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Lecture 12 - 1ELE2110A 2008

Week 12: Output Stages, Frequency Response

ELE 2110A Electronic Circuits

(2 hours only)


Topics to cover z Output Stages

z Amplifier Frequency Response

Reading Assignment: Chap 15.3, 16.1 of Jaeger and Blalock orChap 14.1 14.4 of Sedra & Smith


Multistage Amplifiers

z The first (input) stage is usually required to provide a high input resistance a high common-mode rejection for a differential amplifier

z Middle stages are to provide majority of voltage gain conversion of the signal from differential mode to single-end

mode shifting of the dc level of the signal

z The last (output) stage is to provide a low output resistance in order to

z avoid loss of gain and z provide the current required by the load (power amplifiers)

Practical amplifiers usually consist of a number of stages connected in cascade.


Example

z The input stage (Q1, Q2) is differential-in and differential-out biased by current source Q3

z (Q4, Q5) is a differential-in and single-ended-out stage biased by current source Q6

z Q7 provides additional gain shifting the dc level of the

signalz The output stage Q8 is an

emitter follower


Output Stagesz Function of an output stage is

To provide a low output resistance so that it can deliver the output signal to the load without loss of gain

z Requirements of an output stage: Large input signal range

z b/c it is the final stage of the amplifier, and usually deals with relatively large signals.

z Small-signal approximations and models either are not applicable or must be used with care.

Low distortion High power efficiency


Classification of Output Stagesz Class A: the transistor

conducts for the entire cycle of the input signal

z Class B: the transistor conducts for only half the cycle

z Class AB: conduction cycle is greater than 180o and less than 360o Used for opamp output stage

and audio power amplifiersz Class C: conduction cycle is

less than 180o

Used for radio-frequency (RF) power amplifications (mobile phones, radio and TV)

Class AClass B

Class ABClass C

Collector or Drain current waveformsof different output stages


Class-A Amplifier: Source/Emitter FollowerFor a source follower biased by an ideal current source, vGS is fixed and

+==

n

SSTNIGSIO K

IVvVvv 21

TNDDIGSSS VVvVV ++Input range:

Output range:DDOSS VvV

tDDVov sinThe largest output voltage is

(if VSS=VDD)


Source Emitter with Load

0 +=L

Rov

SSISi

To maintain class A operation, is > 0 at all times:

LSSo RIv tDDVov sinFor largest output amplitude:

We have:

L

DDSS R

VI

tRItV LSSDD allfor sin

LSSDD RIV The lowest value for the LHS occurs when sin t = -1,


Power Efficiency

tDDVov sin

%25/2

)2/(2

2

)2/(2

av

ac

=

==

DDV

LR

DDV

LR

DDV

DDV

SSI

LR

DDV

P

P

- Low efficiency

The largest output voltage is

Efficiency of amplifier is:

( )( ) DDSSSSDDSS

T

DDL

DDSSDDSSav

VIVVI

dtVR

tVVVIT

P

2

sin10

=+=

++=

Average power supplied to the source follower:

Average power delivered to the load:

L

DD

L

DD

ac RV

R

V

P2

2 2

2

=

=

L

DDSS R

VI

(if VSS=VDD)

Lecture 12 - 10ELE2110A 2008

Push-Pull Operation: Class B

Source: B. Putzeys, Digital Audios final frontier, IEEE Spectrum, Mar 2003.

When a push-pull amplifier is operated in Class B, all of the output currentcomes either from the current-sourcing transistor or from the current-sinking device but never from both at the same time.

Lecture 12 - 11ELE2110A 2008

Class-B Amplifierz A complementary pair of source followers

biased at zero source currentz When

neither transistor conductsz No quiescent (DC) current consumption!

TNVIvTPV

Lecture 12 - 12ELE2110A 2008

Class-B Amplifierz When VI > VTN,

M1 turns on and acts as an source follower, vo vI - VTN

M2 offz When VI < VTP,

M2 turns on and acts as an source follower, vo vI VTP

M1 offz Power efficiency is high, can be up to

about 80%

z Disadv.: Output waveform suffers from a dead-zone Large distortion

Lecture 12 - 13ELE2110A 2008

Class AB

Source: B. Putzeys, Digital Audios final frontier, IEEE Spectrum, Mar 2003.

Class AB exhibits less distortion by allowing the transistors to work together when the output signal is near zero, in what is called the crossover region.

Lecture 12 - 14ELE2110A 2008

Class-AB Amplifiers

z Remove dead zone by biasing transistors into conduction but at a low quiescent current level Distortion less than Class-B but worse than Class-A amplifier

z For each transistor, 1800

Lecture 12 - 15ELE2110A 2008

Class-AB AmplifiersBiasing examples:

2

22

= TNVGG

VnKDI

=

TVBRBI

SICI 2exp

DC currents:

Lecture 12 - 16ELE2110A 2008

Topics to cover z Output Stages

z Amplifier Frequency Response

Lecture 12 - 17ELE2110A 2008

Frequency Response of Amplifiers

A typical amplifier:

Block DC and low frequency signals

By-pass high frequency currents

Amplifiers gain is frequency dependent!

Lecture 12 - 18ELE2110A 2008

Typical Amplifier Transfer Function

z In low frequency side, drop in gain is caused by coupling and bypass capacitors

z In high frequency side, drop in gain is caused by transistors parasitic capacitors More on this topic later

z In the mid-band range, no capacitors are in effect: Coupling and bypass capacitors are short circuits Transistor parasitic capacitors are open circuits

Mid-band gain

Lower cut-off frequency Upper cut-off frequency

Lecture 12 - 19ELE2110A 2008

Estimate fL: Short-Circuit Time Constant Method

z Lower cutoff frequency for a network with n coupling and bypass capacitors can be estimated by:

RiS = resistance at terminals of ithcapacitor Ci with all other capacitors replaced by short circuits.

Product RiS Ci is short-circuit time constant associated with Ci.

2/1

1

LLf

n

i iCiSRL

==

Lecture 12 - 20ELE2110A 2008

SCTC: Example

= 100 and VA=75V Q-point: (1.73mA, 2.32V)

AC equivalent with finitecoupling capacitances

=== kmA

mVIVr

B

T 45.1100/73.1

25 =+=+= kmA

VVI

VVrC

CEAo 7.4473.1

32.275

BJT small signal parameters:

Lecture 12 - 21ELE2110A 2008

Time Constant Associated with C1To find the time constant associated with C1:

)()(1 rBRIRRCEinBRIRSR +=+=

(C2 and C3 are short-circuited and set vi = 0)

=+= 2220)1450||7500(10001sRrad/s 225

222.211

11

== FkCR s

Lecture 12 - 22ELE2110A 2008

Time Constant Associated with C2

CRRorCRRRCEoutCRRSR ++=+= 3)(3)(32

=+= kkkkR s 104)7.44||3.4(1002rad/s 1.96

1.010411

22

== FkCR s

To find the time constant associated with C2:


Lecture 12 - 23ELE2110A 2008

Time Constant Associated with C3

1

13

++=

++==

)BRI(RrER

thRr

ERRCCoutERSR

=

+=7.22

1017500||10001450||13003

VkR s

rad/s 4410107.22

11

23

== FkCR s

To find the time constant associated with C3:


Lecture 12 - 24ELE2110A 2008

Lower Cutoff Frequency

rad/s.i iCiSR

L 473044101962253

1

1 =++==

The lower cutoff frequency is:

Hz 7532

== L

Lf

and

In this example the time constant associated with the bypass capacitor C3 is dominant.

Lecture 12 - 25ELE2110A 2008

High Frequency Response

z At high frequency side, drop in gain is caused by transistors parasitic capacitors

Mid-band gain

Lower cut-off frequency Upper cut-off frequency

Lecture 12 - 26ELE2110A 2008

High Frequency Small Signal Model for BJT

B

C: diffusion capacitance of the forward-biased base-emitter junction.C: depletion capacitance of the reverse-biased base-collector junction.

rx: the resistance of the silicon material of the base region between the base terminal and the intrinsic base terminal B that is right under the emitter region.

Model for active mode BJT

Lecture 12 - 27ELE2110A 2008

High-frequency Small Signal Model for MOSFET

ovL

= the capacitance between the Gate and the conducting channel.oxgs WLCC 32=

oxovovgd CWLCC == = the overlap capacitance (very small).

Model for saturation mode MOSFET

Lecture 12 - 28ELE2110A 2008

Open-Circuit Time Constant Method to Determine fH

fH can be estimated by open-circuit time constant method:

2/ ,1

1HHfm

i iCioR

H ==

where Rio is resistance at terminals of ith capacitor Ci with all other capacitors open-circuited.

Lecture 12 - 29ELE2110A 2008

High Frequency Analysis of C-E Amplifier

= 100 and VA=75V Let Q-point be (1.6mA, 3V)

==

2505.0

xrpFC

k10k302||1 == RRBR

=+= 882BRIRBRIR

thR

k3.4k1003 == CRRLR

BRIRBR

ivthv +=

pFC 9.19=

Lecture 12 - 30ELE2110A 2008

High Frequency Small Signal Equivalent

Norton source transformation

xrthR += thvsi

)( xrthRror +=

Lecture 12 - 31ELE2110A 2008

Determine Amid

1531560250882)4120(100 =++=++=

rrR

RAxth

Lomid

xrthR += thvsi )( xrthRror +=

Lsm Rrigv )( 02 =

rrR

rRgrR

rRgvv

xthLm

xthLm

th ++=+=

02

Lecture 12 - 32ELE2110A 2008

OCTC: Time Constant Associated with C

=+=+== 656)250882(||56.1)(|| kxrthRroroR

80 103.1

= RC

To find the time constant associated with C:

(C is open-circuited and set is = 0)

pFC 9.19=

Lecture 12 - 33ELE2110A 2008

OCTC: Time Constant Associated with C

LmxxLLxx RvgiriRiriv )(00 ++=+=

=++=

koR 1786564120)4120(064.01656

90 100.89

=RC

++==

orLR

LRmgoroR 1

xixv

To find the time constant associated with C:

(C is open-circuited and set is = 0)

0riv x=

Lecture 12 - 34ELE2110A 2008

Upper Cutoff Frequency

rad/s 6108.9910898103.111 =+

=+

CoRCoRH

Upper cutoff frequency:

MHz 56.12

== H

Hf

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