btec nc - analogue electronics - principles of analogue circuits

8/2/2019 BTEC NC - Analogue Electronics - Principles of Analogue Circuits

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Brendan Burr BTEC National Certificate in ElectronicsPrinciples of Analogue Circuits

Task 1

1.1 Define the decibel in terms of:

(a) Power Ratio.

PIN POUT

Where:PIN = Power InputPOUT = Power Output

The ratio of IN

OUT

P

P is expressed in Log Form as Decibels (dB).

dB P

P LOG A

IN

OUT

P

= 1010

Where x = Power Ratio in dB’s.

Example

When IN

OUT

P

P = 100, the power ratio in dB would be: dB LOG 2010010 =

The 20dB would express the Power Gain of the Amplifier.

(b) Voltage Ratio.

VIN VOUT

Where:VIN = Voltage InputVOUT = Voltage Output

The ratio of IN

OUT

V

V is expressed in Log Form as Decibels (dB).

dBV

V LOG A

IN

OUT

V

= 1020

Where x = Voltage Ratio in dB’s.

Example

When IN

OUT

V

V = 100, the voltage ratio in dB would be: dB LOG 2010020 =

The 20dB would express the Voltage Gain of the Amplifier.

1

SYSTEM

e.g.

Amplifier

SYSTEM

e.g.

Amplifier




1.2 Using the appropriate formulae convert the following values of gain todecibels:

(a) Power ratio of 0.01.

dB P

P LOG A

IN

OUT

P

= 1010

dB LOG A P 01.010 10=

dB A P

20−=

(b) Voltage ratio of 20.

dBV V LOG A

IN

OUT

V

= 1020

dB LOG AV 2020 10=

dB AV

26=

Answer to 0 decimal place

(c) Current ratio of 10.

dB I

I LOG A

IN

OUT

I

= 1020

dB LOG A I 1020 10=

dB A I

20=

2




1.3 Express an output power level of 10µW in terms of dBm.

As you can see in the chart below, at 0dBm the Power Output equals 1mW.For every increment of 10dBm the Power Output is multiplied by 10. Theopposite is said for every decrement of 10dBm, where the Power Output isdivided by 10.

-20 dBm 0.00001 W

-10 dBm 0.0001 W

0dBm 0.001 W+10 dBm 0.01 W

+20 dBm 0.1 W

It is easily calculated by the following equation:

××

=

=

−

−

3

6

101

101010

110

LOG Input

mW

PowerGain LOG Input

20−=Input

This is confirmed in the chart above, where 10µW equals -20dBm.

3

I/PPower

Gain

(dBm)

10µW




1.4 Explain the decibel in its use as a measurement of Signal to NoiseRatio in electronic systems.

The Signal to Noise Ratio is used to indicate the strength of theunwanted noise present.

It is expressed in Decibels as:

Signal to Noise Ratio = dB NoisePowet

r SignalPowe LOG

10

or

Signal to Noise Ratio = dBage NoiseBVolt

ageSignalVolt LOG

RMS

RMS

20

The higher the Signal to Noise ratio the cleaner the signal will be.

1.5

(a) Identify 3 sources of external noise.

• Radio Frequency Interference

• Radiation (such as the Big Bang)• Other Electrical/Electronic Equipment (in a close proximity such

as a Mains Hum)

(b) Identify 3 types of internal noise.

• Thermal Noise (Johnson Noise)• Components (resistors, capacitors, wiring)

• Shot Noise

4




1.6 Six networks are connected in cascade as follows:

(a) Determine the overall gain/loss in dB’s.

+5-3+12-6+10+2 = +20dB

(b) Determine the output power if the applied input power is 20mW.

( ) OUT

OUT

OUT

OUT

OUT

OUT

P

mW

mW

P mW

mW

P

mW

P AntiLOG

mW

P LOG

mW

P LOG

=××

×=×

=

=

=

=

−3102100

20

20

20100

20100

202

202

201020

W P OUT

2=

(c) Identify the attenuator blocks.

Block 2 = -3dBBlock 4 = -6dB

5

1 2 3 4 5 6

+5

dB-3

dB

+12

dB

-6

dB

+10

dB

+2

dBI/P O/P

20mW+20

dBO/P




Task 22.1

(a) Describe 2 different classes of amplifier.

Class A – This type of amplifier is not very efficient, this is because thedevice is always conducting. So even if the input is on itsnegative part of the cycle, the circuit will still use up thepower, therefore making it 50% efficient at best. With this inmind, if a high output power is required, then the wastedpower is going to become high also.

Even though there is a lack of efficiency in this type of amplifier, it isstill the most common type of amplifier in use for smallsignals. This is because it is cheap, as it consists of onetransistor and a few other components such as resistors andcapacitors.

Class B - This type of amplifier is much more efficient than the Class A, however it only has half of the waveform at the output.They are more efficient than the Class A type amplifier, dueto the circuit not drawing power on the negative part of thecycle. There is, however a large amount of distortionresulting in this type of amplifier not being used in manyapplications.

By using the Class B with another opposing transistor, you can get aClass B Push Pull Amplifier. This allows the whole

waveform to be amplified to the output. It works by the NPNtransistor creating the positive cycle and the PNP transistor creating the negative part of the cycle, so it is constantlyalternating.

6




(b) Identify the biasing points on a typical BJT characteristic input curve.

Image : http://www.electronics-tutorials.ws/transistor/tran_2.html

(c) Give 4 different effects of applying feedback on the performance of anamplifier.

• Increased Stability

• Reduction of Noise

• Reduction of Distortion

• Increased Bandwidth

(d) Summarise the effects on circuit performance in the form of a table.

Type GainInput

ImpedanceOutput

Impedance

CurrentFeedback

ShuntVoltage

Decreased Decreased Decreased

ShuntCurrent

Decreased Decreased Increased

VoltageFeedback

SeriesVoltage

Reduced Increased Decreased

SeriesCurrent

Reduced Increased Increased

7




Task 3

3.1 Design a class A Common Emitter Amplifier with the followingspecification:

DC Power Supply = +12V Load Resistor = 10KΩQuiescent operating point at Vc = 6V, Ic = 1.0mA, hFE=200 AC Input signal of frequency 1 kHz and amplitude of 8V peak to peak.Produce the following:

(a) A circuit diagram Attached

(b) Run a circuit simulation using Croc Tech or Multisim. Attached

(c) All the relevant design calculations (Below)(d) Test results from the simulation showing, typical input and output

waveform voltage levels and frequency.

(c) Design Calculations for the Output Side

6

12

6

=

=

Ve

VccVe

V Ve 2=

V Vc 6=

mA R

Ic

Vc R

L

L

1

6=

=

= K R L

6

8




Design Calculations for the Input Side

200

1

1200

mA Ib

Ib

mA

Ib

Ich FE

=

=

=

A Ib µ 5=

00005.0

10000005.0

10

2

2

2

=×=

×=

I

I

Ib I

A I µ 502=

A A I

Ib I I

µ µ 5501

21

+=+=

A I µ 551=

27.02

2

+=+=

R

E BE R

V

V V V

V V R

7.22=

A R

I V R R

µ 50

7.22

2

22

=

=

= K R 542

7.2121

1

−=−=

R

R

V

VbVccV

V V R

3.91=

A R

I

V R R

µ 55

3.91

1

11

=

=

= K R 1691

9




10




Frequency

Input:

( ) Hz F

F

INPUT

INPUT

1000

101000

16

=×

=−

KHz F INPUT

1=

Output:

( ) ( ) Hz F

F

OUTPUT

OUTPUT

1000

10190101190

166

=×−×

=−−

KHz F OUTPUT

1=

This confirms that the frequency at the input is not changed by the time it getsto get to the output. It is only the amplitude of the waveform that is changed.

Voltage Levels

Input VPK-PK = 4V + 4V = 8 VoltsOutput VPK-PK = 856mV + 10.3V = 11.156 Volts

11




3.2 Design a class B Amplifier with the following specification:

DC Power Supply = +/-12V Load Resistor = 10KΩ AC Input signal of frequency 1 kHz and amplitude of 8V peak to peak.Produce the following:

(a) A circuit diagram Attached

(b) Run a circuit simulation using Croc Tech or Multisim. Attached

(c) All the relevant design calculations

For the Class B Push Pull Amplifier, there are no design calculationsnecessary other that those calculated for the Waveform and todetermine the load resistor.It is critical to ensure the hFE for the PNP and NPN transistors are equaland maintain the same level, as this can affect the gain in thecorresponding half cycle.

(d) Test results from the simulation showing, typical input and outputwaveform voltage levels and frequency.

The waveforms below represent the input and output waveforms for theClass B Push Pull amplifier.

Frequency

Input:

( ) ( ) Hz F

F

INPUT

INPUT

1000

10100101100

166

=×−×

=−−

KHz F INPUT

1=

Output:

( ) ( ) Hz F

F

OUTPUT

OUTPUT

1000

10115101115

166

=×−×

=−−

KHz F OUTPUT

1=

This confirms that the frequency at the input is not changed by the time it getsto get to the output. It is only the amplitude of the waveform that is changed.

Voltage Levels

Input VPK-PK = 4V + 4V = 8 VoltsOutput VPK-PK = 3.4V + 3.4V = 6.8 Volts

12




13




3.3 Compare the results and performance of the two different amplifiers interms of the following:

(a) Output waveforms

The output waveform for the Class A shows that the circuit draws863mV even when the current isn’t flowing through the transistor,making the circuit not very efficient.The waveform is the shape illustrated due to the Vcc restricting theamount of gain available. The waveform is amplified from 8V to justover 11 Volts, however if the Vcc was increased then the waveformwould appear more sinusoidal. As with the Class B Push Pull Amplifier, there is a section in the waveform where the transistor isswitched on. At this point you can see distortion which is the non-linear region in the transistor input characteristic graph.The waveform for this simple amplifier, shows that only half of thewaveform is produced at the output. This is caused by the current notbeing able to flow correctly on the full cycle, therefore indicating adifferent type of amplifier would be better for amplifying somethingrequiring a truer representation of the input waveform.

On the output waveform of the Class B Push Pull Amplifier, it is clear that there is crossover distortion. This is caused by the two transistorshaving to overcome the VBE which turns them on.For these transistors it is around 0.7V, therefore the output voltageremains at 0V until the input voltage is at around 0.7V allowing the

current to flow through the transistor. This is visible on the outputwaveform above, where the red sinusoidal waveform reaches 0V andremains there, until the input voltage is increased to 0.7V or over, onthe positive or negative cycles.Due to this the voltage peak has decreased from 4 Volts to 3.4 Volts,suffering a voltage drop of 0.6V.The flat spot on the waveform lasts for the same time period as theInput waveform takes to go from -0.7V to +0.7V, confirming that this isthe time where the PNP and NPN transistors switch over and reachtheir linear part of their characteristic and turning on.There is no phase shift in the Class B Push Pull Amplifier Waveform

(b) DC Power dissipated in the load

The Power dissipated in the load of the Class A amplifier causes thetransistor to heat up, resulting in the Voltage Gain varying with it.There is high inefficiency in the Class A amplifier due to power constantly being drawn throughout the waveform.For the Class B amplifier the power dissipated is much more efficient.The two transistors should be connected to the same heat sink, this isdue to a difference in the peak voltage of either the positive or negative

cycles on the waveform. If one of the transistors had a higher

14




temperature than the other, one part of the waveform would have ahigher peak voltage, resulting in inaccuracies.

(c) Quiescent operating points

When the Class A amplifier reaches its Quiescent Point it allows thecurrent to flow through the Transistor giving the input signal theamplification the circuit was design for. The Vce for the Transistor inthis circuit is around 0.7V, it isn’t until this voltage is reached that thewaveform begins its half cycle.The point of quiescence controls the point in which the transistor turnson and off, which in turn regulates when the current at the Base of theTransistor can flow. On the Class A amplifier the quiescent point is setto the middle of the DC load line. This results in the Q-point allowingBase current to flow through the transistor throughout the input cycle,this then regulates the current on the Collector which causes outputwaveform to be amplified using the Vcc.

The Class B Push Pull amplifier has two Transistors, this is becausecurrent will not flow for the whole cycle. It requires two transistors,PNP and NPN, this allows the current to flow for each of the half cyclesallowing the current to flow through both of the collectors once eachTransistor reaches its saturation point, determined by the Q-point.

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btec nc - analogue electronics - principles of analogue circuits

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