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LM4861MX bảng dữ liệu(PDF) 11 Page - Texas Instruments |
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LM4861MX bảng dữ liệu(HTML) 11 Page - Texas Instruments |
11 / 20 page LM4861 www.ti.com SNAS095C – MAY 1997 – REVISED MAY 2013 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 needed supply rail to obtain the specified output power. By extrapolating from Figure 11 in Typical Performance Characteristics, the supply rail can be easily found. A second way to determine the minimum supply rail is to calculate the required Vopeak using Equation 5 and add the dropout voltage. Using this method, the minimum supply voltage would be (Vopeak + VOD , where VOD is typically 0.6V. (5) For 1W of output power into an 8 Ω load, the required Vopeak is 4.0V. A minumum supply rail of 4.6V results from adding Vopeak and Vod. But 4.6V is not a standard voltage that exists in many applications and for this reason, a supply rail of 5V is designated. Extra supply voltage creates dynamic headroom that allows the LM4861 to reproduce peaks in excess of 1Wwithout clipping the signal. 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. Once the power dissipation equations have been addressed, the required differential gain can be determined from Equation 6. (6) Rf/Ri = AVD / 2 (7) From Equation 6, the minimum Avd is 2.83: Avd = 3 Since the desired input impedance was 20k Ω, and with a Avd of 3, a ratio of 1:1.5 of Rf to Ri results in an allocation of Ri = 20kΩ, 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. This fact results in a low and high frequency pole of 20Hz and 100kHz respectively. As stated in External Components Description , Ri in conjunction with Ci create a highpass filter. Ci ≥ 1 / (2π*20kΩ*20Hz) = 0.397μF; use 0.39μF. (8) The high frequency pole is determined by the product of the desired high frequency pole, fH, and the differential gain, Avd. With a Avd = 2 and fH = 100kHz, the resulting GBWP = 100kHz which is much smaller than the LM4861 GBWP of 4MHz. This figure displays that if a designer has a need to design an amplifier with a higher differential gain, the LM4861 can still be used without running into bandwidth problems. Copyright © 1997–2013, Texas Instruments Incorporated Submit Documentation Feedback 11 Product Folder Links: LM4861 |
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