chapter vi non-linear applications

1

CONTENTS:

1. Precision rectifiers

2. Peak detectors

3. Logarithmic amplifier

4. The emitter-coupled pair as a simple multiplier

5. Gilbert multiplier circuit

6. Translinear principle

7. RMS-to-DC converters

8. Limiter circuits

Chapter VI –

NON-LINEAR APPLICATIONS

EEL 7300 Chapter VI

2

The Shockley diode equation

6.1 Precision rectifiers

+ vD -

iD

−

= 1

vexpIi D

SD

tn

0.10.0-0.1-0.2-2.0e-15

0.0e+0

2.0e-15

4.0e-15

6.0e-15

8.0e-15

1.0e-14

1.2e-14

Diode Voltage (V)

Dio

de

Cu

rren

t (A

)

IS

Diode behavior near the origin

with IS=10-15 A , n=1 and t = 25 mV

EEL 7300 Chapter VI

/t kT q

n

= - thermal voltage

- ideality factor

3

The ideal diode


A

C

A

C

"On"

C

A

"Off"

"Short" Circuit

"Open" Circuit

F

IdealReal

iD

+

vD

_

iD=0 for vD<0

vD=0 for iD>0

EEL 7300 Chapter VI

4

Precision Half-Wave Rectifier


vy

vI<0

Opamp saturates at negative

limit (close to –VCC) →

Diode OFF → vO=0

( ); exp 1

11 ln 1

y O Oy I O D S

t

t OI O

S

v v vv A v v i i I

n R

n vv v

A A RI

− = − = → − =

= + + +

vI>0

Diode voltage dropR. C. Jaeger and T. Blalock

DC Analysis

EEL 7300 Chapter VI

5

Precision Half-Wave Rectifier• For vI > 0, vO = vI, i > 0, diode is forward-

biased and feedback loop is closed. Rectification is perfect even for small input voltages.

• For vI < 0, diode is cutoff, i = 0, vO = 0.

• Primary sources of error are gain error and offset error due to nonideal op amp.

• For negative input voltages, output voltage vy is saturated at negative limit. Large negative voltages across input can destroy unprotected op amps.

• Response time of circuit is slowed down due to slow recovery of internal circuits from saturation.vO is a rectified replica of vI

without loss of “ON” voltage

drop as in diode rectifier circuit.


vy

R. C. Jaeger and T. Blalock

EEL 7300 Chapter VI

6

Problem

vy

11 ln 1t O

I O

S

n vv v

A A RI

= + + +

vI>0


vI<0 0Ov =

Given that , calculate the difference for the

following conditions:

1. and A=10, 1000, ;

2. and A=10, 1000,

40 mV 1 pAt Sn RI = = I Ov v−

1 VOv =

0.1 VOv =

EEL 7300 Chapter VI

7


For vI > 0, vy is negative (one diode-drop

below zero), D2 is forward biased, current

in R2 is zero, vO = 0, and D1 is reverse

biased. Feedback loop is closed through

D2.

For vI < 0, vy is one diode-drop above

output voltage, diode D1 turns on, and D2

is off. Circuit behaves as an inverting

amplifier with gain - R2 / R1. Feedback

loop is closed through D1 and R2.

Using this scheme, the opamp output

does not saturate

vy


Improved Precision Half-Wave Rectifier

EEL 7300 Chapter VI

8


For vI > 0, vy is negative (one diode-drop

below zero), D2 is forward biased, current

in R2 is zero, vO = 0, and D1 is reverse

biased. Feedback loop is closed through

D2.


vy

EEL 7300 Chapter VI

9


For vI < 0, vy is one diode-drop above

output voltage, diode D1 turns on, and D2

is off. Circuit behaves as an inverting

amplifier with gain - R2 / R1. Feedback

loop is closed through D1 and R2. vy



EEL 7300 Chapter VI

10

6.1 Precision rectifiersApplication to an AC Voltmeter Half-wave rectifier followed by a LP filter

to form the basic ac voltmeter, which needs

a DC voltmeter at the output.

For a sinusoidal input of amplitude VM

and frequency wo, the output is a

rectified sine wave given by its Fourier

series. If the cutoff frequency of the

low-pass filter is wc << wo, the output

consists primarily of the dc voltage

component.

Voltmeter range can be adjusted

through choice of the four resistor

values.

4 2

3 1

MO

R R Vv

R R =


EEL 7300 Chapter VI

Response of the half-wave

rectifier to a sinusoidal input

Average value of v1

11


Full-wave rectification

A. S. Sedra and K. C. Smith

Ideal DA & DB

Exercise: Show the scheme of a full-wave rectifier

using two improved half-wave rectifiers

R1=R2

EEL 7300 Chapter VI

12

6.2 Peak detectors

wikipedia

Peak detectors with

reset switch

Superdiode

Demodulated signalAM (amplitude modulated) signal

EEL 7300 Chapter VI

0 (ideal diode)

outC D

D

dVi i C

dt

i

= =

Ci

http://en.wikipedia.org/wiki/File:PeakDet.svg

http://en.wikipedia.org/wiki/File:PeakDet.svg

13

6.3. The logarithmic & antilogarithmic amplifiers

IR

R

VO

_

+

VI

IC

0; exp 1

ln 1 ln

for

OIR C S

t

I IO t t

S S

IS

VVI I I

R

V VV

RI RI

VI

R

−= = − →

= − + −

What if VI <0 ?

VI >0

VI

VO

VI

IR

R

VO

_

+

IC

0; exp 1

exp

O IR C S

t

IO S

t

V VI I I

R

VV RI

−= = − →

−

VI <0

VI

VO

EEL 7300 Chapter VI

What if VI >0 ?

14

6. 4. The emitter-coupled pair as a simple multiplier

11

22

1 2 1 2

exp

exp

;

BEC S

t

BEC S

t

id BE BE C C EE

VI I

VI I

V V V I I I

= − + =

Simplified analysis:

→; IC is independent of VC

/2 /2

1 2

/2 /2

1 2

1 2

+

tanh2

for 12 2

−

−

− −=

−=

= −

id t id t

id t id t

V V

C C

V V

EE

C C id

EE t

id idC C C EE

t t

I I e e

I e e

I I V

I

V VI I I I

P. R. Gray, P. J. Hurst, S. H. Lewis, and R. G. Meyer

Slope =

t-t

slope for 12 2

idEE

t t

VI

=

EEL 7300 Chapter VI

15

1 2

2

2

for 12 2

( ) /

( ) ( ) /

2

id idC C C EE

t t

EE i

id iC

t

o C C

V VI I I I

I V t R

V t V t RI

V R I

= −

=

=

6. 4. The emitter-coupled pair as a simple multiplier


2 always >0

can be positive or negative

i

id

V

V

Two-quadrant multiplier

EEL 7300 Chapter VI

16

6. 5 Gilbert multiplier circuit

1 23 5 4 6

1 2 1 2

tanh tanh2 2

for , 12 2 2 2

out C C EE

t t

out EE

t t t t

V VI I I I

V V V VI I

− −

= − =


Four-quadrant multiplier

EEL 7300 Chapter VI

The output current can be

converted into an output voltage

by means of two equal resistors

connected to VCC

17

6. 5 Gilbert multiplier circuit

R R

− VO(t) +

V1(t)

V2(t)

Application as a balanced modulator

P. R. Gray, P. J. Hurst, S. H. Lewis, and R. G. MeyerEEL 7300 Chapter VI

18

6. 6 The translinear principle

*B. Gilbert in Chapter 2 of Analogue IC design: the current mode approach (editors:Toumazou, Lidgey, Haigh)

Translinear: transconductance of a BJT is linearly proportional to its collector current*

Translinear principle applied to a 4-diode bridge

ln 1 ln for D DD t t D S

S S

I IV I I

I I

= +

1 3 2 4

31 2 4

1 3 2 4

31 2 4

1 3 2 4

ln + ln = ln + ln

=

D D D D

DD D Dt t t t

S S S S

DD D D

S S S S

V V V V

II I I

I I I I

II I I

I I I I

+ = +

→

1 3 2 4 for equal 'sD D D D SI I II I=

D4

Ia

Ic

Ib

D2

D3

D1

ID1

ID2

ID3

ID4

Translinear

loop

EEL 7300 Chapter VI

19

6.6 The translinear principle

Square root circuit using the translinear principle

3 4

1 2

S So i B

S S

I II I I

I I=

Neglecting the base currents show that

o i BI I I= for identical transistors

EEL 7300 Chapter VI

1 2 3 4BE BE BE BEV V V V+ = +

20

6.6 The translinear principle

Square law circuit using the translinear principle

Neglecting the base currents show that

2 3 4 61

2

2 1 2 5

S S SBo i

B S S S

I I III I

I I I I=

EEL 7300 Chapter VI

21

6.7 RMS-to-DC converter

EEL 7300 Chapter VI

22


1

24 k

=

=

INVI

R

R

2

3 4 1− =I I I1

/2

9 4

/2

22/2 /2

2

3 3/2 /2

1

−

− −

− =

=

IN

T

T

T T

T T

V R I dtT

VIR Rdt dt

T I T I R9

2/2

3 9 9 2

3 /2

/2

2 2

9 ,

/2

/

1

−

−

−= → = →

= =

IN

IN

T

T

T

IN rms

T

VRI V R V dt

TI R

V V dt V VT

Pin 6

to pin 9

EEL 7300 Chapter VI

23


EEL 7300 Chapter VI

6

IN6 IN

6

IN24 24

I

1

N

I

-V /

VI

2I = if V 0

6

I =0 if V 0

I24

I =

+

=

I6

I24

INV

24I

1I

6I

1

24 k

INVI

R

R

=

=

24


2

3 4 1− =I I I

/2

9 4

/2

2/2 /22

1

2

3 3/2 /2

1

IN

T

T

T T

T T

V R I dtT

VIR Rdt dt

T I T I R

−

− −

− =

=

9

2/2

3 9 9 2

3 /2

/2

2 2

9 ,

/2

/

1

−

−

−= → = →

= =

IN

IN

T

T

T

IN rms

T

VRI V R V dt

TI R

V V dt V VT

Pin 9

to pin 6

EEL 7300 Chapter VI

Translinear

loop

( )( )

9

24 1

= =+

V RH j

I RC

ww

R

RCw

-20 dB/decade

1

The LP filter filters out the

fundamental and harmonic

frequencies of I4. The output V9 is

the average value of I4.

1

24 k

INVI

R

R

=

=

LP filter

25

6. 8 Limiter circuits

−ii

EEL 7300 Chapter VI

Finite Rf Infinite Rf

Plot the VTC characteristic of the

limiter above for VZ=5 V, VON=

0.5 V, R1=10 k, R2=100 k

and VCC= 15 V

26

Problem: Find the currents through the diodes for Ia= 1mA,

Ib=2 mA, Ic=3mA . The diodes are identical. Now suppose that

the diodes are replaced with equal value resistors. What are the

currents through the corresponding resistors R1, R2, R3, R4?

D4

Ia

Ic

Ib

D2

D3

D1

ID1

ID2

ID3

ID4

1 2

3 4

1.25 mA; 0.75 mA

2.25 mA; 3.75 mA

= =

= =

D D

D D

I I

I I

Answer:

1 2

3 4

1.5 mA; 0.5 mA

2.5 mA; 3.5 mA

R R

R R

I I

I I

= =

= =

EEL 7300 Chapter VI

27

Problem: What is the purpose of the circuit below? The output is the

differential current I1-I2. Calculate I1,I2 and I1-I2

EEL 7300 Chapter VI

28

A buffered precision

peak rectifier

Problem: Analyze the peak detector shown below

A. S. Sedra and K. C. Smith

2 1 ON, OFF

does not change

I O C

O

v v v D D

v

=

2 1 turns OFF, ON

is charged up to follows

I O C

I O I

v v v D D

C v v v

=

Cv+

−

EEL 7300 Chapter VI

29

Application of log& antilog amplifiers to one-quadrant multiplier/divider

http://pdf1.alldatasheet.com/datasheet-pdf/view/48038/AD/AD538.html

Problem: For the block diagram on the right,

show that the output is given by the equation

on top of the scheme. Assume the differential

amplifier to have gain equal to unity.

EEL 7300 Chapter VI

A. B. Grebene, Bipolar and MOS Analog Integrated Circuit Design, Wiley,

2003.

R. C. Jaeger and T. Blalock, Microelectronic Circuit Design, McGraw-Hill,

New York, any edition.

A. S. Sedra and K. C. Smith, Microelectronic Circuits, any edition.

P. R. Gray, P. J. Hurst, S. H. Lewis, and R. G. Meyer, Analysis and Design

of Analog Integrated Circuits, 4th edition, 2001.

EEL 7300 Chapter VI 30

chapter vi non-linear applications

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