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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.

BASIC COMMON EMITTER AMPLIFIER

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

SCHEMATIC DIAGRAM FOR EXP. 8

PICTORIAL OF EXP. 8 SCHEMATIC DIAGRAM

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.

DIAGRAM FOR EXP. 8, 1ST MODIFICATION

PICTORIAL FOR EXP. 8, 1ST MODIFICATION

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.)

DIAGRAM FOR EXP. 8, 2ND MODIFICATION

PICTORIAL FOR EXP. 8, 2ND MODIFICATION

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).

QUESTIONS?

RESOURCES

Casebeer, J.L., Cunningham, J.E. (2001).

Lesson 1430: Transistors, Part 1.

Cleveland: Cleveland Institute of

Electronics.

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