Section2 - UniMasr.com • Home | Cairo University - … 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

Section2 Chapter 1: Amplifiers Reference : Microelectronic circuits ‘Sedra’ six edition

2012

9/24/2012

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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

For Decibel scale

a. Voltage gain =20 log |Av|

b. Current gain = 20 log |Ai|

c. 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

6. Input power =

= 0.1 mW

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/ = 2.7 , input needed is 2.7/gain = 5.4

m

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)

=

= 30.3 v/v

c)

=

= 333.3 v/v

d) Connect parallel to and satisfy

this equation

6- Frequency response of amplifiers

The frequency response means how the amplifier output respond to

different frequency in frequency.

Frequency response magnitude

Phase

Bode plot relation between gain and w

relation between gain and

Amplifier

In log scale

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Amplifier bandwidth

Example: low pass filter

For K=1

|

| =

gain magnitude

20 log |

|= -10 log (1+

)

Ø = 0 -

gain phase

1 2 3

Gain

Attenuation region 1,2

Useful region 3

S=JW

RC = 𝜏 = 1/𝑤𝑜

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@ w << 20 log |

|= dc gain , Ø=

@ w >> 20 log |

|= attenuation (-ve value)

@ w = 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 = 20 kΩ, = 80 kΩ, and =5pF.

Solution

Why low pass

filter ??

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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