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7/27/2019 Lec2 Frequency Response
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Frequency Response
BJT/MOSFET
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Quiz No 1Group BGroup A
If R1=R2=2kΩ and C=1μFFind ωz & ωp.
ωz = 1/CR1 = 500 rad/sec
ωp = 1/C(R1||R2) = 1000 rad/sec
ωz = 1/CR2 = 500 rad/sec
ωp = 1/C(R1+R2) = 250 rad/sec
15-06-09
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High-Pass Filter
Low Frequency Band
2121
12
1
12
2
12
2 1
1
1
||
)(
)(
R sCR R R
sCR R
sCR
R R
R
sC R R
R
sV
sV
i
o
21
2
21
1
21
2
||1
1
)(
)(
R R
Rk
s
sk
R R sC
sCR
R R
R
sV
sV
p
z
i
o
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Low-Pass Filter
12
21
2
||1
1
)(
)(
R RC
s R R
R
sV
sV
i
o
121
2
1
1
1||
1
||
)(
)(
pi
o
sk
sC R R
sC R
sV
sV
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Low-Pass Filter
p
z
i
o
s
s
k
R R sC
sCR
R R
R
sC R R
sC R
sV
sV sT
1
1
)(1
1
1
1
)(
)()(
21
2
21
2
21
2
Find ‘k’ k = Vo(s)/Vi(s) while C is short circuited
ωZ while Vo(s)=0 s = - 1/CR2
ωp while Vi(s)=0 s = - 1/C(R1+R2)
7/27/2019 Lec2 Frequency Response
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Complete Analysis (Mid-band Frequency)
D C Analysis.
– Suppress ac independent sources – External & Internal Capacitors be Open circuited
– Determine DC operating Point I C /I D
– Calculate small signal model parameters gm, r pi, r e
Mid-Band Analysis AM.
– Suppress DC independent sources
– External Capacitors be short circuited – Internal Capacitors be Open circuited
– Replace BJT with small signal Model
– Analyze the resulting circuit to find voltage gainsGv Ai, input & output resistance
t
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omp ete na ys s ow-FL, g requency FH
Low Frequency Analysis FL.
– Suppress DC independent sources
– Internal Capacitors be Open circuited
– Replace BJT with small signal Model
– Keep the External Capacitors in the circuited – Analyze the resulting circuit to find Req for each capacitor by
using Short Circuit Time Constant Method.
High Frequency Analysis FH.
– Suppress DC independent sources
– External Capacitors be Short circuited
– Replace BJT with small signal Model
– Keep the Internal Capacitors in the circuited – Analyze the resulting circuit to find Req for each capacitor by
using Open Circuit Time Constant Method after using Miller theorem.
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Gain Function A(s)
A(s) = AMFL(s)FH(s)
• FL(s) & FH(s) are gain function in low & high
frequency bands and are dependent on
frequency
• At mid-band, the gain for ω L<< ω <<ωH
A(s) = AM
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Gain Function A(s)• Mid-band gain is determined by analyzing amplifier
equivalent circuit with the assumption• Coupling and bypass capacitors are acting as a perfect short
circuit• Internal capacitors of the amplifier are acting as a perfect open
circuit
• The low frequency function FL(s) is determined by
analyzing amplifier equivalent circuit • Including Coupling and bypass capacitor but assuming
Internal capacitors of the amplifier as a perfect open circuit
• The High frequency function FH(s) is determined byanalyzing amplifier equivalent circuit
• Including Internal capacitors of the amplifier but assumingCoupling and bypass capacitor as a perfect short circuit
121
21
.......
....... )(
p p p
z z L
s
s
s s
s s s F
1
1
.......11
.......11
)(
121
21
p p p
z z
H
s s s
s s
s F
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Dominant-Pole Approximation
ωP1 >> ωP2 , ωP3 ……..ωZ1 ,ωZ2
• Dominant-Pole Approximation can bemade if the highest frequency pole is
separated from the nearest pole or zero by
at least two octave ( a factor of four)
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Complete Analysis Amplifier
AMFL(s)FH(s)
DC Analysis Mid-band Analysis Low frequency Analysis High frequency Analysis
AM FL(s) FH(s)Suppress Independent AC
sources
Suppress Independent DC
sources
Suppress Independent DC
sources
Suppress Independent DC
sources
Open Circuit all capacitors Suppress all capacitors Open Circuit Internal capacitors Short Circuit External capacitors
Calculate Node Voltages & IC Draw Small Signal Model Draw Low Frequency Model Draw High Frequency Model
Find out Mode of operation &
SSM parameters
Find out Gain, Input & output
Impedance
Using Short Circuit Time
Constant, Find out Dominant
pole Frequency
Using Open Circuit Time
Constant, Find out Dominant pole
Frequency
......2...
.......
....... )(
22
12
22
12
121
21
z z p p L
p p p
z z L
s
s
s s
s s s F
.....11
2.........11
1)(
1
1
.......11
.......11 )(
22
12
22
12
121
21
z z p p
H
p p p
z z
H
s
s s s
s s s F
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Normalized High frequency Response
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Low Frequency Response
fL(s)
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Reading Material
Low Frequency Response
• Microelectronics Sedra & Smith
– Section 4.9 Ed 5
– Section 5.9 Ed 5 – Appendix ‘E’ Ed5
– Section 7.1 thru 7.3 Ed 4
• Microelectronics Millman
• Electric Cct Franco
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Low Frequency Response
k r
r
o
X
100
50
100
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DC analysis
External & Internal Capacitors open Circuit
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DC analysis
mA
R R
I
E B
E 1
1
7.04
100
k r r
I
V r
AmV I
V g
e
E
T e
E
T m
5.21
25
/25
V R I V V C C CC C 6
V R I V V B B BB B 4
4.0 BC V V Mode Active for Check
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Mid-band (dc) Gain AM
Suppress DC independent sources
External Capacitors be short circuitedInternal Capacitors be Open circuited
Replace BJT with small signal Model
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Mid-band (dc) Gain AM
sin
in
x
LC om
sig
o
R R
R
r r
r R Rr g
v
v
||||
sig
B
B
o
sig
o
v
v
v
v
v
v
v
v
V V v
v
sig
o /5.22
)(|| r r R R x Bin
+
-
vಗ
vB
Rin
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Low Frequency Response : Equivalent CircuitSuppress DC independent sources
Internal Capacitors be Open circuited
Replace BJT with small signal ModelKeep the External Capacitors in the circuitedAnalyze the resulting circuit to find Req for each capacitor by using Short Circuit TimeConstant Method.
Short Circuit Time Constant Analysis : C
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sec/1891
11
1
111
rad
RC
RC
C C
p
C C p
k r r R R R x BS eqC 3.5)(||1
Find Req for the CC1 by inspection
Suppress the independent voltage source.
Short-Circuit Time Constant Analysis : CC1
Keep CC1 and short cct all other Capacitors
01 zCC
Short Circuit Time Constant Analysis : C
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Short-Circuit Time Constant Analysis : CC2
Keep CC2 and short cct all other Capacitors
k R Rr R LC oC 66.9||2
sec/1031
22
2
222
rad
RC
RC
C C
p
C C p
Find Req for the CC2 by inspection
Suppress the independent voltage source.
02 zCC
or rcu me ons an na ys s :
7/27/2019 Lec2 Frequency Response
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or - rcu me ons an na ys s : E
(Zero Frequency )
sec/4.4
1
011
1||
3 Rad RC
R sC R sC
RC
E E
z
E E
E E
E E
or rcu me ons an na ys s : E
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or - rcu me ons an na ys s : E
Keep CE and short cct all other Capacitors
5.40
1
||||
s B xe E CE
R Rr r R R
sec/2470
13
3
rad RC
RC
CE CE
p
CE CE p
Find Req for the CE by inspectionSuppress the independent voltage source.
r o is neglected to simplify the analysis
Ie /(β+1) Ie(β+1)/β
Ie
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Low frequency Response
sin
in
x
LC om M R R
Rr r
r R Rr g A
||||
11
1
1
C C
p RC
01 zCC )(||1 r r R R R x BS eqC
E E
z RC
13
1
||||
s B x E CE
R Rr r R R
CE CE
p RC
13
LC oC R Rr R ||2
22
2
1
C C
p RC
02 zCC
.......
....... )( 1)( )()(
21
21
p p
z z L H L M
s s
s s s F s F s F A s A
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Low Frequency : Dominant Pole
Hz rad
Hz rad
Hz rad
Hz Rad Hz rad
Hz rad
p L
z
z z
p
p
p L
393sec/2470PoleDominanatApprox
394sec/24792470103189PoleDominantExact
5.0sec/4.4
0
393sec/247015sec/103
30sec/189
3
222
3
21
3
2
1
Hz f frequencylow L 393 3
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Low Frequency Response CE : CE, CC1 & CC2 Included
393 hZ30 hZ
16 hZ
30dB
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Designing of Coupling Capacitors
Design the coupling capacitors so that a dominant low
frequency pole at 100 Hz is obtained.Design so that the contribution of the capacitors are in
the ratio of 8:1:1.
pF C C
antpoledo f E
E
L 39.05.40
1min2
10
8
pF C C
poleother f c
C
L 30103.5
1 2
10
113
1
L L L f Hz f 2100
pF C C poleother f cC
L 5.161066.9
1
210
1
232
k RC eqC
3.51
5.40 E C
eqR k RC
C eq 66.92
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Low Frequency Response : CC Configuration
β =120
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DC Anaysis : CC Configuration
β =120
Mode ActieveV V
V R I V V V
mA R R
V V I
BC
E E BC
B E
BE CC E
4.0
3,5
595.0
1
k r r
I
V r
e
E
T e
084.51
42
Mid b d G i A
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Mid-band Gain AM
Configuration is CC so ‘T’ Model
V V v
v
k R Rr RR
R R
R
R Rr
R R
R R
R
R Rr
R Rr g
v
v
R R
R
R Rr
r R R g
v
v
v
v
v
v
v
v
sig
o
L E e Bin
sig in
in
L E e
L E
sig in
in
L E e
L E em
sig
o
sig in
in
L E e
e L E m
sig
i
i
be
be
o
sig
o
/922.0
4.147||1||
||
||
||
||
||
||
Low Frequency Response : C
7/27/2019 Lec2 Frequency Response
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Low Frequency Response : Cc1
(β+1)IB
A B
k Rr R R R
R R R
Le B sig eq
in sig eq
4.157)')(1(||1
1
3111
1 104.157
11
ceqc p C RC
IB
01
C zC
βIB
Low Frequency Response : C
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I
Low Frequency Response : Cc2
k R R
r R R Rsig B
e E Leq 72.5
)1(
||||2
3
222
2
1072.5
11
ceqc
p
C RC
I/β+1
Iβ /β+1
02
C zC
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Designing of Coupling Capacitors
Design the coupling capacitors so that a dominant low
frequency pole at 10 Hz is obtained.Design so that the contribution of the capacitors are in
the ratio of 10:1 and the total capacitance is minimized
F C C
antpoledo f c
C
L 06.31072.5
1min211
1023
2
F C C poleother f cC
L 11.11042.157
1
211
1
131
L L Lf Hz f 210
k RC eqC 72.5
2
k RC eqC 4.157
1
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Standard Capacitors
Design requires
CC1 = 1.11μF & CC2 =3.06 μF
Standard Capacitor Available commercially are
CC1 = 1.0μF & CC2 =3.3 μF
So selecting Standard capacitor
Dominant Low frequency pole @ f L=9.44 Hz
L f C 1 & C 2
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Low frequency response : Cc1 & Cc2
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High Frequency Response
fH(s)
Frequency Response : Exact Method
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Frequency Response : Exact Method
F R E t M th d
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Frequency Response : Exact MethodReference Chapter 6-6.3
Hi h F R
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High Frequency Response
Mid band Gain
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Mid-band Gain
' L M
in sig
in M R g
R R
R A
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Open Circuit Time Constant for f H• If poles and zeros can be determined easily so
calculate f H
• If not simple to find poles & zero then use opencircuit time constant method.
• Consider one capacitor at a time while reducing
all other capacitance to zero – (open circuit) andsignal source to zero
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Open Circuit Time Constant for f H
2
2
2
1
2
2
2
1
112
11
1
z z p p
H
•Determine resistance Rieq seen by Ci –
•Repeat process for other capacitors
•Compute b1 by Summing individual time constant – called open
circuit time constant
R for C Open Cct Time Constant
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Req for Cgs Open Cct Time Constant
R
1
R
||R R
0V(ii)
circuitopenis (i)as||R sees
gs
p1
gsgs
siggs
sig
sig
gs
gs
in
gd in gs
C
C
R
C RC
To find ωz1,
Find the value of ‘s’ when Vo(s) = 0
when 1/sCgs= 0, Causes Vgs=0 -- gmVgs= 0
so ωz1=∞
To find ωp1,
R for C Open Cct Time Constant
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Req for Cgd Open Cct Time Constant
To find ωz1,Find the value of ‘s’ that causes Vo(s) = 0
Vo(s) will reduce to zero, if current equal
to gmVgs flow through Cgd from left to
right, allowing no current thru RL.
m gd
z gs gd gsmC
so
gsm gd so gsC
g C
sV sC V g I
V To have
V g sC V V I
gd
gd
1
1
0
2
)(
)(
gmVgs
R f C
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Req for Cgd
circuitopenC(i) gs gd C
There is one dependent current source so apply test current IXFind Vtest, & independent source to be suppressed
To find ωp1,
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• @ node G
||where'
sig in
'
X gs R R R R I V
)(
1
)2(
''''
'
''
'
''
''
'
b RC
R R g R R I
V R
R
R R g I
R
V
R
V
R
R I R I g I
R
V V V g I
gd gd gd
Lm L X
X
L
m X
L
X
L
X
L
X L X m X
L
X gs gsm X
gd
@ Node D
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||
' R I V
R R R
x gs
sig in
'
gd gd
p gd gd gd
L Lm Lm L
x
x gd
Lm L x L xm L x x
L gsm L x x
RC RC
R R g R R R g R R I
V R
R R g R R I R R I g R R I V
RV g R R I V
1
1'''
''''
'
2
''''
''''
''
gsm s V g I I
Frequency Response
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q y p
2
2
1
1
1
1
1
)(
)(
p
z
p
M
sig
o
s
s
s A sV
sV
L sig inm L sig in gd
m
gd
sig in gs
Lm
sig in
in
sig
o
R R R g R R R sC
g C
s
R R sC R g
R R
R
sV
sV
'||'||1
1
||1
1'
)(
)(
m
gd
z
g
C 1
12
''2||||
1
L sig inm L sig in gd
p R R R g R R RC
01 z
in gs RC ||R 1sig
p1
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Miller Theorem
Miller effect
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Miller effect• The miller effect is the effective multiplication of a impedance
across a negative gain device.
• Miller effect accounts for an increase in the equivalent inputcapacitance of an inverting voltage amplifier due to amplificationof capacitance between the input and output terminals.
• Although Miller effect normally refers to capacitance, anyimpedance connected between the input and another nodeexhibiting high gain can modify the amplifier input impedance viathe Miller effect.
• This increase in input capacitance is given by
• where Av
is the gain of the amplifier and C is the feedbackcapacitance
• The Miller effect was named after John Milton Miller When Miller published his work in 1920.
v M AC C 1
Miller Effect ACC 1
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Miller Effect
• As most amplifiers are inverting amplifiers (i.e. Av < 0) theeffective capacitance at the input is larger .
• For non-inverting amplifiers, the Miller effect results in anegative capacitor at the input of the amplifier (compareNegative impedance converter ).
• The circuit is also referred as a Capacitance Multiplier.
• The capacitance on the output is often neglected since it sees
C (1 − 1 / A v ) and amplifier outputs are typically low impedance.However if the amplifier has a high impedance output, such as if a gain stage is also the output stage, then this RC can have asignificant impact on the performance of the amplifier. This iswhen pole splitting techniques are used.
Pole splitting causes the pole next in frequency (usually anoutput pole) to move to a higher frequency. This pole movementincreases the stability of the amplifier and improves its stepresponse at the cost of decreased speed
v M AC C 1
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The Miller equivalent circuit.
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q
K
Z
I
V Z
Z
KV V
Z
KV I
Z
V I
K
Z Z
I
V
Z
K V
Z
KV V I
Z
V I
11
00
1
1
2
2
2
11
2
1
2
2
2
1
1
1
1
11
1
1
1
Miller Theorem
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Miller Theorem
• Miller theorem states that bridging impedance Z can be
replaced by two impedances: Z1 connected between node 1 and
ground and Z2 connected between node 2 and ground where
circuit equivalent theobtainto
function gainV
V k where
K
Z
Z K
Z
Z
11
& 1
1
2
21
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• Miller equivalent circuit is valid only as long as theconditions that existed in the network when ‘k’ wasdetermined are not-changed
• Miller theorem is very useful in determining the inputresistance and the gain of the amplifier.
• Miller equivalent circuit cannot be used directly todetermine the output resistance of an amplifier. It is dueto the fact that test source is applied to the outputterminal and input signal is eliminated – obviously itchanges the value of ‘k’.
Miller Theorem
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Miller Theorem
Miller theorem
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Miller theorem
K=-100 V/V, Z = 1 M Ω
Example1 M ohm
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Example
k k
K
Z Z 9.9
1001
100
11
V V R Z
Z
V
V
V
V
V
V
sig sig
O
sig
O /4971001
11
1
M
K
Z Z 99.0
11
2
v
sig
o
A
R R
R
R
V
V
1
2
1
2
11
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OBSERVATIONS
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OBSERVATIONS• The Miller replacement for
a negative feedback resultsin a smaller resistance [bya factor of (1-K)] at theinput.
• The multiplication of afeedback impedance by afactor (1-k) is referred as
Miller Multiplication or Miller Effect
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Application of Miller Effect
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• If the feedback impedance is a capacitor ,thenthe Miller capacitance reflected across the inputterminal is .
• Therefore, connecting a capacitance from theinput to output is equivalent to connecting acapacitance
• Due to Miller effect, a small feedbackcapacitance appears across the input terminalsas a much larger equivalent capacitance with alarge gain (e.g. ).
• At high frequencies, this large capacitance has alow impedance that tends to short out the inputsignal
f C
Application of Miller Effect
)1(1
,
v f
Miller in AC j
Z
f C
)1( v f AC
80|| v A
Analysis using Miller’s Theorem
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Analysis using Miller s Theorem
Since Cgd is small, the current through it will be much smaller
than that of the controlled source gmVgs.
Thus neglect the current through Cgd in determining the output voltage Vo.
‘
‘
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Intrinsic/Cut-off / Unity-gain
Frequency
Gain-Bandwidth Product
Frequency
h Bode Plot
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hfe Bode Plot
Bode Plot
f
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Bode plot for (h fe ) .
84
Cct for hf (s) = I /Ib
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85
Cct for h fe ( s) = I c /I b
Unity Gain Frequency ω
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86
Unity Gain Frequency ωT
p
z
M H M b
c
s
s
s A s F s A I
I
shfe
1
1
)()()()(
0 B
C M
I
I A
r C C s s
s
s F
p
z H
1
1
1
1)(
r C C
p
z
1
pT 0
)(2
C C
g f
C C
g
mT
mT
The Cutoff Frequency
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87
The Cutoff Frequency
sC sC
r
I C C r I V b
b
1||||
V sC g V sC V g I mmC
Common Emitter short circuit gain, as a function of frequency in terms of Hybrid ‘π’
components
@ Output node
@ Input node
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Single Stage BJT Amplifier
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Single Stage BJT Amplifier Small Signal Output
Model Resistance
Pi Model Include
‘T’ Model Include
‘T’ Model Don't
Include
‘T’ Model Dent
Include
Common Emitter (CE)
Common Collector (CC)
Common Emitter (CE)
with Emitter Resistance
Common Base (CB)
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CE with RE
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E E
To analyze this configuration, we first set down the complete nodal equations
Using the relationship , the nodal equations can be rewrite in a more homogeneous form:
CE with RE
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Eliminating vo from the last two nodal equations we find that
and if we substitute this expression into the first nodal equation we find that
E E
CE with RE
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E E
•When but it reduces to
•When this expression reduces to
Finally, substituting these two expressions into the second nodal equation
we find the following expression for the voltage gain: