ECE 352 Electronics II Winter 2003 Ch. 8 Feedback 1 Feedback *What is feedback?Taking a portion of...

Ch. 8 Feedback 1ECE 352 Electronics II Winter 2003

Feedback

* What is feedback? Taking a portion of the signal arriving at the load and feeding it back to the input.

* What is negative feedback? Adding the feedback signal to the input so as to partially cancel the input signal to the amplifier.

* Doesn’t this reduce the gain? Yes, this is the price we pay for using feedback.

* Why use feedback? Provides a series of benefits, such as improved bandwidth, that outweigh the costs in lost gain and increased complexity in amplifier design.

XoXi

Xf

Xs +-

βf


Feedback Amplifier Analysis

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A

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XA

XX

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* Gain desensitivity - less variation in amplifier gain with changes in (current gain) of transistors due to dc bias, temperature, fabrication process variations, etc.

* Bandwidth extension - extends dominant high and low frequency poles to higher and lower frequencies, respectively.

* Noise reduction - improves signal-to-noise ratio* Improves amplifier linearity - reduces distortion in signal due to gain

variations due to transistors

* Cost of these advantages: Loss of gain, may require an added gain stage to compensate. Added complexity in design

Advantages of Negative Feedback

AA

f

LLfHfHf

1

1


* There are four types of feedback amplifiers. Why? Output sampled can be a current or a voltage Quantity fed back to input can be a current or a voltage Four possible combinations of the type of output sampling and input

feedback

* One particular type of amplifier, e.g. voltage amplifier, current amplifier, etc. is used for each one of the four types of feedback amplifiers.

* Feedback factor f is a different type of quantity, e.g. voltage ratio, resistance, current ratio or conductance, for each feedback configuration.

* Before analyzing the feedback amplifier’s performance, need to start by recognizing the type or configuration.

* Terminology used to name types of feedback amplifier, e.g. Series-shunt First term refers to nature of feedback connection at the input. Second term refers to nature of sampling connection at the output.

Basic Types of Feedback Amplifiers


Basic Types of Feedback Amplifiers

Series - Shunt Shunt - Series

Series - Series Shunt - Shunt


* Recognize the feedback amplifier’s configuration, e.g. Series-shunt

* Calculate the appropriate gain A for the amplifier, e.g. voltage gain. This includes the loading effects of the feedback circuit (some

combination of resistors) on the amplifier input and output.

* Calculate the feedback factor f

* Calculate the factor f A and make sure that it is: 1) positive and 2) dimensionless

* Calculate the feedback amplifier’s gain with feedback Af using

* Calculate the final gain of interest if different from the gain calculated, e.g. Current gain if voltage gain originally determined.

* Determine the dominant low and high frequency poles for the original amplifier, but taking into account the loading effects of the feedback network.

* Determine the final dominant low and high frequency poles of the amplifier with feedback using

Method of Feedback Amplifier Analysis

A

AA

ff

1

AA

f

LLfHfHf

1

1


Series-Shunt Feedback Amplifier - Ideal Case

* Assumes feedback circuit does not load down the basic amplifier A, i.e. doesn’t change its characteristics

Doesn’t change gain A Doesn’t change pole frequencies of basic

amplifier A Doesn’t change Ri and Ro

* For the feedback amplifier as a whole, feedback does change the midband voltage gain from A to Af

* Does change input resistance from Ri to Rif

* Does change output resistance from Ro to Rof

* Does change low and high frequency 3dB frequencies

A

AA

ff

1

ARR fiif 1

A

RR

f

oof

1

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f

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1

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Basic Amplifier

Feedback Circuit

Equivalent Circuit for Feedback Amplifier



Midband Gain

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fi

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Input Resistance

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Output Resistance

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1

1

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11

11

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1

1

11

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Low Frequency Pole

High Frequency Pole

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11

11

1

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11

1

11

Low 3dB frequency lowered by feedback.

Upper 3dB frequency raised by feedback.


* Feedback networks consist of a set of resistors Simplest case (only case considered here) In general, can include C’s and L’s (not

considered here) Transistors sometimes used (gives variable

amount of feedback) (not considered here)

* Feedback network needed to create Vf feedback signal at input (desirable)

* Feedback network has parasitic (loading) effects including:

* Feedback network loads down amplifier input Adds a finite series resistance Part of input signal Vs lost across this series

resistance (undesirable), so Vi reduced

* Feedback network loads down amplifier output Adds a finite shunt resistance Part of output current lost through this shunt

resistance so not all output current delivered to load RL (undesirable)

Practical Feedback Networks

Vi

Vf

Vo

* How do we take these

loading effects into account?


* Need to find an equivalent network for the feedback network including feedback effect and loading effects.

* Feedback network is a two port network (input and output ports)

* Can represent with h-parameter network (This is the best for this particular feedback amplifier configuration)

* h-parameter equivalent network has FOUR parameters

* h-parameters relate input and output currents and voltages

* Two parameters chosen as independent variables. For h-parameter network, these are input current I1 and output voltage V2

* Two equations relate other two quantities (output current I2 and input voltage V1) to these independent variables

* Knowing I1 and V2, can calculate I2 and V1 if you know the h-parameter values

* h-parameters can have units of ohms, 1/ohms or no units (depends on which parameter)

Equivalent Network for Feedback Network


* Feedback network consists of a set of resistors

* These resistors have loading effects on the basic amplifier, i.e they change its characteristics, such as the gain

* Can use h-parameter equivalent circuit for feedback network Feedback factor f given by h12 since

Feedforward factor given by h21 (neglected)

h22 gives feedback network loading on output

h11 gives feedback network loading on input

* Can incorporate loading effects in a modified basic amplifier. Basic gain of amplifier AV becomes a new, modified gain AV’ (incorporates loading effects).

* Can then use feedback analysis from the ideal case.

Series-Shunt Feedback Amplifier - Practical Case

fo

f

IV

V

V

Vh

02

112

1

'1

'1

'1'1

'1

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Vf

VVf

AA

A

RRARR

A

AA



* How do we determine the h-parameters for the feedback network?

* For the input loading term h11 Turn off the feedback signal by

setting Vo = 0. Then evaluate the resistance seen

looking into port 1 of the feedback network (also called R11 here).

* For the output loading term h22

Open circuit the connection to the input so I1 = 0.

Find the resistance seen looking into port 2 of the feedback network (also called R22 here).

* To obtain the feedback factor f (also called h12 ) Apply a test signal Vo’ to port 2 of the

feedback network and evaluate the feedback voltage Vf (also called V1 here) for I1 = 0.

Find f from f = Vf/Vo’

Summary of Feedback Network Analysis


* Evaluate modified basic amplifier (including loading effects of feedback network) Including h11 at input Including h22 at output Including loading effects of source resistance Including load effects of load resistance

* Analyze effects of idealized feedback network using feedback amplifier equations derived

* Note Av’ is the modified voltage gain including the

effects of h11 , h22 , RS and RL. Ri’, Ro’ are the modified input and output

resistances including the effects of h11 , h22 , RS and RL.


Summary of Approach to Analysis

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Vf

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AA

A

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Modified Basic Amplifier

Idealized Feedback Network

Practical Feedback Network

Basic Amplifier


* Two stage amplifier

* Each stage a CE amplifier

* Transistor parameters Given: 1= 2 =50, rx1=rx2=0

* Coupled by capacitors, dc biased separately

* DC analysis:

Example - Series-Shunt Feedback Amplifier

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r

VmAV

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r

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2

22

222

1

11

111

DC analysis for each stage can be done separately since stages are isolated (dc wise) by coupling capacitors.



Vi

+_

* Redraw circuit to show Feedback circuit Type of output sampling

(voltage in this case = Vo)

Type of feedback signal to input (voltage in this case = Vf)

Vf

+_ Vo

+

_



Input Loading Effects

Output Loading Effects

KKK

RRhR ff

8.41.07.4

21222

Amplifier with Loading Effects

I1=0

K

KK

RRhR ff

098.0

7.41.0

21111

Vo=0

R1R2


* Construct ac equivalent circuit at midband frequencies including loading effects of feedback network.

* Analyze circuit to find midband gain

(voltage gain for this series-shunt configuration)


R1

R2

R1 R2


Example - Series-Shunt Feedback AmplifierMidband Gain Analysis

dBdBA

V

VA

K

K

KKKK

KKK

RRRR

RRR

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I

VR

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1

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2222

2

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1

1

2

2


Midband Gain with Feedback

* Determine the feedback factor f

* Calculate gain with feedback Avf

* Note f Avo > 0 as necessary for negative feedback

f Avo is large so there is significant feedback. For f Avo 0, there is almost no feedback.

Can change f and the amount of feedback by changing Rf1 and/or Rf2.

NOTE: Since f Avo >> 0

021.07.41.0

1.0

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1

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Input and Output Resistances with Feedback

* Determine input Ri and output Ro resistances with loading effects of feedback network.

* Calculate input Rif and output Rof resistances for the complete feedback amplifier.

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RRRR iBSi

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1


Equivalent Circuit for Series-Shunt Feedback Amplifier

* Voltage gain amplifier

* Modified voltage gain, input and output resistances Included loading effects of

feedback network Included feedback effects

of feedback network Include source resistance

effects

* Significant feedback, i.e. f Avo is large and positive

dBdBA

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A

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21


Low Frequency Poles and Zeros for Series-Shunt Feedback Amplifier

* Six capacitors: Input and output coupling capacitors C1 and C5

Emitter bypass capacitors C3 and C4

Interstage coupling capacitors C2

Feedback coupling capacitor C6

* Analyze using Gray-Searle (Short Circuit) Technique one capacitor at a time

* Find dominant low frequency pole (highest frequency one)


Example - Input Coupling Capacitor’s Pole Frequency

Equivalent circuit for C1

sradFKCR

KKKKRRRR

so

KKKRr

I

RVgIrI

I

VR

where

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VR

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mii

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11

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111

111

11

1

111111

1

11

111

Note that there are some loading effects of the feedback network on this pole frequency. In Ri1 the feedback resistors determine R1

211 ff RRR


Example - Interstage Coupling Capacitor’s Pole Frequency

Equivalent circuit for C2 sradFKCR

so

KKKK

rRRI

VR

xCPL

BCx

xxC

/7.1857.10

11

7.107.04.1910

222

2212

Note: No RE2 since C4 shorts it out.


Example - Feedback Coupling Capacitor’s Pole Frequency


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VR

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Cx

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7.98.49.4

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Example - Emitter Bypass Capacitor’s Pole Frequency


IE1


Example - Emitter Bypass Capacitor’s Pole Frequency


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Example - Output Coupling Capacitor’s Pole Frequency


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4.28.49.4

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Zeros for Series-Shunt Feedback Amplifier Example

* Coupling capacitors C1, C2 and C5 give zeros at = 0 since ZC = 1/sC and they are in the signal line.

* Emitter bypass capacitors C3 and C4 give a zero when the impedance ZCE || RE.

* Feedback capacitor C6 gives a zero when [ZC6

+R2] || RC2 when

sradFKCR

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V

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101

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* Midband Gain Low Frequency Poles Low Frequency Zeros

Series-Shunt Example - Low Frequency

dBdBAA

A

dBdBAA

VfoVfo

Voff

VoVo

7.32)(1.43

4.101021.0

53)(449

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Low 3dB Frequency


* Substitute hybrid-pi model for transistor with C and C

* Short all coupling capacitors and emitter bypass capacitors

* Include loading effects of feedback network R1 and R2

* Find high frequency poles and zeros using Gray-Searle (Open Circuit) Method

Series-Shunt Example - High Frequency


Series-Shunt Example - High Frequency Pole - C1

sradxpFKC

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11

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11

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111

Given: C1 = 15 pF

I1

IS



Given: C1 = 1.2 pF

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1111

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1

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111

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221

1111

221

111

11111

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VR

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63.07.04.1910

822

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VR

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224

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221

22

222

222

2222

2


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C

g

sradxpF

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C

g

mZH

mZH

/101.54.1

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/100.32.1

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10

2

22

10

1

11

Series-Shunt Example - High Frequency Zeros - C1 & C2

For CE amplifier, a high frequency zero occurs when ZH = gm/C

2

20

222

2

2

222

22

2

0

222222

222222

22

2

222

22222

0

11

1

11

getweratio,voltageaasrewritingand

011

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0

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s

g

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V

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R

V

R

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VV

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R

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m

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m

m

C

m

Com

o

C

omo

oo

C

o

o

C

om

I2


* Midband Gain High Frequency Poles High Frequency Zeros

Series-Shunt Example - High Frequency

dBdBAA

A

dBdBAA

VfoVfo

Voff

VoVo

7.32)(1.43

4.101021.0

53)(449

sradxsradx

sradxsradx

PHPH

PHPH

/104.6/102.5

/103.1/100.7

64

72

83

71

sradsradx

sradsradx

ZHZH

ZHZH

//101.5

//100.3

410

3

210

1

sradxsradxAsradx VofPHPHfPH /107.64.10/104.61/104.6 766 High 3dB Frequency

ECE 352 Electronics II Winter 2003 Ch. 8 Feedback 1 Feedback *What is feedback?Taking a portion of...

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Transcript of ECE 352 Electronics II Winter 2003 Ch. 8 Feedback 1 Feedback *What is feedback?Taking a portion of...