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Experiment 8&9

BJT AMPLIFIER

2

BJT AS AMPLIFIER

1. Objective:

1- To demonstrate the operation and characteristics of small signals common emitter

amplifiers.

2- What do we mean by a linear amplifier and why we need it linear

3- Effect of Q point on the shape of output signal .

4- Analyze the dc operation.

5- Stability of Q point

6- See the effect of coupling capacitors

7- Analyze the ac operation.

8- Determine the input resistance.

9- Determine the output resistance.

10- Determining the voltage gain.

11- Understand the model of a transistor

12- Explain the effects of an emitter-bypass capacitor.

13- Demonstrating the effect of a load resistor on the voltage gain.

14- Demonstrating the phase shift.

2. Introduction:

In this experiment, we emphasize the use of the bipolar transistor in linear amplifier applications.

Linear amplifiers imply that, for the most part, we are dealing with analog signals.

A linear amplifier then means that the output signal is equal to the input signal multiplied by a

constant, where the magnitude of the constant of proportionality is, in general, greater than unity.

A linear amplifier prov ides amplification of a signal without any distortion so that the out-put signal

is an exact amplified replica of the input signal. We want the output signal to be linearly

proportional to the input signal so that the output of the speakers is an exact (as much as

possible) reproduction of the signal generated from the compact disc. Therefore, we want the

amplifier to be a linear amplifier.

The transistor is the heart of an amplifier . Bipolar transistors have traditionally been used in linear

amplifier circuits because of their relatively high gain.

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Figure shows the circuit where VBB is a dc voltage to bias the transistor at a particular Q-point

and vs is the ac signal that is to be amplified. To use the circuit as an amplifier, the transistor needs

to be biased with a dc voltage at a quiescent point (Q-point), as shown in the figure, such that

the transistor is biased in the forward-active region.

I f a time-varying (e.g., sinusoidal)signal is superimposed on the dc input voltage, VBB, the output

voltage will change along the transfer curve producing a time-varying output voltage. If the time

varying output voltage is directly proportional to and larger than the time-varying input voltage,

then the circuit is a linear amplifier. From this figure, we see that if the transistor is not biased in the

active region (biased in either cutoff or saturation), the output voltage does not change with a

change in the input voltage. Thus, we no longer have an amplifier. The time-varying signals are

assumed to be small signals, which means that the amplitudes of the ac signals are small enough

to yield linear relations.

3. Effect of biasing on the Q point

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Bias establishes the dc operating point (Q-point) for proper linear operation of an amplifier. If an amplifier is not biased with correct dc voltages on t he input and output, it

can go into saturat ion or cutoff when an input signal is applied. Figure below shows the

effects of proper and improper dc biasing of an invert ing amplifier. In part (a), the output signal is an amplified replica of the input signal except that it is inverted, which means

that it is out of phase with the input. The output signal swings equally above and below the dc bias level of the output,VDC(out). Improper biasing can cause distort ion in the

output signal, as illustrated in parts (b) and (c). Part (b) illustrates limit ing of the posit ive

port ion of the out-put voltage as a result of a Q point (dc operating point) being too close to cutoff. Part (c) shows limit ing of the negative port ion of the output voltage as a

result of a dc operating point being too close to saturat ion.

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6

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4. Effect of Q point on the output signal.

1-connect the below circuit

2-In order to make the input signal, VS, small enough to prevent distort ion of the amplifier output, use the voltage divider circuit shown in Figure below , where VS is the voltage

across the 100Ω resistor. The voltage divider circuit shown in the Figure below divides input

voltage from FGEN by 100:1.Thus, to get VS = 0.02VP-P, set FGEN to provide 2.0VP-P as input to the voltage divider. You may find, it is not possible to see the correct amplitude of

VS on the oscilloscope, but if you are applying 2.0VP-P, 1kHz, at the input of the voltage divider shown in Figure below , you can be assured that VS = 0.02VP-P, 1kHz.

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R2 (kΩ) VO

Shape of the waveform

(e.g.sinusoidal or

distorted)

1

5

10

15

20

25

30

Vs VO

Shape of the waveform (e.g.sinusoidal or

distorted)

1m

2m

5m

15m

20m

25m

50m

Q12N2222

C1

10u

R11k

R268k

83%

RV110k

11%

RV210k

R2(1)

C1(1)

C222u

A

B

C

D+88.8

AC Volts

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5. Common Emitter Amplifier

BJT transistor amplifiers are frequently used in the common-emitter configuration (CE), since this

design gives both a high current gain (AI) and a high voltage gain (AV). This experiment explores

the dc and ac characteristics of the common-emitter circuit and how changing the Q-point

affects circuit performance.

The collector current (IC ) in a BJT circuit depends on the β of the transistor as well as the

transistor’s temperature and the other circuit elements. Good amplifier design requires choosing

biasing resistors such that the quiescent (DC) collector current remains constant, regardless of

whether a transistor with a different β is being used. Selecting the proper biasing network and

resistors keeps the collector current relatively constant. A key element that stabilizes the circuit to

changes in the transistor β is the emitter resistor, RE .

Using a bias network with an emitter resistor is a good way to keep IC relatively constant if the β or

temperature changes. Any increase in IC will cause the feedback voltage drop, VFB, across RE, to

increase, thus lowering the base-emitter voltage, VBE, and causing IB to decrease. This is a

negative feedback effect. Therefore, any changes in β will not cause IC to change significantly,

since IB will scale with 1/β. However, using an emitter resistor lowers the ac voltage gain of the

circuit, since the ac component, ic, must flow through RE to ground. Therefore, RE impedes ic. The

negative feedback voltage, VFB across RE "kills" the ac gain. To get around this undesired ac

response and still maintain the excellent control of ICQ that RE prov ides, a capacitor is placed

across RE in order to

short the ac current, ic, around RE to ground. This capacitor, CE, is called an emitter bypass

capacitor and is generally large, 10μF or greater. This capacitor allows most of the ac current to

flow around RE directly to ground. Note that the dc current, ICQ, still flows through RE, since the

capacitor acts as an open circuit to dc, and the stabilizing effect of RE is maintained. I f RE is

physically removed from the circuit, the negative feedback effect on the gain is taken away The

common emitter amplifier is characterized by moderate input impedance and slightly high output

impedance. Both the voltage and current gain are high resulting in po

wer gain.

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6. Dc analysis

Question:

1-For the circuit below find Q-point and find er mathematically and using Orcad .

2-Change β from 90 to 100 and compute the change in Q point ∆Ic ∆Vce

3-change the transistor itself and find the change in Q point ∆Ic ∆Vce .

Comment the result in your own words

2 CCB

1 2

R VV

R R

E B BEV V V

EE

E

VI

R

CE CC C C E EV V I R I R

eE

26mVr

I

Change Is =2e-15

B=90

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7. Effect of input (coupling ) capacitors

The capacitors at the input and output serve to isolate the signal source and load from the

voltage source VCC it is called coupling capacitors .a coupling capacitor is used to connect two

circuits such that only the AC signal from the first circuit can pass through to the next while DC is

blocked. This technique helps to isolate the DC bias settings of the two coupled circuits.

Use of Coupling Capacitors

Coupling capacitors are useful in many types of circuits where AC signals are the desired signals

to be output while DC signals are just used for prov iding power to certain components in the

circuit but should not appear in the output. For example, a coupling capacitor normally is used in

audio circuits, such as a microphone circuit. DC power is used to give power to parts of the circuit,

such as the microphone, which needs DC power to operate. So DC signals must be present in the

circuit for powering purposes. However, when a user talks into the microphone, the speech is an

AC signal, and this AC signal is the only signal in the end we want passed out. When we pass the

AC signals from the microphone onto the output device, say, speakers to be played or a

computer to be recorded, we don't want to pass the DC signal; remember, the DC signal was

only to power parts of the circuit. We don't want it showing up on the output recording. On the

output, we only want the AC speech signal. So to make sure only the AC passes while the DC

signal is blocked, we place a coupling capacitor in the circuit.

How to Place a Coupling Capacitor in a Cirucit

In order to place a capacitor in a circuit for AC coupling, the capacitor is connected in series with

the load to be coupled.

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Question : Connect the circuit and see the output before and after capacitor .

8. Ac analysis .

Question : Compute :Zo,Zi,Av assume ro=50k mathematically and using orcad compute Av .

1 21 2

1 2

R RR R ||R

R R

eZi R || r

C oZo R ||rC Zo R

ro 10RC |

o C ov

i e

V R ||rA

V r

R1

1k

V

V2

2

V

0

C1

10uV1

FREQ = 1kVAMPL = 10VOFF = 0

Rc

6.8k

0

Q1

Q2N2222

0

0

FREQ = 1kVAMPL = 1m

VOFF = 0

56k

V

8.2k

20u

C1

10u

1.5k

22

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To find

from orcad plot Iin and find the peak to peak value then find Zin

To find

=

from Orcad plot Ic and find the peak to peak value of Ic then plot

Vo and find peak to peak value then find Zo

9. Explain the effects of an emitter-bypass capacitor.

When we remove the emitter-bypass capacitor the voltage gain decrease .

Question :Compute the gain without the capacitor and see the effect in Orcad .

10. Demonstrating the effect of a load resistor on the voltage gain.

When we insert a RL the gain decrease significantly, so we need a matching circuit before

connecting the load.

Quest ion :

1-Give an example of RL in practical

2-Try Rl=50, 10k, 1000k, 1Tera

o C ov

i e

V R ||rA

V r ER

o C ov

i e

V R ||R ||rA

V r

L

C1

10u

0

R1

56k

0

0

Q1

Q2N2222V2

FREQ = 1kVAMPL = 1m

VOFF = 0

R48.2k

R31.5k

C3

1u R5

1k

V1

22

C2

22u

R2

6.8k

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11. AC analysis using Transistor model in Orcad : 1. Insert the dependent current source from the Analog library .

2. Connect the circuit

3. Compute Av using Orcad

12. Compute Zi ,Zo output using Orcad Note that we can not measure Zin and Zo directly as usual impedance because they are an AC

parameter which mean that the transistor must be in the active mode to measure it .

Insert I_test (dc current source equal 1A) and deactivate all the independent sources and short all

the capacitor then measure V_test . _

 _

V testZ

I test

13. Compute Zi,Zo practically In practical, we cannot measure Zi and Zo using the prev ious technique in Orcad but we measure

it as follow. Zin=Iin/Vin and Zo=Vo/Io so we just want to measure In ,Io ,Vin and Vo but sometimes it is

difficult to measure them accurately so we will use this techniques

Zi: Measure the input resistance by inserting a variable resistance in series with the signal

generator and input coupling capacitor and vary the resistance until Vo equal half of the Vo

without this resistance , the value of the variable resistance equal to Rin(why?)

Zo: The output resistance is measured as follows.

a- Measure the output voltage gain at no load.

b- Connect a variable resistance as a load and change its value until half of the outputin

(a)is obtained

c- The value of the variable resistance is equal to Rout (why?)

20u

Bre

1.66kr0

50k

8.2k

C1

10u

0

Rc

6.8k

F1

F

GAIN = 90

1.5k

V

FREQ = 1kVAMPL = 1m

VOFF = 0

0

0

56k22

Bre

1.66k

C1

10u

0

1.5k

0

r0

50k

Rc

6.8k

FREQ = 1kVAMPL = 0VOFF = 0

56k

8.2k

0

F1

F

GAIN = 90

I_test

1

0

20u20u

C1

10ur0

50k

FREQ = 1kVAMPL = 0VOFF = 0

Bre

1.66k

1.5k

F1

F

GAIN = 90

8.2k

0

0

I_test

1

0

0Rc

6.8k

56k

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Lab Report

1- Answer all the quest ions of the experiment (try to just ify and explain all the result s ) ?

2- Summary of your measurement?

3- Bring the following circuit in a breadboard, make sure to choose R2 such that your Q point lay in the middle of the active region (Vce=.5 Vcc) before coming to the

lab ? C1=C2=10u m Ce=22u

Bonus:

use variable resistor and find Zo,Zi in Orcad

Parameter Mathematically Using Orcad

Q-point ∆Ic ∆Vce

Ie

re

Av(NL)

Av(unbypassed)

Av(RL=50)

Av(RL=1T)

Zi

Zo