section 5.6 small signal model & analysis
DESCRIPTION
Section 5.6 Small Signal Model & Analysis. Quiz No 1 DE 27 (CE). 06-03-07. State the purpose and four steps (each) that are to be taken for carrying out DC Analysis (b) Small signal Analysis. The operation of the transistor as an amplifier. - PowerPoint PPT PresentationTRANSCRIPT
Section 5.6
Small Signal Model & Analysis
Quiz No 1 DE 27 (CE)
State the purpose and four steps (each) that are to be taken for carrying out
(a)DC Analysis
(b) Small signal Analysis
06-03-07
The operation of the transistor as an amplifier.
Conceptual circuit with the signal source eliminated .
(vbe =0)
DC Analysis Signal source eliminated
RI-VVV
1
II
II
II
II
0
CCCCCEC
EB
CB
CE
SC
T
BE
VV
be
e
v
Active Mode VerificationVC>VB-0.4 V
The collector Current & Trans-conductance
beBEBE vVv
T
be
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be
T
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BE
Vv
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ionApproximatSignalSmallmVv
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The collector Current & Trans-conductance
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Component Signal Bias DC
beT
CCC v
VIIi
For vbe<< VT, the transistor behaves as a voltage-controlled current device
The trans-conductance of the controlled source is gm
Output resistance is infinity
Linear operation of the transistor under the small-signal condition:
Base Current & Input Resistance at the Base
e
me
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Voltage Gain
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VV
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iRIRViIRV
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DC-AC Models
Large Signal Model Small Signal Model
The amplifier circuit
Figure 5.51 Two slightly different versions of the simplified hybrid- model for the small-signal operation of the BJT.
Figure 5.52 Two slightly different versions of what is known as the T model of the BJT.
Small Signal Analysis
• Coupling Capacitors– Couples the input signal vi to the emitter while
blocks the DC signals
– Don’t let dc biasing established by VCC &VEE be disturbed. when vi is connected
– Capacitor is of very large value –infinite, acts as short circuit at signal frequency of interest.
Application (Steps) : Small Signal Model• Suppress ac independent sources
– ac Voltage Sources be short circuited– ac Current Sources be open circuited– Capacitors be Open circuited
• Determine DC operating Point IC
• Suppress DC independent sources – DC Voltage Sources be short circuited– DC Current Sources be open circuited– Capacitors be short circuited
• Replace BJT with small signal Model
• Analyze the resulting circuit of find voltage gain & input/output resistance
E
T
me
B
T
m
ebbemc
T
Cm
IV
gr
IV
gr
i ivgiVIgCalculate
CE
CB
SC
III
II
II
B
T
BE
VV
eActive Mode Verification VBE > 0.7 VVC> VB-0.4 V
Small Signal Analysis
The Early Effect
• In real world
– (a) Collector current does show some dependence on collector voltage
– (b) Characteristics are not perfectly horizontal line
CEC vi
Figure 5.19 (a) Conceptual circuit for measuring the iC –vCE characteristics of the BJT. (b) The iC –vCE characteristics of a practical BJT.
T
BE
BE
T
BE
vv
SCC
A
CE
C
CEA
atConsvCE
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A
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o
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Figure 5.58 The hybrid-pi small-signal model, in its two versions, with the resistance ro included.
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A
C
CEAo
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IVIV
IV
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10 if neglected becan reducedslightly isgain Thus ||-g
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m
The hybrid- small-signal model, with the resistance ro included.
Problem 5.130
• Find the common-emitter amplifier shown in Fig. P5.130, Let VCC =9V, R1 = 27kΩ, R2 = 15kΩ, RE = 1.2kΩ, and Rc = 2.2kΩ. The transistor has β = 100 and VA = 100 V. Calculate the dc bias current IE. If the amplifier operates between a source for which Rsig = 10 kΩ and a load of 2kΩ replace the transistor with its hybrid-Π model, and find the value of Rm, the voltage gain and the current gain i
o
ii sig
o
vv
Figure P5.130
Figure P5.130
DC AnalysisSuppress the AC (independent Sources)
Short Circuit Voltage Sources
Open Circuit the CapacitorsCalculate DC Node Voltages & Loop Currents
Figure P5.130
DC Analysis
BEvBBV
v9
CI k2.2RI
BR
EIk2.1
β = 100 , α = 0.99VA = 100VIE = ?, Rin = ?, overall gain vo/vsig, io/i1
kR
VkkVV
B
CCBB
64.927152715
21.3152715
Solution P5.130
• DC Values
1
B
E
BEBBE RR
VVI
mAI
mAI
C
E
92.194.199.0
94.1
10164.92.1
7.021.3
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v9
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BR
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3.21 V
9.64 KΩ
1.94 mA
1.92 mA
• Check for ModeVVVV EBEB 33.22.194.17.0
Solution P5.130
BEv
v9
CI k2.2RI
BR
EIk2.1
3.21 V
9.64 KΩ
1.94 mA
1.92 mA
VVV BC 446.233.2776.44.0
VIRVV CCCCC 776.492.12.29
ACTIVE MODE VCB > - 0.4 V
• Small Signal Model
E
kRrRkrRR
Coo
Bi
11.2||15.1||
Solution P5.130sigv
k10
k27 k15r
bevbemvg
or CR LR
ov
IC = 1.92 mAVT = 25 mVβ = 100 , α = 0.99VA = 100 V
kIVr
kg
r
VmAVIg
C
Ao
m
T
Cm
1.52
3.18.76
100
/8.76025.0
94.199.0
E
sigvk10
k27 k15r
bevbemvg
or CR LR
ov
i
o
s
i
s
ov v
vvv
vvA
Solution P5.130
AAR
RRvv
iiA
RRvi
Rvi
L
isig
S
o
i
oi
insig
Si
L
oo /3.45
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o
isig
i
S
i RRrvv
RRR
vv ||||-g
15.11015.1 m
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RA LComisig
iv /13.8||||
E
sigvk10
k27 k15r
bevbemvg
or CR LR
ov
CE with pi Model
CE with ‘T’ Model
Comparison ‘pi’ Vs ‘T’ Model
E
sigvk10
k27 k15r
bevbemvg
or CR LR
ov
Single Stage BJT Amplifier
• Three Configurations
– Common Emitter (CE)• Common Emitter (CE) with Emitter Resistance
– Common Base (CB)– Common Collector (CC)
Figure 5.60 (a) A common-emitter amplifier using the structure of Fig. 5.59.
Amplifiers Configurations
Common Emitter
Amplifiers Configurations Common Emitter
DC AnalysisSuppress Independent ac Source Voltage source ----- Short Cct Current Sources --- Open
Capacitors ---- Open Cct
Redraw the Circuit
Analysis
IE=IIC=αIE
IB=(β+1)IE
VC=VCC-ICRC
VB=-IBRB
VE=VB-VBE
VC
VE
VB
gm=Ic/VT
rл=β/gm
re=α/gm
Amplifiers Configurations Common Emitter
Small Signal AnalysisSuppress Independent DC Source Voltage source ----- Short Cct Current Sources --- Open
Capacitors ---- Short Cct
Redraw the Circuit by replacing BJTWith pi Model
Analysisgm=Ic/VT
rл=β/gm
re=α/gm
Find Rin, Rout, Voltage Gain vo/vi
Common Emitter
Rin=RB||rл
Rout=RC|RL
RRRRRRRg
vv
RRgvv
RRRRR
vv
vv
vv
vvA
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BCLm
i
o
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be
Bsig
B
sig
be
sig
be
be
o
i
ov
||||||
||
||||
vbe
+
-
Short Circuit Current Gain Ais
Ais = ios/iiios=-gmvbevbe=vi=iiRinAis=-gmRin
Summary : CE rrRR Bi ~||
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in
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oi
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Input Resistance
Output Resistance
Open Circuit Voltage Gain
Short Circuit Current Gain
Low to moderate typically a few kilohms
Voltage gain of a few hundred
Current gain equal to β
Output Resistance is relatively low
Figure 5.61 (a) A common-emitter amplifier with an emitter resistance Re.
Quiz No 2 (DE 27 CE)
Redraw the circuit for DC analysis Redraw the circuit for Small Signal pi model analysis
13-03-2007
Figure 5.61 (a) A common-emitter amplifier with an emitter resistance Re.
Figure 5.61 (a) A common-emitter amplifier with an emitter resistance Re.
Small Signal Analysis : CE with Emitter ResistanceInput Resistance
Multiplication by a factor (1+β) is known as the Resistance Reflection Rule.
Analysis
Small Signal Analysis : CE with Emitter ResistanceVoltage Gain
Voltage gain is lower than that of CE because of the additional term (1+β)Re
Small Signal Analysis : CE with Emitter ResistanceCurrent Gain
Small Signal Analysis : CE with Emitter Resistance
Summary
Re introduces negative feedback gives it the name emitter degenerative resistance
Comparison ‘T’ Vs ‘pi’ Model
A common-base amplifier
A common-base amplifier with its T model.
CB has low input resistance
CB is non-inverting amplifier
Small Signal Analysis : CB
• Very Low input resistanceRin=re
• Short Circuit Current Gain is nearly unity
• Open circuit Voltage Gain is equal to CE and is positivegm RC
• Relatively high output resistance (Rc) same as CE
• Excellent high frequency performance
• As short circuit current gain is unity Current Buffer, it accept an input signal current at a low input resistance and delivers equal current at a very high output resistance.
Summary : CB
An emitter-follower circuit : Common Collector
Non-unilateral Amplifier Input Resistance depends upon RL
Output Resistance depends upon Rsig
Common Collector An emitter-follower circuit : T model
An equivalent circuit of the Emitter Follower - CC
An equivalent circuit of the Emitter Follower - CC
Overall Voltage Gain is less than unity:RB>>Rsig, (β+1)(re+(ro||RL))>>(Rsig||RL)
The voltage at the emitter (vo) follows very closely the voltage at the input thus give the circuit the name Emitter Follower
The emitter follower : Reflecting resistance into emitter
For RB>> Rsig & ro >> RL Gain approaches Unity when Rsig/(1+β)<<RL89
Short Circuit Current Gain = 1+β
Common Collector : Output Resistance
Output Resistance is low
Summary : Common Collector
Non-unilateral Amplifier Input Resistance depends upon RL
Output Resistance depends upon Rsig
High Input Resistance
Low out Resistance
Voltage Gain ≈ unity
Relatively Large Current = 1+β
An equivalent circuit of the emitter follower
BJT Configurations
Common Emitter
Common Emitter with Emitter Resistance
Common Base : Current Buffer
Common Collector : Voltage Follower
Summary & Comparison
Comparison of Transistor Configurationsж
Quantity
Common Emitter (CE)
Common Collector (CC)
Common Base (CB)
AI Current Gain High (-50) High (50) Low (0.98)
AV Voltage Gain High (-136) Low (0.99) High (1.4)
Ri Input Resistance Medium (1 kΩ) High (154 kΩ) Low (21 Ω)
Ro
Output Resistance High (∞) Low (80 Ω) High (∞)
ж re = 1.1 kΩ, β = 50, RL= Rs = 3kΩ
Problem 5.135• The amplifier of Fig. P5.135 consists of two identical
common-emitter amplifier connected in cascade. Observe that the input resistance of the second stage, Rin2, constitutes the load resistance of the first stage.– For Vcc = 15V, R1 = 100kΩ , R2 = 47kΩ , RE = 3.9kΩ , Rc = 6.8kΩ ,
and β = 1000, determine the dc collector current and dc collector voltage of each transistor.
– Draw the small-signal equivalent circuit of the entire amplifier and give the values of all its components. Neglect ro1 and ro2
– Find Rin1 and vb1/vsig for Rsig = 5 kΩ– Find Rin2 and vb2/vb1.– For RL = 2kΩ , find vo /vb2– Find the overall voltage gain vo /vsig
Figure P5.135
Solution P5-135DC Analysis
Suppress the AC (independent Sources)Short Circuit Voltage Sources
Open Circuit the Capacitors
Calculate DC Node Voltages & Loop Currents
Solution P5-135DC Analysis
kR
VV
B
BB
32147
10047
8.41547100
47
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mARR
VVI
CCCCC
EC
BE
BEBBE
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97.0
1
VmAVIg
T
Cm /8.3
•β=100, α=0.99
Small Signal Model
Suppress the DC (independent Sources)Short Circuit Voltage SourcesOpen Circuit Current Sources
Short Circuit the Capacitors
Draw the Small Signal Model
Small Signal Model
sigv
SR 1bv
1r
iv
C
1imvg1CR
1BR 2r2imvg
2CR LR
ov
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oo
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21
21
21
21
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ov
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VVvv
vv
vv
vv
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VVvv
RRvgv
VVRR
Rvv
i
o
i
i
S
i
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o
i
oLCimo
i
i
inCimi
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in
sig
i
/1292..
/3.59||
/1.68
||
/32.0
21
21
222
1
2
2112
1
11
Small Signal Model
Figure P5.141 Common Base
For the circuit shown, Assume β=100
(a) Find the input resistance Rin
(b) Find the voltage gain vo/vsig
Figure P5.141 (Common Base)
DC Analysis
Suppress the AC (independent Sources)Short Circuit Voltage Sources
Open Circuit the Capacitors
Calculate DC Node Voltages & Loop Currents
Figure P5.141 (Common Base)
DC Analysis
Calculate DC Node Voltages & Loop Currents
IC
IB
IE
I = IB +IC=IE=0.33 mA
7533.0
25
E
T
E
T
T
Cme I
VI
V
VIg
r
β =100
Figure P5.141 (Common Base)
Small Signal Analysis
Suppress the DC (independent Sources)Short Circuit Voltage SourcesOpen Circuit Current Sources
Short Circuit the Capacitors
Draw the Small Signal Model
e
o
e
e
sig
e
sig
o
iv
vi
vv
vv
VVRr
rvv
sige
e
sig
e /5.0
7511
ee
erv
i
VVvv
sig
o /9
Small Signal Analysis
75ein rR
)5.1||100(99.0)||( kkRRiv
LBe
o
ic=αie
vo
B
C
Eve
ie
Rin
Figure P5.143 Common Collector ( Emitter follower)
For the circuit shown, Assume β=40
(a) Find IE,VE,& VB
(b) Find the input resistance Rin
(c) Find the voltage gain vo/vsig
Figure P5.143 Common Collector ( Emitter follower)
Suppress the AC (independent Sources)Short Circuit Voltage Sources
Open Circuit the Capacitors
Calculate DC Node Voltages & Loop Currents
Figure P5.143 Common Collector ( Emitter follower)
Calculate DC Node Voltages & Loop Currents
mARR
VVIB
E
BECCE 41.2
1
VRIV EEE 41.2
VVVV BEEB 11.3
β=40
37.10E
Te I
Vr
Figure P5.143 Common Collector ( Emitter follower)
Small Signal AnalysisSuppress the DC (independent Sources)
Short Circuit Voltage SourcesOpen Circuit Current Sources
Short Circuit the Capacitors
Draw the Small Signal Model
Small Signal Model P5-143
3.17)||)(1(|| LEeBi RRrRR
b
o
sig
b
sig
o
vv
vv
vv
isig
i
sig
b
RRR
vv
)||(||
lce
Lc
b
o
RRrRR
vv
VVRRr
RRRR
Rvv
lce
Lc
isig
i
sig
o / 621.0)||(
||
Small Signal Model P5-143
Ri
(β+1)ib
vb
vo
Figure P5.144
Problem
Problem
Problem 5.147
• For the circuit in Fig P5.147, called a boot-strapped follower:– Find the dc emitter current and gm, re, and rΠ Use β =
100. – Replace the BJT with its T model (neglecting ro), and
analyze the circuit to determine the input resistance Rin and the voltage gain vo/vsig.
– Repeat (b) for the case when capacitor CB is open –circuited. Compare the results with those obtained in (b) to find the advantages of bootstrapping.
Boot-Strapped Follower
P5-147
Figure P5.147
DC Analysis
Suppress the AC (independent Sources)Short Circuit Voltage Sources
Open Circuit the Capacitors
Calculate DC Node Voltages & Loop Currents
Figure P5.147
DC AnalysisCalculate DC Node Voltages & Loop Currents
V9
V5.4
k10 k10
k2
EV
DC Analysis
mAII
mAk
I
RR
VVI
kRVV
EC
E
BC
BEBBE
B
BB
71.1
73.1
10110102
7.05.41
105.4
V9
V5.4
k10 k10
k2
EV
Solution 5-147
kg
r
gIVr
VmAVI
VIg
m
mC
Te
T
E
T
Cm
46.1
5.145.68
99.073.1
25
/5.68
Figure P5.147
Small Signal Model
Suppress the DC (independent Sources)Short Circuit Voltage SourcesOpen Circuit Current Sources
Short Circuit the Capacitors
Draw the Small Signal Model
B
E
C
αie
Figure P5.147
Small Signal Model
re
B
E
C
αie
B
E
C
αieie
i
Rin
Solution 5-147
B
E
C
αieie
i
Rin
krRrRi
vR
vii
vvi
iii
vR
eEeEb
in
be
bbe
e
bin
148.168))(1(1
)168148()99.01(
)48.1681()166748.14(
)1(
VVRR
RrR
Rvv
vv
vv
insig
in
eE
E
sig
b
bsig
/93.000
Solution 5-147
vo
kkRkR
R
krRrR
iv
R
ib
ibin
eEeEb
ib
12.18)20()20(
203))(1(1
Rib
VVRR
RrR
Rvv
vv
vv
insig
in
eE
E
sig
b
bsig
/64.000
Rin
The value of overall voltage gain and Rin obtained by using Bootstrap capacitor is higher than cct ,without BootstrappingBootstrapping is used to avoid loading of the input cct and to have higher gain.
Solution 5-147Without Boot-Strap Capacitor
Comparison of Transistor Configurations
Quantity
Common Emitter (CE)
Common Collector (CC)
Common Base (CB)
AI Current Gain High High Low
AV Voltage Gain High Low High
Ri Input Resistance Medium High Low
Ro Out Resistance High Low High