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# Exp f1 maycen

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### Exp f1 maycen

1. 1. NATIONAL COLLEGE OF SCIENCE AND TECHNOLOGY Amafel Bldg. Aguinaldo Highway Dasmariñas City, Cavite EXPERIMENT # 1 Class A Power AmplifierBalane, Maycen M. October 11, 2011Signal Spectra and Signal Processing/ BSECE 41A1 Score: Engr. Grace Ramones Instructor
2. 2. Sample Computation:Step 4 VCE=VCEQ + (ICQ)(RC + Re)= 10.273 V + 48.48mA(50Ω+5 Ω)= 12.9394VStep 6Step 7Step 8 VCEQ =VCC - ICQ(RE + Re + RC) VCC - ICQ(RE + Re + RC)= (ICQ)(Rc + Re) VCC (RE + Re + RC)= ( )(Rc + Re)Step 10 VCEQ =20V - ( +5 + )=5.37V VCE=VCEQ + (ICQ)(RC + Re)= 5.37V + (50Ω+5 Ω)= 10.73525VStep 12Step 13Step 14
3. 3. Objectives: 1. Determine the dc load line and locate the operating point (Q-point) on the dc load line for a large- signal class A common-emitter amplifier. 2. Determine the ac load line for a large-signal class A common-emitter amplifier. 3. Center the operating point (Q-point on the ac load for a large-signal class A common-emitter 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 large-signal class A common-emitter amplifier. 5. Observe nonlinear distortion of the output waveshape for a large-signal class A common-emitter amplifier. 6. Measure the large-signal voltage gain of a class A common-emitter amplifier and compare the measured and calculated values. 7. Measure the maximum undistorted output power for a class A amplifier. 8. Determine the amplifier efficiency of a class A amplifier.
4. 4. Datasheet:Materials:One digital multimeterOne function generatorOne dual-trace oscilloscopeOne dc powers supplyOne 2N3904 bipolar transistorCapacitors: two 10 µF, one 470 µFResistors: one 5 Ω, one 95 Ω, two 100 Ω, one 1 kΩ, one 2.4 kΩTheory:A power amplifier is a large-signal amplifier in the final stage of a communication transmitter thatprovides power to the antenna or in the final stage of a receiver that drives the speaker. When an amplifieris biased so that it operates in the linear region of the transistor collector characteristic curve for the full360 degrees of the input sine wave cycle, it is classified as a class A amplifier. This means that collectorcurrent flows during the full sine wave cycle, making class A amplifiers the least efficient of the differentclasses of large-signal amplifiers. In a large-signal amplifier, the input signal causes the operating point(Q-point) to move over much a larger portion of the ac load line than in a small-signal amplifier.Therefore, large signal class A amplifiers require the operating point to be as close as possible to thecenter of the ac load line to avoid clopping of the output waveform. In a class A amplifier, the outputvoltage waveform has the same shape as the input voltage waveform, making it the most linear of thedifferent classes of amplifiers. Most small-signal amplifiers, it is normally used in a low-powerapplication that requires a linear amplifier such as an audio power amplifier or as a power amplifier in alow-power transmitter with low-level AM or SSB modulation. In this experiment you will study a large-signal class A amplifier.For the large –signal class A common-emitter amplifier shown in Figures 14-1 and 14-2, the dc collector-emitter voltage (VCE) can be calculated from VCE = VC – VEThe dc collector current (IC) can be determined by calculating the current in the collector resistor (R C).Therefore, Figure 14–1 Large signal Class A Amplifier, DC Analysis XMM1 R1 Rc V1 2.4kΩ 100Ω 20 V Q1 2N3904 R2 1kΩ Re 5Ω REm 95Ω
5. 5. Figure 14-2 Large-Signal Class A AmplifierThe ac collector resistance (RC) is equal to the parallel equivalent of the collector resistor (RC) and theload resistor (RL). Therefore,The ac load line has a slope of 1/(RC + RL) and crosses the dc load line Q-point. The ac load line crossesthe horizontal axis of the transistor collector characteristic curve plot at V CE equal to VCEQ + (ICQ)(RC +Re), where VcEQ is the collector-emitter voltage at the Q-point and ICQ is the collector current at the Q-point.The amplifier voltage gain is measured by dividing the ac peak-to-peak output voltage (VO) by the acpeak-to-peak input voltage (Vin). The expected amplifier voltage gain for a common emitter amplifier iscalculated fromwhere RC is the collector resistance, re 25 mV/IE(mA), and the Re is the unbypassed emitter resistance.In order to center the Q-point on the ac load line, you must try different values of RE until VCEQ is equal to(ICQ)(Rc + Re), where ICQ IEQ = VE//(RE + Re), VCC - ICQ(RE + Re + RC), Rc is equal to the ac collectorresistance, and RC is equal to the dc collector resistance.The amplifier output power (PO) is calculated as follows:
6. 6. 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)(IS). The current at the source (IS) is determined from IS = I12 + ICQwhere I12 = VCC/(R1 + R2). Note: I12 is the current in resistors R1 and R2, neglecting the base current.Procedure:Step 1 Open circuit file FIG 14–1. Bring down the multimeter enlargement and make sure that V and dc ( ) are selected. Run the simulation and record the dc base voltage (V B). Move the multimeter positive lead to node VE. Run the simulation and record the dc emitter voltage (VE). Move the multimeter positive lead to node VC. Run the simulation and record the dc collector voltage (VC). VB = 5.658V VE = 4.879V VC = 15.152VStep 2 Based on the voltages in Step 1, calculate the dc collector-emitter voltage (VCE) and the dc collector current (IC). IC = 48.48mA VCE=10.273 VStep 3 Draw the dc load line on the graph provided. Based on the calculations in Step 2, locate the operating point (Q-point) on the dc load line. IC(sat) 200 150 100 50 0 5 10 15 20 VCE(V)
7. 7. Step 4 Open circuit file FIG14-2. Bring down the function generator enlargement and make sure that the following settings are selected. Sine wave , Freq = 2 kHz, Ampl = 250 mV, Offset = 0 v. Bring down the oscilloscope enlargement and make sure that the following settings are selected: Time base (Scale = 100 µa/Div, Xpos = 0 Y/T), Ch A (Scale = 200 mV/Div, Ypos = 0, AC), Ch B (Scale = 2 V/Div, Ypos = 0, AC), Trigger (Pos edge, Level = 0, Auto).Based on the value of RC and RL, calculate the ac collector resistance (RC), and then draw the ac load line through the Q-point on the graph in Step 3. VCE= 12.9394VQuestions: Is the operating point (Q-point) in the center of the dc load line? Is it in the center of the ac load line? No, it is not located at the center of the dc load line nor the center of the ac load line.Why is it necessary for the Q-point to be in the center of the ac load line for large signal inputs? To avoid clippings of the output.Step 5 Run the simulation. Keep increasing the input signal voltage on the function generator until the output peak distortion begins to occur. Then reduce the input signal level slightly until there is no longer any output peak distortion. Pause the analysis and record the maximum undistorted ac peak-to-peak output voltage (VC) and the ac peak-to-peak input voltage (Vin) Vin = 303.852 mV Vo = 2.298VStep 6 Based on the voltages measured in Step 5, determine the voltage gain (AV) of the amplifier.Step 7 Calculate the expected voltage gain (AV) based on the value of the ac collector resistance (Rc), the unbypassed emitter resistance (Re), and the ac emitter resistance (re), where re = 25 mV / IE (mA)Questions: How did the measured amplifier voltage gain with the calculated voltage gain? The difference is 22.49% .What effect does unbypassed emitter resistance have on the amplifier voltage gain? What effect does it have on the voltage gain stability? The voltage gain is inversely proportional to the unbypassed emitter resistance When the unbypassed emitter resistance become higher and higher the voltage of an amplifier become smaller and smaller.Step 8 Calculate the value RE required to center the Q-point on the ac load line. Hint: Try different values of Re until VCEQ = (ICQ)(Rc + Re) at the new Q-point. See Theory section for details.
8. 8. Question: Did you need to increase RE to center the Q-point on the ac load line? Explain why. RE needed to decrease. This is to center the Q-pointStep 9 Change RE to the value calculated in Step 8 and repeat the procedure in Step 5. Record the maximum undistorted ac peak-to-peak output voltage (Vo) and the ac peak-to-peak input voltage (Vin) for this centered Q-point. Vin = 547.839 mV Vo = 4.311 VQuestion: How did the maximum undistorted peak-to-peak output voltage measured in Step 9, for the centered Q-point, compare with the maximum undistorted peak-to-peak output voltage measured in Step 5, for the original Q-point that was not centered? It is much higherStep 10 Calculate the new dc values for ICEQ for the new value of RE Locate the new dc load line and the new Q-point on the graph in Step 3. Draw the new ac load line through the new Q-point. VCEQ =5.37V VCE=)= 10.73525VQuestions: What effect did the new Q-point have on the location of the new ac load line? The location of the ac load line changes.What was the location of the new Q-point on the new load line? The Q-point is now located at the center of the ac load line.Step 11 Based on the new centered Q-point and the ac load line, estimate what maximum ac peak-to-peak output voltage (VO) should be before output clipping occurs. Vo = 4.31 VQuestion: How did the maximum undistorted peak-to-peak output voltage measured in Step 9 for the centered Q-point, compared with the expected maximum estimated in Step 11? It is twice.Step 12 Based on the maximum undistorted ac-peak-to-peak output voltage (Vo) measured in Step 9, calculate the maximum undistorted output power (PO) to the load (RL).Step 13 Based on the supply voltage (VCC), the new collector current at the new operating point (ICQ), and the bias resistor current (I12), calculate the power supplied by the dc voltage source (PS).Step 14 Based on the power supplied by the dc voltage source (PS) and the maximum undistorted output power (PO) calculated in Step 12, calculate the percent efficiency (ŋ) of the amplifier.Question: is the efficiency of a class A amplifier high or low? Explain why. The efficiency is low. This is because of the full sine wave conduction of class A.
9. 9. Conclusion: The class A amplifier is one of the classes of amplifier that conducts in full wave.The quiescent point should be in the center of the ac load line so the clipping would not occur.The quiescent point of the dc load line is the intersection of the dc collector-emitter voltage andthe current collector. The ac load line is crossing the dc collector-emitter voltage and thequiescent point. The voltage gain of the class A amplifier is high while the power efficiency islow.