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CEE 321

EXPERIMENT 7

Class A and Class B Output Stages

Objective;

The aim of this experiment is to familisrize students with topologies of basic output stagessuch as class A and B amplifiers .

Equipment:

a)2N2222(npn) and 2N2907 (pnp) transistorsb)Two 1N914 discreate diodes.c)RL=10 and different value of resistors

Prelab:

Part1a)RL=10k Vcc=5 =0.52mA VD=0.5V

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R=(5-0.7)/0.5mA=8.6k

b)Vomax=4.8v

Vo min=-4.8v

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C) There is a gain =1 because its a voltage follower circuit also we can see on the graph ;

gain=1/1=1

PART 2

a) when Q1 is saturate;

5-0.2/RL=4.8/10k=0.48mA

b)Vo min =-5+0.2=-4.8v

Vo max= 5-0.2=4.8v

c)gain is 1 when v=between *0.5 Vcc -Vce+Vbe1]

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INTRODUCTION:

Class A Output Stages

The circuit illustrated in Figure and discussed in the previous section is an example of a ClassA amplifier stage. Class A amplifiers have the general property that the output device(s)always carry a significant current level, and hence have a large Quiescent Current. TheQuiescent Current is defined as the current level in the amplifier when it is producing anoutput of zero. Class A amplifiers vary the large Quiescent Current in order to generate avarying current in the load, hence they are always inefficient in power terms. In fact, theexample shown in Figure is particularly inefficient as it is Single Ended. If you look at Figureagain you will see that the amplifier only has direct control over the current between one ofthe two power rails and the load. The other rail is connected to the output load through a

plain resistor, so its current isnt really under control.

We can make a more efficient amplifier by employing a Double Ended or Push-Pullarrangement. Figure is an example of one type of output stage that works like this. You cansee that this new arrangement employs a pair of transistors. One is an NPN bipolartransistor, the other is a PNP bipolar transistor. Ideally, these two transistors have equivalentproperties e.g. the same current gains, etc except for the difference in the signs of theirvoltages and currents. (In effect, a PNP transistor is a sort of electronic mirror image of an

NPN one.)

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Procedure:

Part 1 :

we construct of figure 2.a lile that circuit;

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we give vi(t)=2sin(2 pi 1000t) with R=9K and we observe the output;

we observe these values ;

volt1=41.6mV

volt2=359mV

=0.4mA

Vopeak =RL*I

Vopeak=0.4mA*10k

Vopeak=4volt

efficency;

n=(4^2/20k)/2*5*0.4mA

20% efficeny here.

gain =there is a 10v/v gain

in prelab this was 1 because of low signal but there is high signal operation

Part 2:

We build that circuit;

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and give vi(t)=2*sin(2*pi*1000t) again.

and we observe that plots on the oscilascop

Volt1=4v

volt2=2v

= 0.08mA

vpeak=0.8v

n=(/4) vpeak/Vcc=0.15=15%

gain=4/2=2v/v

I find at gain is 1 but there is 2 because of high voltage(signal) operation.

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POSTLAB:

frequency responce of DC analysis

analog analysis of DC

These graps are so different from other graps which we find at lab because of operation. Atthe lab we find graphs for high signal operation but there is low signal operation.

Conclusion:

At this lab I learnt ClassA and Class B working principles. Also diffrences between high signal

and low signal operations on A class and B class. Seeing these differences was so useful forme .