NATIONAL COLLEGE OF SCIENCE AND TECHNOLOGY Amafel Bldg. Aguinaldo Highway Dasmariñas City, Cavite EXPERIMENT 2 Class B Push-Pull Power AmplifierPula, Rolando A. October 11, 2011Signal Spectra and Signal Processing/ BSECE 41A1 Score: Engr. Grace Ramones Instructor
Objectives: 1. Determine the dc load line and locate the operating point (Q-point) on the dc load line for a class B push-pull amplifier. 2. Determine the ac load line for a class B push-pull amplifier. 3. Observe crossover distortion of the output waveshape and learn how to estimate it in a class b push-pull amplifier. 4. Determine the maximum ac peak-to-peak output voltage swing before peak clipping occurs and compare the measured value with the expected value for a class B push-pull amplifier. 5. Compare the maximum undistorted ac peak-to-peak output voltage swing for a class B amplifier with the maximum for a class A amplifier. 6. Measure the large-signal voltage gain of a class B push-pull amplifier. 7. Measure the maximum undistorted output power for a class B push-pull amplifier. 8. Determine the amplifier efficiency of a class B push-pull amplifier.
MaterialsOne digital multimeterOne function generatorOne dual-trace oscilloscopeOne dc power supplyOne 2N3904 npn bipolar junction transistorOne 2N3906 pnp bipolar junction transistorTwo 1N4001 diodesCapacitors: two 10 µF, one 100 µFResistors: one 100 Ω, two 5 kΩTheoryA power amplifier is a large-signal amplifier in the final stage of a communications transmitter thatprovides power to the to the antenna or in the final stage of a receiver that drives the speaker. When anamplifier is biased at cutoff so that it operates in a linear region of the collector characteristic curves forone-half cycle of the input sine wave (180o), it is classified as a class B amplifier. In order to produce acomplete reproduction of the input waveshape, a matched complementary pair of transistors in a push-pull configuration, as shown in Figure 15-1,is necessary. In a class B push-pull amplifier, each transistorconducts during opposite halves of the input cycle. When the input is zero, both transistor conductsduring opposite halves of the input cycle. When the input is zero, both transistors are at cutoff (IC = 0).This makes a class B amplifier much more efficient than a class Q amplifier, in which the transistorconducts for the entire input cycle (360o) . the man disadvantage of class B amplifier is that it is not aslinear as class A amplifier. Class b amplifiers are used in high-power applications where a linear amplifieris required, such as high-power audio amplifiers or linear power amplifiers in high-power transmitterswith low-level Am or SSB modulation.Figure 15-1 Class B Push-Pull Amplifier with Crossover Distortion XSC1 Ext T rig + R1 V1 _ XFG1 5kΩ A B 20 V + _ + _ Q1 C1 10µF 2N3904 C3 Q2 C2 100µF R3 100Ω 10µF R2 5kΩ 2N3906In a class B push-pull emitter follower configuration in Figure 15-1, both transistors are biased at cutoff.When a transistor is biased at cutoff, the input signal must exceed the base-emitter junction potential
(VBE) before it can conduct. Therefore, in the push-pull configuration in figure 15-1 there is a timeinterval during the input transition from positive to negative or negative to positive when the transistorsare not conducting, resulting in what is known as crossover distortion. The dc biasing network in figure15-2 will eliminate crossover distortion but biasing the transistors slightly above cutoff. Also, when thecharacteristics of the diodes (D1 and D2) are matched to the transistor characteristics, a stable dc bias ismaintained over a wide temperature range.Figure 15-2 Class B Push-Pull Amplifier – DC Analysis XMM1 + U3 0.000 A DC 1e-009Ohm - R1 5kΩ Q1 V1 D1 2N3904 20 V 1N4001GP D2 1N4001GP Q2 R2 2N3906 5kΩThe dc load line for each transistor in figure 15-2 is a vertical line crossing the horizontal axis at VCE =VCC/2. The load line is vertical because there is no dc resistance n a collector or emitter circuit (slope ofthe dc load line is the inverse of the dc collector and emitter resistance). The Q-point on the dc load linefor each transistor is close to cutoff (Ic = 0). The dc collector-emitter voltages for the two transistors inFigure 15-2 can be determined from the value of VE using the equations VCE1 = VCC – VEand VCE2 = VE – 0 = VEThe complete class B push-pull amplifier is shown in Figure 15-2. Capacitors C1, C2, and C3 are couplingcapacitors to prevent the transistor dc bias voltages being affected by the input circuit or the loadcircuit. The ac load line for each transistor should have a slope of 1/RL (the ac equivalent resistance inthe emitter circuit is RL), cross the horizontal axis at VCC/2, and cross the vertical axis at IC (sat) = VCC/2RL.The Q-point on the ac load line should be close to cutoff (Ic = 0) for each transistor. When one of thetransistors is conducting, its operating point (Q-point) moves up the ac load line. The voltage swing ofthe conducting transistor can go all the way from cutoff to saturation. On the alternate half cycle theother transistor can go all the way from cutoff to saturation. Therefore, the maximum peak-to-peakoutput voltage is equal to 2(VCC/2) = vCC. The amplifier voltage gain is measured by dividing the ac peak-to-peak output voltage (Vo) by the ac peak-to-peak input voltage (Vin). Because the push-pull amplifier infigure 15-3 is an emitter follower configuration, the voltage gain should be close to unity (1). This is not a
problem for large-signal amplifiers because they are used primarily for power amplification rather thanvoltage amplification.Figure 15-3 Class B Push-Pull Amplifier XFG1 XSC1 V1 20 V Ext T rig + R1 _ 5kΩ A B + _ + _ Q1 C2 10µF D1 2N3904 1N4001GP C3 D2 1N4001GP Q2 100µF C1 R3 10µF 100Ω R2 2N3906 5kΩThe amplifier output power (PO) is calculated as follows:The percent efficiency (ŋ) of a large-signal amplifier is equal to the maximum output power (PO) dividedby the power supplied by the source (PS) times 100%. Therefore,where Ps = (VCC)(ICC). The current at the source (IS) is determined fromwhere = VCC/2RL and IC(AVG) is the average value of the half-wave collector current.Note: IRB1 is normally much less than IC(AVG) and can be neglected.Procedure:Step 1 Open circuit file FIG 15-1. Bring down the function generator enlargement. Make sure that the following settings are selected: Time base (Scale = 200 us/Div, Xpos = 0, Y/T), Ch A (Scale =2 V/Div, Ypos = 0, AC), Ch B (Scale = 2 V/Div, Ypos = 0, AC), Trigger (Pos edge, Level = 0, Auto). Run the simulation to four full screen displays, then pause the simulation. You are plotting the amplifier input (red) and the output (blue) on the oscilloscope. Notice the crossover distortion of the output waveshape (blue curve). Draw the waveshape in the space provided and note the crossover distortion.
Step 2 Open circuit file FIG 15-2. Bring down the multimeter enlargement and make sure that V and dc (----) are selected. Run the simulation and record the dc base1 voltage (VB1). Move the multimeter positive lead to node VB2, then node VE, then node A and run the simulation for each reading and record the dc voltages. Also record the dc collector current (IC) VB1 = 10.496 V VB2 = 9.504 V VE = 10.017 V VA =10V IC = 0 AStep 3 Based on the voltages recorded in Step 2, calculate the dc collector-emitter voltage (VCE) for both transistors. VCE1 = 9.983 V VCE2 = 10.017 VStep 4 Draw the dc load line on the graph and locate the operating point (Q-point) on the dc load line based on the data in Step 2 and the calculations in Step 3. 100 75 50 25 0 5 10 15 20Step 5 Open circuit file FIG 15-3. Bring down the function generator enlargement. Make sure that the following settings are selected: Sine wave, freq = 1 kHz, Ampl = 4 V, Offset = 0 V. Bring down the oscilloscope enlargement. Make sure that the following settings are selected Time base (Scale = 200 us/Div, Xpos = 0, Y/T), Ch A (Scale = 2 V/Div, Ypos = 0, AC), Ch B (Scale = 2 V/Div, Ypos = 0, AC), trigger (Pos edge, Level = 0, Auto). Based on the values of VCC and RL, draw the ac load line on the graph in step 4.Questions: Where was the operating point (Q-point) on the dc load line? Where was the operating point on the ac load line? Explain.
Both are located at the cutoff. This is because of the absence of collectors current.What was the relationship between the dc load line and the ac load line? Explain. The dc load line and the ac load line have same quiescent point that is located at the cutoff.Step 6 Run the simulation. Notice that there is hardly any crossover distortion of the output waveshape. Keep increasing the input signal voltage until output peak distortion occurs. Then reduce the input signal level slightly until there is no longer any distortion. Pause the analysis and record the maximum undistorted ac peak-to-peak output voltage (VO) and the ac peak-to-peak input voltage (Vin). Adjust the oscilloscope settings as needed. VO =20.368 V Vin = 10.2 VQuestions: What caused the crossover distortion in Step 1? What does the addition of diodes D1 and D2 accomplish? There is a time interval during the input transition from positive to negative or negative to positive when the transistors are not conducting, resulting in crossover distortion. When the characteristics of the diodes (D1 and D2) are matched to the transistor characteristics, a stable dc bias is maintained over a wide temperature range.How did the maximum undistorted peak-to-peak output voltage for the class B amplifier, measured inStep 6, compare with the maximum undistorted peak-to-peak output voltage for the class A amplifier,measured in Experiment 14, Step 9? The peak voltage of class B is much greater than class A.Step 7 Based on the voltages measured in Step 6, calculate the voltage gain of the amplifier.Questions: How did the measured amplifier voltage gain compare with the expected value for a class B push-pull emitter circuit? They are the sameStep 8 Based on the ac load line and Q-point located on the graph in Step 4, estimate what the maximum ac peak-to-peak output voltage (Vo) should be before output clipping occurs. Record your answer. Vo = 10VQuestion: How did the maximum undistorted peak-to-peak output voltage measured in step 6 compare with the expected maximum estimated in Step 8?
There is only a difference of 0.184 V or 1.81%.Step 9 Based on the maximum undistorted ac peak-to-peak output voltage measured in Step 6, calculate the maximum undistorted output power (PO) to the load (RL).Step 10 Based on the supply voltage (VCC) and the average collector current (IC(AVG)), calculate the power supplied by the dc voltage source (PS).Step 11 Based on the power supplied by the dc voltage source (PS) and the maximum undistorted output power (PO) calculated in Step 9, calculate the efficiency (ŋ) of the amplifier.Questions: How did the efficiency of this class B push-pull amplifier compare with the efficiency of the class A amplifier in Experiment 14? The efficiency of class B amplifier is much larger than the previous experiment.
Conclusion After performing the experiment, I can able to say that class B is conducting at half cyclewave. The quiescent point of class B is located at the cutoff because of the absence of thecollector current. The ac load line crosses the vertical axis saturation current and the cutoff foreach transistor. There is a crossover distortion in the output waveshape if the input voltage isabove the required peak-to-peak input voltage. The voltage gain of the class B amplifier isalmost the same with the unity gain. Finally, power efficiency of class B amplifier is muchgreater than the class A amplifier.