2012
Section2
Chapter 1: Amplifiers
Reference : Microelectronic circuits ‘Sedra’ six edition
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Contents:
1-Introduction
2-What are Amplifiers?
3-Amplifier Saturation
4-Amplifier Biasing
5-Circuit models for amplifiers
6-Frequency response of amplifiers
7-Assignment
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1- Introduction
Problem 23
A particular Signal source produces an output of 30 mV when loaded by a
100kΩ resistor and 10 mV when loaded by 10kΩ resistor, Calculate the
Thevenin voltage, Norton current and source resistance.
Solution
Thus
Therefore
Rs =28.6 kΩ
Vs=38.6 mv
The Norton current
= 1.35
2- What are Amplifiers?



An amplifier is a device for increasing the power of a signal by use
of an external energy source.
Signal amplification is a fundamental signal processing function
that is employed in almost every electronic system.
Signals picked from transducers are said to be weak & posse little
energy so amplification is must to acquire reliable processing
thereafter.
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 Where
Useful laws:
 Voltage gain
 Current gain
 Power gain
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N.B

a.
b.
c.



For Decibel scale
Voltage gain =20 log |Av|
Current gain = 20 log |Ai|
Power gain = 10 log |Ap|
For positive dB value  Amplification  vo>vi
For negative dB value  Attenuation  vo<vi
For real amplifier
Thus
 The dc power delivered to amplifier is
 The balanced power equation is
 The amplifier efficiency is
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Problem 39
Find the voltage, current, and power gains both as ratios and in dB:
a.
= 100mV, = 100µA,
= 10V,
= 100Ω
Solution
1.
20 log 100 =40 dB
2.
20 log 1000 =60 dB
3.
10 log
=50 dB
Problem 40
An amplifier operating from ±3V supplies provides a 2.2 V peak sine
wave across a 100Ω load when provided with a 0.2-V peak input from
which 1.0 mA peak is drawn. The average current in each supply is
measured to be 20 mA, Find the voltage gain, current gain, and power
gain expressed as ratios and in decibels as well as the supply power,
amplifier dissipation, and amplifier efficiency.
Given:
20 mA
v
v
Solution:
1.
20log 11 =20.8 dB
2.
log 22 =26.8 dB
3.
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10log
=50 dB
4. Supply power = 2* * =120 Mw
5. Output power =
=
= 24.2
mW
= 0.1 mW
6. Input power =
7. Amplifier dissipation = supply power – output power = 95.8 mW
8. Amplifier efficiency =
= 20.2 %
3- Amplifier Saturation
Practical the amplifier transfer characteristics remain linear over only a
limited range of input & output voltage, outside this range the o/p is
distorted (clipped).
For linear region
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Problem 41
An amplifier using balanced power supplies is known to saturate for
signals extending within 1.2 V of either supply. For linear operation,
its gain is 500 V/V. What is the rms value of the largest undistorted
sine wave output available, and input needed, with ±5V supplies?
Solution
The largest undistorted sine wave output available is 5 – 1.2 =3.8 v
peak amplitude or 3.8/
m
= 2.7
, input needed is 2.7/gain = 5.4
4- Amplifier Biasing
A dc bias (shift) voltage is added to input signal to shift operation in the
middle of linear region to simplify the work with amplifier with single gain
value.
dc biasing point
Operating point
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5- Circuit models for amplifiers
 For voltage amplifier
Laws



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 Amplifier circuit models :
N.B



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Problem 44
An amplifier with 40 dB of small signal, open-circuit voltage gain, an
input resistance of 1 MΩ, and an output resistance of 10Ω, drives a
load of 100 Ω. What voltage and power gains (expressed in dB) would
you expect with the load connected? If the amplifier has a peak
output-current limitation of 100 mA, what is the rms value of the
largest sine-wave input for which an undistorted output is possible &
what is the corresponding output power available?
Solution
=90.9 v/v
=20 log 90.9 = 39.1 dB
Largest an undistorted output sine-wave peak voltage
= 100mA,*100Ω = 10 v
Largest sine-wave input =
or for rms voltage
Corresponding output power =
Problem 52
A voltage amplifier with an input resistance of 10 kΩ, an output
resistance of 200 Ω, and a gain of 1000 V/V is connected between a
100-kΩ sources with an open-circuit voltage of 10 mV and a 100Ω
load. For this situation:
(a) What output voltage results?
(b) What is the voltage gain from source to load?
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(c) What is the voltage gain from the amplifier input to the load?
(d) If the output voltage across the load is twice that needed and
there are signs of internal amplifier overload, suggest the location and
value of a single resistor that would produce the desired output.
Choose an arrangement that would cause minimum disruption to an
operating circuit (Hint Use parallel rather than series connections.)
Solution
a)
=
=303 mv
b)
=
c)
=
= 30.3 v/v
= 333.3 v/v
d) Connect
parallel to
this equation
and satisfy
6- Frequency response of amplifiers
The frequency response means how the amplifier output respond to
different frequency in frequency.

Amplifier
 Frequency response
magnitude
Phase
 Bode plot
relation between gain and w
In log scale
relation between gain and
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 Amplifier bandwidth
Gain
Attenuation region  1,2
Useful region  3
1
3
2
Example: low pass filter
S=JW
RC = 𝜏 = 1/𝑤𝑜
For K=1
| |=
 gain magnitude
20 log | |= -10 log (1+
Ø=0-
)
 gain phase
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@ w <<
 20 log | |= dc gain , Ø=
@ w >>
 20 log | |= attenuation (-ve value)
@w=
Why low pass
filter ??
 20 log | |= -10 log 2 = -3 dB
Problem 66
Figure P.66 shows a Signal source connected to the input of an amplifier, Here , is
the source resistance. And and , are the input resistance and input capacitance,
respectively, of the amplifier. Derive an expression for , and show that it is of the
low-pass filter type. Find the 3-dB frequency for the case
=5pF.
Solution
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= 20 kΩ,
= 80 kΩ, and
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 Which is the transfer function of low pass filter.
The 3-dB frequency =
=
= 2 MHz
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Biomedical department
2nd year
1st term 2012
Electronics course
Assignment no.2
Student name:
Problem 22
Any given signal source provides an open-circuit voltage
and a short-circuit
current
, Calculate the internal resistance. , the Norton current and the
Thevenin voltage given
Solution:
Problem 45
A 10mV Signal source having an internal resistance of 100 kΩ is connected to an
amplifier for which the input resistance is 10 kΩ, the open-Circuit voltage gain is
1000 V/V. and the output resistance is 1 kΩ. The amplifier is connected in turn to a
100 Ω load. What overall voltage gain results as measured from the source internal
voltage to the load? Where did all the gain go?
Solution
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Problem 50 (bonus)
Design an amplifier that provides 0.5 W of signal power to a 100 Ω load resistance.
The signal source provides a 30-mV rms Signal and has a resistance of 0.5 MΩ. Three
types of voltage-amplifier stages are available:
(a)
(b)
(c)
Design a suitable amplifier using a combination
of these stages. Your design should utilize the minimum number of stages and
should ensure that the signal level is not reduced below 10 mV at any point in the
amplifier chain. Find the load voltage and power output realized.
Solution
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Problem 58 (bonus)
It is required to design an amplifier to sense the short-circuit output current
of a transducer and to provide a proportional current through a load resistor.
The equivalent source resistance of the transducer is specified to vary in the
range of 1 kΩ to 10 kΩ similarly the load resistance IS known to vary over the
range of 1 kΩ to 10 kΩ. The change in load current corresponding to the
specified change in is required to be limited to 10%. Similarly the change in
load current corresponding to the specified change in R should be 10% at
most. Also, for a nominal short-circuit output current of the transducer of 10
.The amplifier is required to provide a minimum voltage across the load of
1 V. Sketch the circuit model of the amplifier and Specify values for its
parameters.
Solution:
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Problem
For the circuit shown:
Prove that it is high pass filter and find its transfer function.
Solution
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