k, R3=10K, C=.33uf

• Display V‐ and Vout on the scope. Set R3=4.7k. Predict what happens to the frequency.

6.101 Spring 2017 34

-+

R3 10K

R1 10K

R2 4.7K

0.33uf

V-

Vout

Lecture 7

High Pass Filter HPF

6.101 Spring 2017 35

RV1 V2

C

1CRsCRs

1CRjCRj

Cj1R

RVVA

1

2v

log f

AV (dB)

-3dB

fLO or f-3dB

slope = 6 dB / octaveslope = 20 dB / decade

0

log f

Degrees

45o

fLO or f-3dB

90o

0o

-45o

PHASE LEAD

Lecture 7

Differentiator Insights


;;11 1

2

1

2 jssCR

sCR

sCR

RAv

at low frequency 11 sCR

12sCRAv

multiplying by s equals differentiation

log f

AV (dB)

-3dB

fLO or f-3dB

slope = 6 dB / octaveslope = 20 dB / decade

0

log f

Degrees

45o

fLO or f-3dB

90o

0o

-45o

PHASE LEAD

differentiation works only at f << flo

Low Pass Filter LPF

6.101 Spring 2017 37

RV1 V2C

11

11

1

1

1

2

sRCA

RCjCj

R

CjXjR

XjVVA

v

C

Cv

log f

AV (dB)

-3dB

fHI or f-3dB

slope = -6 dB / octaveslope = -20 dB / decade

0

log f

Degrees

-45o

fHI or f-3dB

0o

-90o

PHASE LAG

Lecture 7

Integrator Insights


;;1 2

1

2

jssCRR

R

Av

12 sCR

12

12

v sCR1

sCRR

R

A

at high frequency

dividing by s equals integration

log f

AV (dB)

-3dB

fHI or f-3dB

slope = -6 dB / octaveslope = -20 dB / decade

0

log f

Degrees

-45o

fHI or f-3dB

0o

-90o

PHASE LAG

integration works only at f >> fHI 12 sCR

Why R2?

6.101 Spring 2017 39

Without R2, any DC bias current will saturate Vout since the DC gain is the open loop gain

Lecture 7

Frequency Domain Insight


• Integration and differentiation easy to understand in time domain

• In frequency domain, difference between square wave and triangle wave is amplitude and phase –same harmonics.

• Integrator (LPF) rolls off harmonics and phase shift to create a triangle wave

• Differentiator (HPF) amplifies harmonics and phase shift to create a square wave.

Basic OpAmp CircuitsVoltage Follower (buffer)

invoutv

2R

1R

inRRR

out vv1

21

Non-inverting

inv

outv

1invoutv

2inv

1R

1R

2R

2R

121

2

ininRR

out vvv

Differential Inputinout

vv

invoutv

R C

t

inRCout dtvv 1

Integrator

+

-

41

Crossover Distortion (hole)


[a]

vout-

+

LF356

2

34

7

6

vin

RL

2N3904

2N3906

[b]

-

+

LF356

2

34

7

6

vin

2N3904

2N3906

vout

RL

0.1F

0.1F

+15

-150.1F

+15

-15

0.1F

+15

-15-15

+1510k

10k

10k

10k

?k?k

Why is [b] better?

Diode Biasing

6.101 Spring 2017 43

C

+vin

_

RF

2N2219

2N2905

D2 1N914

D1 IN914

RE=5.61/2 watt

RB1

+12 V

RL

-12 V

+12 V

+

vout

_RB2

-12 V

RE=5.61/2 watt

+122

34

7

6

-12

-

+

?opamp

[FromPreamplifier]

1N4001

1N4001

Lecture 7

Documents

L07 Op amps - Massachusetts Institute of Technologyweb.mit.edu/6.101/www/s2017/handouts/L07.pdf · • Op Amps • 741, 356 • Imperfections • Op‐amp applications 6.101 Spring