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

6.101 Spring 2020 Lecture 7 1

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

6.101 Spring 2020 Lecture 7 2

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 2020 3

+15V

2N5459

IDSS = current with VGS=0

VP = pinchoff voltage

iD

2017 Quiz Design Question

6.101 Spring 2020 Lecture 7 4

Design a circuit such that when a momentary push button switch is closed, “switched 9v” will be supplied to your design under test for exactly two minutes.

You have available a 2N7000 (n-channel MOSFET) and a ZVP2106A which is a p-channel complement to the 2N7000.

You have a 100uf capacitor that is exactly 100uf with negligible leakage current.

Based on measurements, your 2N7000 has a VGS(th) 3.0 volts and the ZVP2106A has a VGS(th) -3.0.

The 9V battery is a constant 9V during the test.ln(.3333) = -1.0986ln(.2222) = -1.5041ln(.1111) = -2.1972 [one or more of these constants may be required for your design]

6.101 Spring 2020 5

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 2020 6

LM324

LF353

Lecture 7

356 JFET Input Op‐amp

6.101 Spring 2020 7Lecture 7

JFET Differential  Pair

6.101 Spring 2020 8

Small Signal Model

741 Circuit

6.101 Spring 2020 9

22 transistors, 11 resistors, 1 capacitor, 1 diode

Lecture 7

Discrete 741 

6.101 Spring 2020 10Lecture 7

Discrete 741

6.101 Spring 2020 11Lecture 7

Differential (Emitter Coupled) Pair

6.101 Spring 2020 12

BJT Diff Pair Small Signal Model

Differential Pair – Common Mode Voltage

6.101 Spring 2020 13

Small Signal Model

Differential Pair – Differential Mode Voltage

6.101 Spring 2020 14

MOSFET Differential  Pair

6.101 Spring 2020 15

Small Signal Model

+

6.101 Spring 2020 16

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 2020 17

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 2020 18

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

Idiot proof

Lecture 7

741 Electrical Characteristics

6.101 Spring 2020 19

Almost zero

Lecture 7

LF356 

6.101 Spring 2020 20Lecture 7

not rail to rail

LM6132 – Rail to Rail Output

6.101 Spring 2020 21Lecture 7

Rail to Rail Input

6.101 Spring 2020 22Lecture 7

Decibel (dB)

i

o

VVdB log20

i

o

PPdB log10

6.101 Spring 2020 23

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

6.101 Spring 2020 24Lecture 7

Non‐Inverting Amplifer

6.101 Spring 2020 25

-

++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 2020 26

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

6.101 Spring 2020 27

* comparators have >50 v/μs slew rate

Lecture 7

So Why BJT in Op‐amps?

6.101 Spring 2020 28

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)

6.101 Spring 2020 Lecture 7 29

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

6.101 Spring 2020 30Lecture 7

Input Offset Voltage *

6.101 Spring 2020 Lecture 7 31

* Analog Devices MT-037 Tutorial

741: 6000μv357: 10,000μv

Current technology:10μv

Offset Adjustments

6.101 Spring 2020 32Lecture 7

Input Bias Current *

6.101 Spring 2020 33Lecture 7

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

Current technology:3fA LTC6268

Inverting Amplifier Bias Current Compensation

6.101 Spring 2020 34Lecture 7

-

+

+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

6.101 Spring 2020 35Lecture 7

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

6.101 Spring 2020 Lecture 7 36

AssumptionsInfinite input impedance:

-

+

+15

A

2

34

76

vin

+vout

_

+

_ -15

Rin

Rf

iin if

vin

f

in

out

f

out

in

in

fin

ARR

vv

Rv

Rv

ii

0000

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

6.101 Spring 2020 37Lecture 7

2015 Quiz Question – Digital  Clock Using 60Hz

6.101 Spring 2020 38Lecture 7

The 60 Hz waveform crosses the reference voltage multiple times because of the 200mv noise.

Design a circuit with a Schmitt trigger (“Your design”) using a LM6132 with the thresholds greater than 200mv to fix the problem. A LM7805 voltage regulator IC is used to provide a 5V output to power the digital clock circuit (modeled as a resistor) as well as to supply +5V for the LM6132 in “Your design”.

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 2020 39

-+

R3 10K

R1 10K

R2 4.7K

0.33uf

V-

Vout

Lecture 7

High Pass Filter HPF

6.101 Spring 2020 40

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

6.101 Spring 2020 Lecture 7 41

;;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 2020 42

RV1 V2C

Av V2

V1

j XC

R j XC

1jC

R 1jC

1jRC 1

Av 1

sRC 1

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

6.101 Spring 2020 Lecture 7 43

;;1 2

1

2

jssCRR

R

Av

12 sCR

12

12

v sCR1

sCRR

RA

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 2020 44

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

Lecture 7

Frequency Domain Insight

6.101 Spring 2020 45Lecture 7

• 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

+

-

46

Crossover Distortion (hole)

6.101 Spring 2020 47Lecture 7

[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 2020 48

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

6.101 Spring 2020 49