Analogue ELEK1289 - Electronic systems and practice II - Unit 3 – Amplifier Modelling
Analysis of Amplifier Systems
Amplifier Model
We have seen in the previous chapter, that an amplifier is a device that increases the
magnitude of a signal for use by a load. Amplifiers are complicated arrangements of
transistors, resistors and other components. However a simplified description is all that is
necessary when the requirement is to analyse the source and load behaviour. The amplifier
can be thought of as an interface between the source and the load.
A basic amplifier may be represented as follows:
Gain
Element
Output
Signal
Input
Signal
It is useful to classify amplifiers into four broad categories:
The Voltage Amplifier
 Input signal is voltage VIN ,
 Output signal is voltage VOUT
VOUT
 The voltage gain AV 
VIN
The Transconductance Amplifier
 Input signal is current IIN
 Output signal is voltage VOUT
The Current Amplifier
 Input signal is current IIN
 Output signal is current IOUT
IOUT
 The current gain AI 
IIN
The Transresistance Amplifier
 Input signal is a voltage VIN
 Output signal is current IOUT


The Transconductance gain Gm 
The Transresistance gain Rm 
In this chapter we will focus on the analysis of the voltage amplifier and the current
amplifier.
1
IOUT
VIN
VOUT
IIN
Analogue ELEK1289 - Electronic systems and practice II - Unit 3 – Amplifier Modelling
Non Ideal Amplifiers
Saturation Voltage
An ideal amplifier means that the amplifier introduces no noise or distortion to the signal, i.e.
the output replicates the input exactly as the input signal increases. However, for real
amplifiers the gain will taper off as VIN increases in magnitude. This is known as saturation
of the amplifier.
Non Ideal Characteristic
VOUT (V)
40
Saturation voltage = 30V
30
A=
20
20
10
0
0
1
2
VIN (V)
3
Saturation in Non-Ideal Voltage Amplifier
Frequency Response
A non-ideal amplifier will also have a frequency response in that as the frequency increases
the gain will fall for a particular magnitude of the input signal.
Frequency Response of Non Ideal Amplifier
Gain A
40
30
Ideal
20
Practical
10
0
f (Hz)
0
10KHz
100KHz
1MHz
Frequency Response of Non-Ideal Amplifier
For an ideal voltage amplifier we assume that the gain remains constant as the frequency
increases.
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Analogue ELEK1289 - Electronic systems and practice II - Unit 3 – Amplifier Modelling
Input impedance
For an ideal voltage amplifier we assume that the input impedance is infinite. However, a
real voltage amplifier will also have a finite input impedance.
RIN
VIN
VOUT
Input Impedance
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Analogue ELEK1289 - Electronic systems and practice II - Unit 3 – Amplifier Modelling
Equivalent Circuit Models for Gain Elements
The concept of equivalent circuits, covered in the Electronic Circuits and Devices module to
model voltage and current sources, can be applied to more complicated case of an amplifier.
By drawing an amplifier as an equivalent circuit, you can simplify equations related to its
performance.
The Voltage Amplifier
A basic voltage amplifier may be represented as follows:
VOUT
Transfer Charteristic for Various Gains
100
AV
0
=1
AV
A
V=
20
75
VOUT
VIN
50
5
A V=
25
0
AV 
VOUT
VIN
or
0
5
10
VIN
VOUT  AV .VIN
Voltage Amplifier
The input signal from a source is applied to the input terminals of the amplifier, and an output
is taken from the second set of terminals.
In practice the amplifier input and output voltages will be influenced by the characteristics of
the source, which supplies the input to the amplifier, and the load, which the amplifier output
is required to drive. To analyse this, it is necessary to adopt a model of the interior of the
amplifier.
RS
IOUT
IIN
A
+
VOUT
VIN
VS
RL
GAIN ELEMENT
SOURCE
RS: Source Resistance
VS: Source Voltage
RL: Load Resistance
Source – Gain Element – Load
4
LOAD
Analogue ELEK1289 - Electronic systems and practice II - Unit 3 – Amplifier Modelling
The Voltage Amplifier Equivalent Circuit
The amplifier input and output circuits are characterised using Thevenin’s Theorem to
produce the model for a voltage amplifier.
Gain Element
ROUT
+
VIN
AV VIN
RIN
VOUT
-
Voltage Amplifier Equivalent Circuit Model
As amplifiers are designed to operate from input to output only, VOUT does not influence VIN.
The amplifier’s input terminals present an input resistance to the voltage source. Therefore,
the input circuit is modelled as an input resistance RIN.

RIN is the Thevenin resistance seen at the input terminals and is called the
input resistance of the amplifier.
The output circuit of the amplifier can be modelled as Thevenin source, as shown above:

ROUT is the Thevenin resistance seen at the output terminals and is called the
output resistance of the amplifier

AVVIN is the Thevenin voltage seen at the output of the amplifier where AV is the voltage
gain of the amplifier. Because the magnitude of the Thevenin source is dependent on the
unloaded gain AV and the input voltage VIN., the amplifiers output circuit is said to contain
a dependent source.
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Analogue ELEK1289 - Electronic systems and practice II - Unit 3 – Amplifier Modelling
An amplifier does not exist in isolation and in practice the input will be driven by a non-ideal
source and the output will be connected to a load. The load is assumed to be a resistance
connected across the output terminals as shown below:
Gain Element
Source
Load
RS
ROUT
+
+
VIN
VS
AV VIN
RIN
VOUT
RL
-
-
Voltage Source
Equivalent
Circuit
Voltage Amplifier
Equivalent Circuit Model
Load
Equivalent
Circuit
Source/Amplifier/Load Equivalent Circuits
Analysing the input circuit:
Gain Element
Source
VRS =IIN RS
IIN
+
RS
VS
-
RIN
VIN
Input Circuit
Figure 1: Amplifier Input Circuit
The input resistance RIN affects the input voltage VIN to the amplifier because it forms a
voltage divider with the source resistance RS.
The voltage division rule =>
 RIN 
VIN  VS 

 RIN  RS 
equation 1
So for example, if the source voltage is 3 V, the source resistance is 500  and the input
resistance is 1 k, then the input voltage would be:
VIN = 3 ( 1000 / 1500 ) = 2 V
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Analogue ELEK1289 - Electronic systems and practice II - Unit 3 – Amplifier Modelling
Analysing the output circuit:
Load
Gain Element
VROUT = IOUT ROUT
IOUT
+
VOUT
AV VIN
-
RL
Output Circuit
The presence of RL modifies the behaviour of the amplifier.
The unloaded voltage gain of the amplifier is called AV
AV 
VOUT
VIN
equation 2
RL 
Note AV is independent of the circuit into which the amplifier is connected
RL


VOUT  AVVIN 

 ROUT  RL 
Applying the voltage division rule =>
equation 3
 RIN 
.
 RIN  RS 
But, the input voltage is VIN  VS 
Therefore substituting VIN into the VOUT equation gives =>:
VOUT
RIN 
RL


 AVV

RIN  RS  ROUT  RL 


s


The loaded voltage gain of the amplifier is called AVL and is defined as:
AVL 
VOUT
VS
equation 4
Therefore dividing the VOUT equation above by VS gives an expression for the loaded voltage
gain:
RIN 
RL



 RIN  RS  ROUT  RL





AVL  AV 
equation 5
Looking at this equation, it can be seen the loaded gain AVL must be less than or equal to the
open circuit gain AV:
AVL  AV
In practice the loaded voltage gain is less than the unloaded voltage gain.
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Analogue ELEK1289 - Electronic systems and practice II - Unit 3 – Amplifier Modelling
The loaded voltage gain will be at its maximum if the amplifier is designed such that:
RIN ->  and ROUT -> 0 
Thus,
RIN 
RL


 => AVL --> AV
 RIN  RS  ROUT  RL





AVL  AV 
Ideal voltage amplifier:
Real voltage amplifier
RIN =  , ROUT = 0 
AVL = AV

In reality we require:
RIN
>>

VIN

ROUT <<

VOUT 
=> independent of source and load
RS
VS
RL
AvVS
Example Problems
Attempt the following example problems. The solutions will be covered during lectures.
Exercise 1
A non-ideal voltage amplifier has an input resistance of RIN = 2 k and
output resistance of ROUT = 0 . The gain AV = 20 and the input current is 1 mA. What is
VOUT?
Exercise 2
A voltage amplifier has the following characteristics:
Gain AV = 10, RIN =   and ROUT = 0 .
Calculate VOUT and IOUT if RL = 100 , RS = 1 k, VS = 10 mV.
Exercise 3
Repeat example No. 2 for the two cases:
(a) RS = 500  and RIN = 1.5 k.
(b) RS = 200  and RIN = 800 .
(c) RS = 200  and RIN = 800 . and ROUT = 100 .
(d) RS = 200  and RIN = 800 . and ROUT = 1000 .
Comment on the results obtained
Exercise 4
Calculate the power gain AP and the output power in dBW of a voltage amplifier having RIN =
2 k, AV = 20, VS = 8 V, RS = 200  and RL = 200 . Assume ROUT of the amplifier to be
ideal.
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Analogue ELEK1289 - Electronic systems and practice II - Unit 3 – Amplifier Modelling
Analysis of Amplifier Systems [continued]
The Current Amplifier
A basic current amplifier may be represented as follows:
IOUT (mA)
Transfer Charteristic for Various Gains
of Current Amplifier
40
IIN
10
IOUT
5
A=
A=
30
AI
2
A=
20
10
0
A is the current gain
AI 
0
2
4
6
8
10
12
IOUT
or IOUT = AIIIN
IIN
Current Amplifier
Current Amplifier
Using Norton’s theorem, the following model can be adopted for a current amplifier:
IOUT
IIN
IS
RS
Current Source
Equivalent
Circuit


AIIIN ROUT
RIN
Current Amplifier
Equivalent Circuit Model
VOUT
RL
Load
Equivalent
Circuit
RIN is the resistance seen at the input terminals and is called the
input resistance of the amplifier.
ROUT is the Norton resistance seen at the output terminals and is called the
output resistance of the amplifier
 AIIIN is the Norton current produced at the output of the amplifier where AI is the
current gain of the amplifier.
9
IIN (mA)
Analogue ELEK1289 - Electronic systems and practice II - Unit 3 – Amplifier Modelling
Analysing the input circuit:
IIN
RS
IS
RIN
Current Source
Equivalent Circuit
Current Amplifier
Input Equivalent Circuit
Amplifier Input Circuit
The input resistance RIN affects the input current IIN to the amplifier as follows:
The current division rule =>
 RS 
IIN  IS 

 RIN  RS 
equation 6
Analysing the output circuit:
IOUT
AIIIN ROUT
VOUT
RL
Load
Equivalent Circuit
Current Amplifier
Output Equivalent Circuit
The zero-loaded current gain of the amplifier is called AI
AI 
IOUT
IIN
equation 7
RL  0
Note AI is independent of the circuit into which the amplifier is connected
 ROUT 

 ROUT  RL 
Applying the current division rule => IOUT  AIIIN 
10
equation 8
Analogue ELEK1289 - Electronic systems and practice II - Unit 3 – Amplifier Modelling
 RS 
I
IN

I
S


But, the input current is
 RIN  RS 
Therefore substituting IIN into the IOUT equation gives =>:
RS  ROUT 

IOUT  A I

RIN  RS  ROUT  RL 


I s


The loaded current gain of the amplifier is called AIL and is defined as:
AIL 
IOUT
IS
equation 9
Therefore dividing the IOUT equation above by IS gives an expression for the loaded current
gain:
RS  ROUT 

AIL  A

RIN  RS  ROUT  RL 


I


equation 10
Looking at this equation, it can be seen the loaded gain AIL must be less than or equal to the
open circuit gain AI:
Ideal current amplifier:
Real current amplifier
Ideally:
In reality we require:
RIN
<<
RS
RIN = 0 , ROUT =  
=> AIL =AI => independent of source and load

IIN

IS
resistance
ROUT >>
RL


IOUT 
VOUT 
AIIS
AIISRL
Example Problems
Exercise 5
A non-ideal current amplifier has an input resistance of RIN = 100 ,
output resistance of ROUT =  . The gain AI = 10 and the input voltage is 2 V.
What is IOUT if the load resistance of RL = 10 k.?
Exercise 6
A current amplifier has the following characteristics:
Gain AI = 5, RIN =0  and ROUT =  .
Calculate VOUT and IOUT if RL = 100 , RS = 2 k, IS = 10 mA.
Exercise 7
Repeat previous example for the two cases:
(a) RS = 2 k  and RIN = 2 k.
(b) RS =2 k  and RIN = 100 .
(c) RS =2 k , RIN = 100  and ROUT =100 .
(d) RS =2 k , RIN = 100  and ROUT =10 k .
Comment on the results obtained
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Analogue ELEK1289 - Electronic systems and practice II - Unit 3 – Amplifier Modelling
Amplifiers in Cascade
We can see in the previous examples that the amplifier equivalent circuit models are useful
for analysis of source and load effects. These models are also useful to analysis the internal
loading when two or more stages are cascaded to form a single amplifier. Consider the twostage cascaded voltage amplifier shown below. The overall gain is affected by loading
effects from each of the three loops. The loops are simple series circuits, so the voltages can
easily be calculated with the voltage division rule.
Stage 1
RS
Stage 2
ROUT1
+
+
VS
-
Load
Two-Stage Cascaded Amplifier
Source
VIN1
RIN1
-
ROUT
+
AV1 VIN1
VIN2
RIN2
-
AV2 VIN2
R
VOUT2 L
.
Two-Stage Cascaded Voltage Amplifier
Consider the two-stage cascaded current amplifier shown below. Again, the current can easily
be calculated with the current division rule.
Two-Stage Cascaded Current Amplifier
Stage 1
IIN1
IS
RS
Stage 2
IOUT2
RIN1
AI1IIN1 ROUT1
RIN2
AI2IIN2 ROUT2
Two-Stage Cascaded Current Amplifier
Sample problems covered in class
12
RL