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LM4766 Datasheet(PDF) 17 Page - National Semiconductor (TI) |
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LM4766 Datasheet(HTML) 17 Page - National Semiconductor (TI) |
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17 / 19 page ![]() Application Information (Continued) AUDIO POWER AMPLlFIER DESIGN Design a 30W/8 Ω Audio Amplifier Given: Power Output 30Wrms Load Impedance 8 Ω Input Level 1Vrms(max) Input Impedance 47k Ω Bandwidth 20Hz−20kHz ±0.25dB A designer must first determine the power supply require- ments in terms of both voltage and current needed to obtain the specified output power. V OPEAK can be determined from Equation (4) and I OPEAK from Equation (5). (4) (5) To determine the maximum supply voltage the following conditions must be considered. Add the dropout voltage to the peak output swing V OPEAK, to get the supply rail at a current of I OPEAK. The regulation of the supply determines the unloaded voltage which is usually about 15% higher. The supply voltage will also rise 10% during high line conditions. Therefore the maximum supply voltage is obtained from the following equation. Max supplies ≈ ± (V OPEAK +VOD) (1 + regulation) (1.1) For 30W of output power into an 8 Ω load, the required V OPEAK is 21.91V. A minimum supply rail of 25.4V results from adding V OPEAK and VOD. With regulation, the maximum supplies are ±32V and the required I OPEAK is 2.74A from Equation (5). It should be noted that for a dual 30W amplifier into an 8 Ω load the I OPEAK drawn from the supplies is twice 2.74A PK or 5.48APK. At this point it is a good idea to check the Power Output vs Supply Voltage to ensure that the required output power is obtainable from the device while maintaining low THD+N. In addition, the designer should verify that with the required power supply voltage and load impedance, that the required heatsink value θ SA is feasible given system cost and size constraints. Once the heatsink issues have been addressed, the required gain can be de- termined from Equation (6). (6) From Equation (6), the minimum A V is: A V ≥ 15.5. By selecting a gain of 21, and with a feedback resistor, R f = 20k Ω, the value of R i follows from Equation (7). R i =Rf (AV − 1) (7) Thus with R i =1k Ω a non-inverting gain of 21 will result. Since the desired input impedance was 47k Ω, a value of 47k Ω was selected for R IN. 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 re- sults in a low and high frequency pole of 4Hz and 100kHz respectively. As stated in the External Components sec- tion, R i in conjunction with Ci create a high-pass filter. C i ≥ 1/(2π *1kΩ * 4Hz) = 39.8µF; use 39µF. The high frequency pole is determined by the product of the desired high frequency pole, f H, and the gain, AV. With a A V = 21 and fH = 100kHz, the resulting GBWP is 2.1MHz, which is less than the guaranteed minimum GBWP of the LM4766 of 8MHz. This will ensure that the high frequency response of the amplifier will be no worse than 0.17dB down at 20kHz which is well within the bandwidth requirements of the design. www.national.com 17 |
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