lecture no. 1 · • of course, if you use negative feedback, overall gain of the amplifier is...

1 Linear electronic

Lecture No. 1

2 Linear electronic

3 Linear electronic

4 Linear electronic

5 Linear electronic

Lecture No. 2

6 Linear electronic

7 Linear electronic

8 Linear electronic

9 Linear electronic

10 Linear electronic


Lecture No. 3


Lecture No. 4 Example: find Frequency response analysis for the circuit shown in figure below. Where

RS=4kRB1=8kRB2=4k RE=3.3kRC=6k RL=4k

VCC = 12V, IE 1 mA, o = 100, C= 13.9 pF, Cμ = 2 pF,

ro =100 krπ = 2.56kCin= Cout = 1F, CE=100F.

1. Mid-Band Gain Calculation

k4.2

64

64R '

L

But REF = 0

Cin

Cout


75.93k56.2

k4.2100Av

13.23k31.1k4

k31.175.93AVs

2. Low-Frequency Response

15.9Hz

10k4k62

1f

6out

Hz30

10k31.1k42

1f

6in


2. High-Frequency Response • From here up in frequency, we will start to worry about the details of what’s inside the BJT. Now those junction capacitances matter,

so one needs to use the “full” Hybrid- model.

pF5.208k4.2039.0123.19Ctotal

987

k4k31.1

k4k31.1

RR

RRR

sin

sin's

MHz5.773105.2089872

1f

121h

Also, we will discuss other examples and questions


Lecture No. 5 Multi-stage "Cascaded" Amplifiers

Most practical amplifiers required more gain than can be obtain from a single stage. Consequently, it is common practice to feed the output of one amplifier stage into the input of the next stage. Gain and bandwidth considerations in cascaded amplifiers Most of a cascaded amplifier stages are used to obtain either voltage gain or a current gain. However, in most cascaded amplifiers, ultimately the power gain is important. If a voltage gain is required, we can calculate the total gain by using the equation for voltage gain of one stage. Thus, the voltage gain for stage1 is:

1in

1o1v

V

VA

In addition, the voltage gain for stage 2 is:

2in

2o2v

V

VA

The gain for additional stages can be written in a similar manner. Then, the total amplifier voltage gain Atotal for n-cascaded stages is:

1

11

2

3

1

2

1

1

in

)n(o

on

)n(o

o

o

o

o

in

o

V

V

V

V...

V

V

V

V

V

V

n321total A...AAAA

In same manner, we can derive the current gain or power gain

for the cascaded amplifier. Let f1 and f2 is the lower and upper cutoff frequencies,

respectively. The voltage or current gain of one stage has been reducing to 0.707 of its reference value at these frequencies. Now, if we have an amplifier with two identical stages of amplification, the voltage gain at f1 will be reduced by factor of 0.707 in each stage. Thus, the amplifier gain at f1 (and also f2) will be:

0.707 Av1 0.707 Av2 = 0.5 Atotal


For n-identical cascaded stages of amplification, the gain at f1 and f2 will be (0.707)n Av. The voltage gain per stage is given for the low frequency by:

1o

j

jAA

1

o

j1

1AA

The magnitude of above equation is:

21o

)(1

1AA

If there are n-cascaded stages, the magnitude of the total gain amplifier is:

n

21

nototal

)(1

1AA

Now , if ω is to be equal to ωL , the term multiplying Ao must be equal to .0707 or 1/(2)1/2 . Then

2/n2

L

12/1 12

2

L

1n/1 12

Then,

12 n/1

1L

The voltage gain per stage is given for the low frequency by:


jAA

2

2o

As similar

12 n/12H

From above equations, ωL will be greater than ω1 if n is greater

than one and ωH will be less than ω2 if n is greater than one. Thus, the bandwidth of the amplifier decreases as the number of cascaded stages increases. On the other hand, if the amplifier bandwidth is to remain constant, the stage bandwidth must increase as the number of cascaded stages increases.

Example: If the f2 =32MHz, Atotal=1000, and fH= 12MHz,

Calculate mid-band gain "Aon" for n=3 & 7 cascaded stages.

Determine the gain per stage for each amplifier.

ω2 = 2π32106 rad/sec

ωH = 2π12106 rad/sec

n

2

2

H

nototal

)(1

1AA

1. For n=3 3

2

6

6

3o

10322

101221

1A1000

Ao3 =1218.2

Ao=10.68

2. For n = 7


Ao7 =1584.98

Ao=2.865


Lecture No. 6 PRACTICAL CASCADE AMPLIFIER CIRCUIT

• In the above figure, let us work out the gain assuming nothing about the Rin and Rout of each stage, looking at them as voltage dividers between each stage and between the last stage and the load.

• The equation reduces to the ideal case of AV = A2 for two identical stages if we let the Ro's go to 0 and the Ri's go to infinity.

• For example: let Ro = 100and Rin = 1M, what is the gain with

two stages of gain A in series? (Assume RL = 1Mtoo).


• By the time you reached that point, other effects would have caused much more trouble (for example, the fact that noise from each successive stage is add to the noise coming into that stage and amplified... on down the line!). Cascode Amplifier Design Example Specifications: DC power dissipation: PD < 25 mW Power Supply: 12 VDC Voltage Gain: -50X

Load: Resistive, 50 K

Assume RS = 0 Must use 2N2222A Transistors (NPN, ß = 150 measured)

DESIGN PROCESS: 1) Pick Vcc unless specified.


Vcc= 12 VDC 2) Calculate IMax.

3) Select an IC < IMAX and solve for gm2. Let IC = 1.8 mA

4) Let RS = 0 and solve for RC:

RC = 730 5) Use a 1/4, 1/4, 1/4, 1/4 biasing rule to set up bias resistors. Let VCE of the transistors = Vcc/4 = 3V and solve the following for RE.

6) The required base current is:

7) Solve for the biasing resistors.


Lecture No. 7 FEEDBACK • There are two types of feedback: regenerative (positive feedback) and degenerative (negative feedback). • Unless you want your circuit to oscillate, we usually use NEGATIVE FEEDBACK... • This idea came about in the late 1920’s when they were able to build amplifiers with reasonable gains, but with gains that were difficult to control from amplifier to amplifier... PROPERTIES OF NEGATIVE FEEDBACK • Nonlinear distortion can be reduced. • The effects of noise can be reduced (but not the noise itself). • The input and output impedances of the amplifier can be modified. • The bandwidth of an amplifier can be extended. • Of course, if you use negative feedback, overall gain of the amplifier is always less than the maximum achievable by the amplifier without feedback.

THE BASIC FEEDBACK CIRCUIT

• With an input signal xs, an output signal xo, a feedback signal xf,

and an amplifier input signal xi, let us look at the basic feedback circuit illustrated above.


• The amplifier has a gain of A and the feedback network has a

gain of ... • The input to the amplifier is,

xi = xs - xf

• The output of the amplifier is, xo = A xi

• We can obtain an expression for the output signal in terms of the input signal and the feedback gain...

xo = A xs - xf = A xs - xo

• Re-arranging,

xo = A xs - Axo xo (1 + A= A xs • From which we obtain the negative feedback equation by solving for the overall gain

• For positive feedback, you only need to change the “+” sign in the denominator to a “-” sign.

• It is easy to obtain the equation for the feedback signal, xf,

• If the amplifier gain and the loop gain are large (i.e. A>> 1),

then the feedback signal xf becomes nearly an identical copy of the input signal xs.

• Here we have assumed that there was an input “comparator” or

“mixer” and an output “sampler” that provided us with a copy of the output signal for use as a feedback signal.

• The form these devices take depends upon whether the

amplifier’s input and output are current or voltage based...

• We refer to a given feedback amplifier in terms of the “MIXING - SAMPLING” feedback, where MIXING and SAMPLING are either SHUNT or SERIES... • There are four possible types... (For sampling think of how you would measure V or I in the lab... series for current and shunt for voltage).


TYPES OF MIXER

TYPES OF SAMPLER


Lecture No. 8 INPUT-OUTPUT (MIXING - SAMPLING) SERIES-SHUNT series (voltage) mixing, voltage-sampling V-V SHUNT-SERIES shunt (current) mixing, current-sampling, I-I SERIES-SERIES series (voltage) mixing, current-sampling V-I SHUNT-SHUNT shunt (current) mixing, voltage-sampling I-V

•Let's consider a familiar example.... the common-emitter amplifier.

• Since we sample the output current and generate a voltage feedback signal, this is a series-series feedback topology.

• Considering the output current to be the output signal (e.g. io = ic) and the input to be vs (for simplicity, assume that RB1 and RB2 are very large), the units of the basic amplifier are,

• We know that the feedback voltage is given by Ohm's Law as,

Vf=ioRE so the feedback network gain, = RE (NOTE: don't get

confused! This is NOT the transistor's !)

Transistor's current gain = io/ vi gm in -1

• The output current is given by:


io = gm (vs - vf) = gm(vs - ioRE)

• Combining these equations to find the overall gain for the amplifier, Gm, we end up with an equation we have seen before!

Now that:

SERIES-SHUNT FEEDBACK (SERIES [VOLTAGE] MIXING, VOLTAGE-SAMPLING)

• Two examples are shown below. On the left, the common emitter amplifier connect to feedback circuit consist of (R1 and R2).


• On the right, the non-inverting operation amplifier configuration using an ideal op-amp (infinite input impedance, zero output impedance)...

• Notice how you can re-draw the two feedback resistors as a feedback network of the form we are discussing.

• The feedback network gain can be obtained directly by voltage division,

C1

RL

Vo

Vin Q1NPN R1

R2

RE

Vcc

R2

R1


Lecture No. 9 • This can be plugged into the feedback gain equation to find the overall gain,

• Continuing with the series-shunt case, but including the input and

output resistance terms (Ri and Ro). • We can obtain an expression for the equivalent input and output resistance...


• The output resistance can also be obtained by the same method

we used previously: • Reduce the input signal (Vs) to zero and apply a test voltage Vt at

the output.

• Starting with the definition,

• Therefore,


Lecture No. 10 Power Transistor Amplifier

Amplifiers are used to increase the level of a signal and, depending on the increase required; stages are often cascaded to increase the gain. The last stage of the cascade may be required to drive same form of load, for example a loudspeaker, a servomechanism or a coaxial cable for RF applications.

Power amplifiers are classified by the nature of the collector current waveform into class A, class B and class C. This classification is explained in the following figure, where a typical transfer characteristic of a transistor amplifier is used. CLASS-A AMPLIFIER The simplest possible circuit of a class-A amplifier is shown in Figure below where RL is the load resistance and RB is the biasing resistance.


We already know that for the maximum undistorted peak-to-peak output voltage swing the Q-point should be selected on the basis of the following equation:

In this equation, RDC is the dc resistance and RAC is the ac resistance in the collector-emitter circuit. Thus,

RDC = RAC = RL (In circuit above, there is only one resistance)

The collector-to-emitter voltage at the operating point is:

When the input signal goes negative, it causes a decrease in

the collector current and a corresponding increase in the collector-to-emitter voltage. The ac component of the output voltage is now positive. This situation continues until the collector current reaches zero and the collector-to-emitter voltage becomes equal to VCC.


This is the maximum value of the collector-to-emitter voltage ignoring the nonlinear operation in the cut-off region.

From the above discussion, it should become clear that the

output voltage is zero when the input signal is zero and the collector voltage is ½ VCC. At that time, the collector current is ICQ.

As the collector current becomes 2ICQ, its ac component has a maximum value of ICQ, and the collector-to-emitter voltage is zero. Thus, the ac component of the output voltage has a minimum value of -½VCC. When the collector current becomes zero, the ac component of the collector current has a minimum value of – ICQ, and the ac component of the output voltage attains a maximum value of ½ VCC. With this understanding, we can write the time-domain expressions for the total collector current and total collector-to-emitter voltage as:

The

instantaneous power dissipated by the transistor is:

- Where the first term, ½ ICQ VCC, is the power dissipation at the Q-

point when there is no input signal. - The second term, ½ I VCC sin2 (ωt), is due to the input signal.

- Figure below shows a plot of the instantaneous power dissipation by the transistor.


Since the average value of a sin2 (wt) function is ½, the average power dissipated by the transistor, is:

The instantaneous power delivered to the load is:

Thus, the average power delivered to the load is:

We can substitute for ICQ RL = ½ VCC, and obtain an expression for the maximum value of the average power delivered to the load as:

The power supplied by the source is:

We can determine the efficiency of the class-A amplifier as:

Did you notice that the average power dissipated by the transistor at its Q-point is twice as much as the average power output? For this reason, class-A configuration should be used only when the power output is less than or equal to 1 W.


Lecture No. 11 Example: Design a common-emitter amplifier that delivers 0.5W power to a

100 resistor. Use a transistor that has a maximum current rating of 500mA, collector-to-emitter saturation voltage of 0.5V, breakdown voltage of 40V, and the common-emitter current gain of 100.

Solution: Let us first design the common-emitter amplifier. We have selected a four-resistor bias circuit as shown in Figure below because of its stable operation.

The average power supplied to the 100-W load resistor is 0.5 W. then;

The maximum current through the transistor is expected to be

twice as much, i.e. 200mA during the positive excursion of the collector current. Since the transistor can supply a maximum current of 500mA, it is safe to use this transistor. The undistorted

maximum output voltage swing is 10 V (100mA100 ). Let us add a 10% safety factor to the current in order to keep

the swing from entering the saturation region on one hand and the


cut-off region on the other. This will help keep the distortion to its minimum. Therefore, let us select, ICQ = 110mA. - We can now determine the supply voltage as:

- At the Q-point, the base current is:

mA

mAI

ICQ

BQ11

100

110.

- Let us select a current in R2 that is nearly equal to 10 times the base current.

64101110

7032

.

.R

Thus, the current through R1: mAIII

BQ111111021 ..

We can now compute R1:

k

I

VV

Rbecc 91

10111

70223

11 .

.

.

The power supplied by the dc source, is:

WmAVIIPccCQs

662221111101 .).()(

But: PL=0.5W

Then, %...

.8181880

662

50

S

L

P

P

WmAIVPCQCCMAXT

211110112

1.)(

In order to determine the voltage gain, current gain, power gain and input resistance, we can represent the circuit in the mid-frequency range by its model as shown in Figure below.


Since the base current is 1.1mA, the equivalent resistance in the base circuit is:

72211

25.

.

BQ

T

I

V

r

The base current can now be computed as:

Where; is the equivalent resistance in the emitter circuit. The collector current is:

The output current is:

440101722101

100100

)/.(in

o

v

V

V

A


in

R 16.6

Thus, the current gain:

AI = -73.2 Finally, the power gain is:

AP = AV AI

AP=(-73.2) (-440) = 32208


Lecture No. 12 CLASS B AMPLIFIER

One of the major disadvantages of a class-A amplifier is that it

dissipates maximum power at its Q-point. The amplifier is

consuming power even when there is no signal. In fact, the power

consumption goes down only when the input signal is present. We

can reduce the power consumption just by biasing the transistor at

the cutoff point. At the cutoff point, the voltage drop across the

transistor is at its maximum value while the current though it is

zero. When an amplifier is biased at its cutoff point, it is called

the class-B amplifier.

Let us consider the situation when the input voltage begins its

positive cycle. For the circuit shown, the transistor will not begin to

conduct until the input signal is equal to its base-to-emitter voltage

drop VBE. As soon as the input voltage goes above VBE, the

transistor turns on and the conduction process begins. The output

voltage will simply be VIN – VBE. As the input voltage increases, the

output voltage increases also and so does the current in the

collector.


The average value of the half-wave collector (load) current is:

Thus, the power supplied by the dc voltage source is:

The effective (rms) value of the load current is:

The power supplied to the load is:

It is clear that the efficiency is proportional to the amplitude of


the collector current. Thus, efficiency would be maximum when IO(Peak) = IACM.

Then; R

VI

L

CCACM

Thus, the maximum power supplied by the dc voltage source is:

The maximum power delivered to the load is:

We can now compute the maximum efficiency of the class-B amplifier as:

The power dissipated by the transistor can be obtained by:

Differentiating this equation with respect to IO(Peak) and setting it equal to zero, we obtain the peak value of the collector current that results in maximum power dissipation in the transistor.

Example A class-B amplifier of the type shown in Figure below drives a

load of 100 . It operates from a 15-V dc supply. Assume that the

base-to-emitter turn on voltage is essentially zero, is very large, and the output voltage is basically sinusoidal. 1- What is the maximum power it can deliver to the load when its

collector-to-emitter saturation voltage is 0.5 V? 2- Efficiency. 3- The average and maximum power dissipated by the transistor.


Solution: When the base-to-emitter turn on voltage (VBE) is neglected, the transistor turns on as soon as the input voltage becomes positive. Taking into account the collector-to-emitter saturation [VCE(SAT)] voltage of 0.5 V into account, the maximum possible output voltage is:

The maximum value of the load (collector) current is:

The average and rms values of the load current are:

The power supplied by the 15-V dc source is:


The efficiency of the class-B amplifier is:

The average power dissipated by the transistor is:

The maximum power dissipation by the transistor is:


Lecture No. 13

CLASS-B PUSH-PULL AMPLIFIER

As explained in the previous section, the class-B amplifier is better than the class-A amplifier because it has high efficiency. However, the class-B amplifier can only amplify the positive-half of the input signal. In order to be able to amplify the entire signal, we can use two complementary transistors as shown in Figure below.

When the input signal is positive, the NPN transistor Q1 turns ON, the PNP transistor Q2 is OFF, and the output voltage is positive. The NPN transistor is pushing the current into the load resistor during the positive cycle of the input voltage. The output voltage can be expressed as:

The maximum value of the output voltage is obtained during the

positive cycle when the NPN transistor just begins to saturate.

That is,


And

As the input voltage falls below –VBE, the PNP transistor Q2

begins to conduct and it pulls the current from the load. The NPN

transistor Q1 pushes the current into the load and the PNP

transistor Q2 pulls the current through it. This is why this circuit

configuration is referred to as the Push-Pull amplifier.

As the PNP transistor, Q2 turns ON, the NPN transistor Q1cuts

off and the output voltage is negative. The saturation voltage of the

PNP transistor limits the minimum value of the output voltage.

Since we are using the two complementary transistors, we expect

the magnitude of the saturation voltage for both transistors to be

the same. That is VEC(SAT) of the PNP transistor is equal to VCE(SAT)

of the NPN transistor. Thus, the minimum value of the output

voltage is:

And

Let us denote the amplitude (the magnitude of the maximum possible swing) of each output waveform as:

The maximum current through each transistor is:

The average current through each transistor is:


Example

A class-B push-pull amplifier drives a load of 100. It operates from a dual ± 15V dc supply. Assume that the base-to-emitter turn

on voltage is zero, is very large, and the output voltage is essentially sinusoidal. 1- What is the maximum power it can deliver to the load when the collector-to-emitter saturation voltage for each transistor is 0.5 V? 2- Efficiency of power transistor amplifier. Solution:

The maximum, unclipped peak-to-peak output voltage that we can obtain from the push-pull amplifier is 29 V.


The total power supplied by the 15-V and the –15 V dc sources is:

The efficiency of the class-B amplifier is:


Lecture No. 14 Logic gates A logic gates is an electronic circuit which make decisions. It has one

output and one or more inputs.

1. NOT-Gate

It is so called because its output is NOT the same as its input. It is also

called inverter because it inverts the input. It has one input and one output

as shown in figure below.

Figure below


2. OR-gate

The output of this circuit is logic "1" when either one input is logic "1" as

shown in figure below.

Figure

Figure


AND-gate

lecture no. 1 · • of course, if you use negative feedback, overall gain of the amplifier is...

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