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OBJECTIVES
1. To measure the voltage, current, and
power gains of the CE amplifier
2. To measure the input and output
impedance of the CE amplifier
CE = Common Emitter
INTRODUCTION
• In the previous experiment, we
bypassed the emitter resistor with a
capacitor to avoid degenerative
feedback.
• The resulting voltage, current and power
gains were high.
• In this experiment, we will be leaving the
emitter resistor unbypassed.
•
We want to observe how that affects the
performance of the amplifier
• The following slide shows a basic CE
amplifier with an unbiased emitter resistor.
• The input signal is applied between the
base and ground.
CE AMP WITH A UNBYPASSED RE
• This causes two AC voltage drops: one
is between the base and emitter, and
the other is across RE.
• As the base voltage swings positive,
the base current increases.
• This causes the collector-to-emitter
current to increase.
• The emitter current increases as the
base swings positive, and the voltage
across RE rises.
• Thus the AC signal on the base is in
phase with the AC signal on the emitter.
• This reduces the AC signal between the
base and the emitter, thus reducing the
gain of the amplifier.
REQUIRED PARTS
2
1
2
3
1
1200W, ½ W resistors (brown-red-red)
6800W, ½ W resistor (blue-grey-red)
10kW, ½ W resistors (brown-black-orange)
10mF electrolytic capacitors
MPSA-20 (NPN) silicon transistor
PROCEDURE
1. Construct the circuit on the following
slide.
2. Turn the trainer on and adjust the
positive power supply to 15V.
a) Adjust the FREQUENCY control so the
output is 1 kHz.
COMMON EMITTER AMP FOR EXP. 9
PICTORIAL OF THE CE AMP FOR EXP. 9
3. Measure the DC voltage on the emitter,
base and collector, and check them
against the values shown Schematic
Diagram for Experiment 9 on the
previous slide.
4. Adjust potentiometer R1 until the
output voltage is 1.5 V rms. (AC)
5. Measure the input voltage Vi going into
the amplifier.
a) Record your voltage measurement.
6. Use the formula AV = VO/Vi, and
calculate the voltage gain.
a) Record your gain calculation.
7. Use the input circuit/Exp. 9, modification
circuit for Step 7, and complete this step.
The modification circuit is on the
following slide.
a) Turn the knob on the 100 KW pot fully
counterclockwise (thus making R7 = 0 W.)
EXP. 9 MODIFICATION CKT. FOR STEP 7
EXP. 9 MODIFICATION CKT. FOR STEP 7
b) Adjust the 1 kW potentiometer R1 so that
the output is 1.5 V rms.
c) Adjust the 100 kW potentiometer R7 so
that the output is 0.75 V rms.
d) Turn the trainer off and measure the
resistance between terminals 1 and 2 of
the 100 kW pot R7.
1. Record this input impedance.
8. Use the following schematic diagram /
Exp. 9, 2nd Modification Circuit for Step
8.
a) Disconnect (lift) the ground wire on the
100 kW potentiometer R7 at pin 2.
b) Adjust the 1 kW potentiometer R1 so
that the output voltage is 1.5 V.
EXP. 9 MODIFICATION CKT. FOR STEP 8
PICTORIAL FOR EXP. 9 FOR STEP 8
c) Reconnect the ground wire on the 100
kW potentiometer, and adjust it so the
output is 0.75 V rms.
d) Turn the trainer off and measure the
resistance between terminals 1 and 2
of the 100 kW pot R7.
1. Record this output impedance
measurement.
9. Use the formula IO = VO/RL, and the
values of VO (1.5 V) and RL (10 kW) to
calculate the output current.
a) Record your current calculation
b) Use the formula Ii = Vi /Zi, and the
values of Vi (step 5) and Zi (step 7) that
you recorded earlier, calculate the input
current of the amplifier.
1. Record your current calculation
c) Use the formula AI = IO/Ii, to calculate
the current gain of the amplifier.
1. Record your current gain calculation
10. Use the formula AP = AIAV, using AV (step
6) and AI (step 9) and calculate the power
gain of the amplifier.
a) Record your power gain calculation
CIE RESULTS
5.
6.
7.
8.
9.
0.50 V
3.0…AV = 1.5 V/0.50V = 3.0
1300W
5000W
0.15 mA; 0.35 mA and 0.39
IO = 1.5 V/104 W = 1.5 x 10-4 A
Ii = 0.50 V / 1300W = 0.38 mA
AI = IO/Ii = 0.15 x 10-3 A/0.38 x 10-3 A
AI = 0.39
10. 1.17… AP = AVAI = 3.0 x 0.39 = 1.17
FINAL DISCUSSION
• It was not necessary to use a voltage
divider at the input, since the gain of the
amplifier was so low.
•We measured the input directly (step 5).
•We measured the input and output
impedances (steps 7 and 8).
•(Note that these were fairly close to the input
and output impedances of the common
emitter amplifier using the bypass capacitor
in the last experiment)
• We calculated the current gain in step 9
and the power gain in step 10.
•(Note that the voltage, current ad power
gains of the amplifier were much less than
the amplifier with the bypass capacitor)
•The difference between the two amplifiers
is that the one with emitter feedback has
much less distortion.
•We cannot measure distortion without
special test equipment, but the
improvement in performance is often worth
adding a second stage of amplification to
get the required gain.
• The value of emitter resistance chosen
for a particular amplifier depends on a
compromise between opposite
requirements.
•Good temperature stability requires a high
value of emitter resistance, while good AC
gain requires a low value of emitter
resistance.
•However, to reduce distortion, the value of
the emitter resistance must be increased.
• In practice, a compromise is often made
between these conflicting requirements.
•The emitter resistance is often broken up
into two parts, RE1 and RE2, as shown in the
following schematic diagram.
EMITTER RESISTANCE AS RE1 AND RE2
•The DC bias current sees RE1 and RE2 in
series which results in good temperature
stability, since RE is seen by DC bias as a
large value.
•The AC signal current only sees RE1. (RE2 is
bypassed by a CE) Thus the AC current
doesn’t see R2.
The value of RE1 is chosen t provide the
required compromise between gain and
distortion; while the value of CE is
chosen to provide a low reactance at the
lowest signal frequency to be
encountered.
QUESTIONS?
RESOURCES
Casebeer, J.L., Cunningham, J.E. (2001).
Lesson 1430: Transistors, Part 1.
Cleveland: Cleveland Institute of
Electronics.
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