eces 352 winter 2007ch. 7 frequency response part 51 comparison of amplifier configurations midband...
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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
rr
r
rRR
rRRRg
V
V
V
V
V
VA
Vo
Vo
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s
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2.40
/4.1025.094.0218
sradxpFKCRR
sradxpFCRRr
LCPH
sEePH
ZHZH
/101.73.105.1
11
/105.2174.2
11
,
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101
21
sradFKCRR
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sradFKCR
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1
321
dBdBA
VVA
RrrR
Rrr
rr
rRRg
V
V
V
V
V
V
V
VA
Vo
Vo
Bxs
Bx
xCLm
s
i
i
o
s
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R
sradFKCRR
sradFKCrrRR
sradFKCR
EBsx
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PL
CCLPL
CxBSPL
EEZPZPZP
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1
1
1
/3332.10
11
/71427.0
11
/2521233.0
110
43
22
11
321
sradxpFK
CRRrrRR
gRR
sradxpFKCRRrr
sradxpF
VmA
C
g
SBxLC
mLC
PH
SBx
PH
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
RrrR
Rrr
rr
rRRg
V
V
V
V
V
V
V
VA
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
VmA
C
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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
RrRR
RrR
Rr
R
RRr
RR
V
V
V
V
V
V
V
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Vo
Vo
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sradKFRrRRC
CeELPL
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/25695.12
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0
22
11
21
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CRRrRRRRgr
sradxpFK
RRgC
R
sradxpFKCr
rg
BSxLELEmPH
LEmxC
PH
mZHZH
/101.13.107.0
1
1
1
/100.126.0386.0
1
1
1
/102.11797.0
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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.
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EF CE
Ch. 7 Frequency Response Part 5 8ECES 352 Winter 2007
Cascade Amplifier - DC analysis
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Small Signal Parameters
IE1
IB2
IRE1
IB1
Ch. 7 Frequency Response Part 5 9ECES 352 Winter 2007
Cascade Amplifier - Midband Gain Analysis
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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.
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LPLPLP
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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
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Ch. 7 Frequency Response Part 5 12ECES 352 Winter 2007
Cascade Amplifier - Analysis of Low Frequency Poles Gray-Searle (Short Circuit) Technique
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CLC
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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
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KKKK
r
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IE2
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re1
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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
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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
K
KK
K
KKKK
rgr
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r
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_
Pole for Capacitor Cπ1 = 13.9 pF
sradxpFKPH /100.8
9.1309.0
1 81
Ch. 7 Frequency Response Part 5 17ECES 352 Winter 2007
Cascade Amplifier - High Frequency Poles
K
RRRR SS
7.3
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KKKK
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Ie1
+
_
Pole for Capacitor Cμ1 = 2 pF
sradxpFKPH /104.1
26.3
1 82
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
CC
pFKVmApFpF
RgCCCCC LmT
Using Miller’s Theorem, replace Cμ2 by two capacitors.
CLmCLmo RRg
V
RRVg
V
V
V
VK 2
2
22
21
2
22
2222
222221
11
11
1)1(
CRRg
CK
CC
CRRgCKCC
CLm
CLm
Ch. 7 Frequency Response Part 5 19ECES 352 Winter 2007
Cascade Amplifier - High Frequency Poles
KRRRR SS 7.3' 21
KKKKrRrR
KKK
rg
Rr
rgI
RrI
I
Vr
EeX
m
S
m
S
e
ee
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211
11
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111
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11
Vπ1
IxIe1
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_
Pole for Capacitor CT = 152 pF
sradxpFKPH /100.1
152063.0
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re1
Iπ1
Ve1
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sradxpFKCR
KKKR
RRRR
xCPH
xC
LCLxC
/105.222
11
244
'
8
24
gm2Vπ2 VX+_
Ch. 7 Frequency Response Part 5 20ECES 352 Winter 2007
Cascade Amplifier - High Frequency Zeroes
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rg
so
rC
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or
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me
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11
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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
VmA
C
g
so
C
gs
or
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mZH
m
moC
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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
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105.21
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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
VVA
V
V
V
V
V
V
V
VA
Vo
Vo
S
i
i
oVo
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1
2
2
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sradxpFK
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PH
PH
PH
ZHZH
ZHZH
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84
83
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sradFKCRR
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V
V
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o
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R
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PL
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11
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22
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22
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2321
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CRRrRR
gRR
sradxpFKCRRr
sradxpF
VmA
C
g
SBLC
mLC
PH
SB
PH
mZHZH
/100.42125
1
111
1
/108.49.135.1
11
/100.22
/6.40,
6
222
2
7
22
1
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
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11
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11
Pole for Capacitor CT = 152 pF
amplifier. CE for the /100.4 versus
cascade for the /100.1152063.0
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6
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KKKRRR
pFKVmApFpFRgCCC
CLL
LmT
244'
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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).
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pFpFpFKKVmApFpFRRgCCCso
rg
rrR
bygivenisRwhereamplifierCBofRbecomesamplifierCEstagefirstforRamplifiercascodeFor
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by limited is eperformanc frequency high theamplifier CEFor
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111
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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 Amplifiers
CE 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.
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Ch. 7 Frequency Response Part 5 30ECES 352 Winter 2007
Differential Amplifier Analysis
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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
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High frequency performance is very similar to CE amplifier.
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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.