Lab 5 & 6 - Christian Andrade

docx

School

The City College of New York, CUNY *

*We aren’t endorsed by this school

Course

322

Subject

Electrical Engineering

Date

Jan 9, 2024

Type

docx

Pages

16

Uploaded by AmbassadorScorpionPerson998

Report
City College University of New York Department of Electrical Engineering EE 32200 Saurabh Sachdeva Laboratory Report #5 & #6 Audio Amplifier (Part II & Part III) Christian Andrade Due date: 12/17/2022
Semester: Fall 2022 Introduction An amplifier is an electrical component which increases the input signal, usually voltage, by a specific proportionality constant. This constant is called the gain of the amplifier and it’s derived from the design of the circuit based on the input and output relationship of the signal. It’s controlled by the size of the resistors between the base, collector and emitter (BJT Transistor). It’s necessary to choose the resistors within operating ranges, otherwise effects such as clipping can occur. In this lab, we will build a Power Amplifier that supply a constant Power to the output of the circuit. Objectives We are required to design a power amplifier with the following specifications: The power and the efficiency will be calculated with the following formulas: P o = V RMS 2 8 [ W ] P i avg = 2 V ICC peak Vcc π [ W ] ɲ = Po P i avg 100 [ % ] #5: Audio Amplifier (Part II) Figure 1: Specifications of the Power Amplifier Parameter Min. Typical Max. Units Output Power 0.7 1.0 - Watts Gain 0.6 - - V/V Zin 2000 - - Efficiency (n) 40 50 - % Supply (Vcc) - 12 - Volts
Figure 2: Design 1b Prelab Circuit Topology In order to get the best efficiency from our Power Amplifier, four circuits were tested for the best power & efficiency: i. Design 1b: Figure 3: Icc and VL of design 1b
Your preview ends here
Eager to read complete document? Join bartleby learn and gain access to the full version
  • Access to all documents
  • Unlimited textbook solutions
  • 24/7 expert homework help
Figure 4: Design 1c P o = ( 3.18 0.707 ) 2 8 ¿ 0.63 W P i avg = 2 0.39 12 π = 2.98 W ɲ = Po P i avg = ¿ 21% GAIN = Vo Vin = 3.18 1.3 =− 2.44 ii. iii. Figure 5: Icc and VL of design 1c
P o = ( 3.34 0.707 ) 2 8 ¿ 0.7 W P i avg = 2 0.2 12 π = 1.53 W ɲ = Po P i avg = 46% GAIN = Vo Vin = 3.34 4 = 0.84 The most efficient design is design 1d: push-pull configuration Push-Pull Power Amplifier design i. Figure 7: Icc and VL of design 1d Figure 8: Push-Pull PA with Vd = 0v
*The best choice is option number 2 with an efficiency of 29% ii. This table compares the designs with different Re: Figure 9: Push-Pull PA with Vd = 0.6v Figure 10: Push-Pull PA with Vd = 0.6v Re [Ω] Temp. [C] ´ Icc [A] VL (p-p) [V] Efficiency 0 27 0.2 6.65 28 0 127 0.2 4.37 13 1 27 0.18 6.25 29 1 127 0.4 6.53 14 10 27 0.10 3.32 14 10 127 0.11 3.5 13 V D ´ Icc V L (peak) Efficiency Cross-Over Distortion 0V 0.18A 2.97V 26% Yes 0.6V 0.20A 3.33V 29% No 0.75V 0.20A 2.19V 12% No
Your preview ends here
Eager to read complete document? Join bartleby learn and gain access to the full version
  • Access to all documents
  • Unlimited textbook solutions
  • 24/7 expert homework help
Figure 13: Vin / Vout of Design 2b From the following table we can see that the best configuration is with Re = 1Ω iii. In this part, we have to find the best value for Rd, using the following circuit: After few simulations with different Rd, the best value is Rd = 1.2kΩ Figure 12: Design 2b. Used to find Rd Figure 11: Push-Pull design comparison
Figure 14: Icc of Design 2b iv. Input impedance: Z 3 = Reβ + β ( ℜ+ 8 ) = 1 100 + 100 ( 1 + 8 ) = 1 kΩ 1 100 + 100 ( 1.2 k | | 1 k ) ] | | ( 1 100 + 100 1.2 k ) = 54.64 k | | 120.1 k = 37.5 k Reβ + β ¿ ¿ ( Reβ + βRd )= ¿ Z ¿ = ¿ v. Final design:
Figure 16: Operating points of the final design Figure 17: Vin and Vout of the final design Figure 15: Final PA design
Your preview ends here
Eager to read complete document? Join bartleby learn and gain access to the full version
  • Access to all documents
  • Unlimited textbook solutions
  • 24/7 expert homework help
Figure 18: Icc of the final design P o = ( 3.37 0.707 ) 2 8 ¿ 0.71 W P i = 0.2 12 = 2.4 W ɲ = Po P i avg = 30% GAIN = Vin Vo = 6.75 7.96 = 0.84 Laboratory Results i. Quiescent Voltages: Quiescent Voltage or current Simulated Measured Vb5 6.05V 6.01V Vbe4 0.30V 0.60V Vbe5 -0.50V -0.58V Vbe2 0.50V 0.60V Vbe3 -0.49V -0.58V Ve2 6.05V 6.00V Ve3 6.05V 6.01V Icc 207mA 220mA Figure 19: Measured Q-Voltages
P o = V RMS 2 8 = ( 3.37 0.707 ) 2 8 ¿ 0.71 W P i = V Icc Vcc = 0.21 12 = 2.52 W ɲ = Po P i avg = 28% GAIN = Vin Vo = 6.8 9.12 = 0.74 Conclusions This experiment is very important for Electrical Engineers. Every digital and electronic system nowadays has this component, and we think that it’s a fundamental thing to know how to design and build such a system. It wasn’t easy to implant the design, and we did have a few problems on the way. However, we can defiantly conclude that the amplifier design successfully met the specified criteria. #6: Audio Amplifier (Part III) Introduction: In this experiment we will design a public address (PA) amplifier. A public address amplifier takes a small signal from a microphone (in our case 20 mV), amplifies it to several volts, and
then drives a speaker with its output. The circuits used in the P.A. amplifier will be the circuits designed in Audio Amplifier Part I and Part II. Circuit Design: The P.A. amplifier will consist of three amplifiers: the two amplifiers built in part 1 and part 2, as well as a new amplifier very similar to the one built in part 1. All three of these amplifiers will be placed in series with the part 3 amplifier first followed by the amplifier from part 2 and finally the amplifier from part 3 will be last. The small equivalent circuit of the P.A. amplifier is shown below in figure 1 The design specifications of the P.A. Amplifier are given below in table 1. Parameter Minimum Typical Maximum Units Supply voltage 12 Volts Supply current 0.3 Amps (avg) Output power (8-ohm load) 0.5 0.7 Watts Input Impedance 400 Gain 334.65 V/V Freq. res. (-3dB) 60. - to -15k Hz
Your preview ends here
Eager to read complete document? Join bartleby learn and gain access to the full version
  • Access to all documents
  • Unlimited textbook solutions
  • 24/7 expert homework help
To calculate the minimum required gain of the P.A. amplifier in the table above, we took the minimum voltage required to provide 0.7 W to an 8 Ω resistor found previously, 6.692 V, and divided it the input voltage 20 mV, yielding 6.693/0.02=334.65 . The rest of the values in the table were provided. To design the amplifier 1 in figure 1, we had to calculate the gain necessary using the following calculation: Vo/vi= (600/ (600+600)) A(1k/(1k+1k))20(Zin/(1k+Zin)) (8/9) =334.65 substituting Zin= 36.08 kΩ as calculated previously, and solving for A, we get A=77.38, which means that’s the gain we need in amplifier 1. Using the fact that for the first amplifier, Zin=600Ω , and that we need V B = 1.5 V, we found that for the first amplifier, R 1 =680 Ω and R 2 =4.8 kΩ. This led us to our design of the P.A. amplifier circuit, shown below in figure 2. To produce a higher gain, capacitor C1 has been placed so that it acts as an AC short, which only leaves r E in the denominator of the gain. Laboratory :
We first tested the distortion of the newly designed amplifier 1 (because the other two were tested in previous labs). C1 was placed so that it only shorted R E2 resistor. We set the function generator to a 7 kHz triangular wave. Then the input was adjusted until the output was measured to be 4 V pp , which was measured on R6, with the rest of the other amplifiers disconnected. As shown below in figure 4, the two waveforms almost exactly overlap, so there is virtually no distortion. Then, we connected C1 so that R E1 and R E2 were shorted and made the same measurements as before. As shown below in figure 5, there is now a great deal of distortion.
Setting the function generator to 1 kHz, we measured the gain of amplifier 1, and found the value to be around 70 V/V, which was very close to the desired gain. We then connected the other two amplifiers as shown in the schematic in figure 3 in order to test out the total P.A. amplifier and observe an amplification of the input from the microphone. The results were successful, as there was a clear amplified sound from the speaker. Conclusion:
Your preview ends here
Eager to read complete document? Join bartleby learn and gain access to the full version
  • Access to all documents
  • Unlimited textbook solutions
  • 24/7 expert homework help
Because the amplifier worked so perfectly, the project can be deemed a great success. There were no distortions in the output over the frequencies that were put in. Even the music with high pitches and bass was undistorted in the output. The combination of three stages of amplifiers was used successfully to create an Audio Amplifier. Each stage completed a different task required to obtain the final undistorted output. The first stage gave an enormous gain so that the very small signal coming from the microphone could be sent to the second stage. The second stage brought the gain up even further so that the final stage could amplify the power. Finally, the third stage was able to deliver enough power to the drive the speaker, resulting in an amplified sound.