bipolar sub circuits - users.jyu.fiusers.jyu.fi/~loberg/fyse400slides/lecture3fyse400.pdf ·...

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FYSE400 ANALOG ELECTRONICS

LECTURE 3

Bipolar Sub Circuits

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BIPOLAR SUB CIRCUITS Bipolar Current Sinks and -Sources

Transistor operates in forward-active region.

maxCECN

satCE VVV << max

CEBNCNBN VVVV +<<

Current sink Current source

N

N N

Current sink with higher output resistance

©Loberg University of Jyväskylä

3

NPN-Current Sink

+==

A

CEV/VESFCOUT V

V1eIII TBEα

N

VCC

IOUT

VCN

VBE

+

-

Collector characteristics IOUT

VCN

TBE V/VESF eIα



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PNP-Current Source

Collector characteristics IOUT

VCN

TBE V/VESF eIα

N

VCC

IOUT VCN

VBE +

-

+

-

VCC


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NPN-Current Sink with emitter resistance

N

VCC

IOUT

VCN

VBE

+

-

VBN

RE

E

BEBNFCOUT R

VVII −== α


6

BIPOLAR SUB CIRCUITS Bipolar Diode Connected Voltage Source


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BIPOLAR SUB CIRCUITS

VCC

IREF

OUTi

BiOUTv+

-

1iCi

)1(iandIiwhen FOUTREFB >><< β

+=

ESF

OUTREFTOUT I

iIlnVvα

OUTi

=

ESF

REFTOUT I

IlnVvα

REFI−

OUTv

V7.0vV6.0 OUT <<

Bipolar Diode Connected Voltage Source


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VCC

IREF

2OUTv

1OUTv

Bipolar Diode Connected Voltage Source


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RVV

2I BECC

F

F1C

−+

=ββ If βF>>1 then

RBECC

1C IR

VVI =−

≈

Current Mirror

Identical tansistors

BECECE VVV =≈ 21

Assumptions


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BIPOLAR SUB CIRCUITS Current Mirror

VBE depends on the ambient temperature :

( ) 1VV1

VV

II

BECCBE

BE

R

R

−∝∆∆Sensitivity of reference current IR

is low if we have VCC >>VBE

-2.2mV/°C

Resistor R can be replaced by constant current source IREF

Reference current IR depends on the VCC , R and VBE .

Constant current sink


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If VCE1 > VCE2

+

+=

A

1CER

F

F1C V

V1I2

Iββ

V7.0V 1CE >>


12


Transistors have different emitter areas A1 and A2

2

1

2ES

1ES

AA

II

=

( )( )A2CE

V/V2ESF

A1CEV/V

1ESF

2C

1C

VV1eIVV1eI

II

TBE

TBE

++

=αα

If we assume that 1andVV FCEA >>>> β REF2

1OUT I

AAI ≈


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Identical transistors

F

INOUT 1N1

II

β+

+=

All output transistors has same base current.

Difference between reference and output current is proportional to the number of outputs.

)N1(III BOUTIN +=−

(Multiple output current sink)

Multiple Output Current Mirror


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Emitter follower is used to supply base current.

( )1N11

II INOn

++

+=

ββ

Reduced difference between reference and output current.

β++

=−1

N1III BOUTIN


Modified Multiple Output Current Mirror BIPOLAR SUB CIRCUITS


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Multiple Collector BJT

Different values of sink currents IOn.

Output current depends on the effective value of collector area.

IN

ON

IN

2O

IN

2O

IN

1O

II

II

II

II

10II

minO

maxO ≈


Multiple Output Current Mirror


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( ) TBET1BE2BE V/VV/VV

1C

2C eeII ∆== −

≈

+

=1C

2C

1C

T

1C

2C

1C

T

F

FE I

IlnIV

IIln

IV

1R

ββ

2CE1CE VV ≈Assumptions Early voltage VA is high.

RVV

1I 2BECC

F

F2C

−

+

=β

β

Low output currents with practical resistance values.

High output resistance Ro

( )Emoo Rg1rR +≈

mV60V10II

BE1C

2C =⇒= ∆

mV120V100II

BE1C

2C =⇒= ∆

Widlar Current Source BIPOLAR SUB CIRCUITS


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1E

2E

2C

1C

RR

II

∝

>

Widlar Current Source RE1>RE2

Current Mirror RE1=RE2

Widlar Current Source


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Wilson Current Source

( )( )

( )( ) R

V2V22

2I22

2I BECC

FF

FFIN

FF

FF1C

−++

+=

+++

=ββββ

ββββ

( )( )

( )( ) R

V2V21

1I21

1I BECC

FF

FFIN

FF

FF1C

−++

+=

+++

=ββββ

ββββ

See Modified Multiple Output Current Sink N=1

High output resistance Ro

Wilson Current Source


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BJT or FET

(Emitter-coupled pair)

−+ = VVPower Supply typically

Identical pair

Generic Differential Stage

1Iv 2Iv

IBIAS is Constant Current Source

Sub Circuit of Operational Amplifiers Emitter Coupled Logic (ECL)

Assumption: Input currents iI1=iI2=0

BIPOLAR SUB CIRCUITS Differential Stage

1IV

1iv

2IV

2iv


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1i1I1I vVv += 2i2I2I vVv +=Input signals:

1L

1ON1 R

VVI −=

+

2L

2ON2 R

VVI −=

+


2Iv1Iv1L11ON RIVV −= +

2L22ON RIVV −= +

Quisent values

0III BIAS21 =−+

2I1I2I1I VVvv ===(Ac-component is zero)

Summ of currents I1 and I2 is constant. ( ) 0IIRVVV 21L2ON1ONOD =−−=−=

1IV

1iv

2IV

2iv


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Differential Stage BIPOLAR SUB CIRCUITS

If VON1 is output of differential stage then:

VIN1 is inverting input VIN2 is noninverting input


2Iv1Iv

180° phase difference

Differential mode

2IN1IN VV =2IN1IN VV ∆∆ −=

Common mode

2IN1IN VV =2IN1IN VV ∆∆ =

1IV

1iv

2IV

2iv


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Arbitrary input voltage consists of common mode and differential mode voltages.

DMCMIN VVV +=1 DMCMIN VVV −=2

2IN1IN VV +

2IN1IN VV −( )2IN1IN VV

21

−

( ) CM2IN1IN VVV21

=+

1INV2INV

( ) CM2IN1IN VVV21

=+

( ) DM2IN1IN VVV21

=−

Where



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Common Mode Gain ACM

Differential Mode Gain ADM

Linear system and superposition

CMCMDMDM1ON VAVAV +=

CMCMDMDM2ON VAVAV +−=

Arbitrary output voltages:

Common Mode Rejection Ratio CMRR

CM

DM

AACMRR = In Ideal Case the ACM is zero

Differential Amplifier



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+=

CMRRVVAV CM

DMDM1ON

−−=

CMRRVVAV CM

DMDM2ON

Arbitrary output voltages



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T1BE V/VESF1C eII α=

T2BE V/VESF2C eII α=

Collector currents

( ) TDT21 V/VV/VV

2C

1C eeII

== −

IE1 IE2

Node E : 0III EE2E1E =−+F

2C

F

1CEE

IIIαα

+=

21D VVV −=

BJT Differential Stage BIPOLAR SUB CIRCUITS

(a)

(b)


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TD V/VEEF

1C e1II −+

=α

TD V/VEEF

2C e1II

+=

α

TD V/VCEEF

CC1O e1RIVV −+

−=α

TD V/VCEEF

CC2O e1RIVV

+−=α

+−

+−=−= − TDTD V/VV/VCEEF2O1OOD e1

1e11RIVVV α

DV

2OV1OV

TV2− TV2

"Linear" input voltage region TD V2V <

V

CCV

CEEFCC RIV α−

TV4


(a) and (b)


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9.0V

-9.0V

9.0V

+9.0V

BSX20 BSX20

BSX20 BSX20

Q1 Q2

Q3 Q4

3.8kΩ 3.8kΩ

3.3kΩ

22Ω

1.2kΩ 1.2kΩ

VCQ1 = 6.26V VCQ2 = 6.64V

VEQ = -0.69V

vIN1 vIN2

vIN2

vIN1

vC1 vC2

vE

Amplitude : 50mV or 300mV

Practical example

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vC1

vC2


Collector voltages in fully differential mode

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vC1

vC2

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Input amplitude : 50mV

vC1

vE Input amplitude : 300mV vE

vC1


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The End of Part 3


bipolar sub circuits - users.jyu.fiusers.jyu.fi/~loberg/fyse400slides/lecture3fyse400.pdf ·...

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