eces 352 winter 2007ch. 7 frequency response part 41 emitter-follower (ef) amplifier *dc biasing ●...
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Ch. 7 Frequency Response Part 4 1ECES 352 Winter 2007
Emitter-Follower (EF) Amplifier* DC biasing
● Calculate IC, IB, VCE
● Determine related small signal equivalent circuit parameters Transconductance gm
Input 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 analysis● Gray-Searle (Open Circuit)
Technique Determine pole frequencies ωPH1, ωPH2, ... ωPHn
● Determine zero frequencies ωZH1, ωZH2, ... ωZHn
High and Low Frequency AC Equivalent Circuit
Ch. 7 Frequency Response Part 4 2ECES 352 Winter 2007
EF Amplifier - DC Analysis (Nearly the Same as CE Amplifier)* GIVEN: Transistor parameters:
● Current gain β = 200● Base resistance rx = 65 Ω● Base-emitter voltage VBE,active = 0.7 V● Resistors: R1=10K, R2=2.5K, RC=1.2K, RE=0.33K
* Form Thevenin equivalent for base; given VCC = 12.5V● RTh = RB = R1||R2 = 10K||2.5K = 2K● VTh = VBB = VCC R2 / [R1+R2] = 2.5V ● KVL base loop
IB = [VTh-VBE,active] / [RTh+(β +1)RE] IB = 26 μA
* DC collector current IC = β IB
IC = 200(26 μ A) = 5.27 mA* Transconductance gm = IC / VT ; VT = kBT/q = 26 mV
gm = 5.27 mA/26 mV = 206 mA/V * Input resistance rπ =
β / gm = 200/[206 mA/V]= 0.97 K* Check on transistor region of operation
● KVL collector loop● VCE = 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.
Ch. 7 Frequency Response Part 4 3ECES 352 Winter 2007
EF Amplifier - Midband Gain Analysis
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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!
Ch. 7 Frequency Response Part 4 4ECES 352 Winter 2007
Analysis of Low Frequency Poles Gray-Searle (Short Circuit) Technique
* Draw low frequency AC circuit● Substitute 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= 0● Keep source resistance RS in circuit (do not remove)
* Consider the circuit one capacitor Cx at a time ● Replace all other capacitors with short circuits● Solve remaining circuit for equivalent resistance Rx
seen by the selected capacitor● Calculate 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
xx
PnPPPxLP CR
1...21
Ch. 7 Frequency Response Part 4 5ECES 352 Winter 2007
Emitter Follower - Analysis of Low Frequency Poles Gray-Searle (Short Circuit) Technique
Input coupling capacitor CC1 = 2 μF
srad
xRC
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sec109.395.12
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1
Ri
Vi
Iπ
IX
Ch. 7 Frequency Response Part 4 6ECES 352 Winter 2007
Emitter Follower - Analysis of Low Frequency Poles Gray-Searle (Short Circuit) Technique
srad
PLPLPL
/29337256
21
* Output coupling capacitor CC2 = 3 μF
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* Low 3 dB frequency
Iπ
Ve
Ie IX
re
So dominant low frequency pole is due to CC1 !
Ch. 7 Frequency Response Part 4 7ECES 352 Winter 2007
Emitter Follower - Low Frequency Zeros
* What are the zeros for the EF amplifier?
* For CC1 and CC2 , we get zeros at ω = 0 since ZC = 1 / ωC and these capacitors are in the signal line, i.e. ZC at ω = 0 so Vo 0.
ss
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11
11)(
21
21
21
21
Ch. 7 Frequency Response Part 4 8ECES 352 Winter 2007
Emitter Follower - Low Frequency Poles and ZerosMagnitude Bode Plot
22
2
1
2
1
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1998.0)(
dBdBA
dBdBA
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PL
PL
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Ch. 7 Frequency Response Part 4 9ECES 352 Winter 2007
Emitter Follower - Low Frequency Poles and ZerosPhase Shift Bode Plot
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11
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Ch. 7 Frequency Response Part 4 10ECES 352 Winter 2007
Analysis of High Frequency Poles Gray-Searle (Open Circuit) Technique
* Draw high frequency AC equivalent circuit● Substitute AC equivalent circuit for transistor
(hybrid-pi model for transistor with Cπ, Cμ)
● Consider coupling and emitter bypass capacitors CC1 and CC2 as shorts
● Turn off signal source, i.e. set Vs = 0
● Keep 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 circuits
● Solve remaining circuit for equivalent resistance Rx seen by the selected capacitor
● Calculate pole frequency using● Repeat process for each capacitor
* Calculate the final high frequency pole using
xxPHx CR
1
xxPHnPHPHPxPH
CR
11...
1111
21
1
Ch. 7 Frequency Response Part 4 11ECES 352 Winter 2007
Emitter Follower - Analysis of High Frequency Poles Gray-Searle (Open Circuit) Technique
* Redrawn High Frequency Equivalent Circuit
sourcecurrent dependent to
due resistance equivalent'
'
Vg
VRwhere
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Ch. 7 Frequency Response Part 4 12ECES 352 Winter 2007
Emitter Follower - Analysis of High Frequency Poles Gray-Searle (Open Circuit) Technique
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Modified Equivalent Circuit
Replace thiswith this.
ZB’
ZB’
zπ =1/yπ
Looks like a resistor in parallel with a capacitor.
Ch. 7 Frequency Response Part 4 13ECES 352 Winter 2007
Emitter Follower - Analysis of High Frequency Poles Gray-Searle (Open Circuit) Technique
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RxCπ
* Pole frequency for Cπ =17 pF
Ch. 7 Frequency Response Part 4 14ECES 352 Winter 2007
Emitter Follower - Analysis of High Frequency Poles Gray-Searle (Open Circuit) Technique
sradx
sxpFKCR
KKKK
KKKKKKKVmAK
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PH
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102
112
* Pole frequency for Cμ =1.3 pF
Ch. 7 Frequency Response Part 4 15ECES 352 Winter 2007
Emitter Follower - Analysis of High Frequency Poles Gray-Searle (Open Circuit) Technique
* Alternative Analysis for Pole Due to Cπ
sradxpFKCR
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We get the same result here for the high frequency pole associated with Cπ as we did using the equivalent circuit transformation.
Ch. 7 Frequency Response Part 4 16ECES 352 Winter 2007
Emitter Follower - Analysis of High Frequency Poles Gray-Searle (Open Circuit) Technique* Alternative Analysis for Pole Due to Cµ
sradxpFKCR
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We get the same result here for the high frequency pole associated with Cµ as we did using the equivalent circuit transformation.
Ch. 7 Frequency Response Part 4 17ECES 352 Winter 2007
Emitter Follower - High Frequency Zeros
* What are the high frequency zeros for the EF amplifier?* Voltage gain can be written as
* When Vo/Vπ = 0, we have found a zero.
* For Cμ , we get Vo 0 when ω since the node B’ will be shorted to ground and Vπ = 0 .
* Similarly, we get a zero from Cπ when yπ + gm = 0 since we showed earlier that
* Also, can see this from
s
o
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oV V
V
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Ch. 7 Frequency Response Part 4 18ECES 352 Winter 2007
Emitter Follower - High Frequency Poles and ZerosMagnitude
2
10
2
10
2
10
102
101
102
1
1010
10
1010
10
101.11log10
100.11log10
102.11log101.0)(
1.0987.0log20)(
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/102.1
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102.11
987.0)(
101.11
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1102.1
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xj
xj
xj
A
x
s
x
s
s
x
s
sA
V
Vo
PH
PH
ZH
ZH
Ch. 7 Frequency Response Part 4 19ECES 352 Winter 2007
Emitter Follower - High Frequency Poles and ZerosPhase Shift
10
110
110
1
102
101
102
1
1010
10
1010
10
101.1tan
100.1tan
102.1tan)(
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101.11
100.11
102.11
987.0)(
101.11
100.11
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xxx
sradx
sradx
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xj
xj
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xs
sxs
sA
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PH
ZH
ZH
Ch. 7 Frequency Response Part 4 20ECES 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
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R
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sradFKCR
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E
PL
CCLPL
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/107.112005.0
1
1
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11
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11
/2521233.0
110
43
22
11
321
sradxpFK
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sradxpF
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C
g
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mLC
PH
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mZHZH
/100.53.14.15
1
111
1
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11
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7
2
81
1121
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11
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sradxpFK
CRRrRRRRgr
sradxpFK
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R
sradxpFKCr
rg
BSxLELEmPH
LEmxC
PH
mZHZH
/101.13.107.0
1
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1
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20111
10
2
10
'
1
1021
Better low frequency response !
Much better high frequency response !
Ch. 7 Frequency Response Part 4 21ECES 352 Winter 2007
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 C by the gain is avoided in EF.
* Current gain● For CE amplifier, current gain is high = Ic/Ib
● For EF amplifier, current gain is also high Ie/Ib = +1 !
● Frequency dependence of current gain similar to voltage gain.
* Input and output impedances are different for the two amplifiers!