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поискавой системы для электроныых деталей |
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поискавой системы для электроныых деталей |
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LM4898MMBD датащи(PDF) 14 Page - Texas Instruments |
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LM4898MMBD датащи(HTML) 14 Page - Texas Instruments |
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14 / 23 page  LM4898, LM4898MMBD SNAS216D – MAY 2004 – REVISED MAY 2004 www.ti.com Tolerance Rf1 Rf2 Vo2-Vo1 ILOAD 20% 0.8R 1.2R -0.5V 62.5mA 10% 0.9R 1.1R -0.250V 31.25mA 5% 0.95R 1.05R -0.125V 15.63mA 1% 0.99R 1.01R -0.025V 3.125mA 0 R R R 0 Similar results would occur if the input resistors were not carefully matched. Adding input coupling capacitors in between the signal source and the input resistors will eliminate this problem, however, to achieve best performance with minimum component count it is highly recommended that both the feedback and input resistors matched to 1% tolerance or better. AUDIO POWER AMPLIFIER DESIGN Design a 1W/8 Ω Audio Amplifier Given: • Power Output 1W • Load Impedance 8 Ω • Input Level 1Vrms • Input Impedance 20k Ω • Bandwidth 100Hz–20kHz ± 0.25dB A designer must first determine the minimum supply rail to obtain the specified output power. The supply rail can easily be found by extrapolating from the Output Power vs. Supply Voltage graphs in the Typical Performance Characteristics section. A second way to determine the minimum supply rail is to calculate the required Vopeak using Equation 7 and add the dropout voltages. Using this method, the minimum supply voltage is (Vopeak +(VDO TOP+(VDO BOT )),where VDO BOT and VDO TOP are extrapolated from the Dropout Voltage vs. Supply Voltage curve in the Typical Performance Characteristics section. (7) Using the Output Power vs. Supply Voltage graph for an 8 Ω load, the minimum supply rail just about 5V. Extra supply voltage creates headroom that allows the LM4898 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 8. (8) Rf / Ri = AVD From Equation 8, the minimum AVD is 2.83. Since the desired input impedance was 20kΩ, a ratio of 2.83:1 of Rf to Ri results in an allocation of Ri = 20kΩ for both input resistors and Rf= 60kΩ for both feedback resistors. The final design step is to address the bandwidth requirement which must be stated as a single -3dB frequency point. Five times away from a -3dB point is 0.17dB down from passband response which is better than the required ±0.25dB specified. fH = 20kHz * 5 =100kHz The high frequency pole is determined by the product of the desired frequency pole, fH , and the differential gain, AVD .With a AVD = 2.83 and fH = 100kHz, the resulting GBWP = 150kHz which is much smaller than the LM4898 GBWP of 10MHz. This figure displays that if a designer has a need to design an amplifier with a higher differential gain, the LM4898 can still be used without running into bandwidth limitations. 14 Submit Documentation Feedback Copyright © 2004, Texas Instruments Incorporated Product Folder Links: LM4898 LM4898MMBD |
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