Download frequency response

FREQUENCY
RESPONSE
BJT AMPLIFIER
EMT112
1
Scope

Discuss frequency response of BJT
amplifier

Express the gain in decibels (dB)

Analyze the frequency response of a BJT
amplifier.
EMT112
2
Basic Concept

Most amplifiers have a finite range of frequencies
in which it will operate.

Frequency Response is how the capacitor limits
the passage of certain frequencies.

Coupling capacitors limit the passage of very low
frequencies.

Emitter bypass capacitor will have high reactance
to low frequencies & limiting the gain.
EMT112
3
The Decibel (dB)

A common unit of measurement of voltage gain & frequency
response.

A logarithmic measurement of the ratio of power or voltage

Power gain is expressed in dB by the formula:

AP  10 log AP
where Ap is the actual power gain, Pout/Pin

Voltage gain is expressed by:

If Av is greater than 1, the dB is positive, and if Av is less than 1,
the dB gain is negative value & usually called attenuation
AV ( dB)  20 log Av
EMT112
4
Critical Frequency
Also known as cut-off frequency
 The frequency at which output power
drops by 3 dB. [one half of it’s midrange
value.
 An output voltage drop of 3dB represents
about a 0.707 drop from the midrange
value.
 Power is often measured in units of dBm.
This is decibels with reference to 1mW of
power. [0 dBm = 1mW]

EMT112
5
Amplifier Gain Vs Frequency
Midband
Gain falls of due to the
effects of C and C
Gain falls of due to the
effects of CC and CE
EMT112
6
Definition





Frequency response of an amplifier is the graph of its gain versus
the frequency.
Cutoff frequencies : the frequencies at which the voltage gain
equals 0.707 of its maximum value.
Midband : the band of frequencies between 10fL and 0.1fH. The
voltage gain is maximum.
Bandwidth : the band between upper and lower cutoff frequencies
Outside the midband, the voltage gain can be determined by
these equations:
A
Amid
A
1  fL / f 
2
Amid
1  f / fH 
2
Above midband
Below midband
EMT112
7

Gain-bandwidth product : constant value of the
product of the voltage gain and the bandwidth.

Unity-gain frequency : the frequency at which
the amplifier’s gain is 1
fT  Amid BW
EMT112
8
Low Frequency

At low frequency range, the gain falloff
due to coupling capacitors and bypass
capacitors.

As signal frequency decrease, the XC
increase - no longer behave as short
circuits.
EMT112
9
Short-circuit Time-constant Method
(SCTC)

To determine the lower-cutoff frequency having
n coupling and bypass capacitors:
n
1
L  
i 1 RiS Ci
RiS = resistance at the terminals of the ith capacitor Ci with all
the other capacitors replaced by short circuits.
EMT112
10
Common-emitter Amplifier
Given Q-point values (1.73 mA, 2.32 V)
VCC = 12V
 = 100, VA = 75 V
R1
30 k
RS
RC
4.3 k
C2
0.1 F
C1
RL
100 k
1 k 2 F
vS
R2
10 k
vO
RE
1.3 k
C3
10 F
Therefore, r = 1.45 k, ro =44.7 k
EMT112
11
Low-frequency ac Equivalent Circuit
C2
vo
RS
C1
RC
vs
RL
RB
RE
EMT112
C3
12
Circuit For Finding R1S
RinCE
RS
R1S
RC
RB
RL
Replacing C2
and C3 by
short circuits
RB  R1 R2  7500 
R1S  RS  RB RinCE   RS  RB r   1000  7500 1450  2220 
1
1

 225 rad / s
R1S C1 2.22 k 2.00F 
EMT112
13
Circuit For Finding R2S
RoutCE
Replacing C1
and C3 by
short circuits
R2S
RC
RS
RL
RB
R2 S  RL  RC RoutCE   RL  RC ro   100 k  4.3 k 44.7k   104 k
1
1

 96.1 rad / s
R2 S C2 104 k 0.100F 
EMT112
14
Circuit For Finding R3S
RTH
Replacing C1
and C2 by
short circuits
RC||RL
RS
RB
R3S
RoutCC
RTH  RS RB  882 
R3S  RE RoutCC  RE
RE
r  RTH
1450  882
 1300
 22.7 
 1
101
1
1

 4410 rad / s
R3S C3 22.7 10F 
EMT112
15
Estimation of L
3
1
L  
 225  96.1  4410  4730 rad / s
i 1 RiS Ci
L
fL 
 753 Hz
2
EMT112
16
Common-base Amplifier
RS
100 
vS
4.7 F
1 F
vO
C2
C1
RE
RC
43 k
22 k
-VEE
RL
75 k
+VCC
Given Q-point values (0.1 mA, 5 V) and  = 100, VA = 70 V
Therefore, gm = 3.85 mS, ro = 700 kr = 26 
EMT112
17
Low-frequency ac Equivalent Circuit
RS
vs
vo
C2
C1
RC
RE
EMT112
RL
18
Circuit for finding R1S
RS
Replacing C2
by short circuit
R1S
RE
RC || RL
RinCB

r 
  100  4300 0.26  100 
R1S  RS  RE RinCB   RS   RE

1




1
1

 2.13 10 3 rad / s
R1S C1 100  4.7 F 
EMT112
19
Circuit For Finding R2S
Replacing C1
by short circuit
RoutCB
RC
RS || RE
R2S
RL
R2S  RL  RC RoutCB   RL  RC  75 k  22 k  97 k
1
1

 10.309 rad / s
R2 S C2 97 k 1F 
EMT112
20
Estimation of L
2
1
L  
 2.13 103  10.309  10.309 rad / s
i 1 RiS Ci
L
fL 
 1.64 Hz
2
EMT112
21
Common-collector Amplifier
Given Q-point values (1 mA, 5 V) and
+VCC
 = 100, VA = 70 V
RS
C1
1 k
0.1 F
RB
100 k
vS
C2
100 F
RE
3 k
vO
RL
47 k
-VEE
Therefore, r = 2.6 k, ro =70 k
EMT112
22
Low-frequency ac Equivalent Circuit
RS
C1
C2
vs
vo
RB
RE
EMT112
RL
23
Circuit For Finding R1S
RS
R1S
RinCC
Replacing C2
by short circuit
RB
RE || RL


R1S  RS  RB RinCC   RS  RB r    1ro RE RL 
 74.43k
1
1

 136.18 rad / s
R1S C1 74.43 k 0.1F 
EMT112
24
Circuit For Finding R2S
Replacing C1
by short circuit
RoutCC
R2S
RTH = RS || RB
RE
RL

RTH  r 
R2 S  RL  RE RoutCC   RL   RE
ro 
 1


 47.038 k
1
1

 0.213 rad / s
R2 S C2 47.038 k 100 F 
EMT112
25
Estimation of L
2
1
L  
 136.18  0.213  136.393 rad / s
i 1 RiS Ci
L
fL 
 21.7 Hz
2
EMT112
26
Example
Determine the total low-frequency
response of the amplifier where Q-point
values (1.6 mA,4.86 V) and  = 100,
VA = 70 V
R1
62 k
RS
VCC = 10V
RC
2.2 k
C2
0.1 F
C1
RL
10 k
600 0.1 F
R2
22 k
vS
vO
RE
1.0 k
C3
10 F
Therefore,r = 1.62 k, ro = 43.75 k, gm = 61.54 mS
EMT112
27
Low Frequency Due To C1 And C2 C3
Low frequency due to C1
R1S  RS  RB r   600  16.24k 1.62k   2.07 k
RB  R1 R2  16.24 k
f C1 
1
1

 768.86 Hz  769 Hz
2R1S C1 2 2.07 k 0.1F 
Low frequency due to C2
R2S  RL  RC ro   10 k  2.2k 43.75k   12.09 k
f C2 
1
1

 131.64 Hz  132 Hz
2R2 S C2 2 12.09 k 0.1F 
EMT112
28
Low Frequency Due To C3
Low frequency due to C3
R3S  RE
r  RTH
1.62 k  0.58 k
 1k
 21.32 
 1
101
RTH  RS RB  0.58 k
f C3 
1
1

 746.5 Hz  747 Hz
2R3S C3 2 21.32 10F 
EMT112
29
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 = Cbc ------ 2 pF ~ 50 pF
C = Cbe ------ 0.1 pF ~ 5 pF
EMT112
30
Basic data sheet for the 2N2222 bipolar transistor
Cob = Cbc
Cib = Cbe
 Output capacitance
 Input capacitance
EMT112
31
Miller’s Theorem

simplifies the analysis of feedback
amplifiers.

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.
EMT112
32
Miller Equivalent Circuit
I1

V1  V2
Z
V2

 A V1
I1

V1 (1  A)
Z
Z
I1
I2
I2

V2  V1
Z
V2

 A V1
-A

V1
 Z 


1

A


V1
V2
I2
I2
EMT112

1
V2 1  
 A 

Z

V2

 Z

1  1
 A







33
I2
I1

V1
 Z 
1  A 
V1
I1
 Z 
 

1

A


ZM1
 Z
 
1  A

-A
V1
ZM1
ZM2
V2
V2
I2
EMT112

 Z

1  1
 A










 Z 
 1
1  
 A


 Z

1  1
 A




ZM 2
V2
34






Miller Capacitance Effect
C
I1
ZM1

I2
Z
1 A


Z
1
1
A
X CM 2

XC
1
1
A
1
 CM 2

1
CM 2

-A
V1
X CM 1
ZM 2
V2
XC
1 A
1
 CM 1

1
 C (1  A )
CM 1

C (1  A)
-A
V1
CM1
CM2
EMT112
 C (1 
C (1 
V2
35
1
)
A
1
)
A
High-frequency Hybrid- Model
C
B
C
+
r
V
C
gmV
ro
-
C = Cbc
E
EMT112
C = Cbe
36
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


Cin  C  CMi
Cout  CMo
EMT112
A : midband
gain
37
High-frequency In CE Amplifier
Given  = 125, Cbe = 20 pF, Cbc = 2.4 pF, VA = 70V, VBE(on) = 0.7V.
Determine,
VCC = 10V
(a) Upper cut-off frequencies
(b) Dominant upper cut-off frequency
R1
22 k
RS
RC
2.2 k
vO
C2
10 F
C1
RL
2.2 k
600  10 F
vS
EMT112
R2
4.7 k
RE
470 
C3
10 F
38
High-frequency Hybrid- Model
With Miller Effect For CE Amplifier
RS
vs
vo
R1||R2
C
CMi
r
ro
gmV
 R1 R2 r 
 r R R  56.36
A  gm 
 RS  R1 R2 r  o C L




CMi  Cbc 1  A  2.4 p 57.36  137.66 pF
CMo
CMo
RC||RL
 midband gain
 Miller’s equivalent
capacitor at the input
 Miller’s equivalent
 1
 Cbc 1    2.4 p 1.018  2.44 pF capacitor at the output
 A
EMT112
39
Ri  RS R1 R2 r 600 22k 4.7k 1.55k  389.47
Ro  RC RL ro  2.2k 2.2k 47.62k  1.08k
f Hi
resistance at the output
 total input capacitance
 total output capacitance
1
1


 2.59MHz
2Ri Cin 2 389.47 157.66 p 
f Ho 
resistance at the input
 Thevenin’s equivalent
Cin  Cbe  CMi  20 p  137.66 p  157.66 pF
Cout  CMo  2.44 pF
 Thevenin’s equivalent
1
1

 60.39MHz
2RoCout 2 1.08k 2.44 p 
EMT112
 upper cutoff frequency
introduced by input
capacitance
 upper cutoff frequency
introduced by output
capacitance
40
Determine The Dominant
Frequency

The lowest of the two values of upper cut-off
frequencies is the dominant frequency.

Therefore, the upper cutoff frequency of this
amplifier is
f H  2.59MHz
EMT112
41
TOTAL AMPLIFIER FREQUENCY
RESPONSE
A (dB)
ideal
Amid
actual
-3dB
f (Hz)
fC1
fC2
fC4
fC3
fL
fC5
fH
EMT112
42
Total Frequency Response of CE
Amplifier
Given  = 120, Cbe = 2.2 pF, Cbc = 1 pF, VA = 100V, VBE(on) = 0.7V
Determine :
VCC = 5V
i- lower and upper cutoff frequencies
ii- midband gain
R1
33 k
RS
vO
RC
4 k
C2
2 F
C1
RL
5 k
2 k 1 F
vS
R2
22 k
EMT112
RE
4 k
C3
10 F
43
Q-point Values
VBB  VBE (on)
IB 
 2.615A
RB    1RE
VBB
R2

 VCC  2V
R1  R2
R1  R2
RB 
 13.2 k
R1  R2
I CQ  I B  0.314 mA
EMT112
44
Transistor Parameters Value
r 
VT
I CQ
 9.94 k
VA
ro 
 318.47 k
I CQ
gm 
I CQ
VT
 12.08 mS
EMT112
45
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 
EMT112
46
Lower Cutoff Frequency
Due to C1
1
1 
 130.38 rad / s
R1S C1
R1S  RS  RB r  7.67 k
Due to C2
1
2 
 55.87 rad / s
R2 S C2
R2S  RL  RC ro  8.95 k
Due to C3
1
3 
 1060.9 rad / s
R3S C3
R3S  RE
r  R
S
RB 
 1
 94.26 
3
SCTC
method
 L   1 2  3  1247.15 rad / s
Lower cutoff frequency
i 1
L
fL 
 198.49 Hz
2
EMT112
47
Upper Cutoff Frequency
Miller Capacitance
CMi  Cbc 1  A  1 p 20.47  20.47 pF
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
EMT112
48
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
EMT112
49