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LM4862M bảng dữ liệu(PDF) 8 Page - National Semiconductor (TI) |
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LM4862M bảng dữ liệu(HTML) 8 Page - National Semiconductor (TI) |
8 / 10 page Application Information (Continued) AUDIO POWER AMPLIFIER DESIGN Design a 500 mW/8 Ω Audio Amplifier Given: Power Output 500 mWrms 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 Per- formance Characteristics section, the supply rail can be easily found. A second way to determine the minimum sup- ply rail is to calculate the required V opeak using equation 3 and add the dropout voltage. Using this method, the mini- mum supply voltage would be (V opeak +(2*VOD)), where V OD is extrapolated from the Dropout Voltage vs Supply Volt- age curve in the Typical Performance Characteristics sec- tion. (3) Using the Output Power vs Supply Voltage graph for an 8 Ω load, the minimum supply rail is 4.3V. But since 5V is a stan- dard supply voltage in most applications, it is chosen for the supply rail. Extra supply voltage creates headroom that al- lows the LM4862 to reproduce peaks in excess of 500 mW without clipping the signal. At this time, the designer must make sure that the power supply choice along with the out- put 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 Equa- tion 4. (4) R f/Ri = AVD/2: (5) From equation 4, the minimum A VD is 2; use AVD = 2. Since the desired input impedance was 20 k Ω, and with a A VD of 2, a ratio of 1:1 of Rf to Ri results in an allocation of Ri = R f = 20 kΩ. The final design step is to address the band- width requirements which must be stated as a pair of −3 dB frequency points. Five times away from a −3 dB point is 0.17 dB down from passband response which is better than the required ±0.25 dB specified. This fact results in a low and high frequency pole of 20 Hz and 100 kHz respectively. As stated in the External Components section, R i in con- junction with C i create a highpass filter. C i ≥ 1/(2π*20 kΩ*20 Hz) = 0.397 µF; use 0.39 µF. The high frequency pole is determined by the product of the desired high frequency pole, f H, and the differential gain, A VD. With an AVD = 2 and fH = 100 kHz, the resulting GBWP = 100 kHz which is much smaller than the LM4862 GBWP of 12.5 MHz. This figure displays that if a designer has a need to design an amplifier with a higher differential gain, the LM4862 can still be used without running into bandwidth problems. www.national.com 8 |
Số phần tương tự - LM4862M |
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Mô tả tương tự - LM4862M |
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