Analog Electronics CircuitsNagamani A N

Lecturer, PESIT, Bangalore – 85

Email – nagamani@pes.edu

General Amplifiers

Cascade connection - FET & BJT

Numerical

Cascode connection

Darlington connection

Packaged Darlington connection

Dc bias of Darlington connection

AC equivalent

ac output impedance of Darlington connection

AC voltage gain

Feedback concept

Feedback connection type

Practical feedback circuits

Practical feedback circuits

Numerical

Cascade connection – FET

• Cascade connection is a series connection with the output of one stage thenapplied as input to the second stage.

• Cascade connection provides a multiplication of the gain of each stage for alarger overall gain.

• Gain of overall cascade amplifier is the product of stage gains AV1 and AV2

Av = Av1AV2 = (-gmRD1) (-gmRD2)

• The input impedance of the cascade amplifier is that of stage 1,

Zi = RG1

• Output impedance is that of stage 2,

Z0=RD2

• The main function of cascading the stages is the larger overall gain achieved.

Numerical

Calculate dc bias, voltage gain, input impedance, output impedance,Also

calculate the load voltage if a 10K Ω load is connected across the output

Data for numerical

• C1=C2=C3=0.05uF

• RG1=RG2=3.3 MΩ

• RS1=RS2=680 Ω

• RD1=RD2=2.4 KΩ

• IDSS=10mA; VP=-4V for both stages

Solution

Step 1: from the dc bias details we can find out VGSQ= -1.9V, IDQ=2.8mA

Step 2: both transistors have

gmo=2 IDSS/ Vp =2(10mA)/4 =5mS

At dc bias point, gm=gmo(1-VGS/VP)

gm =5m(1-(-1.9)/(-4) = 2.6mS

Step 3: the voltage gain of each stage

AV1=AV2=-gm RD=-2.6m x 2.4K = -6.2

• Step 4: Overall gain of cascaded stage is

Av=Av1Av2=-6.2 x -6.2 = 38.4

(output is in phase with input)

• Step 5: output voltage is Vo=Av Vi =384 mV

• Cascade amplifier input impedance is

Zi=RG= 3.3 MΩ

• Output impedance (with rd=very high)

Zo=RD= 2.4 KΩ

• Load voltage if load resistance is 10 KΩ

VL= [RL/(RL+Zo)] Vo

=[10K/(10K+2.4)] 384mV=310mV

Cascade amplifier – BJT

• RC coupled cascade amplifier is taken here for example

• Advantage of cascading is increase in the overall voltage gain.

• Dc bias is obtained by procedure followed for single stage amplifier.

• Gain of each stage: AV= -(RC װ RL )/re

• Amplifier input impedance is that of

stage 1: Zi= R1 װ R2 װ βre

• Output impedance is that of stage 2 :

Zo=Rc װ ro

Numerical

Calculate voltage gain, output impedance, input impedance for cascaded BJT

amplifier of fig above. Calculate output voltage resulting if 10K ohms load is

connected to load.

Given,

R1=15KΩ; R2=4.7KΩ;Rc=2.2KΩ;RE=1KΩ

• C1=C2=C3=10uF

• β=200 for both transistors

• Input voltage vi= 25uV

Solution:

• Dc analysis yields

• VB=4.7V;VE=4.0V;VC=11V; IE=4.0mA

• At bias point, re=VT/IE=26m/4.0m=6.5 Ω

• Voltage gain of stage 1 is then,

• AV1= -{RC װ (R1 װ R2 װ βre)}/re

= -665.2/6.5=-102.3

• AV2= -Rc/re = -2.2K/6.5 =-338.46

• Overall gain of AV=AV1AV2

=-102.3 x -338.46

=34,624

• Output voltage is : Vo=AV Vi=34624 x 25u

=0.866V

• Amplifier input impedance is

• Zi= R1 װ R2 װ βre =4.7K װ 15 K װ 200x6.5

=953.6 ohms.

• VL= {RL/Zo+RL} Vo

={10K/2.2K+10K}0.866 = 0.71 V

Cascode connection

• A cascode connection has one transistor on top in series with another.

• Figure below shows CE stage feeding a CB stage.

• This arrangement is designed to provide a high input impedance with low

voltage gain to ensure that the input Miller capacitance is at a minimum with

the CB stage providing good high frequency operation.

Cascade connection configuration fig:1

Cascade connection configuration fig:2

Numerical

Calculate the voltage gain for the cascode amplifier of fig above..

Solution:

Dc analysis: VB1=4.9V ; VB2=10.8V;

IC1=Ic2=3.8mA

Dynamic resistance of each transistor is then re=26/3.8=6.8 ohms

• Voltage gain of stage 1 is

Av1= -Rc/re= -re/re = -1

• Voltage gain of stage 2 is

Av2=Rc/re =1.8K/6.8 = 265

• Resulting in an overall cascode amplifier gain of Av=Av1 x Av2 =-1 x 265=-265

• CE stage with a gain of -1 provides the higher input impedance of CE stage.

• With gain of -1, miller capacitance is kept very small.

• A large gain is then provided by the CB stage, resulting in large overall gainof -265.

Darlington connection

Popular connection operates as “super beta” transistor is Darlington connection.

• Main feature of the Darlington connection is that the composite acts as asingle unit with a current gains of individual transistors.

• Darlington connection provides a current gain of βo= β1+ β2

• If β1= β2= β then βo= β2

• This configuration provides a transistor having a very large current gain,typically a few thousands.

Packaged Darlington transistor

• Specification information of 2N999 Darlington transistor package

DC bias of Darlington circuits

BEEB

EEE

BDBDE

EDB

BEccB

VVV

RIV

Voltages

III

RR

VVI

)1(

Numerical

Calculate dc bias voltages and currents for the Darlington connection. GivenRB=3.3MΩ;RE=390 Ω;βd=8000;VCC=18V;VBE=1.6V

VV

VVVV

VmRIV

Voltages

mAuIII

uAMRR

VVI

C

BEEB

EEE

BDBDE

EDB

BEccB

18

6.96.18

8)390(48.20

48.20)56.2(8000)1(

56.2)390(80003.3

6.118

AC equivalent circuit

Equivalent model

Input impedance

• The ac base current through ri is

Ib=Vi-Vo/ri

• Since Vo=(Ib+βDIb)RE

• Substituting Ib in Vo expression,

Ibri=Vi-Vo=Vi-Ib(1+ βD)RE

solving for Vi,

Vi=Ib[ri+(1+ βD)RE]=Ib(ri+ βDRE)

• Ac input impedance looking into the transistor base is then

Vi/Ib= ri+ βDRE

Zi=RB װ (ri+ βDRE)

ac output impedance of Darlington connection

This can be determined for ac circuit shown in fig below

Output impedance

• The output impedance can be determined by applying a voltage Vo andmeasuring the current Io with Vs setting to zero.

• Solution for Io yields..

Zo= RE װ ri װ ri/βD

oi

D

iEo

i

oD

i

o

E

oBD

i

o

E

oo

VrrR

I

r

V

r

V

R

VI

r

V

R

VI

11

iDiEo

oo rrRI

VZ

//1/1

1

ac voltage gain

Gain expression

On simplification

Numerical

For the Darlington pair, given RE=390 ohms and β=8000. Calculate gain ifri=5KΩ

EbD

EDEbEbDbo

RIIbIbriVi

RRIRIIV

)(

)()(

1)(

)()(

)(

DREREri

DRERE

Vi

VoAv

RRRRr

VV

RRrIV

EDEEDEi

io

EDEibi

998.0]3908000390[5

3908000390

xK

xAv

Feedback concepts

• Depending on the relative polarity of fed back signal in to the circuit, thereare two types of feedback

> Negative feedback

> Positive feedback

Negative feedback results in Reduced gain

Positive feedback are used in oscillators.

Feedback amplifier

Negative feedback circuits

• Reduces the gain

• Increases input impedance

• Better stabilized frequency response

• Lower output impedance

• Reduced noise

• More linear operation

Feedback connection types

• Voltage series feedback

• Voltage shunt feedback

• Current series feedback

• Current shunt feedback

Here voltage refers to small part of voltage as input to the feedback network

Current refers to tapping some part of output current through feedbacknetwork.

Series refers to connecting feedback signal in series with the input signalvoltage.

Shunt refers to connecting feedback signal in shunt with the input signalvoltage

Series feedback connections increases the input resistance

Shunt feedback connections decreases the input resistance.

Voltage series feedback Af=Vo/Vs

Voltage shunt feedback Af=Vo/Is

Current series feedback Af=Io/Vs

Current shunt feedback Af=Io/Is

Gain with feedback

• Gain without feedback is A

• Feedback factor β

• Gain with feedback is (1+A β)

ParameterVoltageseries

Voltageshunt

Currentseries

Currentshunt

Gain withfeedback

A Vo/Vi Vo/Ii Io/Vi Io/Ii

Feedback factor β Vf/vo If/Vo Vf/Io If/Io

Gain withfeedback

Af Vo/Vs Vo/Is Io/Vs Io/Is

Voltage series feedback

• With zero feedback then Vf=0 the voltage gain of amplifier stage is

A=Vo/Vs=Vo/Vi

• If feedback of Vf is connected then,

Vi=Vs-Vf

Vo=AVi=A(Vs-Vf)=AVs-AVf=A(Vs-A(βVo)

Then, (1+ βA)Vo=AVs

Overall gain with feedback is

Af=Vo/Vi=A/(1+A β)

This shows that gain of feedback has reduced by factor (1+A β)

Voltage shunt feedback

• Af=Vo/Is=A Ii / (Ii+If)=AIi/(Ii+ βAIi)

• Af=A/(1+ βA)

Input impedance with FB

• Ref to fig(1)

Ii=Vi/Zi=(Vs-Vf) / Zi = (Vs- βVo) / Zi

Ii Zi= Vs- βAVi

Vs=Ii Zi+ β A Vi = Ii Zi+ β A Ii Zi

Zif = Vs/Ii=Zi+(βA)Zi=Zi(1+ βA)

Improved circuit features of feedback

• Reduction in frequency distortion

When Aβ» 1, then Af=A/(1+A β)≈1/ β

• Here feedback is completely resistive and thus frequency distortion arisingbecause of varying gain with frequency is considerably reduced.

• Bandwidth variation

• When Aβ» 1, then Af=A/(1+A β)≈1/ β

• Therefore, here we can see that, practical circuits, open loop gain drops athigh frequencies.

• Therefore Aβ no longer » 1, hence Af=1/ β

No longer holds good.

• Here reduction in gain has provided improvement in the Bandwidth.

Product of gain and Bandwidth remains same it’s a tradeoff between gain and BW

• Gain stability for Aβ»1,

• This shows that magnitude of relative change in dAf/A is reduced by thefactor Aβ compared to that without feedback dA/A

Numerical

If a amplifier with gain of -1000 and feedback of β=-0.1 has a gain change of 20%due to temperature, calculate the change in gain of the feedback amplifier.

Solution:

Practical feedback circuits

• Voltage series feedback

A

dA

A

1

dA

dA

A1Af

f

%2.0........

%20)1000(1.0

11

A

dA

AAf

dAf

• Here part of output voltage (Vo) is obtained using a feedback network ofresistors R1 and R2.

• The feedback voltage Vf is connected in series with the source signal Vs.their difference being the input signal Vi.

• Gain without feedback A=Vo/Vi=-gmRL

Where RL=parallel combination of RD,Ro,(R1+R2)

• The feedback network provides a feedback factor or β=Vf/Vo = -R2/R1+R2

• Using values of A and β in above equation, Af is

Numerical:

Calculate the gain without and with feedback for the FET amplifier shown in fig.circuit values are given to be R1=80KΩ,R2=20KΩ,RD=10KΩ and gm=4000uS

Solution :

RL=5K Ω

A=-20

β=-0.2 and Af=-4

2

21

212

1

,1..

)/(11

R

RRAf

thenAif

gRRRR

Rg

A

AAf

mL

Lm

Series feedback connection

• Here gain of op-amp is reduced by factor β=R2/R1+R2

Numerical

If open loop gain of op-amp is 100,000 and feedback resistors are R1=1.8K Ω andR2=200 Ω then calculate the gain with feedback .

Solution

• β=0.1

• Af=9.9999

• Here Aβ>>1, Af=1/ β=1/0.1=10

Emitter follower circuit

• The output voltage Vo is also the feedback voltage in series with the inputvoltage.

• Operation of the circuit without the feedback Vf=0 then,

• The operation with feedback then provides that,

1

)/(

o

f

s

Efe

s

ieEfe

s

Ebfe

i

o

V

V

V

Rh

V

hVsRh

V

RIh

V

VA

)/)(1(1

/

1 ieEfe

ieEfe

s

o

hRh

hRh

A

A

V

VAf

Current series feedback

• Feedback technique is to sample the output current (Io) and return aproportional voltage in series with the input.

• It stabilizes the amplifier gain, the current series feedback connectionincreases the input resistance.

• In this circuit, emitter of this stage has an un bypassed emitter, it effectivelyhas current-series feedback.

• The current through RE results in feedback voltage that opposes the sourcesignal applied so that the output voltage Vo is reduced.

• To remove the current-series feedback, the emitter resistor must be eitherremoved or bypassed by a capacitor (as is done in most of the amplifiers)

1

,1

Af

hfeRE

Rhh

Rh

Efeie

Efe

The fig below shows the equivalent circuit for current series feedback

Gain, input and output impedance for this condition is,

Numerical

Calculate the voltage gain of the circuit..

Efeie

CfeCfc

s

o

s

co

s

ovf

ie

Efecoof

ie

Efeieiif

Eie

feE

iefe

s

of

Rhh

RhRAR

V

I

V

RI

V

VA

Afeedbackwith

h

RhRAZZ

h

RhhAZZ

Rh

hR

hh

A

A

V

IA

;..

1)1(

1)1(

)(1

/

1

With RB=470Ω,RC=2.2KΩ,RE=510 Ω, hfe=120,hie=900Ω.

Solution:

• The factor (1+Aβ) is then,

1+(-0.085)(-510) =44.35

• The gain with feedback is

Af=Vo/vi=A/(1+A β)

=-.085/44.35 = -1.92x10e-3

• Voltage gain with feedback is

Avf=Vo/Vs=AfRC=(-1.92x10e-3)(2.2x10e3)=4.2

• Without feedback (RE=0) the voltage gain is

510

085.0510900

120

REIo

VfREhie

hfe

Vi

IoA

Av=-RC/re=-2.2x10e3/7.5= -293.3

Voltage shunt feedback

• Constant gain op-amp circuit provides voltage shunt feedback.

• Ref to fig below. The input impedance of a ideal op-amp is taken to beinfinite. Hence Ii=0,vi=0 and voltage gain is infinity.

• Ie., A=Vo/Ii=infinity

• And β=If/Vo= -1/Ro

• This is transfer resistance gain.

Voltage shunt negative feedback amplifier

1.Constant gain circuit

2.Equivalent circuit

• Voltage gain with feedback ,

1

0

1

1)(

1 R

R

RRo

V

Is

Is

VoAvf

Voltage shunt feedback using FET

Equivalent circuit

• With no feedback A=Vo/Ii=-gmRDRS

• The feedback factor is β=If/Vo= -1/RF

• With feedback, gain of the circuit is,

Numerical

Calculate voltage gain with and without feedback for the circuit of FET f/b. Withthe values, gm=5mS, RD=5.1KΩ, Rs=1KΩ, RF=20KΩ

Solution :

Use above formulae

• Av=-gmRD=-25.5

• Feedback gain Avf=-11.2

SDmF

FDm

SDmF

FDm

SSDmF

FSDm

s

s

s

ovf

SDmF

SDm

s

of

RRgR

RRg

RRgR

RRg

RRRgR

RRRg

V

I

I

VA

isckwithfeedbagainvoltage

RRgR

RRg

A

A

I

VA

)(

1

,..

))(/1(11