New York City College of Technology
Lab Report
Experiment #9
The Common-Collector Amplifier
Emt 1255L (D712)
Faisal Tariq
06/19/2019
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Table of Contents
1. Objective-------------------------------------------------------- page 3
2.Materials Used------------------------------------------------- page 3
3.Procedure------------------------------------------------------- page 3-4
4.Scematic--------------------------------------------------------- page 5-6
5. Data and Results---------------------------------------------- page 7-9
6. Conclusion------------------------------------------------------ page 10
7. Laboratory Data Sheet------------------------------------(Attached)
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Objective: The Objective of this experiment is to be able to computer
the dc and ac parameters for a common-collector amplifier. Measure
the dc and ac parameters including input resistance and power gain.
Test the effect of different load resistors on the ac parameters.
Materials Used:
Resistors: two 1.0kΩ, One 10kΩ, one 33kΩ
Capacitors: one 1.0µF, 10µF
One 10 kΩ potentiometer
One 2N3906 pnp transistor.
Procedure:
1. First we compared the resistor’s nominal value to their measured values to
their measured values and recorded the collected data onto the data table.
2.Next we analyzed the CC Amplifier configuration and computed the DC
parameters of the circuit. After constructing the circuit, we measured and
collected the DC voltages on a table.
3. After measuring the DC parameters we turned on the function generator to
measure all AC parameters of the circuit. Before making any measurements,
we calculated each AC parameters to check if our measurements are close to
the calculated value when measured. For the calculations we assumed the bac
is equal to 100
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4. Next, we analyzed the CE Amplifier configuration and computed the DC
parameters of the circuit. After constructing the circuit we measured and
collected the DC voltages on a table.
5.Afterwards we computed the AC parameters of the following diagram, Then
we constructed the circuit using a signal generator and measured the AC
parameters.
6. Next when we compare the relationship between input and output
waveforms. Our phase relationship between VIN AND VOUT is no change.
7. In Step 7 we reduce RL to 1.0k Ω. We Observe the ac signal voltage at the
transistor’s base, emitter, and collector and measure the voltage gain of the
amplifier when RL is reduced, we observed that we are getting positive
clipping.
8. In step 8 when we change the RL by playing with the potentiometer, we
observe that as we increase the resistance the clipping comes down.
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Schematic:
Page 5
Page 6
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Data Results:
Table 9-1
Resistor
Table 9-3
R1
Listed
Value
33 k
Measured
Value
31.5 kΩ
R2
10 k
9.70 kΩ
RE
1.0 k
1.0 k
982 Ω
RL
Table 9-2
DC
Parameter
VB
AC
Parameter
Vb
Ve
re
981 Ω
Computed
Value
2.84 V
Measured
Value
3V
VE
3.54 V
3.69 V
IE
8.61 mA
VCE
3.54 V

Computed
Value
1.0 Vpp
Measured
Value
350 mVpp
2.90 Ω
343 mVpp
1.0 Vpp
Av
0.99
0.97
Rin(tot)
6.5 kΩ
7.98 kΩ
6.5
8.13
Ap
3.67 V
Table 9-4
Trouble
VB
DC Predictions
VE
VCE
DC Measurements
VB
VE
VCE
R1 open*
0.69 V 1.39 V 1.39 V
R2 open
12.0 V 12.0 V 12.0 V 12.06 V 12.06 V
12.0 V 12.0 V 12.0 V 12.06 V 12.06 V
2.84 V 0.0 V 0.0 V
2.80 V 0.0 V
R1 shorted
RE open
open collector 10.2 V 10.9 V 10.9 V
open emitter
2.84 V 12.0 V 12.0 V
0.36 V
1.08 V
10.3 V 11.07 V
0V
0V
Effect of
Trouble on
Vout
Little effect on
1.65 V
waveform
12.06 V No output.
12.06 V No output.
0.0 V
No output.
0 V
 V
Output
Reduced
No output.
Sample Calculations
𝑉𝐵 =
R2
9.7kΩ
𝑉𝐸𝐸 =
∗ 12𝑉 = 2.84𝑣
R1 + R2
31.3kΩ + 9.7kΩ
𝑉𝐸 = 2.844 + 0.7V = 3.54V
VRE=VEE-VE=12V-3.54V=8.46V
VRE 8.46𝑉
𝐼𝐸 =
=
= 8.61𝑚𝐴
R
982Ω
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1. In step 6, you observed the phase relationship between the
input and output waveforms. Is the phase relationship you
observed the same for an npn circuit? Explain.
Answer: The input and output are in phase for both types of
transistors since the phase relation is a function of the common
terminal of the transistor in this case it’s a common-collector.
2. Is the statement true if the cc circuit had been constructed
with an npn transistor? Why or why not?
Answer: In the equivalent NPN circuit, positive clipping occurs
when the transistor has maximum conduction, that’s why it is
saturation clipping.
3.The circuit used in this experiment used voltage-divider bias
(a) compared to base bias, what is the advantage?
(b) what disadvantages does it have?
Answer:
(a) Compared to base bias, voltage-divider bias has a more
predictable and stable Q-point
(b) Voltage-divider bias requires an additional resistor.
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4.Common-collector amplifiers do not have voltage gain but
still provide power gain. Explain.?
Answer: The output voltage is always smaller than the input
voltage, but the output current is larger than the input current,
that’s why afterall there is a power gain.
5.Figure 9-4 shows a CC amplifier with voltage-divider bias.
Assume Bac=Bdc=100 Compute the dc and ac parameters listed
below for the circuit ?
Answer:
Dc parameter
Ac parameters
VB=+10.9V
re=29.7
VE=+11.6V
AV=0.994
IE=0.84mA
Rin(tot)=34.3kΩ
VCE=-11.6V
Ap=6.8
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Conclusion:
In this laboratory we were able to test the DC parameters and to
make sure that there were no faults in a common collector
amplifier, and if there were to fix the inputs to make sure that in
the output everything is correct. We also found out how to
visually find the power gain and the AC input resistances. Overall
the lab was easy to build.
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