Ch. 7 Frequency Response Part 5 1ECES 352 Winter 2007
Comparison of Amplifier Configurations
Midband Characteristics*
• These are approximate expressions neglecting the effects of the biasing resistors R1 and R2 and the source resistance RS.
J. Millman and A. Grabel, Microelectronics, 2nd Ed., McGraw Hill, NY (1987), p. 420.
Ch. 7 Frequency Response Part 5 2ECES 352 Winter 2007
Characteristics of Amplifier Configurations
Current gain is large ( β) for CE and EF, but < 1 for CB.
Voltage gain is large for CE and CB, but < 1 for EF.
Input resistance is • Very small (few Ωs) for CB, • Medium (few KΩs) for CE, but • Very large (~ 10’s of KΩs) for EF.
Output resistance is • Very small (few Ωs) for EF, • Very large (~ 100’s of KΩs) for CE and CB.
Ch. 7 Frequency Response Part 5 3ECES 352 Winter 2007
Numerical Comparison of Amplifier Configurationsfor the Same Transistor and DC Biasing
• These are approximate expressions neglecting the effects of the biasing resistors R1 and R2 and the source resistance RS.
Ch. 7 Frequency Response Part 5 4ECES 352 Winter 2007
Comparison of CB to CE Amplifier (with same Rs = 5 Ω)
CE (with RS = 5 Ω) CB (with RS = 5Ω)
Midband Gain
Low Frequency Poles and Zeros
High Frequency Poles and Zeroes
dBdBA
VVA
rrr
rRRrR
RRgVV
VV
VV
A
Vo
Vo
xeEs
eECLm
s
e
e
oVo
2.40
/4.1025.094.0218
sradxpFKCRR
sradxpFCRRr
LCPH
sEePH
ZHZH
/101.73.105.1
11
/105.2174.211
,
82
101
21
sradFKCRR
sradxKF
rgrrRRC
sradKFCRRrgrrR
sradFKCR
CCLPL
mx
EsC
PL
BSEmxBPL
BBZPZPZP
/3332.10
11
/100.5010.02
1
1
1
/83112
11
1
/42122110
23
4
1
2
1
321
dBdBA
VVA
RrrRRrr
rrrRRg
VV
VV
VV
VVA
Vo
Vo
Bxs
Bx
xCLm
s
i
i
o
s
oVo
6.45)191log(20
/19193.094.0218
sradxFK
CRRrr
R
sradFKCRR
sradFKCrrRR
sradFKCR
EBsx
E
PL
CCLPL
CxBSPL
EEZPZPZP
/107.112005.0
1
1
1
/3332.10
11
/71427.0
11
/2521233.0
110
43
22
11
321
sradxpFK
CRRrrRR
gRR
sradxpFKCRRrr
sradxpF
VmACg
SBxLC
mLC
PH
SBxPH
mZHZH
/100.53.14.15
1
111
1
/100.917065.0
11
/106.13.1
/206,
7
2
81
1121
Ch. 7 Frequency Response Part 5 5ECES 352 Winter 2007
Comparison of EF to CE Amplifier (For RS = 5Ω )
CE EF
Midband Gain
Low Frequency Poles and Zeros
High Frequency Poles and Zeroes
dBdBA
VVA
RrrRRrr
rrrRRg
VV
VV
VV
VVA
Vo
Vo
Bxs
Bx
xCLm
s
i
i
o
s
oVo
6.45)191log(20
/19193.094.0218
sradxFK
CRRrr
R
sradFKCRR
sradFKCrrRR
sradFKCR
EBsx
E
PL
CCLPL
CxBSPL
EEZPZPZP
/107.112005.0
1
1
1
/3332.10
11
/71427.0
11
/2521233.0
110
43
22
11
321
sradxpFK
CRRrrRR
gRR
sradxpFKCRRrr
sradxpF
VmACg
SBxLC
mLC
PH
SBxPH
mZHZH
/100.53.14.15
1
111
1
/100.917065.0
11
/106.13.1
/206
7
2
81
1121
dBdBA
VVA
RrRRRrR
RrR
RRrRR
VV
VV
VV
VV
A
Vo
Vo
ixBS
ixB
ix
i
LE
EL
s
b
b
i
i
oVo
1.0
/987.0)998.0(999.0015.066
sradKFCrRR
sradKFRrRRC
CeELPL
ixBsCPL
ZPZP
/3793
11
/25695.12
110
22
11
21
sradxpFK
CRRrRRRRgr
sradxpFK
RRgCR
sradxpFKCr
rg
BSxLELEmPH
LEmxC
PH
mZHZH
/101.13.107.0
1
11
/100.126.0386.0
1
1
1
/102.11797.0
20111
10
2
10
'
1
1021
Ch. 7 Frequency Response Part 5 6ECES 352 Winter 2007
Comparison of Amplifier Configurations
Midband Gain and High and Low Frequency Performance
CE CB EF
Midband Voltage Gain -191 V/V +102 V/V +0.987 V/V45.6dB 40.2dB - 0.1dB
Low 3dB Frequency 1.7x104 rad/s 5.0x104 rad/s 2.6x102 rad/s
High 3dB Frequency 5.0x107 rad/s 7.1x108 rad/s 1.0x1010 rad/s
• Results for all three amplifiers with the smaller (5Ω) source resistance RS.
RS= 5 Ω
Ch. 7 Frequency Response Part 5 7ECES 352 Winter 2007
Cascade Amplifier
* Emitter Follower + Common Emitter (EF+CE)* Voltage gain from CE stage, gain of one for EF.* Low output resistance from EF provides a low source resistance for CE amplifier so
good matching of output of EF to input of CE amplifier* High frequency response (3dB frequency) for Cascade Amplifier is improved over CE
amplifier.
pFCCpFCC
rr xx
29.13
0100
21
21
21
21
EF CE
Ch. 7 Frequency Response Part 5 8ECES 352 Winter 2007
Cascade Amplifier - DC analysis
.
7.86.3)101(
7.087.31
87.3)3.4(899
8999.8)101(1
9.83.4110050
7.05
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12
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112
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111
1
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211
21
21
okayisanalysiseapproximatsoII
AKVVI
RIVVVKARIV
IandVcalculateNow
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ionapproximatfirstaasINeglectingRIIRIVV
QBaseKVL
KKKRRR
VVKKV
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EB
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BE
B
B
EBBThBBETh
Th
CCTh
KVmAg
r
VmA
VA
VI
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KVmAg
r
VmA
VA
VI
VIg
m
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B
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Cm
m
T
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9.2/0.34
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9.2/8.34
100
8.340256.0
)9.8(100
2
22
2222
1
11
1111
10021
Small Signal Parameters
IE1IB2
IRE1
IB1
Ch. 7 Frequency Response Part 5 9ECES 352 Winter 2007
Cascade Amplifier - Midband Gain Analysis
KKKK
rRrgr
IrRIrgrI
R
Em
Emi
178)9.23.4)(101(9.2
1
1
21111
1
2111111
91.0178)100100(4
178)100100(
)()(
016.0)9.23.4)(101(9.2
9.2)(1
8.60)9.23.4(1.351
68
21
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2111111
111
21111
21111
1
2
2
22
2
1
1
2
2
KKKK
KKK
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VV
KKKK
rRIrgrIrI
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VV
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VV
VV
VV
VVA
BS
i
s
i
Emi
EmEm
CLmo
s
i
i
o
s
oVo
dBdBA
VVA
Vo
Vo
6.35)2.60log(20
/2.6091.0016.0)8.60(68
Vπ2
+
_
+
_Vπ1+
Vi
_
Note: Voltage gain is nearly equal to that of the CE stage, e.g. – 68 !
Note: rx1 = rx2 = 0 so equivalent circuit is simplified.Iπ1
Ri
Ch. 7 Frequency Response Part 5 10ECES 352 Winter 2007
Cascade Amplifier - Low Frequency Poles and Zeroes
* Use Gray-Searle (Short Circuit) Technique to find the poles.
● Three low frequency poles● Equivalent resistance may
depend on rπ for both transistors.
* Find three low frequency zeroes.
sss
sss
sssssssF
LPLPLP
LZLZLZ
LPLPLP
LZLZLZL
321
321
321
321
111
111
)(
Ch. 7 Frequency Response Part 5 11ECES 352 Winter 2007
Cascade Amplifier - Analysis of Low Frequency Poles Gray-Searle (Short Circuit) Technique
Input coupling capacitor CC1 = 1 μF
sradxRC
xKFRC
KKKK
RRRR
KKKK
rRrgrIVR
rRrgrIrRVgIrIV
IVRRRR
IVR
xCCPL
xCC
iBsxC
Emi
i
EmEmi
iiiBs
x
xxC
/23sec103.4
11sec103.40.431
0.43178504
1789.23.4)101(9.2
1
1
211
1
211
1
21111
2111112111111
1
Ri
Vi
Iπ1
IX
RE1rπ2
rπ1 Vπ1
Vπ2
+
_
RE1 rπ2
Ch. 7 Frequency Response Part 5 12ECES 352 Winter 2007
Cascade Amplifier - Analysis of Low Frequency Poles Gray-Searle (Short Circuit) Technique
sradFKCR
KKKRRR
CCPL
CLC
/1251811
844
222
2
Vo
Vo
VX
Vπ2
rX2
RC RL
gm2Vπ2 RCRL
RE2 CE
CC2
* Output coupling capacitor CC2 = 1 μF
rπ2
Ch. 7 Frequency Response Part 5 13ECES 352 Winter 2007
Cascade Amplifier - Analysis of Low Frequency Poles Gray-Searle (Short Circuit) Technique
sradFKCR
KKKrRR
KKKK
r
KKKrgRr
rgIRrI
IVr
rgrRr
rgIrRrI
IVr
rRIVR
EExPL
eEEx
e
m
S
m
S
e
Ee
m
eE
m
eE
e
Ee
eEx
xEx
/73447029.0
11
029.0029.06.3
029.0101
065.03.49.2
065.0101
7.39.21
')1()'(
1)1()(
3
22
2
11
1
111
11
1
11
22
112
222
1122
2
22
22
sradPLPLPLPL /88273412523321
Emitter bypass capacitor CE = 47 μF
RE1
r π1 Vπ1
Ie1
Iπ1
Iπ2
VX
Ix
gm2Vπ2rπ2Vπ2
RE2
IE2
VE1
re1
Ie2
VE2
re2
Low 3 dB Frequency
The pole for CE is the largest and therefore themost important in determining the low 3 dB frequency.
gm1Vπ1
K
RRRR SS
7.3
' 21
Ch. 7 Frequency Response Part 5 14ECES 352 Winter 2007
* What are the zeros for the Cascade amplifier?
* For CC1 and CC2 , we get zeros at ω = 0 since ZC = 1 / jωC and these capacitors are in the signal line, i.e. ZC at ω = 0 so Vo 0.
* Consider RE in parallel with CE
* Impedance given by
* When Z’E , Iπ 0, so gmVπ 0, so Vo 0
* Z’E when s = - 1 / RE2CE so pole for CE is at
Cascade Amplifier - Low Frequency Zeros
sss
ssF
sss
s
sss
ssssssssssF
L
ZL
PLPLPL
ZLZLZL
PLPLPL
ZLZLZLL
73411251231
9.51)(
73411251231
10101
111
111)(
3
321
321
321
321
EE
EE
E
EEE
ECEE
CEE
CsRRZ
RCsRsC
RZRZ
ZRZ
E
E
2
2'
2
22'
2'
1
11111
sradFKCR EE
ZL /9.5476.311
23
Ch. 7 Frequency Response Part 5 15ECES 352 Winter 2007
Cascade Amplifier - High Frequency Poles and Zeroes
HPHPHPHP
HZHZHZHZH ssss
ssss
sF
4321
4321
1111
1111)(
* Use Gray-Searle (Open Circuit) Technique to find the poles.
● Four high frequency poles● Equivalent resistance may
depend on rπ for both transistors.* Find four high frequency zeroes.
High Frequency Equivalent Circuit
Ch. 7 Frequency Response Part 5 16ECES 352 Winter 2007
Cascade Amplifier - High Frequency Poles
KRRRR SS 7.3' 21
KK
KKK
KKKKK
rgrrR
rRrRR
IV
rRRIrgrrR
rRV
rRIrgrVR
rVIV
So
IrgIIVgIIandrVI
But
rRIRIIV
loopbasearoundKVL
mES
ESX
X
ESXmES
X
EXmX
SX
XX
XmXmeX
EeSXX
09.02.58
3.5
)101(9.2
9.26.39.27.31/9.26.37.3
1'1/'
'1'1
/IVfor solving and gRearrangin
0)(1'
1
0)(')(
111
21
121
21111
21
1
XX
211111
11111111
1
2111
VX
Ix
Iπ1
Ie1
Ix- Iπ1
+
_
Pole for Capacitor Cπ1 = 13.9 pF
sradxpFKPH /100.8
9.1309.01 8
1
Ch. 7 Frequency Response Part 5 17ECES 352 Winter 2007
Cascade Amplifier - High Frequency Poles
K
RRRR SS
7.3
' 21
K
KKKK
rRrgrR
rRrgrRrRrgrR
IV
rRrgrRVIV
IRVIIorRIIV
Also
rRrgrI
rRIrgrIVSo
rgIVgII
EmS
EmS
EmS
X
X
EmS
XXX
S
XXSXX
Em
EmX
mme
6.3
9.23.4)101(9.27.3
)(1'
)(1')(1'
)(1'
/IVfor solving and gRearranginand abovein for ngSubstituti
''
)(1
)(1
1
21111
21111
21111
21111
XX
1
11
211111
2111111
1111111
VX
Ix Iπ1
Ie1
+
_
Pole for Capacitor Cμ1 = 2 pF
sradxpFKPH /104.1
26.31 8
2
Ix- Iπ1
Ch. 7 Frequency Response Part 5 18ECES 352 Winter 2007
Cascade Amplifier - High Frequency Poles and Zeroes
KRRRR SS 7.3' 21 Simplified Equivalent Circuit
222
222212
152}2)/34(1{29.13
'1
CCpFKVmApFpF
RgCCCCC LmT
Using Miller’s Theorem, replace Cμ2 by two capacitors.
CLmCLmo RRg
VRRVg
VV
VVK 2
2
22
21
2
22
2222
222221
1111
1)1(
CRRg
CK
CC
CRRgCKCC
CLm
CLm
Ch. 7 Frequency Response Part 5 19ECES 352 Winter 2007
Cascade Amplifier - High Frequency PolesKRRRR SS 7.3' 21
KKKKrRrR
KKKrgRr
rgIRrI
IVr
EeX
m
S
m
S
e
ee
063.09.23.4065.0
065.0101
7.39.21
'1
'
211
11
1
111
11
1
11
Vπ1
IxIe1
+
_
Pole for Capacitor CT = 152 pF
sradxpFKPH /100.1
152063.01 8
3
VX
re1
Iπ1Ve1
Pole for Output Capacitor Cμ2 = 2 pF
sradxpFKCR
KKKR
RRRR
xCPH
xC
LCLxC
/105.22211
244
'
8
24
gm2Vπ2 VX+_
Ch. 7 Frequency Response Part 5 20ECES 352 Winter 2007
Cascade Amplifier - High Frequency Zeroes
sradxKpFrC
rgso
rCrgs
orrsCrgII
VsCVgII
mZH
m
me
me
/105.2)9.2(9.13
1011
1
010
9
11
113
11
11
111111
111111
Ie1
* When does Vo = 0?
* When ω → ∞, ZCμ1→ 0, so signal shorted to ground. ωZH1= ∞.
* When ω → ∞, ZCπ2→ 0, so rπ2 shorted, so Vπ2 = 0. ωZH2= ∞.
* For Cπ1 , we get a zero when Ie1 = 0.
Ie1
Ch. 7 Frequency Response Part 5 21ECES 352 Winter 2007
Cascade Amplifier - High Frequency Zeroes
sradxpF
VmACg
so
Cgs
or
VgVsCVVsCI
mZH
m
moC
/107.12
/34 10
2
24
2
2
2222222
I Cμ2
* When does Cμ2 produce a zero, i.e. make Vo = 0?
* For Cμ2 , we get a zero when IRL’ = 0, or ICμ2 = gm2Vπ2 , i.e. the output load resistance RL’ is starved of any current.
IRL’= 0
Zero for Output Capacitor Cμ2 = 2 pF
Ch. 7 Frequency Response Part 5 22ECES 352 Winter 2007
Cascade Amplifier - High Frequency Poles and Zeroes
8888
109
8888
109
4321
4321
104.11
105.21
1011
1081
107.11
105.21
)(
104.11
105.21
1011
1081
107.11
105.2111
1111
1111)(
xs
xs
xs
xs
xs
xs
sF
xs
xs
xs
xs
xs
xsss
ssss
ssss
sF
H
HPHPHPHP
HZHZHZHZH
Ch. 7 Frequency Response Part 5 23ECES 352 Winter 2007
Comparison of Cascade to CE Amplifier
CE* Cascade (EF+CE)
Midband Gain
Low Frequency Poles and Zeros
High Frequency Poles and Zeroes
dBdBA
VVAVV
VV
VV
VVA
Vo
Vo
S
i
i
oVo
6.352.60log20
/2.6091.0016.08.6068
1
1
2
2
sradxpFK
sradxpFK
sradxpFK
sradxpFK
sradxsradx
PH
PH
PH
PH
ZHZH
ZHZH
/105.2221
,/100.1152063.0
1
,/104.126.3
1
,/100.89.1309.0
1
/107.1,/105.2
,,
84
83
82
81
104
93
21
sradKFCrR
sradFKCRR
sradKFCrRrgrRR
sradFKCR
EeEPL
CCLPL
CEmBSPL
EEZPZPZP
/73403.047
1)(
1
/1251811
/234311
11
/9.5476.3110
223
22
1221111
2321
dBdBA
VVAVV
VV
VVA
Vo
Vo
s
o
s
oVo
5.29)30log(20
/3037.02.81
2
2
sradFK
CRRr
R
sradFKCRR
sradFKCrRR
sradFKCR
EBs
E
PL
CCLPL
CBSPL
EEZPZPZP
/59147036.0
1
1
1
/1251811
/15714.4
11
/9.5476.3110
22
3
22
121
2321
sradxpFK
CRRrRR
gRR
sradxpFKCRRr
sradxpF
VmACg
SBLC
mLC
PH
SBPH
mZHZH
/100.42125
1
111
1
/108.49.135.1
11
/100.22
/6.40,
6
222
2
7
221
10
2
221
* CE stage with same transistor, biasing resistors, source resistance and load as cascade.
25 X improvement in bandwidth !
2 X improvement in voltage gain !
Ch. 7 Frequency Response Part 5 24ECES 352 Winter 2007
Comparison of Cascade to CE Amplifier* Why the better voltage gain for the cascade?
● Emitter follower gives no voltage gain!● Cascade has better matching with source than CE.
Cascade amplifier has an input resistance that is higher due to EF first stage.
Versus Ri2 = rπ2 = 2.5 K for CE So less loss in voltage divider term (Vi / Vs )
with the source resistance. * 0.91 for cascade vs 0.37 for CE.
* Why better bandwidth?● Low output resistance re1 of EF stage gives smaller
effective source resistance for CE stage and higher frequency for dominant pole due to CT (including Cμ2)
KKKK
rRrR Ei
1789.23.4)101(9.2
1 2111
KKKKrRrR
KKKrgRr
rgIRrI
IVr
EeX
m
S
m
S
e
ee
063.09.23.4065.0
065.0101
7.39.21
'1
'
211
11
1
111
11
1
11
Pole for Capacitor CT = 152 pF
amplifier. CE for the /100.4 versus
cascade for the /100.1152063.0
1
6
83
sradx
sradxpFKPH
KKKRRR
pFKVmApFpFRgCCC
CLL
LmT
244'
152}2)/34(1{29.13'1 222
re1
Ri1
Ri2
Ch. 7 Frequency Response Part 5 25ECES 352 Winter 2007
Another Useful Amplifier – Cascode (CE+CB) Amplifier
* Common Emitter + Common Base (CE + CB) configuration
* Voltage gain from both stages* Low input resistance from second CB stage provides
first stage CE with low load resistance so Miller Effect multiplication of Cμ1 is much smaller.
* High frequency response dramatically improved (3 dB frequency increased).
● Bandwidth is much improved (~130 X).
sradxpFKK
pFpFpFKKVmApFpFRRgCCCso
rgrrR
bygivenisRwhereamplifierCBofRbecomesamplifierCEstagefirstforRamplifiercascodeFor
sradxx
pFpFpFKKVmApFpFRRgCCC
PHin
CLmin
m
xi
iiL
PHin
CLmin
/103.76.292.1
1
6.19)11(3.1172.1005.0/20613.1171
51
,
/106.5sec108.1
1
30221813.1172.19/20613.1171by limited is eperformanc frequency high theamplifier CEFor
8
111
68
1111
Bandwidth is improved by a factor of 130X over that for the CE amplifier !
Large Miller Effect
Small Miller Effect
Ch. 7 Frequency Response Part 5 26ECES 352 Winter 2007
Example of Cascode (CE +CB) Amplifier
http://www.freescale.com ECTW Conf. Proceedings 2003.
Ch. 7 Frequency Response Part 5 27ECES 352 Winter 2007
Other Examples of Multistage AmplifiersCE CE EF EF
Darlington Pair
Ch. 7 Frequency Response Part 5 28ECES 352 Winter 2007
Other Examples of Multistage AmplifiersPush – Pull Amplifier Amplifier with Npn and Pnp Transistors
Amplifier with FETs and Bipolar Transistors
Ch. 7 Frequency Response Part 5 29ECES 352 Winter 2007
Differential Amplifier* Similar to CE amplifier, but two CE’s operated
in parallel* Signal applied between two equivalent inputs
instead of between one input and ground* Common emitter resistor or current source used* Current shared or switched between two
transistors (they compete)* Analyze using equivalent half-circuit
● 1/2 of signal at input● 1/2 of signal at output● 1/2 of source resistance
* Gain and frequency response similar to CE amplifier for high frequencies
* Advantage: ● Rejects common noise pickup on input● No coupling capacitors so can operate down
to zero frequency.
Vo+ _
Ch. 7 Frequency Response Part 5 30ECES 352 Winter 2007
Differential Amplifier Analysis
dBdBAKKK
KKVmA
Rrr
rV
RVgVV
V
V
VV
V
VA
Vo
Sx
Cms
o
s
os
o
Vo
1.54509log20)(
5095.2065.097.0
97.09/206
22
2
2
2
Midband Gain
Low Frequency Poles and Zeros* Direct coupled so no coupling capacitors and no emitter bypass capacitor* No low frequency poles and zeros* Flat (frequency independent) gain down to zero frequency
High Frequency Poles and ZerosDominant pole using Miller’s Thoerem
sradxx
pFKpFpFKKK
CRgCRrr CmsxPH
/101.5sec1095.1
1
27870.01
3.1201172/5065.097.01
12/1
67
High frequency performance is very similar to CE amplifier.
Vo
Vo /2
Vo /2
Ch. 7 Frequency Response Part 5 31ECES 352 Winter 2007
Summary* In this chapter we have shown how to analyze the high and low frequency
dependence of the gain for an amplifier.● Analyzed the effects of the coupling capacitors on the low frequency response
Found the expressions for the corresponding poles and zeros. Demonstrated Bode plots of magnitude and phase.
● Analyzed the effects of the capacitances within the transistor on the high frequency response.
Found the expressions for the corresponding poles and zeros. Demonstrated Bode plots of the magnitude and phase.
* Analyzed the high and low frequency performance of the three bipolar transistor amplifiers: common emitter, common base and emitter follower.
● Found the expressions for the corresponding poles and zeros.● Demonstrated Bode plots of the magnitude and phase.
* Demonstrated how to find the expressions for the gain and the high and low frequency poles and zeros for multistage amplifiers.
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