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...
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.
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Ch. 8 Feedback 2ECE 352 Electronics II Winter 2003
Feedback Amplifier Analysis
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Ch. 8 Feedback 3ECE 352 Electronics II Winter 2003
* 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
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f
LLfHfHf
1
1
Ch. 8 Feedback 4ECE 352 Electronics II Winter 2003
* 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
Ch. 8 Feedback 5ECE 352 Electronics II Winter 2003
Basic Types of Feedback Amplifiers
Series - Shunt Shunt - Series
Series - Series Shunt - Shunt
Ch. 8 Feedback 6ECE 352 Electronics II Winter 2003
* 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
Ch. 8 Feedback 7ECE 352 Electronics II Winter 2003
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
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Basic Amplifier
Feedback Circuit
Equivalent Circuit for Feedback Amplifier
Ch. 8 Feedback 8ECE 352 Electronics II Winter 2003
Series-Shunt Feedback Amplifier - Ideal Case
Midband Gain
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Output Resistance
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R
VAVI
1
1
0
Ch. 8 Feedback 9ECE 352 Electronics II Winter 2003
Series-Shunt Feedback Amplifier - Ideal Case
of
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11
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Low Frequency Pole
High Frequency Pole
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A
AAthen
sA
AFor
11
11
1
1
1
11
1
11
Low 3dB frequency lowered by feedback.
Upper 3dB frequency raised by feedback.
Ch. 8 Feedback 10ECE 352 Electronics II Winter 2003
* 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?
Ch. 8 Feedback 11ECE 352 Electronics II Winter 2003
* 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
Ch. 8 Feedback 12ECE 352 Electronics II Winter 2003
* 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
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Vf
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oofVfiif
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VVf
AA
A
RRARR
A
AA
Ch. 8 Feedback 13ECE 352 Electronics II Winter 2003
Series-Shunt Feedback Amplifier - Practical Case
* 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
Ch. 8 Feedback 14ECE 352 Electronics II Winter 2003
* 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.
Series-Shunt Feedback Amplifier - Practical Case
Summary of Approach to Analysis
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Modified Basic Amplifier
Idealized Feedback Network
Practical Feedback Network
Basic Amplifier
Ch. 8 Feedback 15ECE 352 Electronics II Winter 2003
* 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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Kg
r
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m
T
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m
T
CmC
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4.1
,/36,94.0
2
22
222
1
11
111
DC analysis for each stage can be done separately since stages are isolated (dc wise) by coupling capacitors.
Ch. 8 Feedback 16ECE 352 Electronics II Winter 2003
Example - Series-Shunt Feedback Amplifier
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
+
_
Ch. 8 Feedback 17ECE 352 Electronics II Winter 2003
Example - Series-Shunt Feedback Amplifier
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
Ch. 8 Feedback 18ECE 352 Electronics II Winter 2003
* 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)
Example - Series-Shunt Feedback Amplifier
R1
R2
R1 R2
Ch. 8 Feedback 19ECE 352 Electronics II Winter 2003
Example - Series-Shunt Feedback AmplifierMidband Gain Analysis
dBdBA
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VA
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K
KKKK
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V
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VR
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2
1
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1
1
1
1
2
2
Ch. 8 Feedback 20ECE 352 Electronics II Winter 2003
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
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1.0
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1
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ff
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R
R
RRA
Ch. 8 Feedback 21ECE 352 Electronics II Winter 2003
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.
KKKK
RRRR iBSi
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KK
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RRR Co
4.28.49.4
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A
RR
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1
Ch. 8 Feedback 22ECE 352 Electronics II Winter 2003
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
A
A
V
VA
Vfo
Vof
Vo
fS
oVfo
7.321.43log20)(
1.431
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RR
KARR
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Vofiif
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dBdBA
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R
R
RR
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A
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ff
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1
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1
2
1
21
Ch. 8 Feedback 23ECE 352 Electronics II Winter 2003
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)
Ch. 8 Feedback 24ECE 352 Electronics II Winter 2003
Example - Input Coupling Capacitor’s Pole Frequency
Equivalent circuit for C1
sradFKCR
KKKKRRRR
so
KKKRr
I
RVgIrI
I
VR
where
RRRI
VR
xCPL
iBSxC
mii
iBSx
xxC
/2.1954.10
11
4.104.68.355
4.6098.0514.11
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
Ch. 8 Feedback 25ECE 352 Electronics II Winter 2003
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.
Ch. 8 Feedback 26ECE 352 Electronics II Winter 2003
Example - Feedback Coupling Capacitor’s Pole Frequency
Equivalent circuit for C6
sradFKCR
so
KKK
RRI
VR
xCPL
Cx
xxC
/6.2057.9
11
7.98.49.4
666
226
Ch. 8 Feedback 27ECE 352 Electronics II Winter 2003
sradFKCR
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rgRRRr
I
V
R
VIR
rg
RRr
R
VIR
rgR
VI
RRr
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soIIIEnodeAt
rgR
VI
I
soVgIIIEnodeAt
RIRRrIV
R
VI
I
VR
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xx
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Bs
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E
xx
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xE
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xE
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E
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1
0
113
1111111
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11
11
11
11
11
111
11111
111
11
11
11
1111
11111
11
Example - Emitter Bypass Capacitor’s Pole Frequency
Equivalent circuit for C3
IE1
Ch. 8 Feedback 28ECE 352 Electronics II Winter 2003
Example - Emitter Bypass Capacitor’s Pole Frequency
Equivalent circuit for C4
sradFKCR
KKK
KK
rg
RRrR
I
V
RRr
rg
RVI
becomesIso
RRr
VIsoRRrIV
rgIR
VrgIII
soVgIIIEnodeAt
R
VI
I
VR
ExEPL
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BCE
x
x
BC
m
Exx
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BC
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mE
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xE
x
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/5.28507.0
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4.19107.07.4
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thatknowalsoweBut
11
0
224
22
2122xC4
212
22
2
21222122
2222
2222
2222
22
IE2
Ch. 8 Feedback 29ECE 352 Electronics II Winter 2003
Example - Output Coupling Capacitor’s Pole Frequency
Equivalent circuit for C5
sradFKCR
so
KKK
RRRR
xCPL
ffCxC
/6.41104.2
11
4.28.49.4
555
2125
Ch. 8 Feedback 30ECE 352 Electronics II Winter 2003
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
sradFKCR
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CsR
R
sCRZR
RZ
EZ
EZ
EEEC
EE
E
EECE
EC
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E
E
/3.4507.4
11
/3.4507.4
11
1
11
1
11
1
424
313
226
622
1
622
6262
1
26
2262
26222
1
1
1
1
1
11
C
C
C
C
CCC
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RRsC
CsRR
CsRR
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RsC
RRZR
RZRgV
V
sradKKFRRC
RRCsorRRsC
Cz
CC
/6.209.48.45
11
101
2266
226226
Ch. 8 Feedback 31ECE 352 Electronics II Winter 2003
* 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
sradsrad
sradsrad
sradsrad
PLPL
PLPL
PLPL
/6.20/100
/6.41/7.18
/5.28/2.19
63
52
41
sradsrad
sradsrad
sradsrad
ZZ
PLZ
ZZ
/6.20/3.4
/0/0
/3.4/0
63
52
41
sradsrad
Asrad
Vof
PLPLfPL /6.9
4.10
/100
1/100
Low 3dB Frequency
Ch. 8 Feedback 32ECE 352 Electronics II Winter 2003
* 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
Ch. 8 Feedback 33ECE 352 Electronics II Winter 2003
Series-Shunt Example - High Frequency Pole - C1
sradxpFKC
KKK
KK
KKKK
KKKK
RrgRRr
RRRr
I
V
RIrgr
VRR
r
VIRIRRIV
Irgr
VIrgIIAlso
r
V
r
VI
r
VIIIIorIII
PH
mBS
BS
x
x
xmx
BSx
xBSsx
xmx
xm
xxxxssx
/100.71595.0
1
R
1
95.049.44.1
)49.4(4.1
)098.0(518.3554.1
098.08.3554.1
)1(R
gRearrangin
)1(
)1()1(
since
7
11xC1
11111
1111xC
1111
11
111
111
1111
11
11
111
Given: C1 = 15 pF
I1
IS
Ch. 8 Feedback 34ECE 352 Electronics II Winter 2003
Series-Shunt Example - High Frequency Pole - C1
Given: C1 = 1.2 pF
22111111
1111
221
1
11111
111
1
221
1111
221
111
11111
11111
11111111
11111
11
11111
1111111
11
01
usingso
0
1
1
1
1
1
rRRRRrgRr
RRrgIVso
rRR
V
RRrgRr
RRIrgI
I
rRR
VIrgI
rRR
VVgI
CNode
RRrgRr
rgRrRRI
rgRrIVVVV
and
RRrgRr
RRIIso
RRIIRRIVVAlso
rgRrIVV
BCSBm
SBmxo
BC
o
SBm
SBxmx
BC
omx
BC
omx
SBm
mSBx
meox
SBm
SBx
SBxSBSe
me
sradxpFKCR
K
KKKK
KKKKKKKKKKK
rgRRRr
rgRRRrrRRrgRrRR
I
V
VV
xCPH
mSB
mSBBCmSB
x
x
ox
/102.5)2.1(1.16
1
1
1
1.16
)51(098.058.354.1
)51(58.35098.04.17.04.1910)51(098.04.158.35
1
11
R
getweforaboveexpressionourwithcombiningSo
7
12
11111
1111122111111
1xC
1
Ch. 8 Feedback 35ECE 352 Electronics II Winter 2003
Given: C2 = 12 pF
Series-Shunt Example - High Frequency Pole - C2
sradx
pFKCR
KKKK
rRRI
VR
xCPH
BCx
xxC
/1032.1
1263.0
11
63.07.04.1910
822
3
2212
Ch. 8 Feedback 36ECE 352 Electronics II Winter 2003
Series-Shunt Example - High Frequency Pole - C2
Given: C2 = 1.4 pF
sradxpFKCR
K
KKVmAKKKKK
RRgrRRRRI
VR
getwecombiningso
rRRIVVAlso
rg
RRVgIV
so
IVVgRR
VIVg
RR
V
CatKCL
xCPH
CmBCCx
xxC
BCxxo
m
Cxmxo
xoxmC
oxm
C
o
/104.64.15.111
11
5.111
8.49.4/7117.04.19108.49.4
1
1
0
6
224
222221222
221
22
222
222
2222
2
Ch. 8 Feedback 37ECE 352 Electronics II Winter 2003
sradxpF
VmA
C
g
sradxpF
VmA
C
g
mZH
mZH
/101.54.1
/71
/100.32.1
/36
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
gRearrangin
0
ngsubstitutiso1
0
2
C
gswhenVSo
RRsC
Cg
s
g
RRsC
sCg
V
V
RRsCVsCgV
R
V
R
VVgVVsC
VVsC
sC
VV
Z
VVIBut
R
V
R
VVgICNode
m
C
m
m
C
m
Com
o
C
omo
oo
C
o
o
C
om
I2
Ch. 8 Feedback 38ECE 352 Electronics II Winter 2003
* 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