Post on 03-May-2020
Applied Electronics II
Chapter 1: Feedback Amplifiers
School of Electrical and Computer EngineeringAddis Ababa Institute of Technology
Addis Ababa University
Daniel D./Getachew T./Abel G.
March 2017
Chapter 1: Feedback Amplifiers () SECE March 2017 1 / 45
Overview
Overview
1 Types of Feedback
2 The General Feedback StructureBasic Feedback Amplifier
3 Feedback Topologies
4 Properties of Negative FeedbackGain DesensitivityNoise/Interference ReductionReduction of Nonlinear DistortionControl of Impedance level & Bandwidth Extension
5 Analysis of Feedback AmplifiersVoltage-Series (Voltage Amplifier) FeedbackMethod of Analysiis of Feedback AmplifiersCurrent-Series (Transconductance Amplifier) FeedbackCurrent-Shunt (Current Amplifier) FeedbackVoltage-Shunt (Transresistance Amplifier) Feedback
6 ExerciseChapter 1: Feedback Amplifiers () SECE March 2017 2 / 45
Types of Feedback
Type of Feeedback
Most physical systems incorporate some form of feedback. Feedbackcan be broadly classified as:
1 Posittive FeedbackA portion of the output signal is added to the input. Positvefeedback is used in the design of oscillator and a number of otherapplications (will be discussed in Chapter 4 and 5).
2 Negative FeedbackA portion of the output signal is subtracted from the inputsignal.The basic idea of negative feedback is to trade off gain forother desirable properties listed below
Desensitize the gainReduce nonlinear distortionReduce the effect of noiseControl the input and output resistancesExtend the bandwidth of the amplifier.
Chapter 1: Feedback Amplifiers () SECE March 2017 3 / 45
Types of Feedback
Negeative Feedback
Example
Introducing resistor at the emitter of BJT common-emitter circuitsstabilizes the Q-point against variation transistor parameters.
V+
IC
IE
V-
VBB
IB +
-
VCE+
-VBE
RC
RB
RE
Solution Apply KCL at B-E loop
VBB = IBRB + VBE(on) + IERE + V − (1)
Assuming active-mode of operation
IE = (1 + β)IB and IC = βIB (2)
As IC increases(due to ↑ in T,aging ) thevoltage drop across RE increase thusopposing the base-emitter voltage.
Chapter 1: Feedback Amplifiers () SECE March 2017 4 / 45
The General Feedback Structure
The General Feedback Structure
Figure 1 show the basic structure of a feedback amplifier, where each ofthe quantities x can represent either a voltage or a current signal.
Figure: 1 General structure of the feedback amplifier, the quantities xrepresent either voltage or current signals.
The relationship between the quantities x is
xo = Axi
xf = βxo
xi = xs − xfChapter 1: Feedback Amplifiers () SECE March 2017 5 / 45
The General Feedback Structure
Feedback Systems
Thus xo = A(xs − βxo)The gain with feedback ,Af
Af =xoxs
=A
1 + βA
The open-loop gain, A represents the transfer gain of the basicamplifier without feedback. Implicit in the description is that thesource, the load, and the feedback network do not load the basicamplifier. That is, the gain A does not depend on any of these threenetworks.In practice this will not be the case.
if |Af | < |A| the feed back is negative or degenerative
if |Af | > |A| the feed back is positive or regenerative
If, as is the case in many circuits, the loop gain Aβ is large, Aβ 1,then it follows that
Af u1
β
Chapter 1: Feedback Amplifiers () SECE March 2017 6 / 45
The General Feedback Structure Basic Feedback Amplifier
Basic Feedback Amplifier
A basic representation of feedback amplifier is show in the Figure below
Signal
Source
Sampling
Network
Basic
Amplifier,
gain A
Comparato
r or Mixer
Network
Feedback
Network b
Ii I
If
Io = IL
RL
+
-
+
-
+
-Vi V Vo
Signal Source : This block is a voltage source Vs with a series resistorRs(Thvenin’s equivalent circuit) or a current source Iswith a parallel resistor Rs (Norton’s equivalent circuit)
Feedback Network : Usually a passive two-port network with reversetransmission β
Chapter 1: Feedback Amplifiers () SECE March 2017 7 / 45
The General Feedback Structure Basic Feedback Amplifier
Basic Feedback Amplifier
Sampling Network : Sampling blocks are shown below
A
b
RL
Sampler
Figure: (a) Voltage or node sampling
A
b
Sampler
RL
Figure: (b) Current or loop sampling
(a) Output voltage is sampled by connecting the feedbacknetwork in shunt across the output.
(b) Output current is sampled by connecting the feedbacknetwork in series with the output.
Chapter 1: Feedback Amplifiers () SECE March 2017 8 / 45
The General Feedback Structure Basic Feedback Amplifier
Basic Feedback Amplifier
Comparator or Mixer Network : Two types a series (loop) and shunt(node). A differential amplifier is often used as mixer.
A
b
Rs
Series
MixerSource
Vs
+
-
+
-
Vf
Vi
Figure: (a) Series Mixing
A
b
Rs
Shunt
MixerSource
Is
Ii
If
Figure: (b) Shunt Mixing
Chapter 1: Feedback Amplifiers () SECE March 2017 9 / 45
The General Feedback Structure Basic Feedback Amplifier
Basic Feedback Amplifier
Basic Amplifier : A could be used to represent
VVi
= AV , Voltage gain
IIi
= AI , Current gain
IVi
= GM , Transconductance
VIi
= RM , Transresistance
They are gain of the basic amplifier without feedback
Chapter 1: Feedback Amplifiers () SECE March 2017 10 / 45
Feedback Topologies
Feedback Topologies
There are four basic feedback topologies, based on the parameters tobe amplified (voltage or current)and the output parameter (voltage orcurrent). They are described by the type of connection at the inputand output of the circuit.
(a) Voltage-Series (Series-Shunt) or Voltage Amplifier
(b) Current-Shunt (Shunt-Series) or Current Amplifier
(c) Current-Series (Series-Series) or Transconductance Amplifier
(d) Voltage-Shunt (Shunt-Shunt) or Transeresistance Amplifier
Chapter 1: Feedback Amplifiers () SECE March 2017 11 / 45
Feedback Topologies
Feedback Topologies
Figure: (a) Series-Shunt
Figure: (b) Shunt-Series
Figure: (c) Series-Series
Figure: (d) Shunt-Shunt
Chapter 1: Feedback Amplifiers () SECE March 2017 12 / 45
Properties of Negative Feedback Gain Desensitivity
Gain Desensitivity
Variation in the circuit gain as a result of change in transistorparameters is reduced by negative feedbackFrom the previous slides the gain with feedback,Af is given as
Af =xoxs
=A
1 + βA
Assuming β is constant and taking the derivative of Af with respect toA,
dAfdA
=1
1 + βA− A
(1 + βA)2β =
1
(1 + βA)2or dAf =
dA
(1 + βA)2
Dividing both sides the gain with feedback yields
dAfAf
=
dA(1+βA)2
A1+βA
=1
1 + βA
dA
A
Chapter 1: Feedback Amplifiers () SECE March 2017 13 / 45
Properties of Negative Feedback Gain Desensitivity
Gain Desensitivity
Hence the percentage change in Af (due to variations in some circuitparameter) is smaller than the percentage change in A by a factorequal to the amount of feedback. For this reason, the amount offeedback, 1 +Aβ, is also known as the desensitivity factor.
Example
The open-loop gain of an amplifier is A = 5× 104V/V exhibits a gainchange of 25% as the operating temperature changes. Calculate thepercentage change if the closed loop gain Af = 50V/V .
dAfAf
=1
1 + βA
dA
A=
A
A(1 + βA)
dA
A=AfA
dA
A=
50
5× 104× 25%
dAfAf
= 0.025%
Chapter 1: Feedback Amplifiers () SECE March 2017 14 / 45
Properties of Negative Feedback Noise/Interference Reduction
Noise/Interference Reduction
Under certain condition feedback amplifiers can be used to reducenoise/interference.
This can be achieved if a preamplifer which is (relatively)noise/interference-free precessed the noise/interference-proneamplifierUnder such conditions the Signal-to-Noise ratio can be improved (compare to noise/interference-prone amplifier without feedback)by the factor of the preamplifier gain
Chapter 1: Feedback Amplifiers () SECE March 2017 15 / 45
Properties of Negative Feedback Reduction of Nonlinear Distortion
Reduction of Nonlinear Distortion
Distortion in the output is due to application of large amplitudeinput signal applied beyond the linear region of operation.
Negative feedback can be implemented to reduce nonlineardistortion by a factor of 1 +Aβ.
Assuming that the open-loop gain Aβ 1, the gain with feedback
Af =A
1 +Aβu
1
β
It implies that Af is independent of the nonlinear properties of thetransistors used in the basic amplifier.
Since the feedback network usually consists of passive components,which usually can be chosen to be as accurate as one wishes.
Chapter 1: Feedback Amplifiers () SECE March 2017 16 / 45
Properties of Negative Feedback Control of Impedance level & Bandwidth Extension
Control of Impedance level & Bandwidth Extension
Control of Impedance level: The input and output impedance can beincreased or decreased with the proper type of negativefeedback circuit.
Bandwidth Extension : The improvement in frequency response andbandwidth extension (Chapter 3)
The advantage of negative feedback is at the cost of gain. Undercertain circumstance, a negative feedback amplifier may becomeunstable and break into oscillation.
Chapter 1: Feedback Amplifiers () SECE March 2017 17 / 45
Analysis of Feedback Amplifiers
Fundamental Assumtions
Some fundamental assumptions are taken in order to analyze the fourfeedback configurations.
Input is transmitted through the amplifier only, not through thefeedback.The feedback signal transmitted feedback network only, notthrough the amplifier.β is independent of the load and source impedance.
+−
Ii
Vs
Rif Rof R’of
Io+
-
Vi Ri
Ro
RL
+
-Vo
+
-
Vo
+- Vf
ßVo
AvoVi
+
−
+
−
Figure: Ideal structure of a Voltage-Series feedback amplifierChapter 1: Feedback Amplifiers () SECE March 2017 18 / 45
Analysis of Feedback Amplifiers Voltage-Series (Voltage Amplifier) Feedback
Voltage-Series (Voltage Amplifier) Feedback
Avo represents the open circuit voltage gain taking Rs into accountInput Impedance: The input impedance with feedback is
Rif =VsIi
Also,
Vs = IiRi + Vf = IiRi + βVo and Vo = AvoViRL
Ro +RL
let Av = AvoRL
Ro+RL, where Av is the voltage gain without feedback
taking the RL into account then
Vs = IiRi + βAvVi = IiRi + βAvIiRi
Rif =VSIi
= Ri(1 + βAv)
Chapter 1: Feedback Amplifiers () SECE March 2017 19 / 45
Analysis of Feedback Amplifiers Voltage-Series (Voltage Amplifier) Feedback
Voltage-Series (Voltage Amplifier) Feedback
+
-
Vi Ri
Ro
Vf
ßVo
AvoVi
+
−
+−
+− Vx
Ix
Figure: Ideal structure of a Voltage-Series feedback amplifier
Output Impedance: To find Rof must remove the external signal (setVs = 0 or Is = 0), let RL =∞, impress a voltage Vx across the outputterminals and calculate the current Ix delivered by the test voltage Vx
Ix =Vx −AvoVi
Ro
Since Vi = −βVxRof =
VxIx
=Ro
1 + βAvo
Chapter 1: Feedback Amplifiers () SECE March 2017 20 / 45
Analysis of Feedback Amplifiers Voltage-Series (Voltage Amplifier) Feedback
Voltage-Series (Voltage Amplifier) Feedback
The output resistance with feedback R′of which includes RL as part of
the amplifier is Rof ‖ RL
R′of =
RofRLRof +RL
=
RoRLRof +RL
1 + βAvoRL
Ro +RL
Taking R′o = Ro ‖ RL
R′of =
R′o
1 + βAv
Voltage gain with feedback: Avf taking the load into account.
Vs = Vi + βVo = VoRo +RLAvoRL
+ βVo
Chapter 1: Feedback Amplifiers () SECE March 2017 21 / 45
Analysis of Feedback Amplifiers Voltage-Series (Voltage Amplifier) Feedback
Voltage-Series (Voltage Amplifier) Feedback
Manipulating the equation
Avf =VoVs
=Avo
RLRo +RL
1 + βAvoRL
Ro +RL
The voltage gain with feedback without the load Avfo is
Avfo =VoVs
=Avo
1 + βAvo
In conclusion
Input Impedance: increased by a factor 1 + βAv
output Impedance: decreased by a factor 1 + βAv
Voltage Gain: decreased by a factor 1 + βAv
Chapter 1: Feedback Amplifiers () SECE March 2017 22 / 45
Analysis of Feedback Amplifiers Voltage-Series (Voltage Amplifier) Feedback
Voltage-Series (Voltage Amplifier) Feedback
In practical case
In practical case, feedback network will not be ideal VCVS.
Actually, it is resistive and will load the amplifier.
Source and load resistances will affect A, Ri, and Ro.
Source and load resistances should be lumped with basic amplifier.
Expressed as two-port network.
How To Solve
1. Identify the feedback network
2. Its loading effect at the input is obtained by short circuiting itsport 2 (because it is connected in shunt with the output).
3. The loading effect at the output is obtained by open-circuitingport 1 of the feedback network (because it is connected in serieswith the input)
4. The gain without feedback A is determined
5. The feed back gain β is determinedChapter 1: Feedback Amplifiers () SECE March 2017 23 / 45
Analysis of Feedback Amplifiers Voltage-Series (Voltage Amplifier) Feedback
Voltage-Series (Voltage Amplifier) FeedbackFigure: Finding the A circuit and β for the Voltage-Series feedback amplifier.
Chapter 1: Feedback Amplifiers () SECE March 2017 24 / 45
Analysis of Feedback Amplifiers Method of Analysiis of Feedback Amplifiers
Method of Analysiis of Feedback Amplifiers
Steps
1. Identify if the mixing or comparison is series or shunt
a) Series mixing : If the feedback signal subtracts from the externallyapplied signal as a voltage
b) Shunt mixing : If the feedback signal subtracts from the appliedexcitation signal as a current.
2. Identify the sampled signal as series or shunt
a) Voltage sampling : Set Vo = 0(RL = 0. If Xf becomes zero, we havevoltage sampling.
b) Current sampling : Set Io = 0(RL =∞. If Xf becomes zero, wehave current sampling.
3. The amplifier without feedback but taking the feedback networkloading into account
1) Find the input circuit.
a) Set Vo = 0 for voltage sampling.b) Set Io = 0 for current sampling.
Chapter 1: Feedback Amplifiers () SECE March 2017 25 / 45
Analysis of Feedback Amplifiers Method of Analysiis of Feedback Amplifiers
Method of Analysiis of Feedback Amplifiers
2) Find the output circuit.
a) Set Vi = 0 for shunt comparison so that no feedback current entersthe amplifier input.
b) Set Io = 0 for series comparison so that no feedback voltage reachesthe amplifier input.
4. Find the feedback network β.
5. Calculate β,A,Ri and Ro.
6. Calculate the closed loop Af , Rif , Rof .
Chapter 1: Feedback Amplifiers () SECE March 2017 26 / 45
Analysis of Feedback Amplifiers Method of Analysiis of Feedback Amplifiers
Voltage-Series (Voltage Amplifier) Feedback
Example
Analyze the amplifier to obtain its voltage gain Vo/Vs, input resistanceRin, and output resistance Rout. Find numerical values for casegm1 = gm2 = 4mA/V , RD1 = RD2 = 10kΩ and R2 = kΩ. Forsimplicity, neglect ro of each of Q1 and Q2.
+−Vs
Q1
Q2
RD1
Vo
R1
R2
Rin
Rout
RD2
The next step is identifying the A andβ circuitWe identify the feedback network asthe voltage divider of (R1, R2)
+− Vo
R2
R1
+
-
Vf
Chapter 1: Feedback Amplifiers () SECE March 2017 27 / 45
Analysis of Feedback Amplifiers Method of Analysiis of Feedback Amplifiers
Voltage-Series (Voltage Amplifier) Feedback
Example (Continued)
The A circuit is
Q1
Q2
RD1
Vo
R1 R2
Rout
RD2
+
-Ri
R2
R1
Vi
Vd1
Calculating A1 and A2
Vd1 = 0− id1RD1
Vi = Vgs1 + id1(R1 ‖ R2)
A1 =Vd1Vi
=−id1RD1
Vgs1 + id1(R1 ‖ R2)
A1 =−RD1
1/gm1 + (R1 ‖ R2)
A1 =−gm1RD1
1 + gm1(R1 ‖ R2)
Chapter 1: Feedback Amplifiers () SECE March 2017 28 / 45
Analysis of Feedback Amplifiers Method of Analysiis of Feedback Amplifiers
Voltage-Series (Voltage Amplifier) Feedback
Example (continued)
From A circuit we have
Vo = 0− id2(RD2 ‖ (R1 +R2)) and Vgs2 = Vd1
A2 =VoVd1
=−id2(RD2 ‖ (R1 +R2))
Vgs2= −gm2(RD2 ‖ (R1 +R2))
The open loop gain is
A =VoVi
= A1A2 =gm1gm2RD1[RD2 ‖ (R1 +R2)]
1 + gm1(R1 ‖ R2)
When evaluated
A =4× 4× 10[10 ‖ (1 + 9)]
1 + 4(1 ‖ 9)= 173.913 V/V
Chapter 1: Feedback Amplifiers () SECE March 2017 29 / 45
Analysis of Feedback Amplifiers Method of Analysiis of Feedback Amplifiers
Voltage-Series (Voltage Amplifier) Feedback
Example (continued)
from β circuit we have
β =R1
R1 +R2=
1
1 + 9= 0.1 V/V
The closed loop gain
VoVs
= Af =A
1 +Aβ=
173.913
1 + 173.913× 0.1= 9.45 V/V
The input resistance is infinite because it is the input resistance ofMOSFET.The output resistance is
Rout = Rf =Ro
1 +Aβ=RD2 ‖ (R1 +R2)
1 +Aβ=
10 ‖ (1 + 9)
1 + 173.913× 0.1= 271.87Ω
Chapter 1: Feedback Amplifiers () SECE March 2017 30 / 45
Analysis of Feedback Amplifiers Current-Series (Transconductance Amplifier) Feedback
Current-Series (Transconductance Amplifier) Feedback
+−
Ii
Vs
Rif Rof R’of
Io+
-
Vi Ri Ro RL
+
-
Vo
+- Vf
+
−
GmVi
IoßIo
Input Impedance:
Rif =VsIi
; Vs = IiRi + βIo ; Io = GmViRo
Ro +RL
Rif =IiRi + βGmIiRi
RoRo+RL
Ii= Ri(1 + βGm
RoRo +RL
)
Chapter 1: Feedback Amplifiers () SECE March 2017 31 / 45
Analysis of Feedback Amplifiers Current-Series (Transconductance Amplifier) Feedback
Current-Series (Transconductance Amplifier) Feedback
Gm = Io/Vi is the short-circuit transconductance, andGM = GmRo/(Ro +RL) is the transconductance without feedbacktaking the load into account.
Rif = Ri(1 + βGM )
Output Impedance: calculated by short-circuiting the source andreplacing the source with a voltage source Vx with a current of Ix
Ix =VxRo−GmVi and Vi = βIx
∴ Rof =VxIx
=Ro(Ix +GmβIx)
Ix= Ro(1 + βGm)
The output impedance taking the load as part of the amplifier is:
R′of = (Rof ‖ RL) = (Ro ‖ RL)
1 + βGm1 + βGM
Chapter 1: Feedback Amplifiers () SECE March 2017 32 / 45
Analysis of Feedback Amplifiers Current-Series (Transconductance Amplifier) Feedback
Current-Series (Transconductance Amplifier) FeedbackFigure: Finding the A circuit and β for the Current-Series feedback amplifier.
Chapter 1: Feedback Amplifiers () SECE March 2017 33 / 45
Analysis of Feedback Amplifiers Current-Series (Transconductance Amplifier) Feedback
Current-Series (Transconductance Amplifier) FeedbackExampleCalculate the closed loop voltage gain, output resistance and input resistancefor the circuit below. The output is taken from emitter current of Q3. Thevalues of RC1 = 9kΩ, RC2 = 5kΩ, RC3 = 600Ω, RE1 = 100Ω, RE3 = 100Ωand RF = 640Ω. Assume that the bias circuit, which is not shown, establishesIC1 = 0.6mA, IC2 = 1mA, and IC3 = 4mA. Also assume that for all threetransistors, hfe = 100 and ro =∞.
Q1
Q2
Q3
+
-
Vs
Vo
RE1 RE3
RF
RC1
RC2
RC3
Io
The β circuit.
+
-
RE1 RE3
RF
IoVf
β =VfIo
=[(RF +RE1) ‖ RE2]Io
RE1
RF+RE1
Io
β =RE1 ×RE2
RF +RE1 +RE2= 11.9Ω
Chapter 1: Feedback Amplifiers () SECE March 2017 34 / 45
Analysis of Feedback Amplifiers Current-Series (Transconductance Amplifier) Feedback
Current-Series (Transconductance Amplifier) Feedback
Example (continued)
The A circuit.
Q1
Q2
Q3
Vo
RE1 RE3
RF
RC1
RC2
RC3
Io
RE3 RE1
RF+−
Vi
Ri
Ro
When Aβ 1
Af u1
β=
1
11.9Ω= 84mA/V
lets check by determining eachtransistor gain
A1 =Vc1Vi
=−ic(RC1 ‖ rπ2)
ie(re1 + [RE1 ‖ (RE3 +RF )])
A1 =−α(RC1 ‖ rπ2)
(re1 + [RE1 ‖ (RE3 +RF )])
Since Q1 is biased at 0.6mA,re1 = 41.7Ω. Q2 is biased at 1mA; thusrπ2 = hfe/gm2 evaluating A1
A1 = −14.92V/VChapter 1: Feedback Amplifiers () SECE March 2017 35 / 45
Analysis of Feedback Amplifiers Current-Series (Transconductance Amplifier) Feedback
Current-Series (Transconductance Amplifier) Feedback
Example (continued)
The gain of Q2
A2 =Vc2Vb2
=−ic[RC2 ‖ (hfe + 1)[re3 + (RE3 ‖ (RF +RE1))]]
Vb2
A2 = −gm2[RC2 ‖ (hfe + 1)[re3 + (RE3 ‖ (RF +RE1))]]
re3 = 25/4 = 6.25Ω and substituting the other values
A2 = −131.2V/V
The gain of Q3
A3 =IoVc2
=Ie3Vb3
=1
re3 + (RE3 ‖ (RF +RE1))
when evaluatedA3 = 10.6mA/V
Chapter 1: Feedback Amplifiers () SECE March 2017 36 / 45
Analysis of Feedback Amplifiers Current-Series (Transconductance Amplifier) Feedback
Current-Series (Transconductance Amplifier) Feedback
Example (continued)
The gain without feedback
A = A1A2A3 = −14.92×−131.2× 10.6× 10−3 = 20.7A/V
The gain with feedback
Af =A
1 +Aβ=
20.7
1 + 20.7× 11.9= 83.7mA/V
We can note that it is very close to approximate value. The input resistance
Rin = Rif = Ri(1 +Aβ)
Ri = (hfe + 1)[re1 + (RE1 ‖ (RF +RE2))] = 13.65kΩ
∴ Rif = 13.65(1 + 20.7× 11.9) = 3.38MΩ
Chapter 1: Feedback Amplifiers () SECE March 2017 37 / 45
Analysis of Feedback Amplifiers Current-Series (Transconductance Amplifier) Feedback
Current-Series (Transconductance Amplifier) Feedback
Example (continued)
The output resistanceRof = Ro(1 +Aβ)
Ro = [RE3 ‖ (RF +RE1)] + re3 +RC2
hfe + 1
When evaluated Ro = 143.9Ω
∴ Rof = 143.9(1 + 20.7× 11.9) = 35.6kΩ
Chapter 1: Feedback Amplifiers () SECE March 2017 38 / 45
Analysis of Feedback Amplifiers Current-Shunt (Current Amplifier) Feedback
Current-Shunt (Current Amplifier) Feedback
Ii
Rif
Rof R’of
Io+
-
Vi Ri Ro RL
+
-
Vo
IoßIo
IsAiIi
Ai is the short-circuit current gain taking Rs into accountInput Resistance:
Is = Ii + βIo ; Io = AiIiRo
RL +Ro
taking AI = Ai(Ro/(Ro +RL)), where AI is current gain withoutfeedback taking the load into account.
Chapter 1: Feedback Amplifiers () SECE March 2017 39 / 45
Analysis of Feedback Amplifiers Current-Shunt (Current Amplifier) Feedback
Current-Shunt (Current Amplifier) Feedback
Rif =ViIs
=RiIi
Ii + βAIIi=
Ri1 + βAI
Output Resistance: making Is = 0 and replacing the load with asource
Ix =VxRo−AiIi ; Ii = −If = −βIo = βIx
Ix =VxRo− βAiIx ;
VxRo
= Ix(1 + βAi)
∴ Rof =VxIx
= Ro(1 + βAi)
The output resistance with load
R′of = Rof ‖ RL = (Ro ‖ RL)
1 + βAi1 + βAI
Chapter 1: Feedback Amplifiers () SECE March 2017 40 / 45
Analysis of Feedback Amplifiers Current-Shunt (Current Amplifier) Feedback
Current-Shunt (Current Amplifier) FeedbackFigure: Finding the A circuit and β for the Current-Shunt feedback amplifier.
Chapter 1: Feedback Amplifiers () SECE March 2017 41 / 45
Analysis of Feedback Amplifiers Voltage-Shunt (Transresistance Amplifier) Feedback
Voltage-Shunt (Transresistance Amplifier) Feedback
Ii
Rif
Rof R’of
Io+
-
Vi Ri
Ro
RL
+
-
VoIs
+
− RmIi
ßVo
Rm is the open-circuit transresistance gain taking Rs into accountInput Resistance:
Is = Ii + βVo ; Vo = RmIiRL
RL +Ro
taking RM = Rm(RL/(Ro +RL)), where RM is transresistance gainwithout feedback taking the load into account.
Chapter 1: Feedback Amplifiers () SECE March 2017 42 / 45
Analysis of Feedback Amplifiers Voltage-Shunt (Transresistance Amplifier) Feedback
Voltage-Shunt (Transresistance Amplifier) Feedback
Rif =ViIs
=Vi
Ii + βRMIi=
Ri1 + βRM
Output Resistance: making Is = 0 and replacing the load with asource
Ix =Vx −RmIi
Ro; Ii = −If = −βVo = −βVx
Ix =Vx +RmβVx
Ro;
VxRo
=Ix
(1 + βRm)
∴ Rof =VxIx
=Ro
1 + βAi
The output resistance with load
R′of = Rof ‖ RL =
Ro ‖ RL1 + βRM
Chapter 1: Feedback Amplifiers () SECE March 2017 43 / 45
Analysis of Feedback Amplifiers Voltage-Shunt (Transresistance Amplifier) Feedback
Voltage-Shunt (Transresistance Amplifier) FeedbackFigure: Finding the A circuit and β for the Voltage-Shunt feedback amplifier.
Chapter 1: Feedback Amplifiers () SECE March 2017 44 / 45
Exercise
Exercise
The following questions in the text book are exercises to be done forthe tutorial session.
10.36
10.52
10.57
10.65
Chapter 1: Feedback Amplifiers () SECE March 2017 45 / 45