Frequency Response
Lesson #12
Small Signal Equivalent Circuits for
the BJT
Section 8.4-8.8
BME 373 Electronics II –
J.Schesser
22
Frequency Response
•
•
•
•
•
The gain of an amplifier is affected by the capacitance associated with its circuit. This
capacitance reduces the gain in both the low and high frequency ranges of operation.
The Bode Plot may look something like this where there is a low frequency band, a
midfrequency band and a high frequency band.
The reduction of gain in the low
frequency band is due to the coupling
and bypass capacitors selected. They
are essentially short circuits in the
mid and high bands.
The reduction of gain in the high
frequency band is due to the internal
capacitance of the amplifying device,
e.g., BJT, FET, etc.. This capacitance
is represented by capacitors in the
small signal equivalent circuit for
these devices. They are essentially
open circuits in the low and mid
bands.
First, let’s continue to study the small
signal equivalent circuits.
Mid-Frequency Band
Low Frequency Band
BME 373 Electronics II –
J.Schesser
High Frequency Band
23
Small Signal Equivalent Circuits and Parameters for the BJT
rπ-β Model
•
When the AC Portion of the input is small around the Q point (<<VT in
value) then we can approximate the operation of transistor by an
equivalent circuit consisting of a resistor, rπ=VT/IBQ and a current
source, βib, where ib is the small signal component of the base current:
o
B
o
C
βib
rπ
o
E
•
A more thorough Equivalent Circuit may be needed to specify the
performance of the BJT
BME 373 Electronics II –
J.Schesser
24
Two-Port Devices and the Hybrid Model
• Generalized model for two-port devices
+
v1
i1
i2
+
Two-port
Device
i2  h21i1  h22 v2
-
11  input relationship; h11 
v1  h11i1  h12 v2
v2
v1
i1
22  output relationship; h22 
 input resistance with output shorted-circuited
v2  0
i2
v2
 output conductance with input open-circuited
i1  0
21  forward transfer relationship; h21 
i2
i1
 forward current transfer (or gain) with output short-circuited
v2  0
(short-ciruit current gain)
12  reverse transfer relationship; h12 
v1
v2
 reverse voltage transfer (or gain) with input open-circuited
i1  0
(reverse-open-circuit voltage gain)
BME 373 Electronics II –
J.Schesser
25
Hybrid-Parameter Model for the Common
Emitter BJT
B
ic
ib
+ hie
vbe
-
+
hrevce
-
C
+
hfeib
hoe
vce
-
vbe  hieib  hre vce
ic  h feib  hoe vce
E
E
The parameters defined by this equivalent circuit as
usually provided by transistor manufacturers to describe
the performance of the BJT. For example,  and hfe are
typically given in BJT data sheets.
BME 373 Electronics II –
J.Schesser
26
Hybrid-π Model for the BJT
• Another model typically specified by BJT
manufacturers and is used for frequency analysis
• Includes
– Resistance to model the base-emitter junction, the base
to collector junction, and the collector to emitter path
– Capacitance to model base-emitter junction and the
base to collector junction
– A dependent (forward) current source in the collector
BME 373 Electronics II –
J.Schesser
27
Hybrid-π Model for the BJT (Continued)
B
rx
Cμ
+
vπ
E
•
•
•
•
•
•
•
•
-
C
rπ
rμ
Cπ
gmvπ
ro
E
rx called the base spreading resistance and represents the resistance of the base-emitter junction
rπ represents the dynamic resistance for small signal analysis and depends on the Q-point of the
design - rπ=VT/IBQ
rμ represents the feedback from the collector to the base and is related to the hybrid parameter
hre= rπ /(rπ+ rμ)
ro represents the resistance from the collector to the emitter and is related to the hybrid parameter
hoe ≈ 1/ ro and is also related to the EARLY Voltage by VA/ICQ
Cμ is the depletion capacitance of the collector-base junction
Cπ is the capacitance of the base-emitter junction and depends on the Q-point
gmvπ is the amplification factor and is equal to β ib.
Transition frequency, ft = β /[2πrπ(Cμ+Cπ)], when |Ic/Ib| is unity when the collector is grounded
for ac.
BME 373 Electronics II –
J.Schesser
28
Example
• The Hybrid-π parameters of a
2N2222A:
–
–
–
–
–
–
–
–
gm 
r 
I CQ 10m

 0.385 S,
26m
VT

 225  585 ,
g m .385
Q-point => ICQ=10 mA, VCEQ= 10V
r
Assume VT = 26 mV
r    585 -4  1.5 M,
hre 4  10
Average β => 225
ro  1  5 k ~ 40 k 
hre => 4x10-4
hoe
hoe => 25 mS ~ 200 mS
ro AVG  22.5 k,
Cμ => 8 pF

225
 C 
 8pF
C 
6
2
r
f

ft => 300 MHz (transition frequency)
2  585  300 10
 t
 196 pF
Collector-base time constant:
12
12
150

10
150

10
rxCμ=150 x 10-12

 19 
rx 
ft 
C

2 r (C  C )
BME 373 Electronics II –
J.Schesser
8
29
Analysis of CE at High Frequencies
The Hybrid-π parameters :
VCC 15V
RC
510
C2
1uF
C1
1uF
Rs
50
+
RE
-
0
L
RB
10k
vn
Vs
1V
+-
Q1
Q2N2222A
CE
100uF
1.3k
0
0
VEE –15V
RS
B
rx
+
vs
vin
RB
+
vπ
-
E
–
–
–
+
R
–
510
vout
–
0
–
–
–
–
Cμ
Q-point => ICQ=10 mA, VCEQ= 10V
Assume VT = 26 mV
gm=.385 S
rx=19 Ω
rπ=585 Ω
ro=22.5k Ω
rμ=1.5 M Ω
Cπ=196pF
Cμ=8pF
C
rπ
rμ
Cπ
gmvπ
ro
RC
RL
E
BME 373 Electronics II –
J.Schesser
30
Example Using the Miller Effect
8p
B 19
50
vs
+
+
vin
vπ
10k
r in Miller
r out Miller
C in Miller
C out Miller
+
vo
-
C
+
vin
vπ
-
510
.385vπ
196p
+
10k
510
E
B 19
50
C
1.5 M
22.5k
-
-
vs
585
-
585
.385vπ
+
22.5k
510
510
196p
vo
-
E
BME 373 Electronics II –
J.Schesser
31
Example Calculation of the Miller Parameters
and the Midband Gain
B 19
50
vs
C
+
+
vin
vπ
-
10k
.385vμ
585
22.5k
r in Miller
510
510
r out Miller
-
E
R’L=22.5k||510||510||rμOUTMILLER=252
R’s=[10k||(19+rπ||rμINMILLER)]=550
Miler Effect Parameters
Calculations
rμout MILLER=1.5M*Av /(Av-1) ≈ 1.5M
rπ||rμin MILLER = 563
Av= vo /vπ=-gmR’L=-.385*252=-97
rμin MILLER=1.5M/(1-Av )=1.5M/(1+gmR’L ) ≈ 15.k
vπ /vin =rπ||rμin MILLER / ((rπ||rμin MILLER)+rx) =.97
vin/vs=R’s/(R’s+50)=0.92
Avs= vo/vs= (vo /vπ)(vπ /vin) (vin /vs)=-97(.92)(.97)=-86.1
BME 373 Electronics II –
J.Schesser
32
Example Calculation of the Break Frequencies
r out Miller
r in Miller
B 19
50
vs
+
+
vin
vπ
10k
C
.385vπ
585
-
-
C out Miller
C in Miller
22.5k
510
510
196p
E
R’s=[(50||10k)+19]||585||rμINMILLER=61.3
R’L=22.5k||510||510||rμOUTMILLER=252
Miler Effect Parameters
rμin MILLER=15.k
rμout MILLER= 1.5M
Cμin MILLER=8p*(1-Av ) ≈ 784 pF
Cμout MILLER=8p*(Av-1) /Av ≈ 8.08 pF
Calculations
CT=196+CμinMiller= 980 pF
fb in=1/(2πR’sCT)=1/(2π*61.3*980x10-12)=2.65 MHz
fb out=1/(2πR’LCμout Miller)=1/(2π*252*8.08x10-12)=78.1 MHz
BME 373 Electronics II –
J.Schesser
33
Example Alternative Method Using Circuit
Analysis - Output Circuit
B 19 B’
50
vs
+
+
vin
v
10k π
-
E
CT=Cμinmiller +Cgs
+
vo
.385vπ
-
RB’=rμINMILLER|| rπ = 15k||585
=563
C
-
R’L=252 Cμoutmiller=8p
=784p+196p=980pf
Output Pole Frequency
1
vo
j 8 1012
 g m R' L || ZC
 .385
outmiller
1
v
252 
j 8 10 12
252
252
 .385

.385
1 j 8 1012  252
1 j 2.03 109
252 
f bout 
1
 78.1MHz
2 2.03 10 9
BME 373 Electronics II –
J.Schesser
34
Example Alternative Method Using Circuit Analysis - Input Circuit
50
B 19 B’
vs
+
+
vin
v
10k π
-
C
RB’=rμINMILLER|| rπ = 15k||585=563
Input Pole Frequency
E
CT=Cμinmiller +Cgs
RB ' || Z CT
v

vB RB ' || Z CT  rx
1
RB ' 
jCT
RB '


1
1  jCT RB '
RB ' 
jCT
RB '
v
1  jCT RB '
RB '


RB '
vB
 rx rx (1  jCT RB ' )  RB '
1  jCT RB '

-
R’L=252 Cμoutmiller=8p
=784p+196p=980pf
v v vB
 
vS vB vs
RB ' || Z CT
vo
.385vπ
-
+
RB '
RB '
1

rx  RB '  jCT RB ' rx rx  RB ' 1  jC RB ' rx
T
rx  RB '
Input Pole Frequency
RB || (rx  RB ' || Z CT )
vB

vs RB || ( rx  RB ' || Z CT )  RS
(rx  RB ' || Z CT )  rx 
r (1  jCT RB ' )  RB ' rx  RB '  jCT RB ' rx
RB '
 x

1  jCT RB '
1  jCT RB '
1  jCT RB '
rx  RB '  jCT RB ' rx
r  RB '  jCT RB ' rx
1  jCT RB '

RB || (rx  RB ' || Z CT )  RB || x
r  RB '  jCT RB ' rx
1  jCT RB '
RB  x
1  jCT RB '
RB 
RB ' rx
)
RB  (rx  RB '  jCT RB ' rx )
(rx  RB ' )


RB (1  jCT RB ' )  (rx  RB '  jCT RB ' rx ) RB  rx  RB '  jCT ( RB ' RB  RB ' rx )
RB (rx  RB ' )(1  jCT
BME 373 Electronics II –
J.Schesser
35
CT=Cμinmiller +Cgs
=784p+196p=980pf
50
+
R’L=252
B 19 B’
C
+
10k
vs
Example Input Circuit Cont’d
vin
vπ
-
-
+
vo
.385vπ
E
Cμoutmiller=8p
RB ' rx
)
(rx  RB ' )
RB || (rx  RB ' || Z CT )
RB  rx  RB '  jCT ( RB ' RB  RB ' rx )
vB


RB ' rx
vs RB || (rx  RB ' || Z CT )  RS
RB (rx  RB ' )(1  jCT
)
(rx  RB ' )
 RS
RB  rx  RB '  jCT ( RB ' RB  RB ' rx )
RB’=rμINMILLER|| rπ = 15k||585=563
RB (rx  RB ' )(1  jCT

RB (rx  RB ' )(1  jCT
RB ' rx
(rx  RB ' )
RB ' rx
)
(rx  RB ' )
RB ' rx
)
(rx  RB ' )

)  RS ( RB  rx  RB '  jCT ( RB ' RB  RB ' rx )) RB (rx  RB ' )  jCT RB ' rx RB )  RS ( RB  rx  RB '  jCT ( RB ' RB  RB ' rx ))
RB (rx  RB ' )(1  jCT
RB (rx  RB ' )(1  jCT
RB ' rx
)
(rx  RB ' )

RB (rx  RB ' )  RS ( RB  rx  RB ' )  jCT ( RB ' rx RB  RS ( RB ' RB  RB ' rx ))
RB (rx  RB ' )(1  jCT
BME 373 Electronics II –
J.Schesser
36
Example Input Circuit Cont’d
B 19 B’
50
vs
+
+
vin
v
10k π
-
C
RB’=rμINMILLER|| rπ = 15k||585
=563 Input Pole Frequency
vo
.385vπ
-
E
CT=Cμinmiller +Cgs
+
-
R’L=252 Cμoutmiller=8p
=784p+196p=980pf
v v vB
RB '
1
  
vS vB vs rx  RB ' 1  jC RB ' rx
T
rx  RB '
RB ' rx
)
(rx  RB ' )

RB (rx  RB ' )  RS ( RB  rx  RB ' )  jCT ( RB ' rx RB  RS ( RB ' RB  RB ' rx ))
RB (rx  RB ' )(1  jCT
RB RB '
RB rx  RB RB '  RS RB  RS rx  RS RB '
RB ' RB


RB ' rx RB  RS RB ' RB  RB ' rx RS
RB (rx  RB ' )  RS ( RB  rx  RB ' )  jCT ( RB ' rx RB  RS ( RB ' RB  RB ' rx )) 1  jC (
))
T
RB rx  RB RB '  RS RB  RS rx  RS RB '
fbin 
1
 2.65MHz
2 6 108
BME 373 Electronics II –
J.Schesser
37
Example Input Circuit using Thevenin’s
Theorem
B 19 B’
50
vs
+
+
vin
v
10k π
RB’=rμINMILLER|| rπ = 15k||585
E
CT=Cμnmiller +Cgs
+
vo
.385vπ
-
=563
C
-
R’L=252 Cμoutmiller=8p
=784p+196p=980pf
Thevenin's method for Input Pole
RT  ( RS || RB  rx ) || RB '
(
RS RB
R R r R r R
 rx ) || RB '  ( S B x S x B ) || RB '
RS  RB
RS  RB
RS RB  rx RS  rx RB
) RB '
RS  RB
RS RB RB '  rx RS RB '  rx RB RB '


R R r R r R
( S B x S x B )  RB ' RS RB  rx RS  rx RB  RB ' RS  RB ' RB
RS  RB
(
fbin 
1
 2.65MHz
2 CT RT
BME 373 Electronics II –
J.Schesser
38
Emitter Follower
Cμ
B rx
Rs
C
+
vs
RB
rπ
vπ
rμ
gmvπ
Cπ
ro
E
-
+
VCC 15 V
RE
Rs
C1
510
1uF
Q1
Q2N2222A
RB
10k
100uF
RE
1.3k
RL
50
0
VEE –15 V
vo
-
C2
Vs
1V
0
RL
0
BME 373 Electronics II –
J.Schesser
39
Emitter Follower (Continued)
Rs
Cμ
B rx
C
+
vs
RB
rπ
vπ
rμ
ro
gmvπ
Cπ
E
-
+
=>
R’s=Rs||RB+rx
R’L=RL||RE||ro
RE
Cπ
R’s
v’s
b’
+ vπ Cμ
BME 373 Electronics II –
J.Schesser
vo
-
E
rπ
rμ
RL
+
R’
L
gmvπ
vo
-
C
40
Emitter Follower (Continued)
Cπ
R’s
v’s
b’
+ vπ -
rμ
I fIN 
Vf
v
vb '
  
Z f Z f Z f (1  g m R 'L )
Z in Miller  Z f (1  g m R'L )
vo
L
gmvπ
C
Miller Effect Calculations for the Input
vb '  v∂  vo  (1  g m R' L )v∂
+
R’
Cμ
vo  g m R ' L v ∂
E
rπ
R’s
rπ(1+gmR’L) Cπ / (1+ gmR’L)
b’
rμ
v’s
Cμ
RT
BME 373 Electronics II –
J.Schesser
CT
41
Example
The Hybrid-π parameters :
–
–
–
–
–
–
–
–
–
–
Q-point => IEQ=10.6 mA,
VCEQ= 15V
Assume VT = 26 mV
β=225
gm=.385 S
rx=19 Ω
rπ=585 Ω
ro=22.5k Ω
rμ=1.5 M Ω
Cπ=196pF
Cμ=8pF
The Circuit parameters :
–
–
–
–
RS=510
RB=10 k
RE=1.3 k
RL=50
Calculations:
–
–
–
–
–
R’s=Rs||RB+rx= 510||10k+19=504
R’L=RL||RE||ro= 50||1.3k||22.5k=48
RT=R’S||rμ||rπ(1+gmR’L)=483
CT=Cμ+Cpπ /(1+gmR’L)= 18.1 pF
fb=1/(2πRTCT)=18.2 MHz
Recall for an emitter follower:
– Av=(β+1)R’L/[rπ+(β+1)R’L]=.949
– Rin=RB||[rπ+(β+1)R’L]= 5.33 kΩ
– Avs=AvRin /(Rin+RS)= 8.66
BME 373 Electronics II –
J.Schesser
42
Low Frequency Response of RC-Coupled
Amplifiers
• Coupling Capacitors
– To couple the various stages of a multi-stage amplifier
• For AC performance essentially a short circuit and AC current flows
from one stage to the next stage
– To support the biasing of each stage individually:
• For DC performance: open circuit and no biasing current flows from
one stage to another
• ByPass Capacitors
– To support the addition of a resistor for biasing purposes only
• For DC performance: open circuit and current flows through the
biasing resistor
– Short-circuit the biasing resistor for AC performance.
• For AC performance: short circuit and no current flows through the
resistor (shorted out/bypassed)
BME 373 Electronics II –
J.Schesser
43
Capacitors
VCC 15V
RC
510
C2
+
1uF
C1
Q1
1uF
Rs
50
Vs
+
-
+
510
vout
0
vn
RB
10k
-
RE
1V
0
RL
Q2N2222A
CE
1.3k
0
100uF
0
VEE –15V
Coupling Capacitors
Bypass Capacitors
BME 373 Electronics II –
J.Schesser
44
Coupling Capacitors
Rs
VS
Avs 
+
-
C1
Ro
+
VX
-
VO VX VY VO



VS VS VX VY
+
Rin
-
VY
C2
+
Vo
-
RL
VY=Avo VX
VX
Rin
j ( f / f1 )

,

VS Rin  RS 1  j ( f / f1 )
f1 
1
2 ( RS  Rin )C1
20log|Avsmid|
VX
RL
j( f / f2 )
,


VS RL  Ro 1  j ( f / f 2 )
f2 
1
,
2 ( Ro  RL )C2
VX
 Avo ,
VS
Avs 
Rin
j ( f / f1 )
RL
j( f / f2 )

 Avo 

Rin  RS 1  j ( f / f1 )
RL  Ro 1  j ( f / f 2 )
Avs  Avsmid 
Avsmid 
j ( f / f1 )
j( f / f2 )

1  j ( f / f1 ) 1  j ( f / f 2 )
Rin
RL
 Avo 
Rin  RS
RL  Ro
BME 373 Electronics II –
J.Schesser
f2
f1
45
Bypass Capacitors
• The value of a bypass capacitor is chosen to
provide a short circuit at a frequency sufficiently
low than the band pass of amplifier design
• For a emitter follower, it can be shown
f1=1/(2π R’E CE )
where
R’E is the equivalent resistance reflected into the
emitter circuit
BME 373 Electronics II –
J.Schesser
46
Homework
• Hybrid-
– Problems: 8.28,31
• Common Emitter
– Problems: 8.40-41
• Emitter Follower
– Problems: 8.56-7
• Low Frequency Response
– Problems: 8.60-2
BME 373 Electronics II –
J.Schesser
57