For more courses visit

Transcription

1 For more courses visit

2 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

3 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.

4 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 C 2 to prevent degenerative feedback. The name common emitter comes from the emitter being at AC ground.

5 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.

6 BASIC COMMON EMITTER AMPLIFIER

7 REQUIRED PARTS W, ½ W resistor (brown-black-red) W, ½ W resistors (brown-red-red) W, 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

8 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

9 SCHEMATIC DIAGRAM FOR EXP. 8

10 PICTORIAL OF EXP. 8 SCHEMATIC DIAGRAM

11 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.) R 1 until the output voltage V O is 2V rms.

12 5. Measure the AC voltage V i going into the voltage divider which consists of R 2 and R 3. 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.

13 6. Calculate the input voltage V i of this circuit, using the following voltage divider formula: V i = V i [R 3 /(R 2 + R 3 )] a. Record your calculation. 7. Calculate the voltage gain using the formula: A v = V O / V i a) Record your calculation.

14 8. Replace R 2 and R 3 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 R 1 so the output is 2V rms. c) Then adjust the 100kW pot until the output is 1V rms.

15 DIAGRAM FOR EXP. 8, 1 ST MODIFICATION

16 PICTORIAL FOR EXP. 8, 1 ST MODIFICATION

17 d) Turn the trainer power off, and measure the resistance between terminals 1 and 2 of the 100kW pot. e) Record the measurement

18 9. Replace R 8 with the 100kW pot as shown in the following schematic diagram a) Insert R 2 and R 3 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.)

19 DIAGRAM FOR EXP. 8, 2ND MODIFICATION

20 PICTORIAL FOR EXP. 8, 2ND MODIFICATION

21 d) Adjust R 1, 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

22 g) Record this value 10. Current Gain is the ratio of output current to input current. a) Output current is V O /R L and the input current is V i /Z i. b) Use the value of V O from step 4 (2V) and the value of R L (18kW) to calculate the output current.

23 c) Record your current calculation I O d) Use the value of V i that you measured in step 6 and the input impedance (Z i ) you measured in step 8 to calculate the input current. e) Record your current calculation I i f) Use your values of I O and I i and calculate and record the current gain A i

24 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: A P = A V x A i b) Record your gain calculation A P

25 CIE RESULTS V V 2V x (1kW/(1kW+10kW)) =.2V x = 0.018V V/ W W

26 10. I O = 2V / 18kW = 0.1 ma I i = 0.018V/18kW = ma = 16mA A i = 0.1 ma/0.016 ma = x 6.25 = 694

27 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.

28 This consisted of the 10kW resistor (R 2 ) and the 1kW resister (R 3 ) and we calculated the input voltage V i. As in step 7, we calculated the voltage gain A V by dividing the output voltage V O by the input voltage V i.

29 In step 8, we measured the input impedance Z i 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.

30 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.

31 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

32 In step 11, we calculated the power gain A P, 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!

33 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 R 7 or 1.2kW).

34 QUESTIONS?

35 RESOURCES Casebeer, J.L., Cunningham, J.E. (2001). Lesson 1430: Transistors, Part 1. Cleveland: Cleveland Institute of Electronics.

36 THE END Developed and Produced by the Instructors in the CIE Instruction Department. Copyright 10/2012 All Rights Reserved / Oct. 2012