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LM4871 Datasheet(PDF) 12 Page - Texas Instruments |
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LM4871 Datasheet(HTML) 12 Page - Texas Instruments |
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12 / 30 page ![]() LM4871 SNAS002F – FEBRUARY 2000 – REVISED MAY 2013 www.ti.com AUDIO POWER AMPLIFIER DESIGN Design a 1W/8 Ω Audio Amplifier Given: Power Output 1 Wrms Load Impedance 8 Ω Input Level 1 Vrms Input Impedance 20 k Ω Bandwidth 100 Hz–20 kHz ± 0.25 dB A designer must first determine the minimum supply rail to obtain the specified output power. By extrapolating from the Output Power vs Supply Voltage graphs in the Typical Performance Characteristics section, the supply rail can be easily found. A second way to determine the minimum supply rail is to calculate the required Vopeak using Equation 4 and add the output voltage. Using this method, the minimum supply voltage would be (Vopeak + (VODTOP + VODBOT)), where VODBOT and VODTOP are extrapolated from the Dropout Voltage vs Supply Voltage curve in the Typical Performance Characteristics section. (4) Using the Output Power vs Supply Voltage graph for an 8 Ω load, the minimum supply rail is 4.6V. But since 5V is a standard voltage in most applications, it is chosen for the supply rail. Extra supply voltage creates headroom that allows the LM4871 to reproduce peaks in excess of 1W without producing audible distortion. At this time, the designer must make sure that the power supply choice along with the output impedance does not violate the conditions explained in the POWER DISSIPATION section. Once the power dissipation equations have been addressed, the required differential gain can be determined from Equation 5. (5) Rf/Ri = AVD/2 (6) From Equation 5, the minimum AVD is 2.83; use AVD = 3. Since the desired input impedance was 20k Ω, and with a AVD impedance of 2, a ratio of 1.5:1 of Rf to Ri results in an allocation of Ri = 20kΩ and Rf = 30kΩ. The final design step is to address the bandwidth requirements which must be stated as a pair of −3dB frequency points. Five times away from a −3dB point is 0.17dB down from passband response which is better than the required ±0.25dB specified. fL = 100Hz/5 = 20Hz fH = 20kHz * 5 = 100kHz As stated in the External Components Description section, Ri in conjunction with Ci create a highpass filter. Ci ≥ 1/(2π*20kΩ*20Hz) = 0.397µF; use 0.39µF The high frequency pole is determined by the product of the desired frequency pole, fH, and the differential gain, AVD. With a AVD = 3 and fH = 100kHz, the resulting GBWP = 150kHz which is much smaller than the LM4871 GBWP of 4MHz. This figure displays that if a designer has a need to design an amplifier with a higher differential gain, the LM4871 can still be used without running into bandwidth limitations. 12 Submit Documentation Feedback Copyright © 2000–2013, Texas Instruments Incorporated Product Folder Links: LM4871 |
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