experiment 8&9 - الصفحات الشخصية | الجامعة الإسلامية...
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Experiment 8&9
BJT AMPLIFIER
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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