Transcript Section 5.6.4 - Harispk's Blog | Just another WordPress

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
(a)DC Analysis
(b) Small signal Analysis
The operation of the transistor as an
amplifier.
Conceptual circuit with the signal source eliminated .
(vbe =0)
DC Analysis
Signal source eliminated
vbe  0
I C  ISe
IE 
IC
IB 
IC
VBE
VT


IE
IB 
 1
VC  VCE  VCC - I C R C
Active Mode Verification
V >V -0.4 V
The collector Current & Trans-conductance
vBE  VBE  vbe
iC  I S e
iC  I C e
v BE
VT
 ISe
VBE
VT
e
vbe
VT
vbe
VT
vbe  VT

vbe 
iC  I C 1 

V
T 

Valid for vbe  10m V
Sm all Signal
Approxim at
ion
The collector Current & Trans-conductance
IC
iC  I C  vbe
VT
DC Bias
Signal Component
IC
ic  vbe
VT
ic  g m vbe
IC
g m  transconduc tance 
DirectlyProportion
al to I C
VT
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 smallsignal condition:
Base Current & Input Resistance at the Base
IC
iC  I C  vbe
VT
iC
iC  g m vbe
IC
IC
iB  

vbe
  VT
iB  I B  ib
IB 
IC

I v
g v
ib  C be  m be
VT

r input resist ancebetween Base & Emit ter
Lookingint o t hebase
v
r  be
ib

V
r 
 T
gm I B
ib  g m
vbe

iC   ib
vbe

r 

ib
gm
ie 
g m vbe

vbe 
re 

ie
gm
r    1re
Emitter Current & Input Resistance @ Emitter
iE 
iC


IC


iC

iE  I E  ie
ie 
iC


IC
I
vbe  E vbe
VT
VT
re Small resistancebetween base & Emittter,
lookinginto theEmitterKnown as emitt erresistance
re 
vbe VT

1



ie
I E gm gm
vbe  ie re
vbe  ib r  ie re    1i re
r 
iC
re    1re
ib
ib    1ie
rπ    1re
Voltage Gain
vC  VCC  RC iC
 VCC  RC I C  ic 
 VCC  RC I C   RC ic
vC  VC  RC ic
vc   RC ic   g m RC vbe
vc
IC
Av 
  g m RC   RC
vbe
VT
I C RC
VRC
Av   g m RC  

VT
VT
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
Active Mode Verification
VBE > 0.7 V
VC> VB-0.4 V
I C  ISe
IB 
VBE
VT
IC

IE  IB  IC
Calculate
gm 
IC
VT
ic  g m vbe  ib  ie
r 
re 

gm

gm


Small Signal Analysis
•
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
VT
IB
VT
IE
The Early Effect
• In real world
– (a) Collector current does show some
dependence on collector voltage
– (b) Characteristics are not perfectly
horizontal line
iC  vCE
Figure 5.19 (a) Conceptual circuit for measuring the iC –vCE
characteristics of the BJT. (b) The iC –vCE characteristics of a
practical BJT.
The Early Effect
 vCE 

iC  I S e 1 
vA 

Nonzero slopeindicatesthatoutput resistanceis not infinite
v BE
vT
 iC 

and defined as ro  


 vCE  vBE Cons tan at 
1
v A  vCE
o 
IC
I C & vCE are the valuesat operatingpoint
vA
o  '
IC
I 'C  I S e
v BE
vT
Figure 5.58 The hybrid-pi small-signal model, in
its two versions, with the resistance ro included.
The hybrid- small-signal model, with the resistance ro
included.
VA  VCE VA
ro 

IC
IC
VBE & I B and VCE & I C are DC bias values
 vo  -gm vbe ro || RC  T husgain is slightly reduced
 ro can be neglectedif ro  10RC
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 vo
v sig
and the current gain io
ii
Figure P5.130
Figure P5.130
DC Analysis
Suppress the AC (independent Sources)
Short Circuit Voltage Sources
Open Circuit the Capacitors
Calculate DC Node Voltages & Loop Currents
Figure P5.130
9v
DC Analysis
IC
2.2k
 IE
1.2k
IR
VBB
RB 
vBE
β = 100 , α = 0.99
VA = 100V
IE = ?, Rin = ?, overall gain vo/vsig, io/i1
VCC  15k
VBB 
 3.21V
27  15k
15 27
RB 
 9.64k
15  27
Solution P5.130
• DC Values
IE 
9v
VBB  VBE
R
RE  B
 1
IC
IR
2.2k
1.92 mA
9.64 KΩ
3.21 0.7
 1.94m A
9.64
1.2 
101
I C  0.99 1.94  1.92m A
IE 
RB 
v BE
3.21 V

1.94 mA
IE
1.2k
Solution P5.130
• Check for Mode
9v
VB  VBE  VE  0.7  1.941.2  2.33V
IC
IR
VC  VCC  RC I C  9  2.2 1.92  4.776V
1.92 mA
9.64 KΩ
RB 
v BE
VC  VB  0.4  4.776 2.33  2.446V
3.21 V
ACTIVE MODE VCB > - 0.4 V
2.2k

1.94 mA
IE
1.2k
Solution P5.130
• Small Signal Model
v sig
10k
27k
15k
r

vbe
vo
g m vbe
ro

RC
RL
E
IC = 1.92 mA
VT = 25 mV
β = 100 , α = 0.99
VA = 100 V
gm 
I C 0.991.94

 76.8m A/ V
VT
0.025

100
r 

 1.3k
g m 76.8
ro 
VA
 52.1k
IC
Ri  RB || r  1.15k
Ro  ro || RC  2.11k
Solution P5.130
v sig
10k
27k
15k
r

vbe

vo
g m vbe
ro
E
vo vi vo
Av   
vs vs vi
vi
Ri
1.15


vS Rsig  Ri 10  1.15
Av  
io 
vo
 -gm ro || RC || RL 
vi
Ri
 g m ro || RC || RL   8.13V / V
Rsig  Ri
vo
RL
ii 
vS
Rsig  Rin
Ai 
io
v Rsig  Ri
 o 
 45.3 A / A
ii
vS
RL
RC
RL
CE with pi Model
v sig
10k
27k
15k
r

vbe

E
vo
g m vbe
ro
RC
RL
CE with ‘T’ Model
Comparison ‘pi’ Vs ‘T’ Model
v sig
10k
27k
15k
r

vbe

E
vo
g m vbe
ro
RC
RL
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 Analysis
Suppress Independent ac Source
Voltage source ----- Short Cct
Current Sources --- Open
Capacitors ---- Open Cct
Redraw the Circuit
VC
Analysis
VB
IE=I
IC=αIE
IB=(β+1)IE
VC=VCC-ICRC
VB=-IBRB
VE=VB-VBE
VE
gm=Ic/VT
rл=β/gm
re=α/gm
Amplifiers Configurations Common Emitter
Small Signal Analysis
Suppress Independent DC Source
Voltage source ----- Short Cct
Current Sources --- Open
Capacitors ---- Short Cct
Redraw the Circuit by replacing BJT
With pi Model
Analysis
gm=Ic/VT
rл=β/gm
re=α/gm
Find Rin, Rout, Voltage Gain vo/vi
Common Emitter
+
vbe
-
Av 
vo vo vbe


vi vbe vsig

vbe
RB || R 

vsig Rsig  RB || R 
Rin=RB||rл
Rout=RC|RL
vbe
  g m RL || RC 
vsig
Short Circuit Current Gain Ais
Ais = ios/ii
ios=-gmvbe
vbe=vi=iiRin
Ais=-gmRin
RB || R 
vo
  g m RL || RC 
vi
Rsig  RB || R 
Summary : CE
Ri  RB || r ~ r
Input Resistance
Low to moderate typically a few kilohms
Output Resistance
Ro  ro || RC ~ RC
Output Resistance is relatively low
Open Circuit Voltage Gain
Av  
Ri
 g m Ro || RL  ~ g m ro || RC 
Rsig  Ri
Voltage gain of a few hundred
Short Circuit Current Gain
Ai 
io
g v
  m be   g m r  
vbe
ii
Rin
Current gain equal to β
Figure 5.61 (a) A common-emitter amplifier with an
emitter resistance Re.
Quiz No 2 (DE 27 CE)
13-03-2007
Redraw the circuit for DC
analysis
Redraw the circuit for Small
Signal pi model analysis
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 Resistance
Input Resistance
Multiplication by a factor (1+β) is known as the
Resistance Reflection Rule.
Analysis
Small Signal Analysis : CE with Emitter Resistance
Voltage Gain
Voltage gain is lower than that of CE
because of the additional term (1+β)Re
Small Signal Analysis : CE with Emitter Resistance
Current 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.
Small Signal Analysis : CB
CB has low input resistance
CB is non-inverting amplifier
Summary : CB
• Very Low input resistance
Rin=re
• Short Circuit Current Gain is nearly unity
• Open circuit Voltage Gain is equal to CE and is positive
gm 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.
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
Common
Collector
Base
(CC)
(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 r = 1.1 kΩ, β = 50, R = R = 3kΩ
e
L
s
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 P5-135
DC Analysis
Suppress the AC (independent Sources)
Short Circuit Voltage Sources
Open Circuit the Capacitors
Calculate DC Node Voltages & Loop Currents
•β=100, α=0.99
Solution P5-135
DC Analysis
 47 
VBB  
15  4.8V
 100 47 
 47 100
RB  
  32k
 147 
VBB  VBE
IE 
 0.97m A
RB
RE 
 1
I C  I E  0.99 0.97  0.96m A
VC  VCC  RC I C  15  0.96 6.8k  8.5V
gm 
IC
 3.8mA / V
VT
Small Signal Model
Suppress the DC (independent Sources)
Short Circuit Voltage Sources
Open Circuit Current Sources
Short Circuit the Capacitors
Draw the Small Signal Model
Small Signal Model
RS
ig
vb1
r 1

vi

Rin1
vo
C
RC1
g m vi 2
R B1
g m vi1
Rin 2
r 2
RC 2
RL
Small Signal Model
vb1
RS
v sig
r 1

vi

RC1
g m vi 2
R B1
r 2
RC 2
g m vi1
Rin1
Rin 2
RB1  RB 2  32k
r 1  r 2 
vo
C

gm
 2.6k
RC1  RC 2  6.8k
ro1  ro 2  
Rin1  Rin 2  RB1 || r  2.4k
vi1  Rin1 
  0.32V / V
 
vsig  Rin1  RS 
vi 2   g m vi1 RC1 || Rin2 
vi 2
 68.1V / V
vi1
vo   g m vi 2 RC 2 || RL  
vo
 59.3V / V
vi 2
vo vi1 vi 2 vo
 . .
 1292V / V
vS vS vi1 vi 2
RL
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
β =100
I = IB +IC=IE=0.33 mA
re 

gm


IC
VT

 VT
IE


VT
25

 75
I E 0.33
IB
IE
IC
Figure P5.141 (Common Base)
Small Signal Analysis
Suppress the DC (independent Sources)
Short Circuit Voltage Sources
Open Circuit Current Sources
Short Circuit the Capacitors
Draw the Small Signal Model
Small Signal Analysis
C
Rin  re  75
vo
v
i v
 e  e o
vsig vsig ve ie
ve
re

 0.5V / V
vsig re  Rsig
ie
 1 1
re
75
ve
ic=αie
B
ie
ve
E
vo
  ( R B ||R L )  0.99(100k || 1.5k)
ie
vo
 9V / V
vsig
Rin
vo
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
β=40
IE 
VCC  VBE
 2.41mA
RB


RE  

   1
VE  I E  RE  2.41V
VB  VE  VBE  3.11V
VT
re 
 10.37
IE
Figure P5.143 Common Collector ( Emitter follower)
Small Signal Analysis
Suppress the DC (independent Sources)
Short Circuit Voltage Sources
Open Circuit Current Sources
Short Circuit the Capacitors
Draw the Small Signal Model
Small Signal Model P5-143
Small Signal Model P5-143
Ri  RB || re  (  1)(RE || RL )  17.3
vo
vb vo


vsig vsig vb
vb
vb
Ri

vsig Rsig  Ri
vo
vo
Rc || RL

vb re  ( Rc || Rl )
vo
Ri
Rc || RL


 0.621 V / V
vsig Rsig  Ri re  ( Rc || Rl )
(β+1)ib
Ri
Figure P5.144
Problem
Problem
Problem 5.147
• For the circuit in Fig P5.147, called a bootstrapped 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
Calculate DC Node Voltages & Loop Currents
DC Analysis
9V
10k
10k
VE
4.5V
2k
Solution 5-147
9V
DC Analysis
10k
10k
VBB  4.5V
RB  10k
V  VBE
I E  BB
R
RC  B
 1
4.5  0.7
IE 
 1.73mA
10  10
2k 
101
I C  I E  1.71mA
VE
4.5V
2k
I C I E
gm 

 68.5mA / V
VT VT
re 
r 
VT
25
 0.99



 14.5
I C 1.73 g m 68.5

gm
 1.46k
Figure P5.147
Small Signal Model
Suppress the DC (independent Sources)
Short Circuit Voltage Sources
Open Circuit Current Sources
Short Circuit the Capacitors
Draw the Small Signal Model
Figure P5.147
Small Signal Model
C
αie
B
re
E
Solution 5-147
C
B
αie
E
i
B
E
ie
Rin
αie
C
i
B
Solution 5-147
E
ie
vo
αie
C
vb
i
i  (1   )ie
Rin 
Rin
vb
vb

(14.48  1667) (1681.48)
vb
i  (1  0.99)ie 
(168148)
v
R r
Rin  b  E e  (   1)( RE  re )  168.148k
i
1
ie 
v0 v0 vb  RE   Rin 
 


  0.93V / V
vsig vb vsig  RE  re   Rsig  Rin 
Solution 5-147
Without Boot-Strap Capacitor
Rin
Rib
vb RE  re

 (   1)( RE  re )  203k
i
1
R  (20k)
Rin  ib
 18.12k
Rib  (20k)
Rib 
v0 v0 vb  RE   Rin 
 


  0.64V / V

vsig vb vsig  RE  re   Rsig  Rin 
The value of overall voltage gain and Rin obtained by using Bootstrap
capacitor is higher than cct ,without Bootstrapping
Bootstrapping is used to avoid loading of the input cct and to have higher gain.
Comparison of Transistor Configurations
Quantity
Common
Emitter
(CE)
Common
Common
Collector
Base
(CC)
(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