Laboratory Report
EEX6450
ANALOG ELECTRONIC SYSTEMS &
INSTRUMENTATION
Lab 02
By
C.J.AMARASINGHE
612265044
Submitted to
Department of Electrical and Computer Engineering
Faculty of Engineering Technology
The Open University of Sri Lanka
At
Colombo Regional Center
DESIGN OF A WIDE BAND AMPLIFIER
AIM
To design a Wide Band Amplifier
OBJECTIVES
To become familiar with the procedure of amplifier with emphasis on design consideration at
high frequencies
APPARATUS







Resistors
Capacitors
Transistors
Zener diode
DC power supply
Oscilloscope
Signal Generator
THEORY
An amplifier is a device where a small input signal is used to control a large output signal. This
signal may be current, voltage or power. An amplifier functions by taking power from a power
supply and controlling the output to match the input signal shape but with a larger amplitude.
In this sense, an amplifier modulates the output of the power supply based upon the properties
of the input signal. An amplifier is effectively the opposite of an attenuator: while an amplifier
provides gain, an attenuator provides loss. By considering the range of signal frequency over
which the amplifier has useful gain amplifiers can be classified as
 Audio low frequency amplifiers
 Radio frequency amplifiers
 Wide band amplifiers
 DC amplifiers
Frequency Response curve:
A graph of amplifier gain (in dB) against signal frequency (in logarithmic scale) is called
Frequency Response Curve.
The basic frequency Response Curve for a Wide band Amplifier is shown in the following
graph.
CALCULATIONS
AC Equivalent Circuit
𝛽
𝛽
𝑅𝜋1 = 𝑔 1 ,
𝑅𝑎 = (𝑅1 ∥ 𝑅2 ) ∥ (𝑅𝑓 + 𝑅3 )
𝑅𝜋2 = 𝑔 2
𝑚1
𝑚2
𝑅𝑥 = 𝑅3 ∥ 𝑅𝑓
𝑅𝑖𝑛2 = 𝑅𝜋2 + (𝛽 + 1)𝑅𝑥
Let
𝑉
𝐾1 = 𝑉2
𝐾2 =
1
𝑉𝑜𝑢𝑡 = 𝑔𝑚2 (𝑉2 − 𝑉3 )𝑅𝐿
𝑉𝑜𝑢𝑡
𝑉2
------- (1)
(𝑉2 − 𝑉3 )
𝑉3 = [
+ 𝑔𝑚2 (𝑉2 − 𝑉3 )]𝑅𝑥
𝑅𝜋2
1
𝑉3 = (𝑉2 − 𝑉3 )[𝑅 + 𝑔𝑚2 ]𝑅𝑥
𝜋2
-------- (2)
𝑉2 = −𝑔𝑚1 𝑉1 (𝑅𝐶 ∥ 𝑅𝑖𝑛2 )
𝑉2
= −𝑔𝑚1 (𝑅𝐶 ∥ 𝑅𝑖𝑛2 ) = 𝐾1
𝑉1
From equation (2)
1
1
𝑉3 [1 + (𝑔𝑚2 +
) 𝑅𝑥 ] = 𝑉2 [𝑔𝑚2 + ]𝑅𝑥
𝑅𝜋2
𝑅𝜋
1
[𝑔𝑚2 + 𝑅 ]𝑅𝑥
𝑉3
𝜋
=
= K3
𝑉2 1 + (𝑔 + 1 )𝑅
𝑚2
𝑥
𝑅
𝜋2
𝑉
𝐾3 = 𝑉3
2
From equation (1)
𝑉𝑜𝑢𝑡
= (𝑉2 − 𝑉3 )
𝑔𝑚2 𝑅𝐿
𝑉3 = 𝑉2 −
𝑉𝑜𝑢𝑡
𝑔𝑚2 𝑅𝐿
By both sides divided by 𝑉2
𝑉𝑜𝑢𝑡
𝑉3 𝑉2 − 𝑔𝑚2 𝑅𝐿
=
𝑉2
𝑉2
By eliminating from K3
𝑉
𝑉2 − 𝑔 𝑜𝑢𝑡
𝑅
𝑚2 𝐿
𝑉2
=
1
[𝑔𝑚2 + 𝑅 ]𝑅𝑥
𝜋2
1
1 + (𝑔𝑚2 + 𝑅 )𝑅𝑥
𝜋2
1
[𝑔𝑚2 + 𝑅 ]𝑅𝑥
𝑉𝑜𝑢𝑡
𝜋2
1−
=
𝑔𝑚2 𝑅𝐿 𝑉2 1 + (𝑔 + 1 )𝑅
𝑚2
𝑥
𝑅
𝜋2
1−
1
[𝑔𝑚2 + 𝑅 ]𝑅𝑥
𝜋2
1
1 + (𝑔𝑚2 + 𝑅 )𝑅𝑥
=
𝑉𝑜𝑢𝑡
𝑔𝑚2 𝑅𝐿 𝑉2
𝜋2
𝑉𝑜𝑢𝑡
𝑔𝑚2 𝑅𝐿
=
= 𝐾2
1
𝑉2
1 + (𝑔𝑚2 + 𝑅 )𝑅𝑥
𝜋2
𝐴𝑖𝑜𝐿 =
𝑖𝑜𝑢𝑡
−𝑔𝑚2 (𝑉2 − 𝑉3 )
=
𝑉1
𝑖𝑖𝑛
⁄(𝑅 ∥ 𝑅 ∥ 𝑅 )
𝑠
𝑎
𝜋1
Using K3 , eliminating V3 from 𝐴𝑖𝑜𝐿 ;
−𝑔𝑚2 𝑉2 (1 −
𝐴𝑖𝑜𝐿 =
1
[𝑔𝑚2 + 𝑅 ]𝑅𝑥
𝜋2
)
1
1 + (𝑔𝑚2 + 𝑅 )𝑅𝑥
𝜋2
𝑉1
⁄(𝑅 ∥ 𝑅 ∥ 𝑅 )
𝑠
𝑎
𝜋1
(−𝑔𝑚2 )(−𝑔𝑚2 )(𝑅𝐶 ∥ 𝑅𝑖𝑛2 )(𝑅𝑠 ∥ 𝑅𝑎 ∥ 𝑅𝜋1 )
1
1 + (𝑔𝑚2 + 𝑅 )𝑅𝑥
𝜋2
Obtain the transfer function
𝐴𝑖𝑜𝐿 =
𝑉𝑜𝑢𝑡 = −𝑔𝑚2 (𝑉2 − 𝑉3 )(𝑅𝐿 ∥
1
)
𝐶3 𝑆
1
]𝑅
𝑅𝜋 𝑥
1
1+(𝑔𝑚2 +
)𝑅
𝑅𝜋2 𝑥
From K3 ; 𝑉3 = 𝑉2 [
[𝑔𝑚2 +
]
Converting current source I2 to Voltage source 𝑉3
𝑍1 = (
𝑍1 =
1
∥ 𝑅𝑎 ∥ 𝑅𝜋1 )
𝐶1 𝑆
𝑅𝑎 𝑅𝜋1
𝑅𝑎 + 𝑅𝜋1 + 𝑆𝑅𝑎 𝑅𝜋1 𝐶1
By voltage divider rule
𝑉1 = (
𝑍1
)𝑉
𝑍1 + 𝑅𝑆 𝑆
Since, 𝑍1 = 𝑅
𝑅𝑎 𝑅𝜋1
𝑎 +𝑅𝜋1 +𝑆𝑅𝑎 𝑅𝜋1 𝐶1
Substituting for 𝑍1 we get,
𝑅𝑎 𝑅𝜋1
𝑅𝑎 + 𝑅𝜋1 + 𝑅𝑎 𝑅𝜋1 𝐶1 𝑆
𝑉1 = [
]𝑉𝑆
𝑅𝑎 𝑅𝜋1
+
𝑅
𝑆
𝑅𝑎 + 𝑅𝜋1 + 𝑅𝑎 𝑅𝜋1 𝐶1 𝑆
𝑉1 =
𝑅𝑎 𝑅𝜋1 𝑉𝑆
𝑅𝑎 𝑅𝜋1 + 𝑅𝑎 𝑅𝑆 + 𝑅𝑆 𝑅𝜋1 + 𝑅𝑎 𝑅𝜋1 𝑅𝑆 𝐶1 𝑆
Also, 𝑉2 = (𝑍2 ∥ 𝑅𝑖𝑛2 )(−𝑔𝑚1 𝑉1 )
− − − −(3)
1
; 𝑍2 = 𝑅𝐶 ∥ 𝐶
2𝑆
𝑍2 ∥ 𝑅𝑖𝑛2 = 𝑅𝐶 ∥ 𝑅𝑖𝑛2 ∥
𝑉2 =
1
𝑅𝐶 ∥ 𝑅𝑖𝑛2
=
𝐶2 𝑆 1 + (𝑅𝐶 ∥ 𝑅𝑖𝑛2 )𝐶2 𝑆
(−𝑔𝑚1 𝑉1 )(𝑅𝐶 ∥ 𝑅𝑖𝑛2 )
1 + (𝑅𝐶 ∥ 𝑅𝑖𝑛2 )𝐶2 𝑆
− − − − (4)
From (3) & (4)
𝑉2 =
(−𝑔𝑚1 )(𝑅𝐶 ∥ 𝑅𝑖𝑛2 )(𝑅𝑎 𝑅𝜋1 )𝑉𝑆
[1 + (𝑅𝐶 ∥ 𝑅𝑖𝑛2 )𝐶2 𝑆] [𝑅𝑎 𝑅𝜋1 + 𝑅𝑎 𝑅𝑆 + 𝑅𝑆 𝑅𝜋1 + 𝑅𝑎 𝑅𝜋1 𝑅𝑆 𝐶1 𝑆]
From transfer function
𝑉3
𝑅𝐿
𝑉𝑜𝑢𝑡 = −𝑔𝑚2 𝑉2 (1 − )(
)
𝑉2 1 + 𝑅𝐿 𝐶3 𝑆
𝑉𝑜𝑢𝑡 = −𝑔𝑚2 𝑉2 (1 −
1
[𝑔𝑚2 + 𝑅 ]𝑅𝑥
𝑅𝐿
)(
)
1
1
+
𝑅
𝐶
𝑆
𝐿
3
1 + (𝑔𝑚2 + 𝑅 )𝑅𝑥
𝜋2
𝜋
Eliminating 𝑉2 with 𝑉𝑆 , We get
𝑉𝑜𝑢𝑡
𝑉𝑆
=
𝑔𝑚1 𝑔𝑚2 (𝑅𝐶 ∥𝑅𝑖𝑛2 )𝑅𝑎 𝑅𝜋1 𝑅𝐿
1
[1+(𝑔𝑚2 +
)𝑅 ] [1+𝑅𝐿 𝐶3 𝑆]
𝑅𝜋2 𝑥
[1+(𝑅𝐶 ∥𝑅𝑖𝑛2 )𝐶2 𝑆]
[𝑅𝑎 𝑅𝜋1 +𝑅𝑎 𝑅𝑆 +𝑅𝑆 𝑅𝜋1 +𝑅𝑎 𝑅𝜋1 𝑅𝑆 𝐶1 𝑆]
Where the 3 poles in the transfer function
ω1 =
R a R π1 + R a R S + R 3 R π1
R a R π1 R S C1
ω2 =
R C + R in2
R C R in2 C2
ω3 =
1
R L C3
DISCUSSIONS