EE 359 Electronic Circuits October 15, 2002 0 Differential and Multistage Amplifiers Examples: BJT...

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Transcript of EE 359 Electronic Circuits October 15, 2002 0 Differential and Multistage Amplifiers Examples: BJT...

October 15, 20021

EE 359 Electronic Circuits

Differential and Multistage Differential and Multistage AmplifiersAmplifiers

Examples:

BJT Differential Pair

Small Signal Model

Biasing

Current Mirror

October 15, 20022

EE 359 Electronic Circuits

• HW due Friday (10/18)

• 6.39,6.61,6.71,6.80

October 15, 20023

EE 359 Electronic Circuits

Example: For BJT differential pair configureation, find vd such that:

a) iC1=99% I

b) iC1=95% I

c) iC1=9 iC2

October 15, 20024

EE 359 Electronic Circuits

Use equivalenceFor 100, iC=iE,

a)

b)

c)

TdTbb

Tbb

TbTb

VVVVVE

VVVEEE

VVSE

VVSE

e

I

e

Ii

e

IiIii

eIieIi

//)(2

/)(121

/2

/1

11

1

,

21

12

21

mVVVv

e

eIii

TTd

Vv

VvEC

Td

Td

115595.4)0101.0ln(

0101.0199.0/1

99.01

199.0

/

/22

mVVVv

eIii

TTd

VvEC

Td

7494.2)0526.0ln(

0526.0195.0/195.0 /11

mVVVv

eIiiii

TTd

VvEECC

Td

5520.2)1.0ln(

1.019.0/110/999 /2121

October 15, 20025

EE 359 Electronic Circuits

Biasing in BJT Integrated Ckts

•For what value of would current mirror have a gain error 1%, 0.1 %

• Imperfection due to base current diverted from reference current IREF

October 15, 20026

EE 359 Electronic Circuits

Effect of on BJT Curr. Mirror

/21

1

2

1

2

1

REF

O

EREF

EO

I

I

Ii

Ii

/21 REF

O

II

•Analysis of the current mirror taking into account the finite of the BJTs

998999.02

9899.02

October 15, 20027

EE 359 Electronic Circuits

Current Steering Circuits

•Generation of a number of cross currents

Q5 and Q6 in parallel, combination forms mirror with Q1. I3=2IREF

Q4 forms mirror with Q2.

R

VVVVI BEEBEECE

REF21

October 15, 20028

EE 359 Electronic Circuits

A current mirror with base-current compensation.

Improve dependence of I0 on why?•Error in mirror gain ckt as a result•How? Q3’s emitter supplies base currents of Q1 and Q2

EO

EREF

II

II

1

)1(

2

1 2

October 15, 20029

EE 359 Electronic Circuits

The Wilson current mirror.

332

32

2

2

2

2

3

2

2

3

3

2

112222111

2222

2232

2112113

333

3

2

)1(21

1

112

11

1

2

1

1

1)1()1(

)1(

)1(

11

REF

O

BBBEBE

BEC

BBREFC

BBBBEE

BEO

B

I

I

iiiiii

iii

iiii

iiiiii

iii

i

By adjusting we can set output current to be a closer to IREF

October 15, 200210

EE 359 Electronic Circuits

Wilson current mirror features

a) Q1 is diode connected, 1 does not matter

b) Q2 with low 2, increases IO

c) For high 3, , IO increases

d) If one uses, 1= and 2=(1-k) and

3=(1+k) where k=1/2, IO equals IREF

October 15, 200211

EE 359 Electronic Circuits

The Widlar current source.

Emitter resistance added to current mirror

O

REFTEO

O

REFTBEBE

S

OTBE

S

REFTBE

EOBEBE

I

IVRI

I

IVVV

I

IVV

I

IVV

RIVV

ln

ln

ln,ln

21

21

21

October 15, 200212

EE 359 Electronic Circuits

Example 6.2

• For constant current IO=10uA, determine values of resistors in collector (and emitter for Widlar source). Let VBE=0.7V at 1mA.

• current source ckt:

• Widlar ckt:

kA

R

VmA

AVV TBE

94210

58.010

58.01

10ln7.0

1

1

kR

A

mARx

kR mA

5.11

10

1ln025.01010

3.9

3

36

17.010

2

October 15, 200213

EE 359 Electronic Circuits

Example 6.3/ Opamp Ckt.

DC analysis shown on fig.

(start w/ Q9)

•Rid

•Av

•RO

October 15, 200214

EE 359 Electronic Circuits

Method

• For input resistance look at 1st stage.

• 1st stage gain, consider input resistance of second stage.

kR

kxrr

rr

rrR

id

ee

id

25

1.10100101

10025.0

25

21

21

21

kxrrrr

VVrr

RRR

v

vA

ee

ee

i

id

53.225101,25

/4.22)(||

5454

21

212011

October 15, 200215

EE 359 Electronic Circuits

A differential amplifier with an active load.

Load is Q3 and Q4

2000

2/)2/(

2/

|| 0402

T

Aom

omvodmo

oo

o

V

Vrg

rgArvgv

rR

rrR

October 15, 200216

EE 359 Electronic Circuits

Small-signal model of the differential amplifier

October 15, 200217

EE 359 Electronic Circuits

The differential form of the cascode amplifier

Cascode amp. differential half circuit.

October 15, 200218

EE 359 Electronic Circuits

•A cascode differential amplifier with a Wilson current-mirror active load

October 15, 200219

EE 359 Electronic Circuits

MOSFET differential pair.

October 15, 200220

EE 359 Electronic Circuits

Normalized plots of the currents in a MOSFET differential pair. Note that VGS is the gate-to-source voltage when the drain current is

equal to the dc bias current (I/2).

MOSFET differential pair

October 15, 200221

EE 359 Electronic Circuits

(a) basic, (b) cascode, (c) Wilson, (d) modified Wilson.

MOS current mirrors

October 15, 200222

EE 359 Electronic Circuits

(a) bipolar; (b) MOS; (c) BiCMOS obtained by cascoding Q1 with a

BJT, Q2; (d) BiCMOS

double cascode.

Basic active-loaded amplifier stages

October 15, 200223

EE 359 Electronic Circuits

Voltage gain of the active-loaded common-source amplifier versus the bias current ID. Outside the subthreshold region, nCox = 20 A/V2,

= 0.05 V-1, L = 2 m and W = 20 m.