Download lecture 5:bjt frequency response

LECTURE 4:
BJT AMPLIFIER
FREQUENCY RESPONSE
By:
Syahrul Ashikin Azmi
PPKSE
Lecture’s content
Objectives:
– Discuss the effect of circuit capacitances on
frequency response of an amplifier
– Analyze low and high-frequency response of
amplifiers
Topic to be covered:
–
–
–
–
–
Basic concepts
The Decibel
Low-frequency amplifier response
High-frequency amplifier response
Total amplifier frequency response
Basic concepts
Assumption: Coupling capacitors and bypass capacitors
act as short circuits to the signal voltages and open
circuits to dc voltages.
 Capacitors do not change instantaneously from a
short circuit to an open circuit as the frequency
approaches zero.
Assumption: transistors are ideal in that output signal
respond instantaneously to input signals.
 There are internal capacitances in bipolar transistor.
 Frequency response of amplifier circuits due to circuit
capacitors and transistor capacitances.
The Decibel
The decibel is a common unit of measurement of
voltage gain and frequency response. It is a logarithmic
measurement of the ratio of one power to another or one
voltage to another.
The formulas below are used for calculation of decibels
for power gain and voltage gain.
Ap(db) = 10 log Ap
Av(db) = 20 log Av
If Av is > 1, dB gain is +ve. If Av is < 1, dB gain is –ve
and usually called attenuation.
The Decibel
-0 dB reference0 dB (actual voltage gain =1)  used as a reference
gain with which to compare other values of gains.
Maximum gain occurs in range of frequencies between
upper and lower critical frequencies; is called midrange
gain and assigned as 0 dB value.
Any value of gain below midrange is expressed as a –ve
dB value.
If midrange Av is 100, gain at a certain frequency below
midrange is 50, this reduced Av is expressed as
20log(50/100)= -6 dB. This indicates 6 dB below 0 dB
reference.
Halving o/p voltage for a steady i/p voltage is always 6
dB reduction in gain.
Doubling o/p voltage is always 6 dB increase in gain.
The Decibel
-0 dB referenceFigure shows a normalized gain vs frequency curve for
several dB points. Normalized means midrange voltage
gain is assigned a value of 1 or 0 dB.
The Decibel
-Critical frequency & power measurement in
dBmThe critical frequency, also known as the cutoff
frequency or corner frequency, is the frequency at
which the output power drops by 3 dB, which represents
one-half of its midrange value.
Ap(dB) = 10 log(0.5) = -3dB
An output voltage drop of 3 dB represents about a 70.7%
drop from the midrange value.
Av(dB) = 20 log(0.707) = -3dB
Power is often measured in units of dBm. This is
decibels with reference to 1 mW of power. This means
that 0 dBm = 1 mW.
Each 3dBm increase corresponds to doubling of power,
3 dBm decrease corresponds to a halving of the power.
Basic concepts cont..
-Amplifier gain vs frequencyCc and CE short cct & Cπ
and Cμ open cct.
Gain almost constant.
Midband
Gain falls of due to the
effects of C and C
Gain falls of due to the
effects of CC and CE
-Amplifier gain vs frequencyLOW FREQUENCY RANGE
 gain decreases as the frequency decreases due to
coupling and bypass capacitor effects.
HIGH FREQUENCY RANGE
 gain decreases as the frequency increases due to stray
capacitance and transistor capacitance effects.
MIDBAND RANGE
 gain is almost constant – coupling and bypass
capacitors act as short circuits and stray and transistor
capacitances act as open circuits.
Definitions
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 10f1
and 0.1f2. The voltage gain is maximum.
Bandwidth : the band between upper and lower
cutoff frequencies
Basic concepts cont..
-Low Frequency RangeAt low frequency range, the gain falloff due to
coupling capacitors and bypass capacitors.
As signal frequency  , the XC  - no longer
behave as short circuits.
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.
Basic concepts cont..
-Low Frequency RangeCoupling capacitors C1 and C3 limit the passage of very low
frequencies. Emitter bypass C2 capacitor will have high
reactance to low frequencies as well, limiting the gain. Also
the capacitance causes a phase shift of the signal.
Low frequency amplifier response
Assume CC and CE are short-cct, midrange Av is:
 R1 R2 r 
Vo
ro RC RL 
Av    g m 
 R1 R2 r  RS 
Vs


Low frequency amplifier response cont..
At the low frequency ac equivalent circuit of a capacitor
coupled amplifier, we can see there are three RC circuits
that will limit low frequency response. The input at the
base, the output at the collector, and the emitter.
Input RC Circuit
Output RC Circuit
Bypass RC Circuit
Low frequency amplifier response cont..
-Input RC Circuit-




Rin

V
 in

Vbase
As frequency decreases,
2
2
R

X
in
C1
XC1 increases  less base
voltage due to more voltage
drop across C1. Thus,
voltage gain is reduced.
A critical point occurs when
o/p voltage is 70.7% of its
midrange value. This
condition occur when
XC1=Rin.


 Rin 
Rin


Vin  0.707Vin
Vbase 
Vin  
 2R 
 R2  R2 
in 

 Vbase 
in 
 in
  20 log( 0.707 )  3dB
20 log 
 Vin 
Low frequency amplifier response cont..
-Input RC CircuitThe frequency at which the gain is down by 3 dB is
called the lower critical frequency (fcl). This frequency
can be determined by the formula below.
X C1 
1
2f cl ( input )C1
 Rin
1
1
f cl ( input ) 
@
2RinC1 2( RS  Rin )C1
Rin @ Ri  R1 R2 r
Resistance of input
source taken into
account
fcl also known as lower cutoff frequency, lower corner
frequency or lower break frequency.
Low frequency amplifier response cont..
-Input RC CircuitThe decrease in voltage gain with frequency is
called the roll-off.
A ten times change in frequency is called a
decade.
For each ten times reduction in frequency
below fc, there is a 20 dB reduction in
voltage gain.
The attenuation measured in dB at each decade
is dB/decade.
Low frequency amplifier response cont..
-Input RC CircuitThis typical plot of dB Av
vs frequency is called
Bode plot.
From Bode plot, it is flat
(0dB) down to critical
frequency, at which point
gain drop at -20dB/
decade. Above fc are the
midrange frequencies.
Sometimes roll-off is
expressed in dB/octave,
which is a doubling or
halving of the frequency.
Midrange
Low frequency
Low frequency amplifier response cont..
-Input RC CircuitInput RC circuit also causes an increasing in phase shift
through amplifier.
At midrange, phase shift is approximately zero since
XC1=0Ω.
At lower frequencies, higher values of XC1 cause a
phase shift and o/p voltage of RC circuit leads i/p
voltage.
1  X C 1 

Phase shift in input RC circuit is:   tan 
Rin 

At critical frequency, XC1=Rin, so:
1  Rin 
  tan 1 ( 1 )  450
  tan 
 Rin 
Low frequency amplifier response cont..
-Input RC CircuitAs phase shift approach
90ο, frequency approaches
zero.
The input RC circuit causes
base voltage to lead input
voltage below midrange by
an amount equal to circuit
phase angle, θ.
Low frequency amplifier response cont..
-Output RC CircuitThe output RC circuit affects the response similarly to
the input RC circuit. The formula below is used to
determine the cutoff frequency of the output circuit.
f cl ( output )
1

2( RC ro  RL )C3
Low frequency amplifier response cont..
-Output RC CircuitPhase angle in o/p RC circuit is:
 X C3 

  tan 
 RC  RL 
1
θ≈0o for midrange frequencies and approaches 90o as
frequency approaches zero (XC3 ∞).
At critical frequency, phase shift =45o.
Low frequency amplifier response cont..
-Bypass RC CircuitAt low frequencies, XC2 is in parallel with RE creates an
impedance that reduces the voltage gain.
Example 1
VCC = 12V
Given :
Q-point values :
0.121mA, 11.27V
R1
93.7k
 = 100, VA = 100V
RS
Therefore,
r = 21.49k,
ro =826k
0.5 k
vS
RC
C2
6 k
C1
vO
0.1 F
RL
2 F
R2
6.3k
100 k
RE
0.4 k
C3
10 F
Low-frequency ac equivalent circuit
C3
vo
RS
C1
RC
vs
RB
RE
C2
RL
Circuit for finding R1S
RinCE
RS
R1S
RC
RL
Replacing C2
and C3 by
short circuits
RB
R1 S  RS  RB RinCE   RS  RB r   500  5900 2740  2371 
RB  R1 R2  5900 
1
1

 211 rad / s
R1 S C1 2.371 k 2.00F 
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  6k 105k   106 k
1
1

 94.63 rad / s
R2 S C 2 106 k 0.100F 
Circuit for finding R3S
RTH
Replacing C1
and C2 by
short circuits
RC||RL
RS
RB
RE
RoutCC
R3 S  RE RoutCC  RE
r  RTH
2740  460.94
 400
 29.37 
 1
101
RTH  RS RB  460.94
1
1

 3405 rad / s
R3 S C 3 29.37 10F 
R3S
Estimation of L
3
L  
i 1
1
 211  94.63  3405  3710.63 rad / s
RiS C i
L
fL 
 590.57 Hz
2
Total low-frequency response of amplifier
Each RC circuit has a critical frequency determined by R
and C values.
If one of RC circuits has a critical frequency higher than
the other two  it is called dominant RC circuit.
The dominant circuit determines the frequency at which
the overall gain begins to drop at -20dB/decade.
The other circuits each cause an additional 20dB/decade roll-off below their respective critical
frequencies.
Total low-frequency response of amplifier
Refer to figure below, input RC circuit is dominant and
bypass RC circuit has the lowest fc.
The overall response is shown in blue line.
Example 2
VCC = 10V
Given :
 = 100, VA = 70 V
R1
62 k
Therefore,
RS
r = 1.62 k, ro = 43.75 k,
gm = 61.54 mS
2.2 k
C1
C2
vO
0.1 F
RL
600 0.1 F
vS
R2
22 k
Determine the total lowfrequency response of the
amplifier.
RC
10 k
RE
1.0 k
C3
10 F
Example 2
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 
Example 2
Low frequency due to C3
Low frequency due to C3
r  RTH
1.62 k  0.58 k
R3S  RE
 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 
-High Frequency Range Internal capacitance of BJTs and FETs comes into play
at high frequencies limiting the gain. Remember
reactance is low at high frequencies.
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.
Cμ = Cbc ------ 2 pF ~ 50 pF
Cπ = Cbe ------ 0.1 pF ~ 5 pF
High frequency BJT amplifier response
High-frequency response is limited by
internal capacitances of the transistors.
These act like shunts around the
transistor.
Miller’s theorem allows us to view the
internal capacitances as external
capacitors for better understanding of the
effect they have on the frequency
response.
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.
Miller theorem
General case of Miller input and output capacitances.
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
Input

1
V2 1  
 A 

Z

V2

 Z

1  1
 A

Output






Miller equivalent circuit
I2
I1

V1
 Z 
1  A 
V1
I1
 Z 
 

1

A


ZM1
 Z
 
1  A
Input

-A
V
1
ZM1
ZM2
V
2
V2
I2

 Z

1  1
 A










 Z 
 1
1  
 A


 Z

1  1
 A




ZM 2
V2
Output






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)
Input
-A
V1
CM1
CM2
V2
 C (1 
C (1 
Output
1
)
A
1
)
A
High-frequency hybrid- model
C
B
C
+
r
V
C
gmV
ro
-
C = Cbe
E
C = Cbc
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
Example 3
Given :
VCC = 10V
 = 125, Cbe = 20 pF, Cbc = 2.4 pF,
VA = 70V, VBE(on) = 0.7V
RC
Determine :
R1
i-Upper cutoff frequencies
ii- Dominant upper cutoff
frequency
vS
22 k
RS
2.2 k
C1
C2
vO
10 F
RL
600  10 F
2.2 k
R2
4.7 k
RE
470 
C3
10 F
Example 3
High-frequency hybrid- model
with Miller effect for CE amplifier
Ro
Ri RS
vs
vo
R1||R2
C
CMi
r
ro
gmV
CMo
RC||RL
 R1 R2 r 
 r R R  56.36  midband gain
A  gm 
 RS  R1 R2 r  o C L




CMi  Cbc 1  A  2.4 p 57.36  137.66 pF
CMo
 Miller’s equivalent
capacitor at the input
 1
Miller’s equivalent
 Cbc 1    2.4 p 1.018  2.44 pF 
capacitor at the output
 A
Example 3
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
resistance at the input
 Thevenin’s equivalent
resistance at the output
Cin  Cbe  CMi  20 p  137.66 p  157.66 pF
Cout  CMo  2.44 pF
 Thevenin’s equivalent
 total input capacitance
 total output capacitance
1
1
f cu ( input ) 

 2.59MHz
2RiCin 2389.47157.66 p 
 upper cutoff frequency
introduced by input
capacitance
1
1
upper cutoff frequency
fcu ( output ) 

 60.39MHz introduced
by output
2RoCout 21.08k 2.44 p 
capacitance
Total high-frequency response of BJT amplifier
The lowest of the two values of upper cutoff frequencies
is the dominant frequency.
Therefore, the upper cutoff frequency of the amplifier in
previous example is:
f H  2.59 MHz
Refer to Bode plot in next slide, at fcu(input) voltage gain
begins to roll-off at -20dB/decade. At fcu(output), gain
dropping to -40dB/decade because each RC circuit
provide a -20dB/decade roll-off.
Total high-frequency response of BJT amplifier
The Bode plot of the high frequency response shown
shows the combined effects of each internal
capacitance.
Total Amplifier Frequency Response
-3 break points at lower critical
frequencies (fc1,fc2 and fc3)
produced by 3 low-frequency
RC circuits formed by Cc and CE
-2 break points at upper frequencies
fc4 and fc5 produced by 2 highfrequency RC circuits formed by
transistor’s internal capacitance
Total Amplifier Frequency Response
-Bandwidth
The range of frequencies
lying between fcl(dom) and
fcu(dom)  bandwidth of
amplifier.
Only dominant cutoff
frequency appear in
response curve because
they determine the
bandwidth.
The amplifier’s bandwidth
expressed in hertz as:
BW  f cu ( dom )  f cl ( dom )
Example 4
Total Frequency Response of CE Amplifier
VCC = 5V
Given :
 = 120, Cbe = 2.2 pF, Cbc = 1
pF, VA = 100V, VBE(on) = 0.7V
R1
33 k
Determine :
i-lower and upper cutoff
frequencies
RS
2 k
ii- midband gain
vS
RC
C2
4 k
C1
2 F
RL
1 F
R2
22 k
vO
5 k
RE
4 k
C3
10 F
Example 4
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
Example 4
Transistor parameters value
r 
VT
I CQ
 9.94 k
VA
ro 
 318.47 k
I CQ
gm 
I CQ
VT
 12.08 mS
Example 4
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 
Example 4
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
3
SCTC
method
S
RB 
 1
 L   1 2  3  1247.15 rad / s
i 1
Lower cutoff frequency
L
fL 
 198.49 Hz
2
 94.26 
Example 4
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
Example 4
upper cutoff frequency
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
Summary
The coupling and bypass capacitor of amplifier affect the
low-frequency response.
The internal capacitances affect high-frequency
response.
Each RC circuit causes voltage gain to drop at a rate of
20dB/decade.
For low frequency RC circuit, the highest critical
frequency is the dominant critical frequency.
For high frequency RC circuit, the lowest critical
frequency is the dominant critical frequency.
The bandwidth of amplifier is the range of frequencies
between dominant lower critical frequency and dominant
upper critical frequency.