Download 1E6 Electricity and Magnetism

2. The Practical, Non-Ideal Amplifier Part I
The Real Amplifier:
power supply rail
Ri  
RO
A
RO  0
RS
signal
source
Vi

Ri
VO
VL
RL
load
Vs
Properties:
 Finite Input Resistance, Ri  
The input resistance is finite and not unlimited.
This means that the source must deliver a current (or power) to the input of the
amplifier.
The input signal level present at the input to the amplifier depends on the relative
values of the source resistance and amplifier input resistance.
There can be attenuation of the signal generated by the source on reaching the
input of the amplifier.
Power may be dissipated by both the source resistance and the amplifier input
resistance.
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 Finite Ouput Resistance, RO  0
The ouput resistance of the amplifier is not zero but is finite
There is no limiting factor on the output current available from the amplifier. The
signal present at the load and the power delivered to the load only depends on the
load resistance.
The current or power which can be delivered to the load is limited by the output
resistance.
Considerable power can be dissipated in the output resistance of the amplifier and
this can generate heat. For high power applications means of cooling the amplifier
may be required.
 Limited Bandwidth, BW  
Bandwidth is not infinite but limited.
There is a maximum frequnecy which is passed by the amplifier with full gain.
There is a limit to the rate of change in the signal which the amplifier can handle.
Frequency components outside of the bandwidth of the amplifier are attenuated
and this can lead to distortion f the signal.
AV
fMAX
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 Gain Variation
In simple amplifers manufacturing variations in the semiconductor
fabrication process give rise to large fluctuations in the amplifying properties of the
devices used to construct the amplifier. This gives rise to varition in the amplifer
gain, which can only be overcome by more advanced design techniques such as the
use of feedback.
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Practical Amplifier Limitations
 Input Attenuation
In practice, because of the finite input resistance of the amplifier, not all of
the source signal ‘potentially available’ is transferred to the input of the
amplifier.
RS
signal
source

VS
VS
Under open-circuit conditions (i.e. no load present) the full signal voltage VS
appears at the terminals of the source as no input current flows.
Ii
A
RS
signal
source

VS
Vi
Ri
When the source is connected to the input of the amplifier, the voltage
appearing at the terminals, Vi, is less than the open-circuit voltage, VS.
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Using Ohm’s Law
Ii 
VS
RS  Ri
Then the potential developed across the input resistor, Ri is:
Vi  Ii R i 
VS
Ri
Ri 
VS
RS  Ri
RS  Ri
The input attenuation factor is defined as the fraction of the open-circuit or
‘potentially available’ source signal which is actually transferred to the inut
of the amplifier.
The Attenuation Factor, i is the ratio of the input voltage, Vi to the source
voltage VS. Then
αi 
Vi
Ri

VS R S  R i
Rarely do we want this attenuation factor to exceed 0.8 to 0.9 or 80% to 90%
This requires
R i  R S
Ideally
Ri  
or
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RS  0
Case Study 1:
A transducer has a source resistance of 10kΩ. The input resistance of the
associated amplifier is 100kΩ. If the transducer generates an unloaded signal
voltage of 1V rms, determine the signal present at the amplifier input and the
degree of attenuation present.
Ii
A
RS
signal
source

VS
Vi
Ri
VS
Ri
Vi  Ii R i 
Ri 
VS
RS  Ri
RS  Ri
105
1
Vi  I i R i  4
x1 
 0.9V
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1.1
10  10
Vi
Ri
1
αi 


 0.9
VS R S  R i 1.1
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Case Study 2:
A moving coil pick-up, having an internal resistance of 50kΩ, produces an
unloaded signal level of 8mVrms. An audio preamplifier stage must produce an
open-circuit output signal level of at least 0.5V rms with a voltage gain of 100.
Determine the minimum input resistance required in the amplifier and the current
which the pick-up must deliver to the amplifier.
Ii
A
RS
signal
source

VS
VO
AV 
 100
Vi
;
Vi
Ri
VO  .5V
; VS  8mVrms
Then:
VO 500mV
Vi 

 5mVrms
AV
100
Vi 
Ri
VS
RS  Ri

5mV 
8
Ri
x8mV
RS  Ri
Ri
5

8 RS  Ri
Then:
5R S  R i   8R i
5R S  8R i  5R i
5R S  3R i
5
R i MIN  R S
3
R i MIN 
5x50kΩ 250kΩ

 83.3kΩ
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The input current which the pick-up must deliver to the amplifier input is:
VS
8mV
8x10 3
Ii 


R S  R i 50kΩ  83.3kΩ 133.3x10 3
VS
8mV
8x10 3
Ii 


R S  R i 50kΩ  83.3kΩ 133.3x10 3
8x10 6
Ii 
 0.06μ.
133.3
or 60nA
Note: the low level of current associated with this relatively sensitive transducer.
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