MOSFET Configurations

• Common source • Common Drain ‐ Source follower • Common gate • MOSFET Configurations • Op Amps • 741, 356 • Imperfections • Op‐amp applications Acknowledgements: Ron Roscoe, Neamen, Donald: Microelectronics Circuit Analysis and Design, 3rd Edition 6.101 Spring 2014

Lecture 7

1

6.101 Spring 2014

Common Source

Lecture 7

2

Common Source Small Signal

Av  Vo Vi   g m (ro RD ) 6.101 Spring 2014

Lecture 7

3

6.101 Spring 2014

Lecture 7

4

Common Drain ‐ Source Follower Output Impedance

Common Drain ‐ Source Follower

Av 

6.101 Spring 2014

RS ro

Ri ( ) 1 R  R i Si  RS ro gm

Lecture 7

RO 

5

Lecture 7

6.101 Spring 2014

Av 

Lecture 7

6

Common Gate Small Signal 

Common Gate

6.101 Spring 2014

1 RS ro gm

7

6.101 Spring 2014

g m ( RD RL ) 1  g m RSi

Ai 

IO g R RD ( )( m Si ) Ii RD  RL 1  g m RSi

Lecture 7

8

MOSFET Configuration Summary

JFET Application Current Source

+15V

Configuration

Common Source Source Follower Common Gate

Voltage Gain Av > 1

Av ≈ 1

Av > 1

Current Gain __

__

Ai ≈ 1

•

Input Output Resistance Resistance RTH

•

Moderate to high

• • •

RTH

Low

Low

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

Lecture 7

iD

IDSS = current with VGS=0

Moderate to high

VP = pinchoff voltage

 v iD  I DSS 1  GS  VP 6.101 Spring 2014

2N5459

9

  

2

6.101 Spring 2014

10

Op‐Amp Packaging

Op‐Amps

LM324

• Active device: V0 = a(V+-V-); V+ note that it is the difference V Vof 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. o

LF353 6.101 Spring 2014

Lecture 7

11

6.101 Spring 2014

12

356 JFET Input Op‐amp

JFET Differential  Pair Small Signal Model

6.101 Spring 2014

Lecture 7

13

6.101 Spring 2014

14

Differential (Emitter Coupled) Pair

741 Circuit

BJT Diff Pair

Small Signal Model

22 transistors, 11 resistors, 1 capacitor, 1 diode

6.101 Spring 2014

Lecture 7

15

6.101 Spring 2014

16

Differential Pair – Common Mode Voltage

Differential Pair – Differential Mode Voltage

Small Signal Model

6.101 Spring 2014

17

6.101 Spring 2014

18

Virtual Node Analysis

MOSFET Differential  Pair Small Signal Model

+

V2

Vout = a(V2-V1) a

-

V1 Vx

+

a = gain β = feedback or loop function



V1  Vx  Vout

•

If a>>1 and a>> β then

•

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

V1  Vx  aV2  aV1 V1 1  a   Vx  aV2

(not to be confused of β of a BJT)

 1   a     V2  V2 V1  Vx   1  a   1  a  6.101 Spring 2014

19

6.101 Spring 2014

v1 ~ v2

• Lecture 7

a β is the loop gain 20

Op Amps – Virtual Node • •

741 Op Amp Max Ratings

With negative feedback, output will drive the input voltage difference to  zero  => V+ = V‐ Input current = 0

common mode voltage appears at both inputs

Benefits of Feedback

Stabilize 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

Need +Vcc, -Vee for operation

Disadvantages of Feedback Loss of gain; need more stages

Greater tendency  for  instability  (oscillations) Idiot proof

 1   a     V2  V2 V1  Vx   1  a   1  a  6.101 Spring 2014

Lecture 7

21

6.101 Spring 2014

Lecture 7

22

LF356 

741 Electrical Characteristics Almost zero

not rail to rail 6.101 Spring 2014

Lecture 7

23

6.101 Spring 2014

Lecture 7

24

LM6152 – Rail to Rail Output

Decibel (dB) V  dB  20 log o   Vi 

P  dB  10 log o   Pi 

log10(2)=.301

100 dB = 100,000 = 105 80 dB = 10,000 = 104

3 dB point =

6.101 Spring 2014

Lecture 7

25

half power point

60 dB =

1,000 = 103

40 dB =

100 = 102

Lecture 7

6.101 Spring 2014

26

Non‐Inverting Amplifer

741 Open Loop Frequency Gain Rs

+ _

vin

3

+15

+

7

2

4

-

v 

6

A

v+

Zero input current; therefore v  vin +

-15

v-

vout R2

_

R1

R1  vout but v  v R 1 R 2

so vin 

R1 R1  R 2

vout R1  R 2  vin R1 or Av  1 

β (not to be confused with β of a BJT)

 for finite A

6.101 Spring 2014

Lecture 7

27

6.101 Spring 2014

 vout ;

R2 R1

R1 R1  R2

vout A  Av  1  A vin 28

741 Open Loop Frequency Gain Rs

+ _

vin

3

7 6

A

v+ 2

-

4

+

-15

v-

vout R2

R1

741 vs 356 Comparison

Examples at 1 Hz, 1000 Hz, and 10kHz

+15

+

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. Av 

A 105 10 5    100  40dB 1  A 1  10 5.01 10 3

356

Input device

BJT

JFET

Input bias current

0.5uA

0.0001uA

Input resistance

0.3 MΩ

106 MΩ

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

Slew rate*

0.5 v/us

7.5 v/us

A 103 103 1000 Av      90.9  39.2 dB 1  A 1  103.01 1  10 11

Gain Bandwidth product

1 Mhz

5 Mhz

Output short circuit duration

continuous

continuous

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

A 126 126 126     55.8  34.9dB 1  A 1  126.01 1  1.26 2.26

Lecture 7

Identical pin out * comparators have >50 v/us slew rate

β is the loop transfer function aβ is the loop gain 6.101 Spring 2014

741

29

6.101 Spring 2014

Lecture 7

Gain Bandwidth Product = Constant (No free lunch)

So Why BJT in Op‐amps?

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

Gain: 60dB = 103 Bandwidth = 5x103

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.

6.101 Spring 2014

30

Gain Bandwidth product = 5x106

31

6.101 Spring 2014

Lecture 7

32

Input Offset Voltage *

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 * Analog Devices MT-037 Tutorial

6.101 Spring 2014

Lecture 7

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6.101 Spring 2014

Lecture 7

34

Input Bias Current *

Offset Adjustments

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

* Analog Devices MT-038 Tutorial 6.101 Spring 2014

Lecture 7

35

6.101 Spring 2014

Lecture 7

36

Inverting Amplifier Offset Current Compensation I IN  I B  I F  0

RF IIN

IB

RIN +

VIN

_

IB + VOFF

IF 2

V OFF   R1 I B  V OUT V IN  V OFF V  I B  OFF 0 R IN RF

+15 7

A

6

3

VOUT

-15

R1

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

but with no input signal, V IN and we want V OUT  0, so :

+

4

+

Common Mode Rejection Ratio CMRR

 1  1 1  1    - I B  V OFF   ; - I B   R1 I B   RF  RF   R IN  R IN  1    1 1 thus :       as a condtion for no offset at V o RF   R1   R IN

_

_



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.



V OUT   I B R F // R IN  I B R1 AVOL

IB VDIFF

RF//RIN

IB

R1

CMRR often expressed in dB:

+15

-

V OUT  0 if R F // R IN  R1

AVOL

CMRR  20 log

+

+

VOUT

-15

_



V OUT   I B R F // R IN 



I B R1 AVOL

AOL ACM

V OUT  0 if R F // R IN  R1

6.101 Spring 2014

Lecture 7

37

6.101 Spring 2014

Lecture 7

Inverting Amplifer – Virtual Ground Analysis Assumptions Infinite input impedance: i  0;

A  A

Schmitt Trigger

i  0

v  0 because v is grounded .

A 

V



R R

f

Vin

in

if 2

Rin + _

vin

-

+15

3

+

vin  0 0  vout  0 Rin Rf

7

A

6 4

-15

+ vout

6.101 Spring 2014 6.101 Spring 2014

Lecture 7

•

Can be used to reduce false triggering

•

This is NOT a negative feedback circuit.

R1

vout  R f   Av vin Rin

_

Schmitt trigger have different triggers points for rising edge and falling edge.

Vo

R2

iin  i f  0

• VV+

Rf iin

38

39

Lecture 7

40

High Pass Filter HPF

Schmitt Trigger + RC Feedback =  Oscillator

AV (dB)

R3

10K

•

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.

C

R

V1

V2

0 -3dB

slope = 6 dB / octave slope = 20 dB / decade

log f

V-

+

0.33uf

Vout R1 10K

fLO or f-3dB

Av 

V2 R j CR s CR    V1 R  1 j CR  1 s CR  1 j C

Degrees

PHASE LEAD

90o 45o 0o -45o

R2

4.7K

log f fLO or f-3dB

Lecture 7

6.101 Spring 2014

41

Lecture 7

6.101 Spring 2014

Differentiator Insights

42

Low Pass Filter LPF

AV (dB)

AV (dB)

0 -3dB

R

slope = 6 dB / octave slope = 20 dB / decade

V1

C

V2

0 -3dB

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

log f

log f

fLO or f-3dB Degrees

 R2 1 sC

Av 

R1 

at low frequency

Av





 sCR2 ; s  j ; sCR1  1

90

45o 0o

sCR1  1

6.101 Spring 2014

 j XC V2   V1 R   j X C

Av 

1 sRC  1

j C 1  1 j RC  1 j C

R

Degrees

PHASE LAG

0o -45o -90o

-45o

log f fLO or f-3dB

 sCR2 1

multiplying by s equals differentiation

Av  PHASE LEAD

o

fHI or f-3dB

1

integration works only at f 10 x fHI Lecture 7

fHI or f-3dB

log f

sCR2  1 43

6.101 Spring 2014

Lecture 7

44

Integrator Insights

Why R2?

AV (dB)

0 -3dB

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

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

log f fHI or f-3dB Degrees

PHASE LAG

0o o

-45

R2

Av

 

R1 ; s  j ; 1  sCR2

sCR2  1

at high frequency R2 Av





R1 sCR2



-90o



fHI or f-3dB

1 sCR1

log f

integration works only at f 10 x fHI sCR2  1

dividing by s equals integration

6.101 Spring 2014

Lecture 7

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6.101 Spring 2014

Lecture 7

Basic OpAmp Circuits

Frequency Domain Insight

Non-inverting

Voltage Follower (buffer) • •

•

•

vin +

Integration and differentiation easy to understand in time domain

vin 

vout 

Differential Input

vin1

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

vin 2

Lecture 7

47

R1 R1

R1  R2 R1

vin

Integrator

R2

vout

vout 

R2

vout

R1

vout  vin

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



vout

-

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

6.101 Spring 2014

46

vin

R

C

vout

R2 R2 R1

v

in 2

 vin1 

t

1 vout   RC  vin dt 

48

Crossover Distortion (hole)

Diode Biasing +12 V +12 V

10k

+15 2N3904

+15 10k

10k

2N2219

2N3904

+15

+

-

2

6

LF356 3

0.1F

vout

7

4

RL

?k

vin

-15 2N3906

[a]

3

+

D1 IN914

6 4

RL

0.1F

-15

?k

[b]

-15

vout

7

LF356

0.1F

C + vin

2N3906

_

[From Preamplifier]

2

-

6

?opamp 3

+

D2 1N914

+12

Lecture 7

RE=5.6 1/2 watt

7

2N2905

+ vout

RL

4

-12

_

RB2

-15

1N4001

-12 V -12 V

Why is [b] better?

6.101 Spring 2014

1N4001

RE=5.6 1/2 watt

0.1F

-

2

vin

RB1

+15

10k

RF

49

6.101 Spring 2014

Lecture 7

50