Download FREQUENCY RESPONSE

ANALOG ELECTRONIC
CIRCUITS 1
EKT 204
Frequency Response of
BJT Amplifiers (Part 2)
1
HIGH FREQUENCY


The gain falls off at high frequency end due to
the internal capacitances of the transistor.
Transistors exhibit charge-storage phenomena
that limit the speed and frequency of their
operation.
Small capacitances exist between the
base and collector and between the
base and emitter. These effect the
frequency characteristics of the circuit.
C = Cbe ------ 2 pF ~ 50 pF
C = Cbc ------ 0.1 pF ~ 5 pF
reverse-biased
junction
capacitance
forward-biased
junction
capacitance
2
Basic data sheet for the 2N2222 bipolar transistor
Cob = Cbc
 Output capacitance
Cib = Cbe
 Input capacitance
3
Miller’s Theorem


This theorem simplifies the analysis of
feedback amplifiers.
The theorem states that if an impedance is
connected between the input side and the
output side of a voltage amplifier, this
impedance can be replaced by two
equivalent impedances, i.e. one connected
across the input and the other connected
across the output terminals.
4
Miller Equivalent Circuit
Impedance Z is connected between the
input side and the output side of a voltage
amplifier..
I1

V1  V2
Z
V2

I1
V (1  A)
 1
Z
I2

V2  V1
Z
V2

 A V1
Z
I1
 A V1
I2
I2
V2
I2
-A

V1
 Z

1  A



V1

1
V2 1  
 A 

Z

V2

 Z

1  1
 A







5
Miller Equivalent Circuit (cont)
I2
I1

V1
 Z 
1  A 
V1
I1
 Z 
 

1

A


ZM1
 Z
 
1  A




-A
V1
ZM1
ZM2
.. The impedance Z is being
replaced by two equivalent
impedances, i.e. one connected
across the input (ZM1) and the
other connected across the
output terminals (ZM2)
V2
V2
I2
ZM 2
V2

 Z

1  1
 A










 Z 
 1
1  
 A


 Z

1  1
 A





6

Miller Capacitance Effect
C
ZM1
X CM 1
1
 CM 1
CM 1




Z
1 A
XC
1 A
I1
V1
-A
I2
ZM 2

Z
1
1
A
X CM 2

XC
1
1
A
1
 CM 2

1
CM 2

V2
1
 C (1  A )
C (1  A)
CM = Miller capacitance
Miller effect
Multiplication effect of Cµ
 C (1 
C (1 
1
)
A
1
)
A
7
High-frequency hybrid- model
C
B
C
+
r
V
C
gmV
ro
-
C = Cbe
E
C = Cbc
8
High-frequency hybrid- model
with Miller effect
B
C
r
C
CMi
gmV
ro
CMo
E
CMi  C 1  A  Cbc 1  A
1
1


CMo  C 1    Cbc 1  
A
A


Cout  CMo
Cin  C  CMi
A : midband gain
9
High-frequency in Commonemitter Amplifier
Calculation Example
VCC = 10V
Given :
 = 125, Cbe = 20 pF, Cbc = 2.4 pF,
VA = 70V, VBE(on) = 0.7V
R1
22 k
RS
RC
2.2 k
C2
10 F
C1
RL
2.2 k
600  10 F
vS
R2
4.7 k
vO
RE
470 
C3
10 F
Determine :
1. Upper cutoff frequencies
2. Dominant upper cutoff frequency
10
High-frequency hybrid- model
with Miller effect for CE amplifier
RS
vs
R1||R2
vo
C
CMi
r
ro
gmV
CMo
RC||RL
 R1 R2 r 
 r R R  56.36
A  gm 
 RS  R1 R2 r  o C L


 midband gain
CMi  Cbc 1  A  2.4 p 57.36  137.66 pF
 Miller’s equivalent

CMo

 1
 Cbc 1    2.4 p 1.018  2.44 pF
 A
capacitor at the
input
 Miller’s equivalent
capacitor at the
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output
Calculation (Cont..)
Ri  RS R1 R2 r 600 22k 4.7k 1.55k  389.47
 Thevenin’s equivalent
Ro  RC RL ro  2.2k 2.2k 47.62k  1.08k
 Thevenin’s equivalent
resistance at the input
resistance at the output
Cin  Cbe  CMi  20 p  137.66 p  157.66 pF
 total input capacitance
Cout  CMo  2.44 pF
 total output capacitance
f Hi 
f Ho
1
1

 2.59MHz
2Ri Cin 2 389.47 157.66 p 
1
1


 60.39MHz
2RoCout 2 1.08k 2.44 p 
 upper cutoff frequency
introduced by input
capacitance
 upper cutoff frequency
introduced by output
capacitance
12
How to determine the dominant
frequency


The lowest of the two values of upper cutoff
frequencies is the dominant frequency.
Therefore, the upper cutoff frequency of this
amplifier is
f H  2.59MHz
13
TOTAL AMPLIFIER FREQUENCY
RESPONSE
A (dB)
ideal
Amid
actual
-3dB
fC1
fC2
fC3
fL
fC4
fC5
f (Hz)
fH
14
Total Frequency Response of
Common-emitter Amplifier
Calculation Example
VCC = 5V
Given :
 = 120, Cbe = 2.2 pF, Cbc = 1
pF, VA = 100V, VBE(on) = 0.7V
R1
33 k
RS
2 k
vS
RC
4 k
C2
2 F
C1
RL
5 k
1 F
R2
22 k
vO
RE
4 k
C3
10 F
Determine :
1. Midband gain
2. Lower and upper cutoff frequencies
15
Step 1 - Q-point Values
VBB  VBE (on)
IB 
 2.615A
RB    1RE
VBB
R2

 VCC  2V
R1  R2
R1  R2
RB  R1 || R2 
 13.2 k
R1  R2
I CQ  I B  0.314 mA
16
Step 2 - Transistor parameters value
r 
VT
I CQ
 9.94 k
VA
ro 
 318.47 k
I CQ
gm 
I CQ
VT
 12.08 mS
17
Step 3 - Midband gain
Amid   g m
R
R
S
r
B
r 
 RB

r
r 
o
RC RL


RB   9.94k 13.2k  5.67k
RS  r RB   2k  9.94k 13.2k  7.67k
r
o

RC RL  318.47k 2.22k  2.18k

5.67k 
2.18k   19.47
Amid  12.08m
7.67k 
18
Step 4 - Lower cutoff frequency (fL)
Due to C1
Due to C2
Due to C3
SCTC
method
1
1 
 130.38 rad / s
R1S C1
R1S  RS  RB r  7.67 k
1
2 
 55.87 rad / s
R2 S C2
R2S  RL  RC ro  8.95 k
1
3 
 1060.9 rad / s
R3S C3
R3S  RE
r  R
S
RB 
 1
 94.26 
3
 L   1 2  3  1247.15 rad / s
i 1
Lower cutoff
frequency
L
fL 
 198.49 Hz
2
19
Step 5 - Upper cutoff frequency (fH)
CMi  Cbc 1  A  1 p 20.47  20.47 pF
Miller
capacitance
CMo
 1
 Cbc 1    1 p 1.051  1.05 pF
 A
Cin  Cbe  CMi  22.67 pF
Cout  CMo  1.05 pF
Input & output
resistances
Ri  RS R1 R2 r  1.48 k
Ro  RC RL ro  2.18 k
20
Step 5 - Upper cutoff frequency (fH)
Input side
Output side
f Hi
1
1


 4.74 MHz
2Ri Cin 2 1.48 k 22.67 p 
f Ho
1
1


 69.53MHz
2RoCout 2 2.18k 1.05 p 
Upper cutoff frequency
(the smallest value)
f H  4.74MHz
21
Exercise
Textbook: Donald A. Neamen,
‘MICROELECTRONICS Circuit Analysis &
Design’,3rd Edition’, McGraw Hill
International Edition, 2007
 Exercise 7.11

22