General Amplifiers - BookSpar · General Amplifiers ... • Resulting in an overall cascode...

www.bookspar.com | Website for Students | VTU NOTES | QUESTION PAPERS | NEWS | RESULTS


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


Numerical

http://www.bookspar.com/




Cascade connection – FET

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

• Cascade connection provides a multiplication of the gain of each stage for a larger 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 15 װ K 200 װx6.5

=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 gain of -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 a single 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

VVVRIV

VoltagesIII

RRVVI

+==

≈+=+−

=

βββ

)1(





Numerical

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

VVVVVV

VmRIVVoltages

mAuIII

uAMRR

VVI

C

BEEB

EEE

BDBDE

EDB

BEccB

186.96.18

8)390(48.20

48.20)56.2(8000)1(

56.2)390(80003.3

6.118

==+=+=

===

==≈+=

=+−

=+−

=

βββ





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 and measuring 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

rV

rV

RVI

rV

RVI

++=

++=++=

β

ββ

11

iDiEo

oo rrRI

VZ//1/1

1β++

==





ac voltage gain

Gain expression

On simplification

Numerical

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

EbD

EDEbEbDbo

RIIbIbriViRRIRIIV

)()()(

βββ

++=+=+=

1)(

)()(

)(

≈++

+==

+++

=

++=

DREREriDRERE

ViVoAv

RRRRr

VV

RRrIV

EDEEDEi

io

EDEibi

ββ

ββ

β

998.0]3908000390[5

3908000390=

+++

=xK

xAv





Feedback concepts

• Depending on the relative polarity of fed back signal in to the circuit, there are 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 feedback network.

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

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

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)

Parameter

Voltage series

Voltage shunt

Current series

Current shunt

Gain with feedback

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

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

Gain with feedback

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 arising because 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 at high 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 the factor 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:


• Voltage series feedback

AdA

A1

dAdA

A1Af

f

β

β

=

+=

%2.0........

%20)1000(1.0

11

=

−−==

AdA

AAfdAf

β





• Here part of output voltage (Vo) is obtained using a feedback network of resistors 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

RRRAf

thenAifgRRRR

RgA

AAfmL

Lm

+−==

>>++

−=

+=

β

ββ





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 and R2=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 input voltage.

• 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

VV

VRh

VhVsRh

VRIh

VVA

β

)/)(1(1/

1 ieEfe

ieEfe

s

o

hRhhRh

AA

VVAf

+=

+==

β





Current series feedback

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

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

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

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

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

1,1

≈>>

+=

AfhfeRE

RhhRh

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

RhhRh

RARVI

VRI

VVA

AfeedbackwithhRh

RAZZ

hRh

hAZZ

Rhh

R

hhA

AVIA

+

−≅=

===

−

+≅+=

+≅+=

+

−−+

−=

+==

;..

1)1(

1)1(

)(1

/1

β

β

β





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

−=−==

−=+

−=

+−

==

REIoVf

REhiehfe

ViIoA

β





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 be infinite. 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 ,

10

11)(

1 RR

RRo

VIs

IsVoAvf −

=−==





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. With the 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

RRgRRRg

RRgRRRg

RRRgRRRRg

VI

IVA

isckwithfeedbagainvoltageRRgR

RRgA

AIVA

+−=

+−

=

+

−==

−−−−+

−=

+==

)(

1

,..))(/1(11 β



General Amplifiers - BookSpar · General Amplifiers ... • Resulting in an overall cascode...

Documents

Transcript of General Amplifiers - BookSpar · General Amplifiers ... • Resulting in an overall cascode...