for more courses visit 1. to measure the voltage, current, and power gains of the common emitter...
TRANSCRIPT
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OBJECTIVES
1. To measure the voltage, current, and
power gains of the common emitter
amplifier
2. To measure the input and output
impedance of the common emitter
amplifier
INTRODUCTION
• The following is a schematic diagram of
a Common Emitter Amplifier.
• The base is biased through a voltage
divider and the emitter is biased
through a emitter resistor.
• The input signal is injected into the base
and the output signal is taken off the
collector.
• The emitter is at AC ground because the
emitter is bypassed by C2 to prevent
degenerative feedback.
• The name common emitter comes from
the emitter being at AC ground.
• The collector bias current is the same as
it was for the common base amplifier.
• This is because the circuit uses the same
components except for the 100mF emitter
bypass capacitor.
• This gives us a more valid basis of
comparison between the two amplifiers.
REQUIRED PARTS
1 1000W, ½ W resistor (brown-black-red)
2 1200W, ½ W resistors (brown-red-red)
1 6800W, ½ W resistor (blue-grey-red)
2 10kW, ½ W resistors (brown-black-orange)
1 18kW, ½ W resistor (brown-gray-orange)
3 10mF electrolytic capacitors
1 100mF electrolytic capacitor
1 MPSA-20 (NPN) silicon transistor
PROCEDURE
1. Construct the circuit on the following
slide
2. Turn on the trainer and adjust the
positive power supply to +15V
a) Adjust the FREQUENCY control so the
output is 1k Hz
3. Measure the DC voltages on the emitter,
base, and collector and check them
against those on the previous slide.
a) They should be approximately the same
value.
4. Adjust the potentiometer (Pot.) R1 until the
output voltage VO is 2V rms.
5. Measure the AC voltage Vi’ going into the
voltage divider which consists of R2 and
R3.
a) Record your measurement
b) You may have difficulty getting an accurate
reading, using the analog multimeter,
since the deflection may be small, but
make as good an estimate as your can.
6. Calculate the input voltage Vi of this circuit , using the following voltage divider formula:
Vi = Vi’ [R3 /(R2 + R3)] a. Record your calculation.
7. Calculate the voltage gain using the
formula: Av = VO / Vi a) Record your calculation.
8. Replace R2 and R3 with the 100kW pot
as shown in the schematic on the
following slide.
a) Turn the knob on the 100kW pot fully
counterclockwise.
b) Adjust R1 so the output is 2V rms.
c) Then adjust the 100kW pot until the
output is 1V rms.
d) Turn the trainer power off, and measure
the resistance between terminals 1 and
2 of the 100kW pot.
e) Record the measurement
9. Replace R8 with the 100kW pot as
shown in the following schematic
diagram
a) Insert R2 and R3 as in the main circuit
b) Turn your trainer on
c) Lift the ground wire on the 100kW pot.
(Disconnect the wire connecting the
100kW pot to ground.)
d) Adjust R1, the 1kW pot, so that the
output measures 2V rms.
e) Connect the ground wire to the 100kW
pot at pin 2 and adjust the pot so that
the output (across pins 1 and 2) is 2V
rms.
f) Turn the trainer off, measure the
resistance between pins 1&2 of the pot
g) Record this value
10. Current Gain is the ratio of output
current to input current.
a) Output current is VO/RL and the input
current is Vi/Zi.
b) Use the value of VO from step 4 (2V) and
the value of RL (18kW) to calculate the
output current.
c) Record your current calculation IO
d) Use the value of Vi that you measured in
step 6 and the input impedance (Zi) you
measured in step 8 to calculate the
input current.
e) Record your current calculation Ii f) Use your values of IO and Ii and
calculate and record the current gain Ai
11. Use the value of voltage gain (AV) you
measured in step 7 and the current
gain you found in step 10 to calculate
the power gain.
a) Use the following power gain formula:
AP = AV x Ai
b) Record your gain calculation AP
CIE RESULTS
5. 0.2V
6. 0.018V…2V x (1kW/(1kW+10kW))
= .2V x 0.0909 = 0.018V
7. 111…2V/0.018
8. 1100W
9. 5000W
10. IO = 2V / 18kW = 0.1 mA
Ii = 0.018V/18kW = 0.016 mA = 16mA
Ai = 0.1 mA/0.016 mA = 6.25
11. 694…111 x 6.25 = 694
FINAL DISCUSSION
• Again, as in the previous experiment, the
voltage gain was so high, we could not
measure the input voltage directly with
the analog meter.
•We measured the voltage across the
voltage divider.
•This consisted of the 10kW resistor (R2)
and the 1kW resister (R3) and we
calculated the input voltage Vi.
•As in step 7, we calculated the voltage
gain AV by dividing the output voltage VO
by the input voltage Vi.
• In step 8, we measured the input
impedance Zi by inserting a resistance in
series with the input of the amplifier.
•First we set the 100KW pot to 0W,
and then adjusted the input voltage so
that the output voltage was 2V.
•Next we added series resistance with the
100kW potentiometer until the output
voltage fell/decreased from 2V to 1V.
•At this point, half of the input voltage was
dropped across the 100kW pot and half was
dropped across the input impedance of the
amplifier; so, the resistance of the 100kW
pot was equal to the input impedance of the
amplifier.
•We measured the resistance between pins 1 and 2 of the 100KW pot, after disconnecting it, to determine the input resistance of the amplifier.
• In step 9, We used the same procedure as in the previous experiment to determine the output resistance of the amplifier.
•We loaded the output of the amp until the output voltage fell to one half its unloaded value
• In step 11, we calculated the power gain
AP, by multiplying the current gain by the
voltage gain
•We found that the voltage gain was high;
the current gain was moderate and the
power gain was high.
•Your results may vary due to component
tolerances!
• The main difference between the (CB)
common-base and (CE) common-emitter
amplifiers was in the area of input
impedance.
•The input impedance of the CB amp was
low, (about 292W).
•The input impedance of the CE amp was
moderate, about equal to R7 or 1.2kW).
RESOURCES
Casebeer, J.L., Cunningham, J.E. (2001).
Lesson 1430: Transistors, Part 1.
Cleveland: Cleveland Institute of
Electronics.