1 FREQUENCY RESPONSE OF AMPLIFIERS * Effects of capacitances within transistors and in amplifiers * Build on previous analysis of amplifiers from EENG 341 Transistor DC biasing Small signal amplification, i.e. voltage and current gain Transistor small signal equivalent circuit * Use in Bipolar and FET Transistors Amplifiers and their analysis * Build on previous analysis of single time constant circuits Review simple RC, LC and RLC circuits Recall frequency dependent impedances for C and L Review frequency dependence in transfer functions * Magnitude and phase * Examine origins of frequency dependence in amplifier gain Identify capacitors and their origins; find the dominant C Determine equivalent R and determine RC time constant Use to describe approximately the amplifier’s frequency behavior Examine effects of other capacitors * GOAL: Use results of analysis to modify circuit design to improve performance.
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1
FREQUENCY RESPONSE OF AMPLIFIERS* Effects of capacitances within transistors and in amplifiers
* Build on previous analysis of amplifiers from EENG 341Transistor DC biasingSmall signal amplification, i.e. voltage and current gainTransistor small signal equivalent circuit
* Use in Bipolar and FET Transistors Amplifiers and their analysis* Build on previous analysis of single time constant circuits
Review simple RC, LC and RLC circuitsRecall frequency dependent impedances for C and LReview frequency dependence in transfer functions
* Magnitude and phase* Examine origins of frequency dependence in amplifier gain
Identify capacitors and their origins; find the dominant CDetermine equivalent R and determine RC time constantUse to describe approximately the amplifier’s frequency behaviorExamine effects of other capacitors
* GOAL: Use results of analysis to modify circuit design to improve performance.
2
Analysis of Amplifier Performance
DC bias or quiescent point
iB iC
vBE vCE
* Previously analyzed DC bias pointAC analysis (midband gain)
* Neglected all capacitances in the transistor and circuit
* Gain at middle frequencies, i.e. not too high or too low in frequency
3
Frequency Response of Amplifiers* In reality, all amplifiers have a limited
range of frequencies of operationCalled the bandwidth of the amplifierFalloff at low frequencies
* At ~ 100 Hz to a few kHz* Due to coupling capacitors at the
input or output, e.g. CC1 or CC2 Falloff at high frequencies
* At ~ 100’s MHz or few GHz* Due to capacitances within the
transistors themselves.
Equivalent circuit for bipolar transistor
20 logT(ω)
ω
Midband Gain
4
Frequency Response of Amplifiers
* First approximation – describe the amplifier’s high and low frequency responses in terms of that of single time constant (STC) circuits
High frequency falloff –Like that of a low pass filter
* Simple RC equivalent circuit * Shunting capacitor shorts signal at the
output at high frequenciesLow frequency falloff
Like that of a high pass filter* Simple RC equivalent circuit* Series capacitor blocks output signal at
low frequencies (acts like open circuit)* Amplifier frequency analysis
Determine equivalent R for each CCompare and find the most important (dominant RC) combination
Find the dominant one (RC)at high frequencies
Find the dominant one (RC)at low frequencies
Low-Pass Network
High-Pass Network
( )⎩⎨⎧
→∞∞→
→==0 (open)
011ωω
ω asasshort
CjsCZC
Vi
Vi
5
Review of Complex Numbers* Complex numbers
General form a + bj wherea = real part, b = imaginary part and
Magnitude of complex numberPhase of complex numberPhasor form
* Complex number mathMultiplication of two complex numbers a + bj = Me jθ and c + dj =Ne jφ
Reciprocal of a complex number
1−=j
θθ
jj e
MMebja
ORba
bjba
ababja
bjabja
bjabja
−==+
+−
+=
+−
=−−
+=
+
111
11222222
22 babjaM +=+=
⎟⎠⎞
⎜⎝⎛= −
ab1tanθ
θjMebja =+
( )( )
( )( ) ( )( ) ( )ϕθϕθ +==++
++−=++
jjj MNeNeMedjcbjaOR
adbcjbdacdjcbja )()(
6
Amplifier Transfer Function (Gain) - General Form* A (s) = Gain Function (general form of amplifier transfer function)
AM = midband gain (independent of frequency)FH(s) = high frequency function (acts like low pass filter)
FL(s) = low frequency function (acts like high pass filter)
HH s
sFω/1
1)(+
=
)()()( sFsFAsA LHM=
ssF
LL /1
1)(ω+
=
ωHωL
Magnitude
FH(s)FL(s)
AM
7
Amplifier Transfer Function (Gain) - General Form
)()()( sFsFAsA LHM=
Coupling Capacitors
Transistor’s Capacitors
ωHωL
FH(s)FL(s)
AM
8
Frequency Response of MOSFET vs BJT AmplifiersEquivalent circuit for bipolar transistor Equivalent circuit for MOSFET
Similar equivalent circuits
Corresponding amplifier circuits
Similar frequency performance
Common Source Amplifier
Common Emitter Amplifier
Gain Gain
9
Amplifier Transfer Function (Gain) - General Form
)()()( sFsFAsA LHM=
Coupling Capacitors
Transistor’s Capacitors
ωHωL
Now we consider the low frequency behavior.
10
Summary* Examined origin of falloff in amplifier gain at low and high frequencies.
Degradation in magnitude of the gain.Shift in phase of output relative to input.
* Due to presence of capacitors within the amplifier (Create poles and zeros).Coupling capacitors limit gain at low frequencies.Transistor’s capacitances limit gain at high frequencies.
* Examined and quantified the falloff due to single and multiple poles and zeroes. Bode plots of gain and phase shift with frequency.
* Next, we will apply this method of analysis to transistor amplifiers.Multiple capacitors so multiple RC combinations.Investigate how to determine which capacitors are most important and limit the bandwidth.Examine how to change the amplifier to get better frequency performance.
Higher frequency operation before falloff (improved bandwidth).Better low frequency behavior.
11
Analysis of Bipolar Transistor Amplifiers* Single stage amplifiers
Common Emitter (CE)Common Base (CB)Emitter Follower (EF) (Common Collector)
* DC biasingCalculate IC, IB, VCEDetermine related small signal equivalent circuit parameters
Transconductance gmInput resistance rπ
* Low frequency analysisGray-Searle (Short Circuit) Technique
Determine the pole frequencies ωPL1, ωPL2, ... ωPLnDetermine the zero frequencies ωZL1, ωZL2, ... ωZLn
* High frequency analysisGray-Searle (Open Circuit) Technique
Determine the pole frequencies ωPH1, ωPH2, ... ωPHnDetermine the zero frequencies ωZH1, ωZH2, ... ωZHn
12
CE Amplifier Frequency Analysis - Long and Difficult Way* DC analysis: IC , IB , VCE ; gm , rπ* Draw ac equivalent circuit* Substitute hybrid-pi model for
transistor* Obtain KVL equations (at least one
for each capacitor in circuit (5))* Solve set of 5 simultaneous equations
to obtain voltage gain AV = Vo /Vs* Put expression in standard form for
gain AV(ω) = AVo FH(ω) FL(ω)* Identify midband gain AVo* Determine FH(ω) part and factor to
Determine high frequency poles ωPH1 and ωPH2Determine high frequency zeros ωZH1 and ωZH2
* Determine FL(ω) part and factor toDetermine low frequency poles ωPL1, ωPL2 and ωPL3Determine high frequency zeros ωZL1, ωZL2 and ωZL3
* Is there an easier way ? YES⎟⎟⎠
⎞⎜⎜⎝
⎛+⎟⎟
⎠
⎞⎜⎜⎝
⎛+
⎟⎟⎠
⎞⎜⎜⎝
⎛+⎟⎟
⎠
⎞⎜⎜⎝
⎛+
=
21
21
11
11)(
PP
ZZH ss
ss
sF
ωω
ωω
( )( )( )( )( )( )
⎟⎠⎞
⎜⎝⎛ +⎟⎠⎞
⎜⎝⎛ +⎟⎠⎞
⎜⎝⎛ +
⎟⎠⎞
⎜⎝⎛ +⎟⎠⎞
⎜⎝⎛ +⎟⎠⎞
⎜⎝⎛ +
=
++++++
=
sss
sss
sssssssF
PPP
ZZZ
PPP
ZZZL
321
321
321
311
111
111
)(
ωωω
ωωωωωωωωω
13
CE Amplifier - Starting Point is DC Analysis
* Q is quiescent point (DC bias point)* Q needs to be in the active region
IC = β IB
* If Q is in saturation (VCE < 0.3 V) , then IC < β IB and there is little or no gain from the transistor amplifier
* If the transistor is in the cutoff mode, there is virtually no IC so there is no gain, i.e. gm 0.
* Q depends on the choice of R1 and R2 since they determine the size of IB.
* Q point determines the size of the small signal parameters
* Check on transistor region of operation (Find VCE)KVL collector loopVCE = VCC - IC RC - (β +1) IB RE = 4.4 V
(okay since not close to zero volts, i.e. > 0.2V).
R1 = 10KR2 = 2.5KRC = 1.2KRE = 0.33K
15
* Construct amplifier’s small signal ac equivalent circuit (set DC supply to ground)
* Substitute small signal equivalent circuit (hybrid-pi model) for transistor
* Neglect all capacitancesCoupling and emitter bypass capacitors become shorts at midband frequencies (~ 105 rad/s)
Why? Their impedances are negligibly small, e.g. few ohms because CC1, CC2, CE are large,
e.g.~ few µF (10-6F)
Transistor capacitances become open circuits at midband frequencies
Why? Their impedances are very large, e.g. ~ 10’s M Ω because Cπ , Cµ are very small, e.g. ~ pF(10-12 F)
* Calculate small signal voltage gain AVo = Vo /Vs
CE Amplifier - Midband Gain Analysis
Ω== 10)1)(/10(
1~15 FsradC
ZC µω
Ω== 75 10
)1)(/10(1~1
pFsradCZC ω
Hybrid-Pi Model for BJT
16
CE Amplifier - Midband Gain Analysis
( ) ( ) ( )( )
( )[ ]( )[ ]
( )[ ]( )[ ]
( )( )( )( ) dBdBA
VVA
KK
KKKKKKK
RrrRRrr
VV
KKK
rrr
VV
KKVmARRgV
RRVgVV
VV
VV
VV
VVA
Vo
Vo
Bxs
Bx
s
i
xi
CLmCLmo
s
i
i
o
s
oVo
7.27)6.24log(20
/6.2412.094.0218
12.068.568.0
0.2065.097.050.2065.097.0
94.0065.097.0
97.0
21892.1/206
=−=
−=−=
==++
+=
++
+=
=+
=+
=
−=−=−=−
=
==
π
π
π
ππ
π
π
π
π
π
KRKR
KRKR
KRKR
S
E
C
L
55.2
1033.02.1
9
2
1
======
RL||RC
+
-Vi
Io
+
-
Vi
Break voltage gain into a series of voltage ratios
Vs
Vorx
rπ gmVπ
Negative sign means output signal is 180o out of phase with the input signal.
Vπ
17
Analysis of Low Frequency Poles Gray-Searle (Short Circuit) Technique
* Draw AC equivalent circuit at low frequency Include coupling and emitter bypass capacitors CC1,CC2, CE
Substitute AC equivalent circuit for transistor (hybrid-pi for bipolar transistor) Ignore (remove) all transistor capacitances Cπ , Cµ
* Turn off signal source, i.e. set Vs= 0Keep source resistance RS in circuit (do not remove)
* Analyze the circuit one capacitor Cx at a time Replace all other capacitors with short circuitsSolve remaining circuit for equivalent resistance Rx seen by the selected capacitorCalculate pole frequency usingRepeat process for each capacitor finding equivalent resistance seen by it and find corresponding pole frequency
* Determine which is dominant (largest) low frequency pole* Calculate the final, low 3dB frequency using
xxPx CR
1=ω
∑ ∑=+==xx
PnPPPxLP CR1...21 ωωωωω
18
Common Emitter - Analysis of Low Frequency PolesGray-Searle (Short Circuit) Technique
Common Emitter - Analysis of Low Frequency PolesGray-Searle (Short Circuit) Technique
( ) sradFKCR
KKKKK
K
RRrrRR
RRrrrg
R
RRrrrg
RRrrRVI
R
RRrrVrg
RRrrV
RVI
RRrrVrrIV
RRrrVI
RVI
VgIIIIVR
EExPL
BsxEEx
Bsx
m
E
Bsx
m
BsxEx
x
Ex
Bsx
xm
Bsx
x
E
xx
Bsx
x
Bsx
x
E
xe
mexx
xEx
/342,512016.0
11
016.0201
2597.0065.033.0
1
11
111
?
3 ===
=⎟⎟⎠
⎞⎜⎜⎝
⎛ ++=
⎟⎟⎠
⎞⎜⎜⎝
⎛+++
=
+++
+=
+++
+++==
+++
+++=
++−
==
++−
==
−−===
µω
βπ
π
π
π
π
π
π
π
π
π
ππππ
ππ
ππ
sradPLPLPLPL /546353423388321 =++=++= ωωωω
rX
rπ Vπ gmVπ
VX
RS RB
RB
RSrX
rπ Vπ gmVπ RC RL
RE
CC1
CE
CC2
REIe
Ix
Iπ
Emitter bypass capacitor CE = 12 µF
+
Vo
VS
REx
KCL at node E
Recall β = gmrπ
Low 3db frequency
20
Analysis of High Frequency Poles Gray-Searle (Open Circuit) Technique
* Draw AC equivalent circuit at high frequency Substitute AC equivalent circuit for transistor (hybrid-pi model for transistor with Cπ, Cµ )Remove coupling and emitter bypass capacitors CC1, CC2, CE (consider as shorts).Turn off signal source, i.e. set Vs = 0Keep source resistance RS in circuit Neglect transistor’s output resistance ro
* Consider the circuit one capacitor Cx at a time Replace all other transistor capacitors with open circuitsSolve remaining circuit for the equivalent resistance Rx seen by the selected capacitorCalculate the pole frequency usingRepeat process for each capacitor
* Calculate the final high 3dB frequency using
xxPHx CR
1=ω
∑∑ =⎥⎦
⎤⎢⎣
⎡++=⎥
⎦
⎤⎢⎣
⎡=
−−
xxPHnPHPHPxPH
CR
11...1111
21
1
ωωωωω
Cπ comes from Emitter-Base p-n junction.
Cµ comes from Base-Collector p-n junction.
21
Common Emitter - Analysis of High Frequency PolesGray-Searle (Open Circuit) Technique
AC equivalent circuit at High frequency
CE shorts REat high frequencies.
Vo
VS
VoVS Vπ
22
Common Emitter - Analysis of High Frequency PolesGray-Searle (Open Circuit) Technique
* Equivalent circuit for Capacitor Cπ = 17 pF
( ) ( )
( )
sradxxCR
xpFKCR
KKKKKRRrrIVR
xPH
x
SBxX
Xx
/100.1sec100.1
11sec100.11759.0
59.052065.097.0
881
8
===
==
=+=+==
−
−
ππ
ππ
ππ
ω
VoVS Vπ
VX
IX
Note: This frequency is very high due to the very small size of the capacitor.
23
Common Emitter - Analysis of High Frequency PolesGray-Searle (Open Circuit) Technique
* Equivalent circuit for Capacitor Cµ = 1.3 pF
( ) ( )( )
( ) ( )( )
( )
sradxxxCRCR
isfrequencydBfrequencyhighFinal
sradxxCR
xpFKCRK
KKKKKK
VmAKK
RRrrRR
gRRR
getweIVRforsolvingandVforngSubstituti
RRV
RRgV
RVV
RVVVgI
VVVorVVVRV
RVVgICnodeAt
RRrrIVso
RRrV
rVIBnodeAt
xxPH
xPH
x
SBxLC
mLCx
x
xx
LCx
LCm
L
x
C
xmx
xoox
L
o
C
omx
SBxx
SBxx
/106.5sec107.1100.1
11
3
/109.5sec107.1
11sec107.13.1130
130
52065.097.091
2.11/206192.1
111
1111
0Using
11'
678
672
7
=+
=+
=
===
==
=
⎥⎦
⎤⎢⎣
⎡+⎟
⎠⎞
⎜⎝⎛ +++=
⎥⎦
⎤⎢⎣
⎡+⎟⎟
⎠
⎞⎜⎜⎝
⎛+++=
=
⎥⎦
⎤⎢⎣
⎡++⎥
⎦
⎤⎢⎣
⎡++−=
−−
−−−=
−==++−
−−−=
⎥⎦
⎤⎢⎣
⎡
++=
++=
−−
−
−
µµππ
µµ
µµ
πµ
µπ
πππ
π
ππ
π
ππ
π
π
π
ω
ω
B/
Need to solve forX
XX I
VR =µ
VXIX
Vπ
Vπ
C
24
Common-Base (CB) Amplifier* DC biasing
Calculate IC, IB, VCEDetermine related small signal equivalent circuit parameters
Transconductance gmInput resistance rπ
* Midband gain analysis* Low frequency analysis
Gray-Searle (Short Circuit) Technique
Determine pole frequencies ωPL1, ωPL2, ... ωPLn
Determine zero frequencies ωZL1, ωZL2, ... ωZLn
* High frequency analysisGray-Searle (Open Circuit) Technique
* Check on transistor region of operationKVL collector loopVCE = VCC - IC RC - (β +1) IB RE = 4.4 V
(okay since not close to zero volts).
R1 = 10KR2 = 2.5KRC = 1.2KRE = 0.33K
26
* Construct small signal ac equivalent circuit (set DC supply to ground)
* Substitute small signal equivalent circuit (hybrid-pi model) for transistor
* Neglect all capacitancesCoupling and emitter bypass capacitors become shorts at midband frequencies (~ 105 rad/s)
Why? Impedances are negligibly small, e.g. few ohms because CC1, CC2, CE ~ few µF (10-6F)
Transistor capacitances become open circuits at midband frequencies
Why? Impedances are very large, e.g. ~ 10’s M Ωbecause Cπ , Cµ ~ pF (10-12 F)
* Calculate small signal voltage gain AVo = Vo /Vs
CB Amplifier - Midband Gain Analysis
Ω=== 10)1)(/10(
115 FsradC
ZC µω
Ω=== 75 10
)1)(/10(11
pFsradCZC ω
High and Low Frequency AC Equivalent Circuit
27
CB Amplifier - Midband Gain Analysis
( ) ( ) ( )( )
[ ][ ]
[ ][ ]
( )( )( )( ) dBdBA
VVAKK
KKKKK
rRRrR
VV
KKK
rrIIr
VV
KKVmARRgV
RRVgVV
VV
VV
VV
VVA
Vo
Vo
eEs
eE
s
e
xe
CLmCLmo
s
e
e
o
s
oVo
14)20.0log(20
/20.0001.094.0218
001.00050.50050.0
0051.033.050051.033.0
94.0065.097.0
97.0)(
21892.1/206
−==
=−−=
==+
=+
=
−=+
−=
+−=
−=−=−=−
=
==
ππ
πππ
π
π
π
π
π
KRKR
KRKR
KRKR
S
E
C
L
55.2
1033.02.1
9
2
1
======
Equivalent resistance re
KKKrrrg
rrrg
rr
rrrg
rVV
IVr
rrgV
rgVI
rVVgI
EnodeatKCLIVr
x
m
x
m
x
m
e
e
ee
mme
me
e
ee
0051.02001
97.0065.011
11
11
0
=++
=++
=++
=
+⎟⎟⎠
⎞⎜⎜⎝
⎛ +−−=
+−==
⎟⎟⎠
⎞⎜⎜⎝
⎛ +−=⎟⎟
⎠
⎞⎜⎜⎝
⎛+−=
=++
=
βπ
π
π
π
π
π
π
π
π
π
π
ππ
ππ
π
ππ
Ve
+_
Iπ
π
ππ
ππ
ππ
rrr
VV
rrV
rVI
xe
x
e
+−
=
+−
==
re
βIπ
Voltage gain is less than one !
28
What Happened to the CB Amplifier’s Midband Gain?
* Source resistance Rs = 5K is killing the gain.
Why? Rs >> re = 0.0051 Kso Ve/Vs<<1
* Need to use a different signal source with a very low source resistance Rs , i.e. ~ few ohms
* Why is re so low?Vs drives formation of Ve
Ve creates Vπ across rπVπ turns on dependent current sourceGet large Ie for small Veso re =Ve/Ie is very small.
( )( )( )[ ][ ]
[ ][ ]
[ ][ ]
[ ][ ]
( )( )( )dBdBA
VVAand
KKK
KKKKK
rRRrR
VV
RFor
ceresislowwithsourcesignalNew
KK
KKKKK
rRRrR
VV
VVA
Vo
Vo
eEs
eE
s
e
s
eEs
eE
s
e
Vo
2.40)5.102log(20)(
/5.1025.094.0218
5.0005.0005.0
0050.00051.033.0005.0
0051.033.0
5
tan
001.00050.50050.0
0051.033.050051.033.0
/20.0001.094.0218
==
=−−=
=+
=+
=+
=
Ω=
==+
=+
=
=−−=
Ve
+_
re
Voltage gain is now much bigger than one !
29
Analysis of Low Frequency Poles Gray-Searle (Short Circuit) Technique
* Draw low frequency AC circuitSubstitute AC equivalent circuit for transistor (hybrid-pi for bipolar transistor) Include coupling and base capacitors CC1, CC2, CB
Ignore (remove) all transistor capacitances Cπ , Cµ
* Turn off signal source, i.e. set Vs= 0Keep source resistance RS in circuit (do not remove)
* Consider the circuit one capacitor Cx at a timeReplace all other capacitors with short circuitsSolve remaining circuit for equivalent resistance Rx seen by the selected capacitorCalculate pole frequency usingRepeat process for each capacitor finding equivalent resistance seen and the corresponding pole frequency
* Determine the dominant (largest) pole frequency* Calculate the final low pole frequency using
xxPx CR
1=ω
∑ ∑=+==xx
PnPPPxLP CR1...21 ωωωωω
30
Common Base - Analysis of Low Frequency PolesGray-Searle (Short Circuit) Technique
* Base capacitor CB = 12 µF
( )
( )( ) ( )
( )( )( )
( )( )[ ]( )( )[ ]
( )
sradxCR
xKFRC
KKK
KKKKK
RRrgrrRR
RRrgrV
RRgr
VrR
RRgr
VRRVgIVV
rVIce
VVr
rVV
IVR
RrRR
BxCPL
xCB
SEmxBxC
SEm
SEm
i
SEmSEmi
iiii
ixBxC
B
B
B
B
/83102.111
sec102.10.112
0.103.22
005.033.020197.0065.02
1
1
11
11
sin/
21
2
===
==
==
++=
+++=
++=⎥⎦
⎤⎢⎣
⎡⎟⎟⎠
⎞⎜⎜⎝
⎛++
=
⎥⎦
⎤⎢⎣
⎡⎟⎟⎠
⎞⎜⎜⎝
⎛++=++=
====
+=
−
−
ω
µ
ππ
πππ
ππ
π
πππππ
π
ππ
ππ
πππ
Vi
+
_RxCB
Ri
Iπ
Low Frequency AC Equivalent Circuit
Vπ
Vx
Vo
Ix
31
Common Base - Analysis of Low Frequency PolesGray-Searle (Short Circuit) Technique
Input coupling capacitor CC1 = 2 µF
( )( )
( )( )
( )( )
( )
sradxxRC
xKFRC
KKK
KKKK
rgrrRRR
rgrr
IrgrrI
IVr
rrIVIrgVgII
IVrrRR
IVR
C
C
C
C
xCCPL
xCC
m
xEsxC
m
x
m
x
e
ee
xe
mme
e
eeeEs
x
xxC
/100.5sec100.2
11
sec100.2010.02
010.00051.0005.0
20197.0065.033.0005.0
1
11
1
451
2
51
1
1
1
1
===
==
=+=
⎥⎦
⎤⎢⎣
⎡ ++=
⎥⎦
⎤⎢⎣
⎡
++
+=
++
=+−
+−==
+−=+−=−−=
=+==
−
−
ω
µ
π
π
π
π
ππ
ππ
ππ
ππππ
Ve
Ie
Iπ
Ve
+_
re
Vπ
Vo
Ix Rs
Vx
32
Common Base - Analysis of Low Frequency PolesGray-Searle (Short Circuit) Technique
srad
PLPLPLPL
/116,5033000,5083
321
=++=
++= ωωωω
* Output coupling capacitor CC2 = 3 µF
( ) sradFKCR
KKKRRR
CCPL
CLC
/3332.10
112.102.19
223
2
===
=+=+=
µω
* Low 3dB frequency
Vo
RL
VX
RC
Dominant low frequency pole is due to CC1 !
33
Analysis of High Frequency Poles Gray-Searle (Open Circuit) Technique
* Draw high frequency AC equivalent circuitSubstitute AC equivalent circuit for transistor (hybrid-pi model for transistor with Cπ, Cµ) Consider coupling and emitter bypass capacitors CC1, CC2, CB as shortsTurn off signal source, i.e. set Vs = 0Keep source resistance RS in circuit Neglect transistor’s output resistance ro
* Consider the circuit one capacitor Cx at a time Replace all other transistor capacitors with open circuitsSolve remaining circuit for equivalent resistance Rx seen by the selected capacitorCalculate pole frequency usingRepeat process for each capacitor
* Calculate the final high frequency pole using
xxPHx CR
1=ω
∑∑ =⎥⎦
⎤⎢⎣
⎡++=⎥
⎦
⎤⎢⎣
⎡=
−−
xxPHnPHPHPxPH
CR
11...1111
21
1
ωωωωω
34
Common Base - Analysis of High Frequency PolesGray-Searle (Open Circuit) Technique
High frequency AC equivalent circuit
NOTE: We neglect rx here since the base is grounded. This simplifies our analysis,but doesn’t change theresults appreciably.
35
Common Base - Analysis of High Frequency PolesGray-Searle (Open Circuit) Technique
* Equivalent circuit for Ze
βππ
ππ
π
ππ
π
ππ
ππ
π
π
ππ
π
ππ
π
π
π
π
π
+=
+=
=
⎟⎠⎞⎜
⎝⎛
⎥⎦
⎤⎢⎣
⎡+
=
⎥⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢⎢
⎣
⎡
⎟⎠⎞⎜
⎝⎛
+
⎟⎟⎠
⎞⎜⎜⎝
⎛+
=
⎥⎦
⎤⎢⎣
⎡+
+=
⎥⎦
⎤⎢⎣
⎡+
+==
⎥⎦
⎤⎢⎣
⎡+
+=
⎥⎦
⎤⎢⎣
⎡++=−
⎟⎠⎞⎜
⎝⎛
+=
−==
11
11
11
1
1
1
11
1
1
11
rrg
rr
where
ZrZSo
sCrgr
sCrgr
sCr
rgsCr
rgV
VIVZ
sCr
rgV
gsCr
VVg
sC
VrVI
givesEnodeatKCLVVIVZ
me
Cee
m
m
mme
e
e
ee
me
memee
e
ee
ee
Ze
Ze
Replace thiswith this.
Ve
+_
Parallel combination of a resistor and capacitor.
36
* Pole frequency for Cπ =17pF
Common Base - Analysis of High Frequency PolesGray-Searle (Open Circuit) Technique
( ) sradxsxpFCR
KKKKR
KKrg
rr
RRrR
xCPH
xC
me
sEexC
/105.2101.4
1174.211
4.20024.0005.033.00048.0
8.40048.02001
97.01
10111 ==
Ω==
Ω===
Ω==+
=+
=
=
−π
π
π
π
π
π
ω
Turn off signal source when finding resistance seen by capacitor.
37
Common Base - Analysis of High Frequency PolesGray-Searle (Open Circuit) Technique
* Equivalent circuit for Capacitor Cµ = 1.3 pF
* Pole frequency for Cµ =1.3pF
( ) sradxsxpFKCR
KKKR
RRR
xCPH
xC
LCxC
/101.7104.1
13.105.1
11
05.192.1
892 ====
==
=
−µµ
µ
µ
ω
sradxsx
xxCRCR
PH
xCxCPH
/109.61044.11
104.1101.411
89
911
==
+=
+=
−
−−
ω
ωµπ µπ
* High 3 dB frequency
Dominant high frequency pole is due to Cµ !
Rs || RE || r π
= 0
38
Comparison of CB to CE AmplifierCE (with RS = 5K) CB (with RS = 5Ω)
Midband Gain
Low Frequency Poles and Zeros
High Frequency Poles and Zeroes
( ) [ ][ ]
( )( )( )( ) dBdBA
VVA
rrr
rRRrR
RRgVV
VV
VVA
Vo
Vo
xeEs
eECLm
s
e
e
oVo
2.40
/4.1025.094.0218
=
+=−−=
⎟⎟⎠
⎞⎜⎜⎝
⎛
+−
+−==
π
ππ
π
[ ] ( )
( ) ( ) sradxpFKCRR
sradxpFCRRr
LCPH
sEePH
ZHZH
/101.73.105.1
11
/105.2174.211
,
82
101
21
===
=Ω
==
∞=∞=
µ
π
ω
ω
ωω
( )
( )( )[ ] ( )
( )
( ) ( ) 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
==+
=
==
⎪⎭
⎪⎬⎫
⎪⎩
⎪⎨⎧
⎥⎦
⎤⎢⎣
⎡++
+
=
==+++
=
=====
µω
µω
µω
µωωω
ππ
ππ
( )[ ] ( )[ ]( )[ ]
( )( )( )( ) dBdBA
VVA
RrrRRrr
rrrRRg
VV
VV
VV
VVA
Vo
Vo
Bxs
Bx
xCLm
s
i
i
o
s
oVo
7.27)6.24log(20
/6.2412.094.0218
=−=
−=−=
⎥⎥⎦
⎤
⎢⎢⎣
⎡
++
+⎥⎦
⎤⎢⎣
⎡
+−===
π
π
π
ππ
π
( )
( )[ ] ( )
( ) ( )
( ) sradFK
CRRrr
R
sradFKCRR
sradFKCrrRR
sradFKCR
EBsx
E
PL
CCLPL
CxBSPL
EEZPZPZP
/342,512016.0
1
1
1
/3332.10
11
/8827.5
11
/2521233.0
110
3
22
11
321
==
⎥⎥
⎦
⎤
⎢⎢
⎣
⎡
⎟⎟⎠
⎞⎜⎜⎝
⎛
+++
=
==+
=
==++
=
=====
µ
β
ω
µω
µω
µωωω
π
π
( )[ ] ( )
( ) ( )( )
( ) sradxpFK
CRRrrRR
gRR
sradxpFKCRRrr
sradxpF
VmACg
SBxLC
mLC
PH
SBxPH
mZHZH
/109.53.1130
1
111
1
/100.11759.0
11
/106.13.1
/206,
6
2
81
1121
==
⎪⎭
⎪⎬⎫
⎪⎩
⎪⎨⎧
⎥⎦
⎤⎢⎣
⎡+⎟⎟
⎠
⎞⎜⎜⎝
⎛+++
=
==+
=
===∞=
µπ
ππ
µ
ω
ω
ωω
Note: CB amplifier has much better high frequency performance!
39
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
VVA
Vo
Vo
xeEs
eECLm
s
e
e
oVo
2.40
/4.1025.094.0218
=
+=−−=
⎟⎟⎠
⎞⎜⎜⎝
⎛
+−
+−==
π
ππ
π
[ ] ( )
( ) ( ) sradxpFKCRR
sradxpFCRRr
LCPH
sEePH
ZHZH
/101.73.105.1
11
/105.2174.211
,
82
101
21
===
=Ω
==
∞=∞=
µ
π
ω
ω
ωω
( )
( )( )[ ] ( )
( )
( ) ( ) 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
==+
=
==
⎪⎭
⎪⎬⎫
⎪⎩
⎪⎨⎧
⎥⎦
⎤⎢⎣
⎡++
+
=
==+++
=
=====
µω
µω
µω
µωωω
ππ
ππ
( )[ ] ( )[ ]( )[ ]
( )( )( )( ) 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
=−=
−=−=⎥⎥⎦
⎤
⎢⎢⎣
⎡
+++
⎥⎦
⎤⎢⎣
⎡+
−===π
π
π
ππ
π
( )
( )[ ] ( )
( ) ( )
( ) 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
==
⎥⎥⎦
⎤
⎢⎢⎣
⎡⎟⎟⎠
⎞⎜⎜⎝
⎛+++
=
==+
=
==++
=
=====
µ
β
ω
µω
µω
µωωω
π
π
( )[ ] ( )
( ) ( )( )
( ) 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
==
⎪⎭
⎪⎬⎫
⎪⎩
⎪⎨⎧
⎥⎦
⎤⎢⎣
⎡+⎟⎟
⎠
⎞⎜⎜⎝
⎛+++
=
==+
=
===∞=
µπ
ππ
µ
ω
ω
ωω
Note: CB amplifier has much better high frequency performance!
40
Conclusions* Voltage gain
Can get good voltage gain from both CE and CB amplifiers.Low frequency performance similar for both amplifiers.CB amplifier gives better high frequency performance !
CE amplifier has dominant pole at 5.0x107 rad/s.CB amplifier has dominant pole at 7.1x108 rad/s.
* Bandwidth approximately 14 X larger!* Miller Effect multiplication of Cµ by the gain is avoided in
CB configuration.
* Current gainFor CE amplifier, current gain is high AI = Ic / Ib
For CB amplifier, current gain is low AI = Ic / Ie (close to one)!Frequency dependence of current gain similar to voltage gain.
* Input and output impedances are different for the two amplifiers! CB amplifier has especially low input resistance.
41
Emitter-Follower (EF) Amplifier* DC biasing
Calculate IC, IB, VCEDetermine related small signal equivalent circuit parameters
Transconductance gmInput resistance rπ
* Midband gain analysis* Low frequency analysis
Gray-Searle (Short Circuit) Technique
Determine pole frequencies ωPL1, ωPL2, ... ωPLn
Determine zero frequencies ωZL1, ωZL2, ... ωZLn
* High frequency analysisGray-Searle (Open Circuit) Technique
Determine pole frequencies ωPH1, ωPH2, ... ωPHn
Determine zero frequencies ωZH1, ωZH2, ... ωZHn
High and Low Frequency AC Equivalent Circuit
42
EF Amplifier - DC Analysis (Nearly the Same as CE Amplifier)* GIVEN: Transistor parameters:
Current gain β = 200Base resistance rx = 65 ΩBase-emitter voltage VBE,active = 0.7 VResistors: R1=10K, R2=2.5K, RC=1.2K, RE=0.33K
* Form Thevenin equivalent for base; given VCC = 12.5VRTh = RB = R1||R2 = 10K||2.5K = 2KVTh = VBB = VCC R2 / [R1+R2] = 2.5V
* Check on transistor region of operationKVL collector loopVCE = VCC - (β +1) IB RE = 10.8 V (was 4.4 V for CE amplifier) (okay since not close to zero volts).
R1 = 10KR2 = 2.5KRC = 0 KRE = 0.33K Note: Only difference here from CE case is VCE is larger
since RC was left out here in EF amplifier.
43
EF Amplifier - Midband Gain Analysis
( )( ) ( ) ( )
( )( )
( )( ) 998.0
905.19.1
65065.02005.065065.02
999.065065.0
65
015.0661
1
1
1
6697.020133.091
==++
+=
+++
=
=+
=+
=
=+
=+
=+
=
==+
=⎥⎦
⎤⎢⎣
⎡+
=
==
KK
KKKKKKK
RrRRRrR
VV
KKK
RrR
VV
VVVV
VVV
KKK
rrgRR
V
VgrVRR
VV
VV
VV
VV
VV
VVA
ixBS
ixB
s
b
ix
i
b
i
ooi
mEL
mELo
s
b
b
i
i
o
s
oVo
π
π
ππ
π
π
π
ππ
π
π
π
π
Ω=====
==
5005.05.2
1033.0
09
2
1
KRKR
KRKR
KRKR
S
E
C
L
Equivalent input resistance Ri
( ) KKVVr
rV
VVIVR ooi
i 6566197.01 =+=⎟⎟⎠
⎞⎜⎜⎝
⎛+=
⎟⎟⎠
⎞⎜⎜⎝
⎛+
==π
π
ππ
π
π
( )( )( )( )( ) dBdBA
A
Vo
Vo1.0987.0log20)(
987.0998.0999.0015.066−==
==
Vi
+
_
Ri
Vb
+
_
Iπ
KVmAg
r
VmAmVmA
VIg
m
T
Cm
97.0/206
200
/2062627.5
===
===
βπ
DC analysis is nearly the same!IB , IC and gm are all the same. Only VCE is different since RC=0.
VO
NOTE: Voltage gain is only ~1!This is a characteristic of the EF amplifier!Cannot get voltage gain >1 for this amplifier!
44
Analysis of Low Frequency Poles Gray-Searle (Short Circuit) Technique
* Draw low frequency AC circuitSubstitute AC equivalent circuit for transistor (hybrid-pi for bipolar transistor) Include coupling capacitors CC1, CC2
Ignore (remove) all transistor capacitances Cπ , Cµ
* Turn off signal source, i.e. set Vs= 0Keep source resistance RS in circuit (do not remove)
* Consider the circuit one capacitor Cx at a time Replace all other capacitors with short circuitsSolve remaining circuit for equivalent resistance Rxseen by the selected capacitorCalculate pole frequency using
Repeat process for each capacitor finding equivalent resistance seen and corresponding pole frequency
* Calculate the final low 3 dB frequency using
xxPx CR
1=ω
∑ ∑=+==xx
PnPPPxLP CR1...21 ωωωωω
45
Emitter Follower - Analysis of Low Frequency PolesGray-Searle (Short Circuit) Technique
Input coupling capacitor CC1 = 2 µF
( )
( )( ) ( )( )[ ]( )( )
( )
( )
( )
( )
sradxRC
xKFRC
KKKKK
RrRRR
KKKK
RRrgrIVR
RRrgrIRRVgIrIV
IVRRrRR
IVR
xCCPL
xCC
ixBsxC
LEmi
i
LEmLEmi
iiixBs
x
xxC
/256sec109.3
11sec109.395.12
95.165065.02005.0
65933.0)201(97.0
1
1
3111
311
1
1
===
==
=++=
++=
=+=
++==
++=++=
=++==
−
−
ω
µ
πππ
πππππππ
π
Ri
Vi
Iπ
IX
46
Emitter Follower - Analysis of Low Frequency PolesGray-Searle (Short Circuit) Technique
srad
PLPLPL
/29337256
21
=+=
+= ωωω
* Output coupling capacitor CC2 = 3 µF
( )[ ][ ]
( )
( ) sradFKCR
KKKKrRRR
KKKKK
rgRRrr
rgIRRrrI
r
RRrrIV
rgIVgII
IVrrRRR
CxCPL
eELxC
m
SBx
m
SBxe
SBxe
mme
e
eeeELxC
/373005.9
11
005.9005.033.09
005.0201
005.02065.097.0
11
1
222
2
2
===
=+=+=
=++
=
+
++=
+−
++−=
++−=
+−=−−=
=+=
µω
π
π
ππ
ππ
ππ
ππππ
* Low 3 dB frequency
Iπ
Ve
Ie IXre
So dominant low frequency pole is due to CC1 !
47
Analysis of High Frequency Poles Gray-Searle (Open Circuit) Technique
* Draw high frequency AC equivalent circuitSubstitute AC equivalent circuit for transistor (hybrid-pi model for transistor with Cπ, Cµ) Consider coupling and emitter bypass capacitors CC1 and CC2 as shortsTurn off signal source, i.e. set Vs = 0Keep source resistance RS in circuit Neglect transistor’s output resistance ro
* Consider the circuit one capacitor Cx at a time Replace all other transistor capacitors with open circuitsSolve remaining circuit for equivalent resistance Rx seen by the selected capacitorCalculate pole frequency usingRepeat process for each capacitor
* Calculate the final high frequency pole using
xxPHx CR
1=ω
∑∑ =⎥⎦
⎤⎢⎣
⎡++=⎥
⎦
⎤⎢⎣
⎡=
−−
xxPHnPHPHPxPH
CR
11...1111
21
1
ωωωωω
48
Emitter Follower - Analysis of High Frequency PolesGray-Searle (Open Circuit) Technique
* Redrawn High Frequency Equivalent Circuit
sourcecurrent dependent to
due resistance equivalent'
'
=−
=
==
πVgVRwhere
RRRIVZ
m
oS
SLEe
oeq
Ie
( ) ( )
( )
( ) ⎥⎦
⎤⎢⎣
⎡ +=
⎥⎦
⎤⎢⎣
⎡+⎟
⎟⎠
⎞⎜⎜⎝
⎛=
⎥⎦
⎤⎢⎣
⎡+⎟
⎟⎠
⎞⎜⎜⎝
⎛−=+=
=
+−=−=
+=+==
+=⎭⎬⎫
⎩⎨⎧
++=
=−++
π
π
π
π
π
ππ
πππππ
π
ππππ
ππππ
π
ygyRRZ
gyy
RR
gyg
RRRRRRRZ
findcanweRRRZSince
gyRRgVg
VRso
sCrZrz
ydefinewewhere
gyRRgsCr
RRVV
RRVVgVsC
rVEatKCL
mLEeq
mLE
m
m
LELESLEeq
SLEeq
mLEmm
oS
C
mLEmLEo
LE
om
1
11111
1
1111
1
0
'
'
'
Zeq
EIe
zπ =1/yπ
49
Emitter Follower - Analysis of High Frequency PolesGray-Searle (Open Circuit) Technique
( ) ( ) ( )( )
( )
( )
( )
( )
( )( )[ ] ( )[ ]
( ) ( )[ ]( )
( )LEm
C
CLEmLEb
b
LEmLEm
LE
LEm
LEb
LEm
LE
LEmLE
mLEeqb
b
RRgCssC
Zwhere
ZRRgrRRZ
isZso
RRgCs
RRgr
RR
RRgr
CsrRRZ
rCsrsC
ryforngSubstituti
RRgy
RR
yRRg
RRyy
gyRRy
ZZ
ZisgroundandBbetweenimpedanceTotal
+
==
++=
++
+
+=
⎟⎟⎟⎟⎟
⎠
⎞
⎜⎜⎜⎜⎜
⎝
⎛
+
⎥⎦
⎤⎢⎣
⎡ ++=
+=+=
⎟⎟⎠
⎞⎜⎜⎝
⎛
+
+=
++=+⎥
⎦
⎤⎢⎣
⎡ +=+=
1
1'
11
111
1
1
11
11
1
1
111
'
'
''
'
'
'
'
π
π
π
π
π
ππ
π
πππ
ππ
π
πππ
π
π
Modified Equivalent Circuit
Replace thiswith this.
ZB’
ZB’
zπ =1/yπ
Looks like a resistor in parallel with a capacitor.
50
Emitter Follower - Analysis of High Frequency PolesGray-Searle (Open Circuit) Technique
( )[ ] [ ]( )( )( )[ ] [ ]
[ ] [ ]
( )( )( )
sradx
sxpFKKKVmA
pFKRRg
CRCR
KKK
KKKKKKKVmAK
RRRRrRRgrR
PH
LEmxC
xCPH
LEBSxLEmxC
/100.1
1086.91
255.0386.01
933.0/206117386.0
1
1
11
386.039.06.64
933.02005.0065.0933.0/206197.0
1
101
11
'''
1
'
=
==
⎥⎦
⎤⎢⎣
⎡
+
=
⎥⎦
⎤⎢⎣
⎡
+
==
==
+++=
+++=
−
ω
ωπ
πππ
ππ
RxCπ
* Pole frequency for Cπ =17 pF
51
Emitter Follower - Analysis of High Frequency PolesGray-Searle (Open Circuit) Technique
( )[ ]( ) [ ]( )( )( )[ ] ( )( ) [ ]
( )
( )sradx
sxpFKCR
KKKK
KKKKKKKVmAK
RRrRRRRgrR
PH
xCPH
BSxLELEmxC
/101.1
101.91
3.107.011
07.007.032.06.64
2005.0065.0933.0933.0/206197.0
1
102
112
=
===
=+=
+++=
+++=
−
ω
ωµµ
πµ
* Pole frequency for Cµ =1.3 pF
52
Emitter Follower - Analysis of High Frequency PolesGray-Searle (Open Circuit) Technique
* Alternative Analysis for Pole Due to Cπ
( )( ) ( )( )
( )( ) ( )( )
( )( ) ( )[ ] ( )
( )( ) ( )
( )( ) ( )
( )( )
( ) sradxpFKCR
KKKKKK
KKKKKK
RRrRRrgRRrRR
rIIr
IVR
soRRrRRrg
RRrRRII
RRrRRIRRrRRrgI
RRrIIVRRIrgII
sorIVandIIIBut
RRrIIVRRVgI
givesloopbasearoundKVLIIr
IV
IVRandVVNote
xCH
BSxLEm
BSxLE
xx
xxC
BSxLEm
BSxLE
x
BSxLExBSxLEm
BSxxLEmx
xe
BSxxLEme
xxx
xxCx
/100.117006.0
11is frequency pole theSo
006.02005.0065.0933.0201
2005.0065.0933.097.0
1
1
1get wegRearrangin
0
0
:
101P ===
=++
++=
+++++
===
+++++
=
++=+++
=+−+−+−−
=−=
=+−+−+−
====
ππ
ππ
πππ
π
π
ππ
πππππ
ππππ
πππ
πππππ
ω
EIe
Vx
Ix
Iπ
Ix-Iπ
Ie+gmVπ
We get the same result here for the high frequency pole associated with Cπ as we did using the equivalent circuit transformation.
53
Emitter Follower - Analysis of High Frequency PolesGray-Searle (Open Circuit) Technique* Alternative Analysis for Pole Due to Cµ
( )( ) ( )
( )( )
( )[ ] [ ] [ ]
[ ] [ ] ( )( )
( )( ) ( )( )[ ] ( )( )[ ][ ] ( )[ ][ ] [ ]
( ) sradxpFKCR
KKK
KKKKKK
RRrgrRRr
RRrgrRRrRRrgrRRr
RRrIVR
RRrgrrVRRr
rRRrIV
V
RRrrVRRrIRRrIIV
RRrgrrVV
sorRR
rgVRRVgIVV
xCH
LEmBsx
LEmBsxLEm
Bsx
Bsx
x
xxC
LEmxBsxBsxxx
BsxBsxxBsxxx
LEmx
LEmLEmx
/101.13.107.0
11is Cfor frequency pole theSo
07.00.6507.0
933.0)201(97.02005.0065.0
1
111
1
11
get wegRearrangin
11
get wefor ngSubstituti
writealsocan We
1
11
102P ===
==
++=
+++=
+++
+
=
+++
+
+==
⎥⎥⎦
⎤
⎢⎢⎣
⎡
+++−+=
+−+=+−=
⎥⎥⎦
⎤
⎢⎢⎣
⎡
++=
⎥⎦
⎤⎢⎣
⎡++=++=
µµ
µ
ππ
ππππ
µ
ππ
π
π
π
π
ππ
ππ
ππ
ππππππ
ω
VxIx
Iπ
Ix-Iπ
Iπ+gmVπ
E
We get the same result here for the high frequency pole associated with Cµ as we did using the equivalent circuit transformation.
54
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
=−=
−=−=⎥⎥⎦
⎤
⎢⎢⎣
⎡
+++
⎥⎦
⎤⎢⎣
⎡+
−===π
π
π
ππ
π
( )
( )[ ] ( )
( ) ( )
( ) 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
==
⎥⎥⎦
⎤
⎢⎢⎣
⎡⎟⎟⎠
⎞⎜⎜⎝
⎛+++
=
==+
=
==++
=
=====
µ
β
ω
µω
µω
µωωω
π
π
( )[ ] ( )
( ) ( )( )
( ) 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
==
⎪⎭
⎪⎬⎫
⎪⎩
⎪⎨⎧
⎥⎦
⎤⎢⎣
⎡+⎟⎟
⎠
⎞⎜⎜⎝
⎛+++
=
==+
=
====∞
µπ
ππ
µ
ω
ω
ωω
( )[ ][ ]
( )( )
( )( )( )( ) dBdBA
VVA
RrRRRrR
RrR
RRrRR
VV
VV
VV
VVA
Vo
Vo
ixBS
ixB
ix
i
LE
EL
s
b
b
i
i
oVo
1.0
/987.0)998.0(999.0015.066
−=
==
++
+⎟⎟⎠
⎞⎜⎜⎝
⎛
++==
π
π
π
( )[ ] ( )
( )[ ] ( ) sradKFCrRR
sradKFRrRRC
CeELPL
ixBsCPL
ZPZP
/3793
11
/25695.12
110
22
11
21
==+
=
==++
=
==
µω
µω
ωω
( )
( )
( )[ ] [ ]
( ) 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
==
+++=
==
⎥⎦
⎤⎢⎣
⎡
+
=
==⎟⎟⎠
⎞⎜⎜⎝
⎛ +==∞
µπ
ππ
ππ
π
ω
ω
ωω
Better low frequency response !
Much better high frequency response !
55
Conclusions* Voltage gain
Can get good voltage gain from CE but NOT from EF amplifier (AV 1).Low frequency performance better for EF amplifier.EF amplifier gives much better high frequency performance!
CE amplifier has dominant pole at 5.0x107 rad/s.EF amplifier has dominant pole at 1.0x1010 rad/s.
* Bandwidth approximately 200 X larger!* Miller Effect multiplication of Cµ by the gain is avoided in EF.
* Current gainFor CE amplifier, current gain is high β = Ic/Ib
For EF amplifier, current gain is also high Ie/Ib = β +1 !Frequency dependence of current gain similar to voltage gain.
* Input and output impedances are different for the two amplifiers!
56
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
====
≈===
µµ
ππ
ββ
EF CE
57
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
]1[
50100100
510200100
12
2
22222
112
22
111
1
2
12111111
1
211
21
21
okayisanalysiseapproximatsoII
AK
VVI
RIVVVKARIV
IandVcalculateNow
AAIIThen
AKK
VVI
ionapproximatfirstaasINeglectingRIIRIVV
QBaseKVL
KKKRRR
VVKKV
RRRV
EB
B
EBBEB
EEB
BB
BE
B
B
EBBThBBETh
Th
CCTh
<<
=−
=
++===≈
==+=
=++
−≈
−++=−
===
==+
=
µ
βµ
µµβ
µ
β
KVmAg
r
VmA
VA
VI
VIg
KVmAg
r
VmA
VA
VI
VIg
m
T
B
T
Cm
m
T
B
T
Cm
9.2/0.34
100
0.340256.0
)7.8(100
9.2/8.34
100
8.340256.0
)9.8(100
2
22
2222
1
11
1111
===
====
===
====
β
µβ
β
µβ
π
π
10021 == ββ
Small Signal Parameters
IE1IB2
IRE1
IB1
58
Cascade Amplifier - Midband Gain Analysis
( ) ( )[ ]
( )( )
KKKK
rRrgr
IrRIrgrI
R
Em
Emi
178)9.23.4)(101(9.2
1
1
21111
1
2111111
=+=
++=
++=
πππ
π
πππππ
[ ]( )
( )
( )
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
21
2111111
111
21111
21111
1
2
2
22
2
1
1
2
2
=+
=+
=
=+
=++
=
==⎟⎟⎠
⎞⎜⎜⎝
⎛+=
+=
−=−
=
==
KKKKKKK
RRRRRRR
VV
KKKK
rRIrgrIrI
VV
KKVmArRg
rVrRVgI
VV
VRRVg
VV
VV
VV
VV
VV
VVA
BS
i
s
i
Emi
EmEm
CLmo
s
i
i
o
s
oVo
πππππ
πππ
πππ
πππ
π
π
π
π
π
π
π
π
π
( ) ( )( )( ) 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
59
Cascade Amplifier - Low Frequency Poles and Zeroes
* Use Gray-Searle (Short Circuit) Technique to find the poles.
Three low frequency polesEquivalent 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
)(
ωωω
ωωωωωωωωω
60
Cascade Amplifier - Analysis of Low Frequency PolesGray-Searle (Short Circuit) Technique
Input coupling capacitor CC1 = 1 µF
( )( ) ( )( )[ ]( )( )
( )
( )
sradxRC
xKFRC
KKKK
RRRR
KKKK
rRrgrIVR
rRrgrIrRVgIrIVIVRRRR
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
===
==
=+=
+=
=+=
++==
++=++=
=+==
−
−
ω
µ
ππππ
πππππππππ
π
Ri
Vi
Iπ1
IX
RE1rπ2
rπ1 Vπ1
Vπ2
+
_
RE1 rπ2
61
Cascade Amplifier - Analysis of Low Frequency PolesGray-Searle (Short Circuit) Technique
( ) sradFKCR
KKKRRR
CCPL
CLC
/1251811
844
222
2
===
=+=+=
µω
Vo
Vo
VX
Vπ2
rX2
RC RL
gm2Vπ2 RC RL
RE2 CE
CC2
* Output coupling capacitor CC2 = 1 µF
rπ2
62
Cascade Amplifier - Analysis of Low Frequency PolesGray-Searle (Short Circuit) Technique
( ) sradFKCR
KKKrRR
KKKK
r
KKKrg
RrrgI
RrII
Vr
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
===
===
=+
=
=+
=
++
=+−+−
==
++
=+−+−
==
==
µω
π
π
ππ
ππ
π
π
ππ
ππ
sradPLPLPLPL /88273412523321 =++=++= ωωωω
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
=
=
63
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
=−=
−=−=
= π
π
π
π
( )
( )
( )
( ) 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
==
==
==
==
==
∞=∞=
ω
ω
ω
ω
ωω
ωω
( )
( )( )[ ] ( )
( ) ( )
( ) sradKFCrR
sradFKCRR
sradKFCrRrgrRR
sradFKCR
EeEPL
CCLPL
CEmBSPL
EEZPZPZP
/73403.047
1)(
1
/1251811
/234311
11
/9.5476.3110
223
22
1221111
2321
===
==+
=
==+++
=
=====
µω
µω
µω
µωωω
πππ
( )( )( ) dBdBA
VVAVV
VV
VVA
Vo
Vo
s
o
s
oVo
5.29)30log(20
/3037.02.81
2
2
=−=
−=−=
== π
π
( )
[ ] ( )
( ) ( )
( ) 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
==
⎥⎥⎦
⎤
⎢⎢⎣
⎡⎟⎟⎠
⎞⎜⎜⎝
⎛+
+=
==+
=
==+
=
=====
µ
β
ω
µω
µω
µωωω
π
π
[ ] ( )
( ) ( )( )
( ) 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
==
⎪⎭
⎪⎬⎫
⎪⎩
⎪⎨⎧
⎥⎦
⎤⎢⎣
⎡⎟⎟⎠
⎞⎜⎜⎝
⎛+++
=
===
===∞=
µπ
ππ
µ
ω
ω
ωω
* CE stage with same transistor, biasing resistors, source resistance and load as cascade.
25 X improvement in bandwidth !
2 X improvement in voltage gain !
64
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 CESo 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
=+=
++= ππ β
( )( )
KKKKrRrR
KKKrg
RrrgI
RrIIVr
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
===
=+
=
++
=+−+−
==
π
π
π
ππ
ππ
Pole for Capacitor CT = 152 pF
( )amplifier. CE for the /100.4 versus
cascade for the /100.1152063.0
1
6
83
sradx
sradxpFKPH ==ω( )
KKKRRR
pFKVmApFpFRgCCC
CLL
LmT
244'
1522)/34(129.13'1 222
===
=++=++= µπ
re1
Ri1
Ri2
65
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).