Analog Electronics CircuitsNagamani A N
Lecturer, PESIT, Bangalore – 85
Email – [email protected]
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