FREQUENCY RESPONSE OF AMPLIFIERS
*
*
*
*
Effects of capacitances within transistors and in amplifiers
* Build on previous analysis of amplifiers from EENG 341
z Transistor DC biasing
z Small signal amplification, i.e. voltage and current gain
z Transistor small signal equivalent circuit
* Use in Bipolar and FET Transistors Amplifiers and their analysis
Build on previous analysis of single time constant circuits
z Review simple RC, LC and RLC circuits
z Recall frequency dependent impedances for C and L
z Review frequency dependence in transfer functions
* Magnitude and phase
Examine origins of frequency dependence in amplifier gain
z Identify capacitors and their origins; find the dominant C
z Determine equivalent R and determine RC time constant
z Use to describe approximately the amplifier’s frequency behavior
z Examine effects of other capacitors
GOAL: Use results of analysis to modify circuit design to improve performance.
1
Analysis of Amplifier Performance
*
iB
Previously analyzed
z DC bias point
z AC 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
iC
DC bias or
quiescent point
vBE
vCE
2
Frequency Response of Amplifiers
*
20 logT(ω)
Midband Gain
In reality, all amplifiers have a limited
range of frequencies of operation
z Called the bandwidth of the amplifier
z Falloff at low frequencies
* At ~ 100 Hz to a few kHz
* Due to coupling capacitors at the
input or output, e.g. CC1 or CC2
z Falloff at high frequencies
* At ~ 100’s MHz or few GHz
* Due to capacitances within the
transistors themselves.
Equivalent circuit for bipolar transistor
ω
3
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
z High frequency falloff –
Æ Like that of a low pass filter
* Simple RC equivalent circuit
* Shunting capacitor shorts signal at the
output at high frequencies
z Low frequency falloff
Æ 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
z Determine equivalent R for each C
z Compare and find the most important
(dominant RC) combination
ÆFind the dominant one (RC)
at high frequencies
ÆFind the dominant one (RC)
at low frequencies
Low-Pass Network
Vi
High-Pass Network
Vi
ZC =
⎧0 (short ) as ω → ∞
1
1
=
→⎨
sC jωC
⎩ ∞ (open) as ω → 0
4
Review of Complex Numbers
*
Complex numbers
z General form a + bj where
a = real part, b = imaginary part and j = − 1
z
z
z
*
Magnitude of complex number
Phase of complex number
Phasor form
a + bj = Me jθ
M = a + bj = a 2 + b 2
⎛b⎞
⎝a⎠
θ = tan −1 ⎜ ⎟
Complex number math
z Multiplication of two complex numbers a + bj = Me jθ and c + dj =Ne jφ
(a + bj )(c + dj ) = (ac − bd ) + j (bc + ad )
OR
(a + bj )(c + dj ) = (Me jθ )(Ne jϕ ) = MNe j (θ +ϕ )
z
Reciprocal of a complex number
a − bj
a
b
1
1 a − bj
=
= 2
−j 2
= 2
2
2
a +b
a + b2
a + bj a + bj a − bj a + b
OR
1
1
1 − jθ
=
=
e
jθ
a + bj Me
M
5
Amplifier Transfer Function (Gain) - General Form
*
A (s) = Gain Function (general form of amplifier transfer function)
A( s ) = AM FH ( s ) FL ( s )
z
AM = midband gain (independent of frequency)
FH(s) = high frequency function (acts like low pass filter)
FH ( s ) =
z
FL(s) = low frequency function (acts like high pass filter)
FL ( s ) =
z
1
1 + s / ωH
1
1 + ωL / s
Magnitude
AM
FL(s)
FH(s)
ωL
ωH
6
Amplifier Transfer Function (Gain) - General Form
A( s ) = AM FH ( s ) FL ( s )
AM
FL(s)
FH(s)
Coupling
Capacitors
ωL
Transistor’s
Capacitors
ωH
7
Frequency Response of MOSFET vs BJT Amplifiers
Equivalent circuit for MOSFET
Equivalent circuit for bipolar transistor
Similar equivalent circuits
Common Source
Amplifier
Common Emitter
Amplifier
Corresponding
amplifier circuits
Gain
Gain
Similar frequency
performance
8
Amplifier Transfer Function (Gain) - General Form
A( s ) = AM FH ( s ) FL ( s )
Now we consider the
low frequency behavior.
Coupling
Capacitors
ωL
Transistor’s
Capacitors
ωH
9
Summary
*
*
*
*
Examined origin of falloff in amplifier gain at low and high frequencies.
z Degradation in magnitude of the gain.
z Shift in phase of output relative to input.
Due to presence of capacitors within the amplifier (Create poles and zeros).
z Coupling capacitors limit gain at low frequencies.
z Transistor’s capacitances limit gain at high frequencies.
Examined and quantified the falloff due to single and multiple poles and zeroes.
z Bode plots of gain and phase shift with frequency.
Next, we will apply this method of analysis to transistor amplifiers.
z Multiple capacitors so multiple RC combinations.
z Investigate how to determine which capacitors are most important and limit the
bandwidth.
z Examine how to change the amplifier to get better frequency performance.
Higher frequency operation before falloff (improved bandwidth).
Better low frequency behavior.
10
Analysis of Bipolar Transistor Amplifiers
*
Single stage amplifiers
z Common Emitter (CE)
z Common Base (CB)
z Emitter Follower (EF) (Common Collector)
* DC biasing
z Calculate IC, IB, VCE
z Determine related small signal equivalent circuit parameters
Transconductance gm
Input resistance rπ
* Low frequency analysis
z Gray-Searle (Short Circuit) Technique
Determine the pole frequencies ωPL1, ωPL2, ... ωPLn
z Determine the zero frequencies ωZL1, ωZL2, ... ωZLn
* High frequency analysis
z Gray-Searle (Open Circuit) Technique
Determine the pole frequencies ωPH1, ωPH2, ... ωPHn
z Determine the zero frequencies ωZH1, ωZH2, ... ωZHn
11
CE Amplifier Frequency Analysis - Long and Difficult Way
*
*
*
*
*
*
*
*
*
⎛
s ⎞⎛
s ⎞
⎜⎜1 +
⎟⎟⎜⎜1 +
⎟
ωZ 1 ⎠⎝ ωZ 2 ⎟⎠
⎝
FH ( s ) =
⎛
s ⎞⎛
s ⎞
⎜⎜1 +
⎟⎟⎜⎜1 +
⎟⎟
⎝ ω P1 ⎠⎝ ω P 2 ⎠
FL ( s ) =
(s + ωZ 1 )(s + ωZ 1 )(s + ωZ 3 )
(s + ωP1 )(s + ωP 2 )(s + ωP 3 )
⎛ ωZ 1 ⎞⎛ ωZ 2 ⎞⎛ ωZ 3 ⎞
⎜1 +
⎟⎜1 +
⎟⎜1 +
⎟
s ⎠⎝
s ⎠⎝
s ⎠
=⎝
⎛ ω P1 ⎞⎛ ω P 2 ⎞⎛ ω P 3 ⎞
⎜1 +
⎟⎜1 +
⎟⎜1 +
⎟
s ⎠⎝
s ⎠⎝
s ⎠
⎝
*
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
z Determine high frequency poles
ωPH1 and ωPH2
z Determine high frequency zeros
ωZH1 and ωZH2
Determine FL(ω) part and factor to
z Determine low frequency poles
ωPL1, ωPL2 and ωPL3
z Determine high frequency zeros
ωZL1, ωZL2 and ωZL3
Is there an easier way ? YES
12
CE Amplifier - Starting Point is DC Analysis
*
*
*
*
*
Saturation
region
Active
region
*
Q point
Q is quiescent point (DC bias point)
Q needs to be in the active region
z I C = 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
z Transconductance
gm = IC / VT
VT = kBT/q = 26 mV
z Input resistance rπ = β / gm
Cutoff region
13
Example of CE Amplifier - DC Analysis
*
GIVEN: Transistor parameters:
z Current gain β = 200
z Base resistance rx = 65 Ω
z Base-emitter voltage
VBE,active = 0.7 V
z Resistors: R1=10K, R2=2.5K, RC=1.2K, RE=0.33K
*
Form Thevenin equivalent for base; given VCC = 12.5V
z
RTh = RB = R1||R2 = 10K||2.5K = 2K
z
VTh = VBB = VCC R2 / [R1+R2] = 2.5V
z DC Base Current (use 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 (Find VCE)
z KVL collector loop
z VCE = VCC - IC RC - (β +1) IB RE = 4.4 V
(okay since not close to zero volts, i.e. > 0.2V).
*
*
*
R1 = 10K
R2 = 2.5K
RC = 1.2K
RE = 0.33K
*
14
CE Amplifier - Midband Gain Analysis
*
*
*
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 capacitances
z Coupling 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)
ZC =
z
Hybrid-Pi Model for BJT
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)
ZC =
*
1
1
~
= 10 Ω
ωC (105 rad / s )(1µF )
1
1
~
= 107 Ω
5
ωC (10 rad / s )(1 pF )
Calculate small signal voltage gain
AVo = Vo /Vs
15
CE Amplifier - Midband Gain Analysis
Io
rx
Vs
+
+
V
Vii rπ
RL = 9 K
Vo
RC = 1.2 K
RE = 0.33K
gmVπ
Vπ
R1 = 10 K
--
R2 = 2.5 K
RS = 5 K
RL||RC
V
V V V
Break voltage gain into a series of voltage ratios
AVo = o = o π i
Vs Vπ Vi Vs
Vo − g mVπ RL RC
=
= − g m RL RC = − (206 mA / V ) 1.2 K 9 K = −218
Vπ
Vπ
(
)
(
)
Vπ
rπ
0.97 K
=
=
= 0.94
Vi
rx + rπ
0.97 K + 0.065 K
[
]
[
(
]
)
(rx + rπ ) RB
(0.97 K + 0.065 K ) 2.0 K
Vi
0.68 K
=
=
=
= 0.12
Vs Rs + (rx + rπ ) RB
5 K + (0.97 K + 0.065 K ) 2.0 K
5.68 K
Negative sign means output signal is
AVo = (− 218 )(0.94 )(0.12 ) = −24.6 V / V
[
]
[
AVo (dB ) = 20 log( − 24.6 ) = 27.7 dB
]
180o out of phase with the input signal.
16
Analysis of Low Frequency Poles
Gray-Searle (Short Circuit) Technique
* Draw AC equivalent circuit at low frequency
z
Include coupling and emitter bypass capacitors CC1,CC2, CE
Substitute AC equivalent circuit for transistor
(hybrid-pi for bipolar transistor)
z
Ignore (remove) all transistor capacitances Cπ , Cµ
z
* Turn off signal source, i.e. set Vs= 0
z
Keep source resistance RS in circuit (do not remove)
* Analyze the circuit one capacitor Cx at a time
z
z
z
z
Replace all other capacitors with short circuits
Solve remaining circuit for equivalent resistance Rx seen by
the selected capacitor
1
ω
=
Px
Calculate pole frequency using
Rx C x
Repeat 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
ω LP =
∑
ω Px = ω P1 + ω P 2 ...ω Pn =
∑
1
RxC x
17
Common Emitter - Analysis of Low Frequency Poles
Gray-Searle (Short Circuit) Technique
AC equivalent circuit
at low frequency
RS CC1
VS
RB
CC2
rX
rπ
Vπ
gmVπ
RE
*
Input coupling capacitor CC1 = 2 µF
IX
RS
RC1 =
VX
RB
rX
rπ
Vπ
RL
CE
*
Output coupling capacitor CC2 = 3 µF
VX
IX
VX
= RS + RB (rx + rπ ) = 5 K + 2 K (0.065 K + 0.97 K ) ) = 5.7 K
IX
1
1
=
= 88 rad / s
ω PL1 =
RC1CC1 5.7 K (2 µF )
RC
Vo
Vo
RC
RC 2 =
RL
VX
= RL + RC = 9 K + 1.2 K = 10.2 K
IX
ωPL 2 =
1
1
=
= 33 rad / s
RC 2CC 2 10.2 K (3µF )
18
Common Emitter - Analysis of Low Frequency Poles
Gray-Searle (Short Circuit) Technique
V
REx = x =?
Ix
Ie =
Vx
RE
Vπ = I π rπ =
Ix =
KCL at node E
I x = I e − I π − g mVπ
Iπ =
RS CC1
rX
RB
rπ
VS
− Vx
rx + rπ + Rs RB
− rπ Vx
rx + rπ + Rs RB
Vx
Vx
g m rπ Vx
+
+
RE rx + rπ + Rs RB rx + rπ + Rs RB
1 + g m rπ
1
+
RE rx + rπ + Rs RB
Vπ
gmVπ
RE
Recall β = gmrπ
RL
RC
CE
rX
RS
+
RB
rπ
Iπ
Vπ
gmVπ
⎛ r + r + Rs RB ⎞
⎟
REx = RE ⎜⎜ x π
⎟
β +1
⎝
⎠
⎛ 0.065 K + 0.97 K + 5 K 2 K ⎞
⎟ = 0.016 K
= 0.33K ⎜⎜
⎟
201
⎠
⎝
1
1
ω PL 3 =
=
= 5,342 rad / s
REx C E 0.016 K (12 µF )
Vo
Emitter bypass capacitor CE = 12 µF
I
g m rπ
1
1
1
= x =
+
+
REx Vx RE rx + rπ + Rs RB rx + rπ + Rs RB
=
CC2
Ix
Ie
RE
REx
VX
Low 3db frequency
ω PL = ω PL1 + ω PL 2 + ω PL3 = 88 + 33 + 5342 = 5463 rad / s
19
Analysis of High Frequency Poles
Gray-Searle (Open Circuit) Technique
*
*
Cπ comes from Emitter-Base p-n junction.
Cµ comes from Base-Collector p-n junction.
*
Draw AC equivalent circuit at high frequency
z Substitute AC equivalent circuit for
transistor (hybrid-pi model for transistor
with Cπ, Cµ )
z Remove coupling and emitter bypass
capacitors CC1, CC2, CE (consider as shorts).
z Turn off signal source, i.e. set Vs = 0
z Keep source resistance RS in circuit
z Neglect transistor’s output resistance ro
Consider the circuit one capacitor Cx at a time
z Replace all other transistor capacitors with
open circuits
z Solve remaining circuit for the equivalent
resistance Rx seen by the selected capacitor
1
z Calculate the pole frequency using ω PHx =
Rx C x
z Repeat process for each capacitor
Calculate the final high 3dB frequency using
⎡
ω PH = ⎢
⎣
∑
1 ⎤
⎥
ω Px ⎦
−1
⎡ 1
1
1 ⎤
... +
=⎢
+
⎥
ω PHn ⎦
⎣ ω PH 1 ω PH 2
−1
=
1
∑R C
x x
20
Common Emitter - Analysis of High Frequency Poles
Gray-Searle (Open Circuit) Technique
Vo
VS
CE shorts RE
at high frequencies.
AC equivalent circuit at High frequency
VS
Vπ
Vo
21
Common Emitter - Analysis of High Frequency Poles
Gray-Searle (Open Circuit) Technique
VS
Vπ
Vo
* Equivalent circuit for Capacitor Cπ = 17 pF
IX
Rπx =
VX
= rπ (rx + RB RS ) = 0.97 K (0.065 K + 2 K 5 K ) = 0.59 K
IX
Rπx Cπ = 0.59 K (17 pF ) = 1.0 x10 −8 sec
VX
ω PH 1 =
1
1
=
= 1.0 x108 rad / s
−8
Rπx Cπ 1.0 x10 sec
Note: This frequency is very high due to the
very small size of the capacitor.
22
Common Emitter - Analysis of High Frequency Poles
Gray-Searle (Open Circuit) Technique
⎡1
⎤
Vπ
Vπ
1
so Vπ = I x ⎢ +
+
⎥
rπ rx + RB RS
⎣ rπ rx + RB RS ⎦
V
V
At node C I x = − g mVπ − o − o
RC RL
At node B' I x =
Vπ
Using − Vπ + Vx + Vo = 0 or Vo = Vπ − Vx
⎡
⎡1
Vπ − Vx Vπ − Vx
1
1⎤
1⎤
−
= −Vπ ⎢ gm +
+ ⎥ + Vx ⎢ + ⎥
RC
RL
RC RL ⎦
⎣
⎣ RC RL ⎦
V
Substituting for Vπ and solving for Rµx = x we get
Ix
I x = − g mVπ −
*
Equivalent circuit for Capacitor Cµ = 1.3 pF
B/ IX
VX
C
⎡ ⎛
⎤
1
1 ⎞
Rµx = (RC RL )⎢1 + ⎜⎜ g m +
+ ⎟⎟ rπ (rx + RB RS ) ⎥
RC RL ⎠
⎣ ⎝
⎦
(
)
(
)
⎡ ⎛
⎤
1
1 ⎞
= (1.2K 9K )⎢1 + ⎜ 206mA / V +
+
⎟ 0.97K (0.065K + 2K 5K ) ⎥
1.2K 9K ⎠
⎣ ⎝
⎦
= 130K
Vπ
RµxCµ = 130K (1.3 pF ) = 1.7 x10− 7 sec
ωPH 2 =
Need to solve for RµX =
VX
IX
1
1
=
= 5.9 x106 rad / s
−7
RµxCµ 1.7 x10 sec
Final high frequency 3dB frequency is
ωPH =
1
1
=
= 5.6 x106 rad / s
−8
−7
RπxCπ + RµxCµ 1.0 x10 + 1.7 x10 sec
23
Common-Base (CB) Amplifier
*
*
*
*
Input at emitter, output at collector.
DC biasing
z Calculate IC, IB, VCE
z Determine related small signal
equivalent circuit parameters
Transconductance gm
Input resistance rπ
Midband gain analysis
Low frequency analysis
z Gray-Searle (Short Circuit)
Technique
Determine pole frequencies
ωPL1, ωPL2, ... ωPLn
z Determine zero frequencies
ωZL1, ωZL2, ... ωZLn
High frequency analysis
z Gray-Searle (Open Circuit)
Technique
Determine pole frequencies
ωPH1, ωPH2, ... ωPHn
z Determine zero frequencies
ωZH1, ωZH2, ... ωZHn
24
CB Amplifier - DC Analysis (Same as CE Amplifier)
*
GIVEN: Transistor parameters:
z Current gain β = 200
z Base resistance rx = 65 Ω
z Base-emitter voltage
VBE,active = 0.7 V
z Resistors: R1=10K, R2=2.5K, RC=1.2K, RE=0.33K
*
Form Thevenin equivalent for base; given VCC = 12.5V
z
RTh = RB = R1||R2 = 10K||2.5K = 2K
z
VTh = VBB = VCC R2 / [R1+R2] = 2.5V
z 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
z KVL collector loop
z VCE = VCC - IC RC - (β +1) IB RE = 4.4 V
(okay since not close to zero volts).
*
*
*
R1 = 10K
R2 = 2.5K
RC = 1.2K
RE = 0.33K
*
25
CB Amplifier - Midband Gain Analysis
*
*
*
Construct small signal ac equivalent circuit
(set DC supply to ground)
Substitute small signal equivalent circuit
(hybrid-pi model) for transistor
Neglect all capacitances
z Coupling and emitter bypass capacitors become shorts
at midband frequencies (~ 105 rad/s)
Why? Impedances are negligibly small, e.g.
few
-6
ohms because CC1, CC2, CE ~ few µF (10 F)
ZC =
High and Low Frequency AC Equivalent Circuit
z
Transistor capacitances become open circuits at
midband frequencies
Why? Impedances are very large, e.g. ~ 10’s M Ω
because Cπ , Cµ ~ pF (10-12 F)
ZC =
*
1
1
=
= 10 Ω
ωC (105 rad / s )(1µF )
1
1
=
= 10 7 Ω
5
ωC (10 rad / s )(1 pF )
Calculate small signal voltage gain
AVo = Vo /Vs
26
CB Amplifier - Midband Gain Analysis
V
− Ve
Iπ = π =
rπ rx + rπ
Vπ
− rπ
=
Ve rx + rπ
RL = 9 K
RC = 1.2 K
Iπ
re
+
Ve
_
RS = 5 K
Equivalent resistance re
Vo − gmVπ (RL RC )
=
= −gm (RL RC ) = −(206mA/ V )(1.2K 9K ) = −218
Vπ
Vπ
Vπ
rπ Iπ
− 0.97K
=
=
= −0.94
Ve − Iπ (rx + rπ ) 0.97K + 0.065K
[RE re ] = [0.33K 0.0051K ] = 0.0050K = 0.001
Ve
=
Vs Rs + [RE re ] 5K + [0.33K 0.0051K ] 5.0050K
AVo = (− 218)(− 0.94)(0.001) = 0.20 V / V
R1 = 10 K
R2 = 2.5 K
V V V V
AVo = o = o π e
Vs Vπ Ve Vs
AVo (dB) = 20log(0.20) = −14dB
RE = 0.33K
βIπ
Voltage gain is
less than one !
re =
Ve
Ie
KCL at node E
I e + g mVπ +
Vπ
=0
rπ
⎛
1
I e = −Vπ ⎜⎜ g m +
rπ
⎝
⎞
⎛ 1 + g m rπ ⎞
⎟⎟ = −Vπ ⎜⎜
⎟⎟
r
π
⎠
⎝
⎠
⎛ r + r ⎞ rπ
V
V
rπ
re = e = − e
= −⎜⎜ − x π ⎟⎟
Ie
Vπ 1 + g m rπ
rπ ⎠ 1 + g m rπ
⎝
=
rx + rπ
r +r
0.065K + 0.97 K
= 0.0051 K
= x π =
1 + g m rπ
1+ β
1 + 200
27
What Happened to the CB Amplifier’s Midband Gain?
+
Ve
_
re
AVo = (−218 )(−0.94 )(0.001) = 0.20 V / V
[
]
[
]
0.33 K 0.0051K
RE re
Ve
0.0050 K
=
= 0.001
=
=
5 K + 0.33 K 0.0051K
5.0050 K
Vs Rs + RE re
[
]
[
]
New signal source with low resistance
For Rs = 5Ω
[
]
[
]
0.33 K 0.0051K
RE re
Ve
0.0050 K
=
=
=
0.005 K + 0.33K 0.0051K
0.005 K + 0.005 K
Vs Rs + RE re
[
]
[
and
AVo = (− 218 )(− 0.94 )(0.5) = 102.5 V / V
AVo ( dB ) = 20 log(102.5) = 40.2dB
]
* Source resistance Rs = 5K is
killing the gain.
z Why? Rs >> re = 0.0051 K
so 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?
z Vs drives formation of Ve
z Ve creates Vπ across rπ
z Vπ turns on dependent
current source
= 0.5 z Get large Ie for small Ve
so re =Ve/Ie is very small.
Voltage gain is
now much bigger than one !
28
Analysis of Low Frequency Poles
Gray-Searle (Short Circuit) Technique
* Draw low frequency AC circuit
z
z
z
Substitute 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= 0
z
Keep source resistance RS in circuit (do not remove)
* Consider the circuit one capacitor Cx at a time
z
z
z
z
Replace all other capacitors with short circuits
Solve remaining circuit for equivalent resistance Rx seen
by the selected capacitor
1
Calculate pole frequency using ω Px = R C
x x
Repeat 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
ω LP =
∑
ω Px = ω P1 + ω P 2 ...ω Pn =
∑
1
RxC x
29
Common Base - Analysis of Low Frequency Poles
Gray-Searle (Short Circuit) Technique
Low Frequency AC Equivalent Circuit
RxCB = RB (rx + Ri )
Ri =
Vi
Vi
V
=
= rπ i
Iπ Vπ / rπ
Vπ
sin ce Iπ =
Vπ
rπ
⎡ ⎛1
⎤
⎞
Vi = Vπ + (Iπ + g mVπ )(RE RS ) = Vπ ⎢1 + ⎜⎜ + g m ⎟⎟(RE RS )⎥
⎠
⎣ ⎝ rπ
⎦
*
⎡ ⎛1
⎤
⎞
Vπ ⎢1 + ⎜⎜ + g m ⎟⎟(RE RS )⎥
⎝ rπ
⎠
⎦ = r + (1 + g r )(R R )
Ri = rπ ⎣
π
m π
E
S
Vπ
Base capacitor CB = 12 µF
Vx Ix
RxCB = RB [rx + rπ + (1 + g m rπ )(RE RS )]
Iπ
Vo
+
Vπ
Vi
= 2K [0.065K + 0.97K + (201)(0.33K 0.005K )]
= 2K 2.03K = 1.0K
CB RxCB = 12µF (1.0K ) = 1.2 x10−2 sec
ωPL1 =
RxCB
_
1
1
=
= 83 rad / s
RxCB CB 1.2 x10−2
Ri
30
Common Base - Analysis of Low Frequency Poles
Gray-Searle (Short Circuit) Technique
Input coupling capacitor CC1 = 2 µF
V
RxCC1 = x = Rs + RE re
Ix
V
re = e
Ie
I e = − Iπ − g mVπ = −(1 + g m rπ )Iπ
Ve = − Iπ (rx + rπ )
(r + r )
V
− Iπ (rx + rπ )
re = e =
= x π
I e − (1 + g m rπ )Iπ (1 + g m rπ )
⎡ r +r ⎤
RxCC1 = Rs + RE ⎢ x π ⎥
⎣ 1 + g m rπ ⎦
Iπ
Vo
⎡ 0.065 K + 0.97 K ⎤
= 0.005 K + 0.33K ⎢
⎥
201
⎣
⎦
Vπ
Ve
= 0.005 K + 0.0051K = 0.010 K
CC1RxCC1 = 2µF (0.010 K ) = 2.0 x10 −5 sec
1
1
ω PL 2 =
=
= 5.0 x10 4 rad / s
−
5
CC1RxCC1 2.0 x10 sec
Vx
Ie
re
+
Ve
_
Ix
Rs
31
Common Base - Analysis of Low Frequency Poles
Gray-Searle (Short Circuit) Technique
*
Output coupling capacitor CC2 = 3 µF
VX
Low 3dB frequency
ω PL = ω PL1 + ω PL 2 + ω PL3
Vo
RC
*
RL
RC 2 = RL + RC = 9 K + 1.2 K = 10.2 K
1
1
ω PL3 =
=
= 33 rad / s
RC 2CC 2 10.2 K (3µF )
= 83 + 50,000 + 33 = 50,116 rad / s
Dominant low frequency
pole is due to CC1 !
32
Analysis of High Frequency Poles
Gray-Searle (Open Circuit) Technique
*
*
*
Draw high frequency AC equivalent circuit
z Substitute AC equivalent circuit for transistor
(hybrid-pi model for transistor with Cπ, Cµ)
z Consider coupling and emitter bypass
capacitors CC1, CC2, CB as shorts
z Turn off signal source, i.e. set Vs = 0
z Keep source resistance RS in circuit
z Neglect transistor’s output resistance ro
Consider the circuit one capacitor Cx at a time
z Replace all other transistor capacitors with
open circuits
z Solve remaining circuit for equivalent
resistance Rx seen by the selected capacitor
1
z Calculate pole frequency using
ω PHx =
RxC x
z Repeat process for each capacitor
Calculate the final high frequency pole using
⎡
ω PH = ⎢
⎣
∑
1 ⎤
⎥
ω Px ⎦
−1
⎡ 1
1
1 ⎤
... +
=⎢
+
⎥
ω PHn ⎦
⎣ ω PH 1 ω PH 2
−1
=
1
∑R C
x x
33
Common Base - Analysis of High Frequency Poles
Gray-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 the
results appreciably.
34
Common Base - Analysis of High Frequency Poles
Gray-Searle (Open Circuit) Technique
Ze =
Ve
Ie
Ve = −Vπ
KCL at node E gives
⎤
⎡1
Ve
Ve
+
− g mVπ = Ve ⎢ + sCπ + g m ⎥
rπ ⎛ 1
⎞
⎦
⎣ rπ
⎜ sC ⎟
π ⎠
⎝
⎡1 + g m rπ
⎤
= Ve ⎢
+ sCπ ⎥
⎣ rπ
⎦
Ie =
Ze
+
Ve
_
Replace this
with this.
*
1
Ve
=
⎡1 + g m rπ
⎤ ⎡1 + g m rπ
⎤
+ sCπ ⎥ ⎢
+ sCπ ⎥
Ve ⎢
⎣ rπ
⎦ ⎣ rπ
⎦
⎡ rπ ⎤ ⎛ 1
1
⎞
=
=⎢
⎜ sC ⎟
⎥
π ⎠
⎡
⎤ ⎣1 + g m rπ ⎦ ⎝
⎢
⎥
1
1
⎢
⎥
+
⎢ ⎛ rπ
⎞ ⎛ 1
⎞⎥
⎟⎟ ⎜ sC ⎟ ⎥
⎢ ⎜⎜
π ⎠
⎝
Parallel combination
⎣⎢ ⎝ 1 + g m rπ ⎠
⎦⎥
Ze =
Equivalent circuit for Ze
Ve
=
Ie
So Z e = re Z Cπ
of a resistor and
capacitor.
where
Ze
re =
rπ
r
= π
1 + gmr 1 + β
35
Common Base - Analysis of High Frequency Poles
Gray-Searle (Open Circuit) Technique
Turn off signal source
when finding resistance
seen by capacitor.
*
Pole frequency for Cπ =17pF
RxCπ = re RE Rs
re =
rπ
0.97 K
=
= 0.0048 K = 4.8Ω
1 + g m rπ 1 + 200
RxCπ = 0.0048 K 0.33K 0.005 K = 0.0024 K = 2.4Ω
ω PH 1 =
1
RxCπ Cπ
=
1
1
=
= 2.5 x1010 rad / s
2.4Ω(17 pF ) 4.1x10 −11 s
36
Common Base - Analysis of High Frequency Poles
Gray-Searle (Open Circuit) Technique
*
Pole frequency for Cµ =1.3pF
RxCµ = RC RL
RxCµ = 1.2 K 9 K = 1.05 K
ω PH 2 =
*
Equivalent circuit for Capacitor Cµ = 1.3 pF
=0
Rs || RE || r π
*
1
RxCµ Cµ
=
1
1
=
= 7.1x108 rad / s
9
−
1.05 K (1.3 pF ) 1.4 x10 s
High 3 dB frequency
ω PH =
1
1
=
RxCπ Cπ + RxCµ C µ 4.1x10 −11 + 1.4 x10 −9
ω PH =
1
8
=
6
.
9
x
10
rad / s
−9
1.44 x10 s
Dominant high frequency
pole is due to Cµ !
37
Comparison of CB to CE Amplifier
CE (with RS = 5K)
Midband Gain
[
CB (with RS = 5Ω)
]
⎡ r ⎤⎡ (rx + rπ ) RB
V V V V
AVo = o = o π i = − gm RL RC ⎢ π ⎥⎢
Vs Vπ Vi Vs
⎣ rx + rπ ⎦⎢⎣ Rs + (rx + rπ ) RB
AVo = (− 218)(0.94)(0.12) = −24.6 V / V
[
)]
(
[
⎤
⎥
⎥⎦
]
AVo(dB) = 20log(− 24.6 ) = 27.7dB
ωZP1 = ωZP2 =0
Low Frequency
Poles and Zeros
1
1
=
= 88 rad / s
ωPL1 =
RS + RB (rx + rπ ) CC1 5.7K (2µF )
ωPL2 =
ωPL3 =
]
1
⎛ rx + rπ + Rs RB ⎞⎤
⎜
⎟⎥CE
⎜
⎟⎥
β +1
⎝
⎠⎦
ωZH 1 = ∞, ω ZH 2 =
High Frequency
Poles and Zeroes
ω PH 1 =
ω PH 2 =
=
1
= 5,342 rad / s
0.016K (12µF )
g m 206 mA / V
=
= 1.6 x1011 rad / s
Cµ
1.3 pF
1
1
=
= 1.0 x108 rad / s
rπ (rx + RB RS ) Cπ 0.59 K (17 pF )
[
]
1
⎧⎪
⎤ ⎫⎪
⎡ ⎛
1
1 ⎞
⎟⎟ rπ (rx + RB RS ) ⎥ ⎬Cµ
+
⎨(RC RL )⎢1 + ⎜⎜ g m +
RC RL ⎠
⎪⎩
⎦ ⎪⎭
⎣ ⎝
1
=
= 5.9 x106 rad / s
(
)
130 K 1.3 pF
(
ωZP1 = ωZP2 =0
ωPL1 =
)
ωZP3 =
⎞
⎟
⎟
⎠
1
1
=
= 42 rad / s
RBCB 2K(12µF )
1
{RB [rx + rπ + (1+ gmrπ )(RE RS )]}CB
1
ωPL2 =
1
1
=
= 33 rad / s
(RL + RC )CC2 10.2K (3µF )
⎡
⎢RE
⎢
⎣
(
AVo (dB ) = 40.2dB
1
1
=
= 252 rad / s
ωZP3 =
RECE 0.33K (12µF )
[
) [ [ ] ]
RE re
⎛ − rπ
V V V
⎜
AVo = o π e = − g m RL RC
Vπ Ve Vs
Rs + RE re ⎜⎝ rx + rπ
AVo = (− 218)(− 0.94 )(0.5) = +102.4 V / V
=
=
1
= 83 rad / s
12µF(1K )
1
= 5.0x104 rad / s
2µF(0.010K )
⎧⎪
⎡ r + r ⎤⎫⎪
CC1⎨Rs + RE ⎢ x π ⎥⎬
⎪⎩
⎣1+ gmrπ ⎦⎪⎭
1
1
ωPL3 =
=
= 33 rad / s
(RL + RC )CC2 10.2K(3µF)
ω ZH 1 = ∞, ω ZH 2 = ∞
1
1
=
= 2.5 x1010 rad / s
re RE Rs Cπ 2.4Ω(17 pF )
ω PH 1 =
[
ω PH 2 =
1
1
=
= 7.1x108 rad / s
(RC RL )Cµ 1.05K (1.3 pF )
]
Note: CB amplifier has much better
high frequency performance!
38
Comparison of CB to CE Amplifier (with same Rs = 5 Ω)
CE (with RS = 5 Ω)
⎡ r ⎤⎡ [(rx + rπ ) RB ] ⎤
Vo Vo Vπ Vi
=
= [− gm (RL RC )]⎢ π ⎥⎢
⎥
Vs Vπ Vi Vs
⎣ rx + rπ ⎦⎣⎢ Rs + [(rx + rπ ) RB ]⎦⎥
AVo = (− 218)(0.94)(0.93) = −191 V / V
AVo =
Midband Gain
AVo (dB) = 20log(−191) = 45.6dB
ωZP1 = ωZP2 =0
Low Frequency
Poles and Zeros
ωZP3 =
(
ωZP1 = ωZP2 =0
1
1
=
[RS + RB (rx + rπ )]CC1 0.7K(2µF ) = 714 rad / s
ωPL1 =
ωPL2 =
1
1
=
= 33 rad / s
(RL + RC )CC 2 10.2K (3µF )
ωPL2 =
⎡
⎢RE
⎢⎣
⎛ rx + rπ + Rs RB ⎞⎤
⎜⎜
⎟⎟⎥CE
β +1
⎝
⎠⎥⎦
ωZH 1 = ∞, ω ZH 2 =
ω PH 1 =
ω PH 2 =
1
=
= 1.7 x104 rad / s
0.005K (12µF )
g m 206 mA / V
=
= 1.6 x1011 rad / s
Cµ
1.3 pF
1
1
=
= 9.0 x108 rad / s
rπ (rx + RB RS ) Cπ 0.065K (17 pF )
[
⎞
⎟
⎟
⎠
AVo (dB ) = 40.2dB
1
1
=
= 252 rad / s
RECE 0.33K (12µF )
1
) [ [ ] ]
RE re
⎛ − rπ
V V V
⎜
AVo = o π e = − g m RL RC
Vπ Ve Vs
Rs + RE re ⎜⎝ rx + rπ
AVo = (− 218)(− 0.94 )(0.5) = +102.4 V / V
ωPL1 =
ωPL3 =
High Frequency
Poles and Zeroes
CB (with RS = 5Ω)
]
1
⎧⎪
⎤ ⎫⎪
⎡ ⎛
1
1 ⎞
⎟⎟ rπ (rx + RB RS ) ⎥ ⎬Cµ
+
⎨(RC RL )⎢1 + ⎜⎜ g m +
RC RL ⎠
⎪⎩
⎦ ⎪⎭
⎣ ⎝
1
=
= 5.0 x107 rad / s
(
)
15.4 K 1.3 pF
(
)
ωZP3 =
1
1
=
= 42 rad / s
RBCB 2K(12µF )
1
{RB [rx + rπ + (1+ gmrπ )(RE RS )]}CB
1
=
=
1
= 83 rad / s
12µF(1K )
1
= 5.0x104 rad / s
2µF(0.010K )
⎧⎪
⎡ r + r ⎤⎫⎪
CC1⎨Rs + RE ⎢ x π ⎥⎬
⎪⎩
⎣1+ gmrπ ⎦⎪⎭
1
1
ωPL3 =
=
= 33 rad / s
(RL + RC )CC2 10.2K(3µF)
ω ZH 1 = ∞, ω ZH 2 = ∞
1
1
=
= 2.5 x1010 rad / s
re RE Rs Cπ 2.4Ω(17 pF )
ω PH 1 =
[
ω PH 2 =
1
1
=
= 7.1x108 rad / s
(RC RL )Cµ 1.05K (1.3 pF )
]
Note: CB amplifier has much better
high frequency performance!
39
Conclusions
* Voltage gain
z Can get good voltage gain from both CE and CB amplifiers.
z Low frequency performance similar for both amplifiers.
z 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 gain
z For CE amplifier, current gain is high AI = Ic / Ib
z For CB amplifier, current gain is low AI = Ic / Ie (close to one)!
z Frequency dependence of current gain similar to voltage gain.
* Input and output impedances are different for the two amplifiers!
z CB amplifier has especially low input resistance.
40
Emitter-Follower (EF) Amplifier
*
*
*
High and Low Frequency AC Equivalent Circuit
*
DC biasing
z Calculate IC, IB, VCE
z Determine related small signal
equivalent circuit parameters
Transconductance gm
Input resistance rπ
Midband gain analysis
Low frequency analysis
z Gray-Searle (Short Circuit)
Technique
Determine pole frequencies ωPL1,
ωPL2, ... ωPLn
z Determine zero frequencies
ωZL1, ωZL2, ... ωZLn
High frequency analysis
z Gray-Searle (Open Circuit)
Technique
Determine pole frequencies
ωPH1, ωPH2, ... ωPHn
z Determine zero frequencies
ωZH1, ωZH2, ... ωZHn
41
EF Amplifier - DC Analysis (Nearly the Same as CE Amplifier)
*
GIVEN: Transistor parameters:
z Current gain β = 200
z Base resistance rx = 65 Ω
z Base-emitter voltage
VBE,active = 0.7 V
z Resistors: R1=10K, R2=2.5K, RC=1.2K, RE=0.33K
*
Form Thevenin equivalent for base; given VCC = 12.5V
z
RTh = RB = R1||R2 = 10K||2.5K = 2K
z
VTh = VBB = VCC R2 / [R1+R2] = 2.5V
z 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
z KVL collector loop
z VCE = VCC - (β +1) IB RE = 10.8 V (was 4.4 V for
CE amplifier) (okay since not close to zero volts).
*
*
*
R1 = 10K
R2 = 2.5K
RC = 0 K
RE = 0.33K
*
Note: Only difference here from CE case is VCE is larger
since RC was left out here in EF amplifier.
42
EF Amplifier - Midband Gain Analysis
DC analysis is nearly the same!
IB , IC and gm are all the same.
Only VCE is different since RC=0.
gm =
I C 5.27 mA
=
= 206 mA / V
VT
26 mV
+
Vo
=
Vπ
RE = 0.33K
Vi
_
_
Vo Vo Vπ Vi Vb
=
Vs Vπ Vi Vb Vs
L
⎤
⎡V
RE )⎢ π + g mVπ ⎥
⎦ = (R R )1 + g m rπ = (9 K 0.33 K ) (201) = 66
⎣ rπ
L
E
Vπ
rπ
0.97 K
Vπ
Vπ
1
1
=
=
=
= 0.015
Vi Vπ + Vo 1 + Vo 1 + 66
Vπ
Vi
Ri
65 K
=
=
= 0.999
Vb rx + Ri 0.065 K + 65 K
RC = 0 K
Vb
200
=
= 0.97 K
rπ =
g m 206 mA / V
(R
RL = 9 K
Iπ
Ri
β
AVo =
+
2 K (0.065 K + 65 K )
RB (rx + Ri )
Vb
1 .9 K
=
=
=
= 0.998
Vs RS + RB (rx + Ri ) 0.005 K + 2 K (0.065 K + 65 K ) 1.905 K
VO
R1 = 10 K
R2 = 2.5 K
RS = 0.005K = 5Ω
Equivalent input resistance Ri
⎛ V
V
V + Vo
Ri = i = π
= rπ ⎜⎜1 + o
Iπ
⎛ Vπ ⎞
⎝ Vπ
⎜
⎟
⎜r ⎟
⎝ π ⎠
⎞
⎟ = 0.97 K (1 + 66 ) = 65 K
⎟
⎠
AVo = (66 )(0.015)(0.999 )(0.998) = 0.987
AVo (dB ) = 20 log(0.987 ) = −0.1dB
NOTE: Voltage gain is only ~1!
This is a characteristic of the EF amplifier!
Cannot get voltage gain >1 for this amplifier!
43
Analysis of Low Frequency Poles
Gray-Searle (Short Circuit) Technique
* Draw low frequency AC circuit
z
z
z
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
z
Keep source resistance RS in circuit (do not remove)
* Consider the circuit one capacitor Cx at a time
z
z
z
z
Replace all other capacitors with short circuits
Solve remaining circuit for equivalent resistance Rx
seen by the selected capacitor
1
Calculate pole frequency using ω Px =
Rx C x
Repeat process for each capacitor finding equivalent
resistance seen and corresponding pole frequency
* Calculate the final low 3 dB frequency using
ω LP =
∑ω
Px = ω P1 + ω P 2 ...ω Pn =
∑ R 1C
x x
44
Emitter Follower - Analysis of Low Frequency Poles
Gray-Searle (Short Circuit) Technique
Input coupling capacitor CC1 = 2 µF
V
RxC1 = x = Rs + RB (rx + Ri )
Ix
(
V
Ri = i
Iπ
[
)
(
Vi = Iπ rπ + (Iπ + gmVπ ) RE RL = Iπ rπ + (1 + gmrπ ) RE RL
V
Ri = i = rπ + (1 + gmrπ ) RE RL
Iπ
(
(
)]
)
)
= 0.97K + (201) 0.33K 9K = 65K
RxC1 = Rs + RB (rx + Ri )
IX
Iπ
= 0.005K + 2K (0.065K + 65K ) = 1.95K
Ri
CC1RxC1 = 2µF (1.95K ) = 3.9x10−3 sec
ω PL1 =
1
1
=
= 256 rad / s
CC1RxC1 3.9x10−3 sec
Vi
45
Emitter Follower - Analysis of Low Frequency Poles
Gray-Searle (Short Circuit) Technique
*
Output coupling capacitor CC2 = 3 µF
V
re = e
Ie
RxC 2 = RL + RE re
I e = − Iπ − g mVπ = − Iπ (1 + g m rπ )
Ve = − Iπ rπ + rx + RB RS
re
[
]
− Iπ [rπ + rx + RB RS ] rπ + rx + RB RS
=
=
− Iπ (1 + g m rπ )
=
1 + g m rπ
0.97 K + 0.065K + 2 K 0.005K
201
= 0.005K
RxC 2 = RL + RE re = 9 K + 0.33K 0.005K = 9.005K
ω PL 2 =
Iπ
Ve
re Ie
*
1
1
=
= 37 rad / s
RxC 2CC 2 9.005K (3µF )
Low 3 dB frequency
ω PL = ω PL1 + ω PL 2
IX
= 256 + 37 = 293 rad / s
So dominant low frequency pole is due to CC1 !
46
Analysis of High Frequency Poles
Gray-Searle (Open Circuit) Technique
*
*
*
Draw high frequency AC equivalent circuit
z Substitute AC equivalent circuit for transistor
(hybrid-pi model for transistor with Cπ, Cµ)
z Consider coupling and emitter bypass
capacitors CC1 and CC2 as shorts
z Turn off signal source, i.e. set Vs = 0
z Keep source resistance RS in circuit
z Neglect transistor’s output resistance ro
Consider the circuit one capacitor Cx at a time
z Replace all other transistor capacitors with
open circuits
z Solve remaining circuit for equivalent
resistance Rx seen by the selected capacitor
1
z Calculate pole frequency using
ω PHx =
Rx C x
z Repeat process for each capacitor
Calculate the final high frequency pole using
⎡
ω PH = ⎢
⎣
∑
1 ⎤
⎥
ω Px ⎦
−1
⎡ 1
1
1 ⎤
... +
=⎢
+
⎥
ω PHn ⎦
⎣ ω PH 1 ω PH 2
−1
=
1
∑R C
x x
47
Emitter Follower - Analysis of High Frequency Poles
Gray-Searle (Open Circuit) Technique
Z eq =
where
Vo
= R E R L R S'
Ie
R S' =
Vo
= equivalent resistance due
− g mV π
to dependent current source
KCL at E
Ie
*
Redrawn High Frequency Equivalent
Circuit
zπ =1/yπ
E
Zeq
Ie
Vπ
V
+ sCπVπ + gmVπ − o = 0
rπ
RE RL
⎧1
⎫
Vo
= (RE RL )⎨ + sCπ + gm ⎬ = (RE RL ){yπ + gm }
Vπ
⎩ rπ
⎭
1 1
1
1
where we define yπ = = +
= + sCπ
zπ rπ ZCπ rπ
so RS' = −
1
Vo
= − (RE RL ){yπ + gm }
gmVπ
gm
Since Zeq = RE RL RS' we can find
1
1
1
1
=
+ ' =
Zeq RE RL RS RE RL
⎛ 1 ⎞⎡ yπ ⎤
⎟⎢
= ⎜⎜
⎥
⎟
R
R
⎝ E L ⎠⎣ yπ + gm ⎦
⎡ y + gm ⎤
Zeq = (RE RL )⎢ π
⎥
⎣ yπ ⎦
⎛ 1 ⎞⎡ g m ⎤
⎟⎢
− ⎜⎜
⎥
⎟
R
R
⎝ E L ⎠⎣ yπ + gm ⎦
48
Emitter Follower - Analysis of High Frequency Poles
Gray-Searle (Open Circuit) Technique
ZB’
Total impedance between B' and ground is Z b'
zπ =1/yπ
Replace this
with this.
Modified Equivalent Circuit
ZB’
(1 + g m (RE RL ))
⎡ y + gm ⎤ 1
1
= (RE RL )⎢ π
= (RE RL ) +
⎥+
yπ
yπ
⎣ yπ ⎦ yπ
1
= (RE RL ) +
⎞
⎛
yπ
⎟
⎜
⎜ 1 + g (R R ) ⎟
m
E
L
⎠
⎝
1
1 + srπ Cπ
Substituting for yπ = + sCπ =
rπ
rπ
Z b' = Z eq +
Z b' = (RE RL ) +
1
⎛ ⎡1 + srπ Cπ ⎤ ⎞
⎜ ⎢
⎥ ⎟
⎜ ⎣ rπ
⎦ ⎟
⎜ 1 + g (R R ) ⎟
m
E
L ⎟
⎜
⎜
⎟
⎝
⎠
1
= (RE RL ) +
Cπ
1
+s
rπ [1 + g m (RE RL )] [1 + g m (RE RL )]
so Z b' is
Z b' = (RE RL ) + (rπ [1 + g m (RE RL )]) Z C '
where Z C ' =
1
=
sC ' s
1
Cπ
1 + g m (RE RL )
Looks like a resistor in
parallel with a capacitor.
49
Emitter Follower - Analysis of High Frequency Poles
Gray-Searle (Open Circuit) Technique
RxCπ
* Pole frequency for Cπ =17 pF
[
RxCπ ' = [rπ (1 + g m RE RL )] rx + RS RB + RE RL
][
[
]
= 0.97 K (1 + (206 mA / V )(0.33K 9 K )) 0.065K + 0.005K 2 K + 0.33K 9 K
]
= [64.6 K ] [0.39 K ] = 0.386 K
ω PH 1 =
1
=
1
⎡
⎤
Cπ
RxCπ ' ⎢
⎥
⎣1 + g m RE RL ⎦
ω PH 1 = 1.0 x1010 rad / s
RxCπ 'Cπ '
=
1
⎡
⎤
17 pF
0.386 K ⎢
⎥
⎣1 + (206mA / V )(0.33K 9 K )⎦
=
1
1
=
0.386 K (0.255 pF ) 9.86 x10 −11 s
50
Emitter Follower - Analysis of High Frequency Poles
Gray-Searle (Open Circuit) Technique
* Pole frequency for Cµ =1.3 pF
RxCµ = ([rπ (1 + g m RE RL )] + RE RL )
[r + R R ]
= ([0.97 K (1 + (206 mA / V )(0.33K 9 K ))]+ (0.33K 9 K )) [0.065 K + 0.005 K 2 K ]
= (64.6 K + 0.32 K )
ω PH 2 =
1
RxCµ C µ
=
x
S
B
0.07 K = 0.07 K
1
1
=
0.07 K (1.3 pF ) 9.1x10 −11 s
ω PH 2 = 1.1x1010 rad / s
51
Emitter Follower - Analysis of High Frequency Poles
Gray-Searle (Open Circuit) Technique
*
Alternative Analysis for Pole Due to Cπ
Ix-Iπ
Ix
Vx
Iπ
Ie
E
Ie+gmVπ
Note : Vx = Vπ and RxCπ =
Vx Vπ rπ Iπ
=
=
Ix Ix
Ix
KVL around base loop gives
− (I e + g mVπ )(RE RL ) − Vπ + (I x − Iπ )(rx + RS RB ) = 0
But I e = Iπ − I x and Vπ = Iπ rπ so
− (Iπ − I x + g m rπ Iπ )(RE RL ) − Vπ + (I x − Iπ )(rx + RS RB ) = 0
Rearranging we get
Iπ [(1 + g m rπ )(RE RL ) + (rx + RS RB )] = I x (RE RL + rx + RS RB )
RE RL + rx + RS RB
Iπ
=
so
I x (1 + g m rπ )(RE RL ) + (rx + RS RB )
RxCπ =
We get the same result here for the high
frequency pole associated with Cπ as we
did using the equivalent circuit
transformation.
RE RL + rx + RS RB
Vx rπ Iπ
=
= rπ
(1 + g m rπ )(RE RL ) + (rx + RS RB )
Ix
Ix
= 0.97 K
0.33K 9 K + 0.065K + 0.005K 2 K
(201)(0.33K 9K ) + 0.065K + 0.005K 2K
So the pole frequency is
1
1
ωPH 1 =
=
= 1.0 x1010 rad / s
RxCπ Cπ 0.006K (17 pF )
52
= 0.006K
Emitter Follower - Analysis of High Frequency Poles
Gray-Searle (Open Circuit) Technique
Ix-Iπ
*
Alternative Analysis for Pole Due to Cµ
⎡
R R ⎤
Vx = Vπ + (Iπ + gmVπ )(RE RL ) = Vπ ⎢1 + (1 + gmrπ ) E L ⎥
⎣
Vx
Ix
Iπ
E
Iπ+gmVπ
⎦
so
⎤
⎡
rπ
Vπ = Vx ⎢
⎥
⎢⎣ rπ + (1 + gmrπ )(RE RL )⎥⎦
We can also write
Vx = (I x − Iπ )[rx + Rs RB ] = I x [rx + Rs RB ] −
Vπ
[rx + Rs RB ]
rπ
Substituting for Vπ we get
⎤
⎡
rπ
1
[rx + Rs RB ]Vx ⎢
⎥
rπ
⎢⎣ rπ + (1 + gmrπ )(RE RL )⎥⎦
Rearranging we get
Vx = I x [rx + Rs RB ] −
RxCµ =
We get the same result here for the high
frequency pole associated with Cµ as
we did using the equivalent circuit
transformation.
rπ
Vx
=
Ix
rx + Rs RB
1
=
1
1
rx + Rs RB
+
1+
rπ + (1 + gmrπ )(RE RL ) rx + Rs RB rπ + (1 + gmrπ )(RE RL )
= [rx + Rs RB ] [rπ + (1 + gmrπ )(RE RL )]
= [0.065K + 0.005K 2K ] [0.97K + (201)(0.33K 9K )]
= [0.07K ][65.0K ] = 0.07K
So the pole frequency for Cµ is
ωPH 2 =
1
1
=
= 1.1x1010 rad / s
RxCµ Cµ 0.07K (1.3 pF )
53
Comparison of EF to CE Amplifier (For RS = 5Ω )
CE
EF
⎡ r ⎤⎡ [(rx + rπ ) RB ] ⎤
Vo Vo Vπ Vi
=
= [− gm (RL RC )]⎢ π ⎥⎢
⎥ AVo = Vo Vπ Vi Vb = (RL RE ) ⎛⎜ Ri ⎞⎟ RB (rx + Ri )
[
]
(
)
Vs Vπ Vi Vs
r
+
r
R
+
r
+
r
R
⎥
x
B ⎦
π
⎣ x π ⎦⎣⎢ s
Vπ Vi Vb Vs [rπ + [RE RL ]] ⎜⎝ rx + Ri ⎟⎠ RS + RB (rx + Ri )
AVo = (− 218)(0.94)(0.93) = −191 V / V
AVo = (66 )(0.015)(0.999 )(0.998) = 0.987 V / V
AVo =
Midband Gain
AVo (dB ) = −0.1dB
AVo (dB) = 20log(−191) = 45.6dB
ωZP1 = ωZP2 =0
Low Frequency
Poles and Zeros
1
1
=
= 252 rad / s
RECE 0.33K (12µF )
ωZP1 = ωZP2 = 0
1
1
=
ωPL1 =
[RS + RB (rx + rπ )]CC1 0.7K(2µF ) = 714 rad / s
ωPL1 =
1
1
=
= 33 rad / s
ωPL2 =
(RL + RC )CC 2 10.2K (3µF )
ωPL3 =
High Frequency
Poles and Zeroes
ωZP3 =
1
⎡
⎢RE
⎢⎣
ω ZH 1 =∞
ω PH 1 =
ω PH 2 =
⎛ rx + rπ + Rs RB ⎞⎤
⎜
⎟⎥CE
⎜
⎟
+
1
β
⎝
⎠⎥⎦
ωZH 2 =
ωPL2 =
1
=
= 1.7x104 rad / s
0.005K (12µF )
g m 206 mA / V
=
= 1.6 x1011 rad / s
Cµ
1.3 pF
1
1
=
= 9.0 x108 rad / s
rπ (rx + RB RS ) Cπ 0.065K (17 pF )
[
]
1
⎧⎪
⎡ ⎛
⎤ ⎫⎪
1
1 ⎞
⎟⎟ rπ (rx + RB RS ) ⎥ ⎬Cµ
+
⎨(RC RL )⎢1 + ⎜⎜ g m +
RC RL ⎠
⎪⎩
⎣ ⎝
⎦ ⎪⎭
1
=
= 5.0 x107 rad / s
15.4 K (1.3 pF )
(
)
1
1
=
= 256 rad / s
CC1 Rs + RB (rx + Ri ) 2µF(1.95K )
[
]
1
[RL + (RE re )]CC2
=
1
= 37 rad / s
3µF(9K )
Better low frequency response !
⎞ 1
201
⎟⎟
=
= 1.2 x1010 rad / s
(
)
C
K
pF
0
.
97
17
⎠ π
1
1
=
= 1.0 x1010 rad / s
ωPH1 =
(
)
K
pF
0
.
386
0
.
26
⎡
⎤
Cπ
RxCπ ' ⎢
⎥
+
g
1
m RE RL ⎦
⎣
1
ωPH 2 =
[rπ (1 + gm RE RL ) + RE RL ] [rx + RS RB ]Cµ
ωZH1 =∞
=
⎛ 1 + g m rπ
⎝ rπ
ωZH 2 = ⎜⎜
1
= 1.1x1010 rad / s
(
)
0.07K 1.3 pF
Much better high frequency response !
54
Conclusions
* Voltage gain
z Can get good voltage gain from CE but NOT from EF amplifier (AV ¡ 1).
z Low frequency performance better for EF amplifier.
z 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 gain
z For CE amplifier, current gain is high β = Ic/Ib
z For EF amplifier, current gain is also high Ie/Ib = β +1 !
z Frequency dependence of current gain similar to voltage gain.
* Input and output impedances are different for the two amplifiers!
55
Cascade Amplifier
EF
CE
β1 = β 2 = 100
rx1 = rx 2 ≈ 0
Cπ 1 = Cπ 2 = 13.9 pF
C µ1 = C µ 2 = 2 pF
*
*
*
*
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.
56
Cascade Amplifier - DC analysis
VTh1 =
IB1
IE1
R2
100 K
10V = 5V
VCC =
200 K
R1 + R2
RTh1 = R1 R2 = 100 K 100 K = 50 K
IB2
β1 = β 2 = 100
KVL Base Q1
VTh1 − V BE 1 = I B1 RTh1 + {[ β 1 + 1] I B1 − I B 2 }R E 1
Neglecting I B 2 as a first approximat ion
IRE1
I B1 ≈
5V − 0.7V
= 8 .9 µA
50 K + (100 + 1)4.3 K
Then
Small Signal Parameters
I C1 β1 I B1 100(8.9 µA)
mA
=
=
= 34.8
VT
VT
V
0.0256V
g m1 =
rπ 1 =
β1
g m1
gm2 =
rπ 2 =
100
=
= 2.9 K
34.8mA / V
I C 2 β 2 I B 2 100(8.7 µA)
mA
=
=
= 34.0
VT
VT
0.0256V
V
β2
gm2
100
=
= 2.9 K
34.0mA / V
I E 1 = (β 1 + 1)I B1 = (101)8 .9 µA = 899 µA
Now calculate V B 2 and I B 2
V B 2 ≈ I E 1 R E 1 = 899 µA( 4.3 K ) = 3 .87V
V B 2 = V BE 2 + (β 2 + 1)I B 2 R E 2
3.87V − 0.7V
= 8 .7 µ A
(101)3.6 K
<< I E 1 so approximat e analysis is okay .
I B2 =
I B2
57
Cascade Amplifier - Midband Gain Analysis
Iπ1
+
+
Vπ1
_
Vi
_
AVo =
Note: rx1 = rx2 = 0 so equivalent circuit is simplified.
Ri
+
Vπ2
_
Vo Vo Vπ 2 Vπ 1 Vi
=
Vs Vπ 2 Vπ 1 Vi Vs
Vo [− g m 2Vπ 2 ](RL RC )
=
= −68
Vπ 2
Vπ 2
Ri =
Iπ 1
= rπ 1 + (1 + g m1rπ 1 )(RE1 rπ 2 )
⎞
Vπ 2 (I π 1 + g m1Vπ 1 )RE1 rπ 2 ⎛ 1
mA
=
= ⎜⎜ + g m1 ⎟⎟ RE1 rπ 2 = 35.1
(4.3K 2.9 K ) = 60.8
V
Vπ 1
Vπ 1
⎝ rπ 1
⎠
2.9 K
Vπ 1
I π 1rπ 1
=
=
= 0.016
Vi
I π 1rπ 1 + (1 + g m1rπ 1 )I π 1 ( RE1 rπ 2 ) 2.9 K + (101)(4.3K 2.9 K )
(100 K 100 K ) 178 K
( R1 R2 ) Ri
Vi
=
=
= 0.91
Vs RS + RB ( R1 R2 ) 4 K + (100 K 100 K ) 178 K
I π 1rπ 1 + (1 + g m1rπ 1 )I π 1 [(RE1 rπ 2 )]
= 2.9 K + (101)(4.3K 2.9 K ) = 178 K
AVo = (− 68)(60.8)(0.016 )(0.91) = −60.2 V / V
AVo (dB ) = 20 log( − 60.2 ) = 35.6dB
Note: Voltage gain is nearly equal to
that of the CE stage, e.g. – 68 !
58
Cascade Amplifier - Low Frequency Poles and Zeroes
(s + ωZ 1L )(s + ωZ 2 L )(s + ωZ 3L )
FL ( s ) =
(s + ω P1L )(s + ω P 2 L )(s + ω P3 L )
⎛ ω Z 1L ⎞⎛ ω Z 2 L ⎞⎛ ω Z 3 L ⎞
⎟
⎜1 +
⎟⎜1 +
⎟⎜1 +
s ⎠⎝
s ⎠⎝
s ⎠
⎝
=
⎛ ω P1L ⎞⎛ ω P 2 L ⎞⎛ ω P 3 L ⎞
⎟
⎜1 +
⎟⎜1 +
⎟⎜1 +
s
s
s
⎝
⎠⎝
⎠⎝
⎠
*
*
Use Gray-Searle (Short Circuit)
Technique to find the poles.
z Three low frequency poles
z Equivalent resistance may
depend on rπ for both
transistors.
Find three low frequency zeroes.
59
Cascade Amplifier - Analysis of Low Frequency Poles
Gray-Searle (Short Circuit) Technique
Input coupling capacitor CC1 = 1 µF
RxC1 =
Vx
= Rs + RB Ri
Ix
Ri =
Vi
Iπ
rπ1
Vi = Iπ1rπ1 + (Iπ1 + gm1Vπ1 )(RE1 rπ 2 ) = Iπ1[rπ1 + (1+ gm1rπ1 )(RE1 rπ 2 )]
Ri =
Vi
= rπ1 + (1+ gm1rπ1 )(RE1 rπ 2 )
Iπ
RE1
= 2.9K + (101)(4.3K 2.9K ) = 178K
RxC1 = Rs + RB Ri
ωPL1 =
1
1
=
= 23 rad / s
CC1RxC1 4.3x10−2 sec
+
Vπ2
_
rπ2
IX
Iπ1
= 4K + 50K 178K = 43.0K
CC1RxC1 = 1µF(43.0K ) = 4.3x10−2 sec
Vπ1
Ri
Vi
RE1
rπ2
60
Cascade Amplifier - Analysis of Low Frequency Poles
Gray-Searle (Short Circuit) Technique
CC2
rX2
gm2Vπ2
Vπ2
r π2
RE2
*
RC
Vo
RL
CE
Output coupling capacitor CC2 = 1 µF
VX
Vo
RC
RL
RC 2 = RL + RC = 4 K + 4 K = 8 K
ω PL 2 =
1
1
=
= 125 rad / s
RC 2CC 2 8K (1µF )
61
Cascade Amplifier - Analysis of Low Frequency Poles
Gray-Searle (Short Circuit) Technique
Emitter bypass capacitor CE = 47 µF
R Ex
Iπ1
V
= x = R E 2 re 2
Ix
re 2 =
− I π 2 ( rπ 2 + R E 1 re 1 ) rπ 2 + R E 1 re 1
VE 2
=
=
I e2
1 + g m 2 rπ 2
− I π 2 (1 + g m 2 rπ 2 )
re 1 =
r + RS '
− I π 1 ( rπ 1 + R S ' )
V E1
=
= π1
I e1
− I π 1 (1 + g m 1 rπ 1 ) 1 + g m 1 rπ 1
2 .9 K + 3 .7 K
=
= 0 . 065 K
101
2 . 9 K + 4 . 3 K 0 . 065 K
re 2 =
= 0 . 029 K
101
R Ex = R E 2 re 2 = 3 . 6 K 0 . 029 K = 0 . 029 K
ω PL 3 =
1
1
=
= 734 rad / s
R Ex C E
0 . 029 K (47 µ F )
r π1
RS ' = RS R1 R2
= 3.7 K
Vπ1
gm1Vπ1
VE1
Ie1
re1
RE1
Iπ2
rπ2
re2
Vπ2
Ie2
VE2
gm2Vπ2
Ix
VX
RE2
IE2
Low 3 dB Frequency
ω PL = ω PL1 + ω PL 2 + ω PL 3 = 23 + 125 + 734 = 882 rad / s
The pole for CE is the largest and therefore the
most important in determining the low 3 dB frequency.
62
Comparison of Cascade to CE Amplifier
CE*
AVo =
Midband Gain
Cascade (EF+CE)
Vo Vo Vπ 2
=
Vs Vπ 2 Vs
AVo =
AVo (dB ) = 20 log( − 60.2 ) = 35.6dB
AVo (dB) = 20log(− 30) = 29.5dB
Low Frequency
Poles and Zeros
ωZP3 =
1
ωZP1 = ωZP2 = 0 ωZP3 =
1
= 157 rad / s
4.4K (1µF )
ωPL1 =
{
ωPL2 =
1
1
=
= 125 rad / s
(RL + RC )CC2 8K(1µF)
ωPL3 =
1
1
=
= 734 rad / s
(RE2 re2 )CE 47µF(0.03K )
[R
ωPL2 =
1
1
=
= 125 rad / s
(RL + RC )CC 2 8K (1µF )
ωPL3 =
S + RB rπ 2 ]CC1
1
⎛ rπ 2 + Rs RB ⎞⎤
⎜
⎟
⎜ β +1 ⎟⎥CE
⎝
⎠⎥⎦
⎡
⎢RE 2
⎢⎣
ωZH 1 = ∞, ω ZH 2 =
High Frequency
Poles and Zeroes
1
1
=
= 5.9 rad / s
RE 2CE 3.6K (47µF )
ωPL1 =
=
ω PH 1 =
[r
π2
ω PH 2 =
2 X improvement
in voltage gain !
AVo = (− 68)(60.8)(0.016 )(0.91) = −60.2 V / V
AVo = (− 81.2)(0.37) = −30 V / V
ωZP1 = ωZP2 = 0
Vo Vπ 2 Vπ 1 Vi
Vπ 2 Vπ 1 Vi VS
=
1
= 591 rad / s
0.036K (47µF )
g m 2 40.6 mA / V
=
= 2.0 x1010 rad / s
Cµ 2
2 pF
1
1
=
= 4.8 x10 7 rad / s
RB RS Cπ 2 1.5K (13.9 pF )
]
1
⎧⎪
⎤ ⎫⎪
⎡ ⎛
1
1 ⎞
⎟⎟ rπ 2 (RB RS ) ⎥ ⎬Cµ 2
+
⎨(RC RL )⎢1 + ⎜⎜ g m 2 +
RC RL ⎠
⎪⎩
⎦ ⎪⎭
⎣ ⎝
1
=
= 4.0 x106 rad / s
(
)
125K 2 pF
(
)
1
1
=
= 5.9 rad / s
RE2CE 3.6K(47µF )
1
1
=
= 23 rad / s
RS + RB [rπ1 + (1+ gm1rπ1 )(RE2 rπ 2 )] CC1 1µF(43K )
}
ωZH 1 = ∞, ωZH 2 = ∞,
ωZH 3 = 2.5 x109 rad / s, ω ZH 4 = 1.7 x1010 rad / s
1
= 8.0 x108 rad / s,
0.09 K (13.9 pF )
25 X improvement
1
8
ω PH 2 =
= 1.4 x10 rad / s, in bandwidth !
3.6 K (2 pF )
1
ω PH 3 =
= 1.0 x108 rad / s,
0.063K (152 pF )
1
ω PH 4 =
= 2.5 x108 rad / s
2 K (2 pF )
63
ω PH 1 =
* CE stage with same transistor, biasing resistors, source resistance and load as cascade.
Comparison of Cascade to CE Amplifier
*
Why the better voltage gain for the cascade?
z
z
Ri1
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.
Ri1 = rπ 1 + (β + 1)RE1 rπ 2
= 2.9 K + (101)(4.3K 2.9 K ) = 178K
Ri2
*
Pole for Capacitor CT = 152 pF
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?
z 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)
V
− I (r + R ') r + R '
re1 =
re1
CT = Cπ 2 + C µ 2 (1 + g m 2 RL ') = 13.9 pF + 2 pF {1 + (34mA / V )2 K } = 152 pF
RL ' = RL RC = 4 K 4 K = 2 K
e1
I e1
=
π1 π1
S
S
= π1
− I π 1 (1 + g m1rπ 1 ) 1 + g m1rπ 1
2.9 K + 3.7 K
= 0.065 K
101
RX = re1 RE1 rπ 2 =0.065 K 4.3K 2.9 K = 0.063K
=
ω PH 3 =
1
= 1.0 x108 rad / s for the cascade
0.063K (152 pF )
versus 4.0 x106 rad / s for the CE amplifier.
64
Another Useful Amplifier – Cascode (CE+CB) Amplifier
*
*
*
*
For CE amplifier
the high frequency
C in 1 = C π 1 + C µ 1 [1 + g m 1 (R L R C
)] = 17 pF
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).
z Bandwidth is much improved (~130 X).
performanc e is limited
by
+ 1 . 3 pF (1 + 206 mA / V (9 K 1 . 2 K
1
= 5 . 6 x10 6 rad / s
−8
1 . 8 x10 sec
For cascode amplifier , R L for first stage CE amplifier
)) = 17 pF
+ 1 . 3 pF (1 + 218 ) = 302 pF
ω PHin =
Ri =
rx + rπ
= 5Ω
1 + g m rπ
1
(1 . 2 K 9 K )(2 . 6 pF
becomes
R i of CB amplifier
where R i is given by
Small Miller Effect
so C in 1 = C π 1 + C µ 1 [1 + g m (R L R C
ω PHin =
Large Miller Effect
)
)] = 17 pF
+ 1 . 3 pF (1 + 206 mA / V (0 . 005 K 1 . 2 K
= 7 . 3 x10 8 rad / s
)) = 17 pF
+ 1 . 3 pF (1 + 1) = 19 . 6 pF
Bandwidth is improved by a factor of 130X
over that for the CE amplifier !
65
Other Examples of Multistage Amplifiers
CE
CE
EF
EF
Darlington Pair
66
Other Examples of Multistage Amplifiers
Push – Pull Amplifier
Amplifier with Npn and Pnp Transistors
Amplifier with FETs and
Bipolar Transistors
67
Differential Amplifier
*
*
+V
o
_
*
*
*
*
*
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
z 1/2 of signal at input
z 1/2 of signal at output
z 1/2 of source resistance
Gain and frequency response similar to CE
amplifier for high frequencies
Advantage:
z Rejects common noise pickup on input
z No coupling capacitors so can operate
down to zero frequency.
68
Differential Amplifier Analysis
Midband Gain
⎛
⎞
⎟ ⎛ − g mVπ RC ⎞⎜
rπ
⎟⎜
⎟ = ⎜⎜
⎟
Vπ
⎟ ⎝
⎠⎜ rπ + rx + RS
2
⎝
⎠
0.97 K
⎛
⎞
= (− 206mA / V )(9 K )⎜
⎟ = −509
⎝ 0.97 K + 0.065K + 2.5 K ⎠
⎛ Vo
Vo ⎜ 2
2
AVo =
=
=⎜
Vs
Vs ⎜ Vπ
2
⎝
Vo
Vo
Vo /2
⎞⎛
⎟⎜ Vπ
⎟⎜ V
⎟⎜ s
⎠⎝ 2
⎞
⎟
⎟
⎟
⎠
AVo (dB) = 20 log 509 = 54.1dB
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
Vo /2
High Frequency Poles and Zeros
Dominant pole using Miller’s Thoerem
ω PH =
=
=
1
[rπ (rx + Rs / 2)][Cπ + (1 + g m RC )Cµ ]
1
1
=
0.97 K (0.065 K + 5K / 2) [17 pF + 201(1.3 pF )] (0.70 K )(278 pF )
[
]
1
1.95 x10
−7
sec
= 5.1x106 rad / s
High frequency performance is very similar to CE amplifier.
69
Summary
* In this chapter we have shown how to analyze the high and low frequency
dependence of the gain for an amplifier.
z 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.
z 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.
z Found the expressions for the corresponding poles and zeros.
z 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.
70