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NATIONAL COLLEGE OF SCIENCE AND TECHNOLOGY                        Amafel Bldg. Aguinaldo Highway Dasmariñas City, Cavite  ...
Objectives:   1. Determine the dc load line and locate the operating point (Q-point) on the dc load line for a      class ...
Sample Computation:Step 7 – Computation of voltage gainStep 9 Output powerStep 10 Input powerStep 11 Power efficiency3|Page
Data Sheet:Materials:One digital multimeterOne function generatorOne dual-trace oscilloscopeOne dc power supplyOne 2N3904 ...
In a class B push-pull emitter follower configuration in Figure 15-1, both transistors are biased atcutoff. When a transis...
saturation. Therefore, the maximum peak-to-peak output voltage is equal to 2(VCC/2) = vCC. Theamplifier voltage gain is me...
(blue) on the oscilloscope. Notice the crossover distortion of the output waveshape                                  (blue...
V/Div, Ypos = 0, AC), Ch B (Scale = 2 V/Div, Ypos = 0, AC), trigger (Pos edge, Level               = 0, Auto). Based on th...
Questions: How did the measured amplifier voltage gain compare with the expected value for a              class B push-pul...
Conclusion        After performing the experiment about class B, I conclude that class B is much differentfrom class A. Cl...
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SIGNAL SPECTRA EXPERIMENT 2 - FINALS (for CAUAN)

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SIGNAL SPECTRA EXPERIMENT 2 - FINALS (for CAUAN)

  1. 1. NATIONAL COLLEGE OF SCIENCE AND TECHNOLOGY Amafel Bldg. Aguinaldo Highway Dasmariñas City, Cavite EXPERIMENT # 2 Class B Push-Pull Power AmplifierCauan, Sarah Krystelle P. October 11, 2011Signal Spectra and Signal Processing/ BSECE 41A1 Score: Engr. Grace Ramones Instructor1|Page
  2. 2. 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.2|Page
  3. 3. Sample Computation:Step 7 – Computation of voltage gainStep 9 Output powerStep 10 Input powerStep 11 Power efficiency3|Page
  4. 4. Data Sheet:Materials:One 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 kΩTheory:A 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 an amplifier is biased at cutoff so that it operates in a linear region of the collectorcharacteristic curves for one-half cycle of the input sine wave (180o), it is classified as a class Bamplifier. In order to produce a complete reproduction of the input waveshape, a matchedcomplementary pair of transistors in a push-pull configuration, as shown in Figure 15-1,isnecessary. In a class B push-pull amplifier, each transistor conducts during opposite halves of theinput cycle. When the input is zero, both transistor conducts during opposite halves of the inputcycle. When the input is zero, both transistors are at cutoff (IC = 0). This makes a class B amplifiermuch more efficient than a class Q amplifier, in which the transistor conducts for the entire inputcycle (360o) . the man disadvantage of class B amplifier is that it is not as linear as class A amplifier.Class b amplifiers are used in high-power applications where a linear amplifier is required, such ashigh-power audio amplifiers or linear power amplifiers in high-power transmitters with low-levelAm 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 2N39064|Page
  5. 5. In a class B push-pull emitter follower configuration in Figure 15-1, both transistors are biased atcutoff. When a transistor is biased at cutoff, the input signal must exceed the base-emitter junctionpotential (VBE) before it can conduct. Therefore, in the push-pull configuration in figure 15-1 thereis a time interval during the input transition from positive to negative or negative to positive whenthe transistors are not conducting, resulting in what is known as crossover distortion. The dcbiasing network in figure 15-2 will eliminate crossover distortion but biasing the transistorsslightly above cutoff. Also, when the characteristics of the diodes (D1 and D2) are matched to thetransistor characteristics, a stable dc bias is maintained 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 of the dc load line is the inverse of the dc collector and emitter resistance). The Q-point onthe dc load line for each transistor is close to cutoff (Ic = 0). The dc collector-emitter voltages forthe two transistors in Figure 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 arecoupling capacitors to prevent the transistor dc bias voltages being affected by the input circuit orthe load circuit. The ac load line for each transistor should have a slope of 1/RL (the ac equivalentresistance in the emitter circuit is RL), cross the horizontal axis at VCC/2, and cross the vertical axisat IC (sat) = VCC/2RL. The Q-point on the ac load line should be close to cutoff (Ic = 0) for eachtransistor. When one of the transistors is conducting, its operating point (Q-point) moves up the acload line. The voltage swing of the conducting transistor can go all the way from cutoff tosaturation. On the alternate half cycle the other transistor can go all the way from cutoff to5|Page
  6. 6. saturation. Therefore, the maximum peak-to-peak output voltage is equal to 2(VCC/2) = vCC. Theamplifier voltage gain is measured by dividing the ac peak-to-peak output voltage (Vo) by the acpeak-to-peak input voltage (Vin). Because the push-pull amplifier in figure 15-3 is an emitterfollower 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 than voltageamplification.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)divided by 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 output6|Page
  7. 7. (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. 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 A Step 3 Based on the voltages recorded in Step 2, calculate the dc collector-emitter voltage (VCE) for both transistors. VCE1 = VCC – VE = 20V – 10.017 V = 9.983 V VCE2 = VE – 0 = VE = 10.017 V Step 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. IC(sat) 100 AC load line 75 50 25 Q-pointDC load line 0 5 10 15 20 VCE(V) Step 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 7|Page
  8. 8. 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. The operating point on the dc load line is at the cutoff where I C = 0. While the operating points for each transistors in the ac load line is also at the cutoff.What was the relationship between the dc load line and the ac load line? Explain. The dc load line is vertical however the ac load line has a slope of 0.01. The dc and the ac load line have same operating point which 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 = 10.184 V for one transistor 20.368 V for two transistors Vin = 10.2 VQuestions: What caused the crossover distortion in Step 1? What does the addition of diodes D1 and D2 accomplish? There is crossover distortion in Step 1 because when the transistors are not conducting, there is a time interval during the input transition from positive to negative or vise versa. The addition of D1 and D2 minimize the distortion because there is a stable dc biasing is maintained.How did the maximum undistorted peak-to-peak output voltage for the class B amplifier, measuredin Step 6, compare with the maximum undistorted peak-to-peak output voltage for the class Aamplifier, measured in Experiment 14, Step 9? The maximum undistorted peak-to-peak output voltage for the class B amplifier measured in Step 6 is 18 V more than the maximum undistorted peak-to-peak output voltage for the class A amplifier measured in the previous experiment.Step 7 Based on the voltages measured in Step 6, calculate the voltage gain of the amplifier.8|Page
  9. 9. Questions: How did the measured amplifier voltage gain compare with the expected value for a class B push-pull emitter circuit? The measured and expected value of the amplifier voltage gain have a difference of 0.002 therefore, the measured AV is almost what expected value is.Step 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? The difference is 0.184 V or 1.81% difference.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 this class B push-pull amplifier is much higher than the efficiency of the class A amplifier.9|Page
  10. 10. Conclusion After performing the experiment about class B, I conclude that class B is much differentfrom class A. Class B conducts at half cycle unlike class A. To produce full cycle wave class B uses amatched complementary pairs of transistors in a push-pull configuration. Moreover, the quiescentpoint of the class B amplifier is located at the cutoff of each transistor because the value of thecollector current is zero that is why the dc load line appears as a vertical line. The ac load linecrosses the saturation current and the half of the supply voltage. Its slope is the inverse of the loadresistance. In addition, when the input voltage exceeds the maximum ac peak-to-peak output voltageswing, a crossover distortion will occur in the output signal. Furthermore, a crossover distortionoccurs because there are time intervals when the transistors are not conducting. To avoid this,diodes are connected to the network to maintain a stable dc bias. The voltage gain of class B push-pull amplifier is close to the unity gain. Lastly, the power efficiency of class B is much higher thanwith class A amplifier.10 | P a g e

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