Frequency Response of Amplifier Analog Electronics Engineering

10/05/2025 Frequency Response of Amplifier 1
Frequency Response of Amplifier

Low-Frequency Response (BJT) Amplifier
Low Frequency response of CE amplifier and all
other capacitively coupled amplifier is limited by
DC blocking and bypass capacitor.

Low-Frequency Response (BJT) Amplifier

Effect of Cs
• the general form of the R-C configuration is
established by the network of the following
Figure. The total resistance is Rs + Ri.
• The Cutoff frequency:
R = Ri+ Rs
Ri = Rth//rπ

Common Emitter with Coupling Capacitor
Small signal equivalent circuit
C
i
Si
L
C
R
R
f
)
(
2
1




Common Emitter with Coupling
# Corner frequency is f = 1/(2pt) where t = ReqC
Req is resistance seen by capacitor
• Ri = RB // [r π + (1+β) RE ]
• = RB // Rib
• Time constant for the circuit is function of equivalent
resistance seen by capacitor.
• τs = ( Rsi + Ri) Cc
• fL = 1/ (2 π τs )

Effect of CC
Since the coupling capacitor is
normally connected between
the O/P of the active device and
the applied load, the R-C
configuration that determines
the low cut-off frequency due to
CC appears in the Figure.
The total series resistance is now RO + RL and the cutoff frequency due to
CC is determined by:

Effect of Output Coupling Capacitor
For the network of the loaded
BJT amplifier, the ac equivalent network
for the O/P section with Vi = 0 V appears
in the following Figure.
The resulting value of Ro in
the equation of fLC is then simply

Common Source with Output Coupling
Capacitor
C
C
D
L
C
R
R
f
)
(
2
1




Effect of CE
1





 TH
E
outCC
E
e
R
r
R
R
R
R
B
S
TH R
R
R 
RS RB
R3S
RE
RC||RL
RoutCC
RTH
1
2π [𝑅 𝐸‖
[(𝑅 𝑆‖𝑅1‖𝑅2)+𝑟 π]
¿
1+β
¿ 𝐶𝐸

Emitter Bypass Capacitor

Common-emitter Amplifier
30 k
VCC = 12V
10 k
RS C1
2 F
C2
C3
10 F
0.1 F
1 k
1.3 k
4.3 k
R1 RC
RE
R2
vS
vO
RL
100 k
Given :
Q-point values :
1.73 mA, 2.32 V
 = 100, VA = 75 V
Therefore,
r = 1.45 k,
ro =44.7 k

D.C. Analysis
β

Ex 7.3 Electronic Circuits-Neamen
Rsi = 0.1 kΩ, R1 = 20 KΩ, R2 = 2.2 KΩ, RE = 0.1 KΩ, RC = 2 KΩ,
Cc = 47μF, VCC = 10 V
VBE = 0.7 V, β = 200, VA =∞
Determine Corner frequency.

β
RTH = R1//R2 = 2.2// 20 = 1.98 KΩ
VTH = R2/(R1+R2) Vcc = (2.2/(2.2+20)) * 10 =0.991 V
IBQ = = (0.991- 0.7)/ (1.98K+((1+200) *0.1)
= 0.0132 mA
ICQ = β IBQ = 200* 0.0132 = 2.64 mA
rπ = β VT/ ICQ = 200 * 0.026 /2.64 = 1.97 KΩ
g m = ICQ / VT = 2.64 / 0.026 = 101.4 mA/ V
rib = rπ + (( I + β) RE =1.97+(201* 0.1)= 22.1K Ω
RTH = R1//R2 = 1.98 KΩ
ri = RTH // rib = 1.82 KΩ C
i
Si
L
C
R
R
f
)
(
2
1




Common Source
Design the audio amplifier
circuit such that lower cut-
off frequency is 20 Hz.

C
L
D
L
C
R
R
f
)
(
2
1



Cc = 1 / (2 π (RD + RL) = ?
Cc = 0.477 μF

Emitter Follower

Common Base Configuration Amplifier

Common Base Amplifier
FLCC1 =
FLCC2 =
Write Equation for
FLCC1 andFLCC2

• When we say that the base is "grounded" in a
common-base amplifier, we're talking about the AC
small-signal analysis of the circuit. The key concept
is that there is no signal voltage on the base
terminal.
• This does not mean that there's no DC bias on that
terminal, if it is required for the circuit to operate.
But note that the base COULD be directly connected
to ground if the emitter has a negative bias
(assuming NPN transistor).

Common Collector Amplifier
𝑅𝑂=𝑅𝐸‖𝑟𝑂‖{[𝑟 𝜋 +( 𝑅𝑆‖𝑅𝐵 ¿¿ /(¿1+ 𝛽)}¿

Example 7.5 …………………………….
Determine 3dB frequency of an emitter follower with output
coupling capacitor. Β = 100 VA = 120 V Cc2 = 1 micro Farad

EX. 7.5
Determine required value of Cc2 to yield 3dB
frequency =10 Hz for an emitter follower.

• Common Collector
• Common Base
• Common Emitter
2
)
(
2
1
C
L
o
L
C
R
R
f



𝑅𝑂=𝑅𝐸‖𝑟𝑂‖{[𝑟 𝜋 +( 𝑅𝑆‖𝑅𝐵 ¿¿ /(¿ 1+ 𝛽)}¿
FLCC1 =
FLCC2 =
C
i
Si C
R
R
fls
)
(
2
1



1
2π [𝑅 𝐸‖
[(𝑅 𝑆‖𝑅1‖𝑅2)+𝑟 π]
¿
1+β
¿ 𝐶𝐸
FLCE =

Example 7.7
Determine mid-band gain, corner frequencies, and bandwidth of a circuit. VBE= 0.7 V, β = 100 and VA = ∞
Since these values differ a large in
magnitude, Corner frequencies
are far apart.
A circuit with coupling capacitor and
load capacitor.

Example
E
C
i
S
L
R
r
Ri
C
R
R
R
R
f
)
1
(
)]
(
[
2
1
2
1


 




L
L
C
H
C
R
R
f
)
(
2
1


Cutoff frequency due to Cc
Cutoff frequency due to CL

Example
• DC Analysis
• ICQ= 0.99 mA
• gm = ICQ/ VT = 0.99/0.026 = 38.08 mA/V
• rπ=β VT/ICQ = 100 * 0.026 / 0.99 = 2.626 KΩ
• = 2.626 +(101)0.5 = 53.1 KΩ
• Midband gain = gm rπ (RCIIRL )
E
R
r
Ri )
1
( 
 



Example -Mid-band gain
)
( 2
1 i
S
i
R
R
R
R
Vi
I


Ii
R
R
R
R
R
I
i
b )
)
(
(
2
1
2
1


b
I
r
V 
 
)
( L
C
m llR
R
V
g
Vo 


Av =
)
)]
(
[
1
)(
)
(
)(
(
2
1
2
1
2
1
i
S
i
L
C
m
R
R
R
R
R
R
R
R
R
llR
R
r
g
Av


 
E
R
r
Ri )
1
( 
 



Example_cut-off frequency
)
)]
(
[
1
)(
)
(
)(
(
2
1
2
1
2
1
i
S
i
L
C
m
R
R
R
R
R
R
R
R
R
llR
R
r
g
Av



 
= -6.14
E
C
i
S
L
R
r
Ri
C
R
R
R
R
f
)
1
(
)]
(
[
2
1
2
1


 




FL= 3.42 Hz
FH= 3.18 MHz
L
L
C
H
C
R
R
f
)
(
2
1


Bandwidth = 3.18 MHz - 3.42 Hz ≈ 3.18 MHz

Gain –Bandwidth Product
• GB = │AVmax│. Bandwidth
• For fixed load capacitance, above product is
constant.

Expanded Hybrid π equivalent circuit
C,B E terminals are external connections to the transister
and C’, B’, E’ are idealised internal collector ,Base and emitter regions.
gmVπ is transistor collector current controlled by internal base emitter
voltage.
ro is due to early effect.
rμ is of the order of megohm and can be neglected.
Cμ can not be neglected due to Miller Effect.

Short Circuit Current Gain
Determine Ai= Ic/Ib
A
XXXX

From KCL equation at output node,
• [Vπ / (1/jwCμ )] + Ic = gmVπ
• Ic = Vπ (gm - jwCμ )
A

Magnitude of current gain decreases with increasing frequency. At FT ,
which is cutoff frequency, this gain goes to 1.

Example
• Determine 3dB frequency of short circuit
current gain of a bipolar transistor.
• rπ = 2.6K C π = 2 pF and Cμ = 0.1 pF
Fβ = 29.1 MHz

• Calculate Fβand capacitance C πof bipolar
transistor. FT = 500 MHz at Ic = 1mA, = 100
and C
• Fβ=/ β0= 5MHz
• gm = ICQ/ VT = 38.46mA/V
• = Cπ= 11.9 pF

Miller Effect
• Miller capacitance was identified in 1920 in triode vacuum
tubes by John Milton Miller.
• The Miller effect accounts for the increase in the equivalent
input capacitance of an inverting voltage amplifier due to amplification
of the effect of capacitance between the input and output terminals.
The virtually increased input capacitance due to the Miller effect is
given by CM=C(1+AV) where AV is the gain of the amplifier and C is
the feedback capacitance.
• Although the term Miller effect normally refers to capacitance, any
impedance connected between the input and another node exhibiting
gain can modify the amplifier input impedance via this effect.

Z1 = Z/(1 – K) Z2 = Z/ 1- (1/K))
V0 = K Vi

Frequency Response of Amplifier Analog Electronics Engineering

Writing KVL equations at input and output terminals,
Vπ = I1 (1/(jωCμ) + VO ---------------A
VO = I2 (1/(jωCμ) + Vπ ----------------B
Using A & B form two port network. Convert Thevenin
equivalent circuit into Norton equivalent circuit.

• Typical value of Cμ =0.2 pF gm = 50 mA/V
• Calculate frequency at which magnitude of two dependent
current sources are equal.
If ωCμ Vπ = gm Vπ , then
f = gm/(2π Cμ) = 39.8 x109
Hz
Since frequency of operation of bipolar transistors is far less
than 40 GHz, Current source
Isc = jωCμVπ is negligible compared to gmVπ
source.

Approximations
• Calculate frequency at which magnitude of impedance Cμis
equal to RC R
‖ L
• If RC R
‖ L = , then
• If we assume RC = RL = 4 KΩ , Cμ=0.2 pF
= 398 MHz
• If frequency of operation is much smaller than 400 MHz then
will be much greater than (RC R
‖ L ) and can be considered to
be open circuit.
• Using approximations, circuit reduces to
f=
f= 1
2π(0.2 10
∗ −12)[(4 10
∗ −3 ‖4 10
∗ −3)¿
¿

Small signal equivalent circuit, including
approximations
I1 = (Vπ –VO) / ( ) = jωCμ(Vπ–VO)
VO = - gm Vπ)
I1= j ωCμ[ 1+ gm)] Vπ

Small signal equivalent circuit, including
Miller capacitance
• CM is called as Miller capacitance and multiplication
effect of Cμ is the Miller effect.
CM = Cμ[ 1+ gm)]

Example
• Determine 3 dB frequency of the current gain,
both with and without the effect of CM
• == 4KΩ rπ = 2.6 KΩ=200 KΩ Cπ = 4pF Cμ= 0.2pF
gm = 38.5 mA/V

Example
• Ai = IO/IS
• IO= - gm Vπ())
• Vπ=IS [‖ rπ ] = I
‖ ‖ S[]
• Ai = IO/IS =- gm ())[]
• 3 dB frequency is []
• Neglecting effect of CM
• f3dB = [] = 15.5MHz
• Miller capacitance CM = Cμ[ 1+ gm)]= 15.6 pF
• f3dB = [] = 3.16 MHz

Ex 7.11
=200KΩ= 220KΩ=2.2K= 4.7KΩrS= 100
KΩ=5 V
The transistor parameters are---
VBE= 0.7 V, β = 100, VA =∞ Cπ =10pF
Cμ= 2pF using simplified hybrid-π
model calculate Miller capacitance, and
3 dB frequency

= RS R
‖ 1 R
‖ 2
CM = Cμ[ 1+ gm)] = 109 pF
3 dB frequency is [] = 0.506MHz

TYU 7.6
• ICQ, β0, Fβ, are given. Calculate .
• =
• gm = ICQ/ VT
• Fβ=/ β0
• =

TYU 7.8
• r ,β0,FT , are given. Calculate &.
Fβ=/ β0
• =

Model based on inherent resistances and capacitances in the n-channel MOSFET Structure
• and represent interaction between gate and channel inversion charge near source and drain terminals.
• If device is biased in non-saturation region and VDS is small, the channel inversion charge is approximately
uniform –
• )WLCOX where COX (F/cm2
)= OX / tOX
• OX - oxide permittivity = 3.9OO = 8.85 *10 -14
F/cm permittivity of free space
• tOX - oxide thickness in cm

Model based on inherent resistances and capacitances
in the n-channel MOSFET Structure
• However when transistor is biased in the saturation region,
channel is pinched off at drain and inversion charge is no
longer uniform.
• 0 (2/3) WLCOX
• Example: oxide thickness = 500 0
A
Channel length L = 5cm; Channel Width W= 50
0.12pF
Value of
• , are parasitic or overlap capacitance---gate oxide overlaps
source and drain contacts because of fabrication factors.
lowers bandwidth of FET

• – drain to substrate pn junction capacitance
• rS , rdseries resistances of source and drain
terminals
• The internal gate to source voltage controls
small signal channel current through the
transconductance.
Model based on inherent resistances and
capacitances in the n-channel MOSFET Structure

Equivalent circuit of n channel common
source MOSFET
rO is not shown:rO is associated with slope of ID versus
VDS. In the ideal MOSFET biased in saturation region,
ID is independent of VDS . rO is infinite. However rO is
finite in short channel length devices, because of
channel length modulation and is included in
V’gs is the internal gate
to source voltage that
controls the channel
current. Assume
contains parasitic
overlap capacitances.

Equivalent circuit of n channel common
source MOSFET
• Source resistance rS can have significant effect on transistor
characteristics.
• To illustrate figure shows simplified low frequency equivalent circuit
including rS , but not rO. ID = gmV’gs
• Vgs=V’gs+ ( gm V’gs)rS
• = (1+ gm rS)V’gs
• ID = (gm /((1+ gm rS)) Vgs
• = g’mVgs
• source resistance reduces effective transconductance or transistor gain.
Low frequency equivalent circuit

Unity Gain Bandwidth- MOSFET
• Neglecting rS , rd, rOand ; Connect drain to signal ground.
Since input gate impedance is no longer infinite at high
frequencies, define short circuit current gain.
• Write KCL at input node
• Ii = -------A
• = [ j + )]
• Write KCL at output node
• + ID = gmVgsID = Vgs(gm - )
• Vgs=ID/(gm - ) -------B
• Substitute B in A

• Ii = -------A
• Vgs=ID/(gm - ) -------B
• Ii = ID
• Ai = ID / Ii =
• = 0.05 pF, = 1mA/V and maximum frequency f = 100MHz, ,
therefore,
• Ai = FET = =
• Unity gain frequency or bandwidth is parameter of
transistor and is independent of circuit.

Example 7.12
• Determine unity gain bandwidth of FET.
• Consider an N-channel MOSFET with parameters Kn =
0.25 mA/V2
, VTN = 1 V, λ=0, = 0.04 pF assume
transistor is biased at =3 V
• Transconductance gm = 2Kn ( – VTN)
• = 2 *0.25(3-1)= 1mA/V
• = = 6.63 x 108
Hz =663 MHz
• Compared to BJT, high frequency FETs require small
capacitances and small device size.

• Typical values of for MOSFET are in the range of 0.1 to 0.5
pF and values of are from 0.01 to 0.04 pF
• Equivalent circuit is same for MOSFETs, JFETs and MESFETs.
• For JFETs and MESFETs capacitances and are depletion
capacitances rather than oxide capacitances.
• For JFETs and are larger than for MOSFETs while values for
MESFETs are smaller.
• Gallium Arsenide MESFETs are often used in microwave
Amplifier.(Unity gain bandwidth is in the range of tens of
GHz)

Miller Effect- FET
MOSFET high frequency circuit, including
equivalent Miller capacitance
Equivalent high frequency small signal
circuit of MOSFET with RL
CM = C gd[ 1+ gm)]
Ai =
CM = Cμ[ 1+ gm)]
fcutoff= []
For BJT

Example 7.14 Determine upper corner frequency and midband gain of CE circuit.
=40KΩ, = 5.72KΩ, =5 K= 10 KΩ, RS = 0.1 KΩ, = 5 V, = 5 V The transistor
parameters are--- VBE = 0.7 V, β = 150, VA =∞ Cπ =35 pF Cμ = 4pF

Example 7.14
• ICQ= 1.03 mA
• gm = ICQ/ VT = 1.03/0.026 = 39.6 mA/V
• rπ=β VT/ICQ = 150 * 0.026 / 1.03 = 3.79 KΩ
• CM = Cμ[ 1+ gm)]
• Midband gain = gm R’L [] = 126
MHz
C
C
R
R
r
f
M
S
B
H 94
.
2
)
](
[
2
1







Ex 7.14 VBE = 0.7 V, β = 125, VA =200V Cπ =24 pF Cμ = 3pF
Calculate Miller capacitance, upper corner frequency and
midband gain.
RTH = R1//R2 = 10 KΩ
VTH = 0 V
IBQ= (0- 0.7-(-5))/(10+((1+125) *5)= 0.00672mA
ICQ = β IBQ = 0.84 mA
rπ = β VT/ ICQ = 125 * 0.026 /0.84 = 3.87 KΩ
g m = ICQ / VT = 0.84 / 0.026 = 32.3 mA/ V
rO = VA / ICQ = 238 KΩ

CM = Cμ[ 1+ gm)]
MHz
C
C
R
R
r
f
M
S
B
H 21
.
1
)
](
[
2
1






Midband gain = gm R’L []
= 155pF
= 1.21 MHz
= -37.3

Common Base
R1 & R2 are effectively short circuit.
RO is assumed infinite.
Cμ is no longer between i/p and o/p.

Common Base
KCL at emitter
Ie + gm Vπ+ Vπ/(1/ (j)) + (Vπ/rπ) = 0
Vπ= -Ve
= gm + j + (1/ rπ) =
= + j
= + j
One end of Cis tied to ground.
It eliminate Miller multiplication effect. ?????





 C
R
R
r
f
S
E
H
]
))
1
/(
[(
2
1




 C
R
R
f
L
C
H
]
[
2
1

gm(RC[]= AV
Cμ+CL

Example 7.15
Determine upper corner frequency and
midband gain. =40KΩ, = 5.72KΩ, =5 K=
10 KΩ, RS = 0.1 KΩ, =5 V, =-5 V The
transistor parameters are--- VBE = 0.7 V,
β = 150, VA =∞ Cπ =35 pF Cμ= 4pF
DC Analysis:
ICQ= 1.03 mA
gm = ICQ/ VT = 1.03/0.026 = 39.6 mA/V
rπ=β VT/ICQ = 150 * 0.026 /1.03 = 3.79 KΩ

10/05/2025 Frequency Response of Amplifier
MHz
C
R
R
r
f
S
E
H 236
]
))
1
/(
[(
2
1








MHz
C
R
R
f
L
C
H 9
.
11
]
[
2
1





gm(RC[ ]= AV
= 236 MHz
= 11.9 MHz
=25.5

Ex. 7.15
For the common base circuit; β =100, VBE = 0.7 V,
VA =∞ Cπ =24 pF Cμ = 3 pF
Determine upper 3dB frequencies (F Hπ and F Hμ) .
Calculate small signal midband voltage gain.

• IBQ =(10- 0.7) / (100+(101*10) =0.0084mA
• ICQ =0.84 mA
• rπ =β VT/ ICQ = 3.1 K
• gm = ICQ / VT = 32.22 mA/V

Emitter follower Circuit
)
1
](
)
1
(
[
2
1
'
'
L
m
L
m
B
S
H
R
g
C
C
r
R
g
R
R
f









RL’ = RE
VO = (I’b +gmV----------------A
V = I’b/ y -----------------B
Where y= (1/r) +sC
V= V+V
Therefore Z’b = ------C

VO = (I’b +gmV----------------A
V = I’b/ y -----------------B
Where y= (1/r) +sC
V= V+V
Therefore Z’b = ------C
=
= + +
= (1+ ) +
= (1+ ) + (1+ ) +

• Z’b= (1+ ) +
• = +

• Impedance ) in parallel with is large
compared to .
• So neglecting ,
)
1
](
)
1
(
[
2
1
'
'
L
m
L
m
B
S
H
R
g
C
C
r
R
g
R
R
f









7.12

TYU 7.12
Kn = 1 mA/V2
; VTN =0.8 V
Cgs = 2pF; Cgd = 0.2 pF
• DC Analysis
• VG = [(50/200) 10 ]- 5 = -2.5 V
• Kn (VGS - VTN)2
= ID
• VS = VG – VGS
• ID = (Vs - (-5))/Rs
• Kn (VGS - VTN)2
= ID
• =
• =
• VGS = 1.505 V
• gm = 2 Kn (VGS - VTN)
• = 1.41 mA/V

𝑓 𝐻=
1
2 𝜋(𝑅𝑖 ∕ ∕ 𝑅𝐺)(𝐶𝑔𝑠+𝐶𝑀)
= 3.38MHz
Av = - gm = - 4.6
CM = C gd[ 1+ gm)] = 0.2pF[1+ 1.41* 5] = 1.61pF

PMOS Common source
7.58 VTP = -2V Kp = 1mA/V2
λ= 0 Cgs = 15pF
Cgd = 3pF
Determine upper 3dB frequency, miller
capacitance, midband voltage gain

PMOS Common source
VG = () 20 – 10
= 4.67 V
ID =
= (10- Vs) / Rs
=
= KP(VSG +VTP)2
=
KP(VSG +VTP)2
VSG = 3.77V
gm = 2 KP(VSG +VTP)
= 3.54mA/V

• CM = C gd[ 1+ gm(∕ ∕ RL)]
• = 3 pF[1+ 3.54*(∕ ∕ 5)]
• = 18.2pF
• FH =
• Req = Ri∕∕ R1 ∕∕R2
• Ceq=CGS + CM
• FH= 10.4MHz
• VO = - gm VGS (∕ ∕ RL)
• VGS = Vi
• Av = VO / Vi
• = - 4.66

Common Gate
7.61 VTN = 1V Kn = 3mA/V2
λ= 0 Cgs = 15pF
Cgd = 4pF
Determine upper 3dB frequency, miller capacitance,
midband voltage gain

Low frequency….FET
Design circuit such that IDQ = 0.5mA
VDSQ = 4.5 V Rin = 200KΩ FL = 20 Hz
Given data: VTN = 1.5 V
Kn = 0.2mA/V2
λ= 0

7.20
VTP = -1.5V Kp = 1mA/V2
λ= 0
Calculate lower cutoff frequency.

Common Source
Low frequency Equivalent
PMOS

Common Source
High frequency Equivalent

Common Gate

Common Drain

Reference
• Neaman

Frequency Response of Amplifier Analog Electronics Engineering

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