21673105 6 Introduction to Amplifiers

7/29/2019 21673105 6 Introduction to Amplifiers

1/38

Introduction to Amplifiers


2/38

Outline

Amplifier Properties

BJT Amplifier Configurations

Amplifier Classifications Decibels


3/38

Amplification

Amplification

the process of increasing the powerof an ac signal

BJT amplifier, JFET amplifier, OP-AMP amplifier

What is

amplification?


4/38

Part 1.



5/38


Three Fundamental Properties

Gain

Input impedance

Output impedance

outputinput

Zout

Zin A

General amplifier model


6/38

Amplifier Gain

Gain

A multiplier that exists between the input and output

of a circuit.

For example, if the gain of an amplifier is 100, thenthe output signal is 100 times as great as the input

signal under normal operating conditions.

Types of Gain:

Voltage gain, AV Current gain, Ai

Power gain, Ap


7/38

Gain as a Ratio

Gain

Ratio of an output value to its corresponding input

signal

Av = vout

vin

Where

Vout = the ac output voltage from the amplifier

Vin = the ac input voltage to the amplifier

Ai = iout

iin

Ap = Pout

Pin


8/38

The General Voltage Amplifier

Model

outputinput

Zout

Zin Avvin

voltage amplifier model

Zout

Zin Avvin

RS

vSRL

Voltage source


9/38

Amplifier Input Impedance (Zin)

Input impedance (Zin)

The load that an amplifier places on its source.

When an amplifier is connected to a signal source,

the source sees the amplifier as a load. The inputimpedance of the amplifier is the value of this load.

vin = vs Zin

RS + Zin

Amplifier input circuit

Zout

Zin Avvin

RS

vSvin

1.5k

100

2 mV

Example.

Calculate vin.

vin = 2mV (1.5k)/ 1.6k

= 1.88 mV


10/38

Amplifier Output Impedance (Zout)

Output impedance (Zout) The source impedance that an amplifier presents to its load.

When a load is connected to an amplifier, the amplifier acts asthe source for that load. As with any source, there is somemeasurable value of source impedance, in this casr, the outputimpedance of the amplifier.

vL = vout RL

Zout + RL

Amplifier output circuit

Zout

Zin Avvin

RL

vL

1.2k

300

300 mV

Example.

Calculate vL.

vL = 300mV (1.2k)/ 1.5k

= 240 mV


11/38

Combined Effects of the Input and

Output Circuits The combination of the input and output circuits can cause a fairly

significant reduction in the effective voltage gain of an amplifier.

Av (eff) = vL

vS

Zout

Zin vout = AvvinRL

1.2k

250

AV = 340

vin = 15mV (980)/ 1k = 14.7 mV

vout = 340(14.7) mV = 5 v

vL = 5v ( 1.2k)/ 1.45k = 4.14 v

RS

vS

20

15 mV 980

vS = 15mv vL = 4.14v

AV(eff) = 4.14V/15 mV = 276

AV = 340 Av(eff) = 276

reduction

of voltage

gain!


12/38

How do you reduce the effects of the input and

output circuits on an amplifier voltage gain?

1. Increasing the value of Zin

2. Decreasing the value of Zout.

Av (eff) = vL

vS

Zout

Zin vout = AvvinRL

1.2k

20

AV = 340

vin = 15mV (8k)/ 8.02k = 15 mV

vout = 340(15) mV = 5.1 v

vL = 5.1v ( 1.2k)/ 1.22k = 5 v

RS

vS

20

15 mV 8 k

vS = 15mv vL = 5vAV(eff) = 5 V/15 mV = 333

increased

significantly!

Zin and Zout are affected by the choice of active components used as well as the type of biasing circuit and

com onent values. Soon!!!


13/38

The Ideal Voltage Amplifier

1. Infinite gain (if needed).

2. Infinite input impedance.

3. Zero output impedance

Zout

Zin voutRL

1.2k

0 RS

vS

vin vL

no current in

input circuit

vin = vS (ideal)

no voltage divider

in output circuit

vL

= vout

(ideal)Av = AV(eff)


14/38

The Current Amplifier Model

Current Amplifier a circuit designed to provide aspecific value of current gain.

ZoutZin Aiiin

ZoutZin AiiinRL

iSRS


15/38


-Input Circuit-

where:

iin = amplifier input current

iS =the source current

RS|| Zin = the parallel combination of RS and Zin

iin < iS

ZoutZin Aiiin

iin = is RS || Zin

Zin

iS

Did we just use

the current

divider formula?


16/38


-Output Circuit-

where:

iL = amplifier load current

RL|| Zout = the parallel combination of RL and Zout

iout = Ai iin

iL < iout

ZoutZin Aiiin

iL = iout RL || Zout

RL

RL

Combination of Input and Output circuit Effect reduced effective current gain

Ai(eff) = iL / iS solution: Decreasing the value of Zin

Increasing the value of Zout


17/38

The Ideal Current Amplifier

Infinite gain (if needed).

Zero input impedance (Zin = 0).

Infinite output impedance (Zout= )

iin = iS (for ideal current amplifier)

iL = iout (for ideal current amplifier)


18/38


19/38

BJT Amplifier Configurations

common- emitter amplifier

common-collector amplifier

common-base amplifier


20/38

Common-Emitter Amplifier

CE amplifier is the most widely used BJT amplifier The emitter terminal of the transistor is common to both input and

output circuits.

The emitter terminal of the transistor is normally returned to acground(orac common) provided by the bypass capacitor (CB).

The CE amplifier is unique it produces a 180 voltage phase shift

from its input to its output.


21/38

Common-Collector Amplifier

CC amplifier is also known as emitter-follower

This circuit is most commonly used for its current

gain and impedance characteristics.


22/38


23/38

Comparing the BJT Amplifier

Configurations

AP = AV Ai

Common

Emitter

Emitter

Follower

Common

BaseAv

Midrange Less than 1 Midrange

AiMidrange Midrange Less than 1

APHigh Midrange Midrange

ZinMidrange High Low

ZoutMidrange Low High


24/38

Part 3.

Amplifier Classifications


25/38

Amplifier Classifications

Class A amplifier an amplifier with a single transistorthat conducts during the entire input cycle.

Class B amplifier an amplifier with two transistors thateach conduct for approximately half the input cycle.

Class C amplifier an amplifier with one transistor that

conducts for less than 180 of the input cycle. Class AB amplifier an amplifier with two transistors that

each conduct for slightly 180 of the input cycle.

Amplifier Efficiency

Efficiency () the percentage of the power drawn from the dc power

supply than an amplifier actually delivers to its load.

= (PL/ Pdc ) x 100where: = (eta) efficiency of the amplifier, in %

PL

= ac load power

P = dc input power


26/38

Distortion

One of the goals in amplification is to produce an output waveformthat has the same shape as the input waveform.

Distortion any undesired change in the shape of a waveform

Two types of Distortion: Nonlinear distortion

Crossover distortion


27/38

Class A Amplifiers

Characteristics: An active device that conducts during the entire 360 of the input cycle.

An output that contains little or no distortion.

A maximum theoretical efficiency of 25%.

Class A operation is achieved in a BJT amplifier by midpoint biasingthe transistor.

Because of their relatively poor efficiency ratings, class A amps aregenerally used as small-signal(low power) amplifiers.


28/38

Class B Amplifiers

Characteristics: Two transistors that are biased at cutoff (each conducts during one

alternation of the ac input cycle).

An output that contains little or no distortion.

A maximum theoretical efficiency of approximately 78.5%.

The relatively high efficiency rating makes it very useful as a high-

power amplifier.


29/38

Class AB Amplifiers

One variation of the class B amplifier. Class B amplifier an amplifier with two transistors that each

conduct for slightly more than 180 of the input cycle.

Also known as diode-biased amplifier.

This is used to prevent a specific type of distortion that can beproduced by a standard class B amplifier.


30/38

Class C Amplifiers

The BJT in the class C amp is biased deeply into cutoff. The ac input to theamp causes the transistor to conduct for a brief time during the input cycle.

The output waveform is produced by the LC tank in the collector circuit.

Tuned amplifier an amplifier designed to have a specific value of gain overa specified range of frequencies.

Characteristics: A single transistor that conducts for less than 180 of the ac input cycle.

An output that may contain a significant amount of distortion.

A maximum theoretical efficiency rating of approximately 99%.


31/38

Circuit:

Conduction:

Maximum

theoretical

efficiency:

Distortion:


32/38


33/38

Part 4.

Decibels


34/38

Decibels

Decibel (dB) a logarithmic unit used to express theratio of one value to another.

Writing numbers in dB form allows us to easily representvery large gain values as relatively small numbers.

dB Power Gain the ratio of circuit output power to inputpower, equal to 10 times the common log of that ratio.

Ap(dB)

= 10 log AP

= 10 log (Pout

/ Pin

)

Ap =log-1 (Ap(dB) / 10) Inverse log =

antilog=(log-

1

)


35/38

Positive versus negative dB values

Positive dB values represent a powergain, whilenegative dB values represent a powerloss.

Positive and negative decibels of equal magnituderepresent reciprocal gains and losses.

Say what?

Try this examples and compare.

1. Pin = 50 mW and Ap(dB) = 3 dB; Pout?

2. Ap(dB) = -3 dB and Pout = 50 mW, Pin?


36/38

The dBm Reference

This rating tells you that the maximum output power fromthe amplifier is a certain value above 1 mW.

dBm values represent actual power levels, while dBvalues represent powerratios.

Number 1.

.How can

this be?

Try this example.

1. An amplifier has a rating of Ap = 50 dB. Calculate the output power

of the amplifier. Pout?

2. The output rating of an amplifier is given as 50 dBm. Calculate the

output power for the circuit.


37/38

dB Voltage Gain

Av(dB) = 20 log Av = 20 log (vout / vin)

Ap =log-1 (Av(dB) / 20)


38/38

One Final Note on Decibels

1. Decibels are logarithmic representations of gainvalues.

2. Decibel power gain is found as 10 log AP.

3. Decibel voltage gain is found as 20 lof AV.

4. When AV changes by a given number ofdecibels, AP changes by the same number ofdecibels.

5. You cannot use dB voltage and power gain

values as multipliers. For example, if you want todetermine vout, given vin and Av(dB), you mustconvert Av(dB) to standard numeric form beforemultiplying to find vout.

21673105 6 Introduction to Amplifiers

Documents

Transcript of 21673105 6 Introduction to Amplifiers