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LM4861 Datasheet(PDF) 7 Page - National Semiconductor (TI) |
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LM4861 Datasheet(HTML) 7 Page - National Semiconductor (TI) |
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7 / 8 page ![]() Application Information (Continued) tion of R f and Cf will cause rolloff before 20 kHz. A typical combination of feedback resistor and capacitor that will not produce audio band high frequency rolloff is R f = 100 kΩ and C f = 5 pF. These components result in a −3 dB point of ap- proximately 320 kHz. Once the differential gain of the ampli- fier has been calculated, a choice of R f will result, and Cf can then be calculated from the formula stated in the External Components Description section. VOICE-BAND AUDIO AMPLIFIER Many applications, such as telephony, only require a voice- band frequency response. Such an application usually re- quires a flat frequency response from 300 Hz to 3.5 kHz. By adjusting the component values of Figure 2, this common application requirement can be implemented. The combina- tion of R i and Ci form a highpass filter while Rf and Cf form a lowpass filter. Using the typical voice-band frequency range, with a passband differential gain of approximately 100, the following values of R i,Ci,Rf, and Cf follow from the equa- tions stated in the External Components Description sec- tion. R i =10kΩ,Rf = 510k ,Ci = 0.22 µF, and Cf =15pF Five times away from a −3 dB point is 0.17 dB down from the flatband response. With this selection of components, the re- sulting −3 dB points, f L and fH, are 72 Hz and 20 kHz, re- spectively, resulting in a flatband frequency response of bet- ter than ±0.25 dB with a rolloff of 6 dB/octave outside of the passband. If a steeper rolloff is required, other common bandpass filtering techniques can be used to achieve higher order filters. SINGLE-ENDED AUDIO AMPLIFIER Although the typical application for the LM4861 is a bridged monoaural amp, it can also be used to drive a load single- endedly in applications, such as PC cards, which require that one side of the load is tied to ground. Figure 3 shows a com- mon single-ended application, where V O1 is used to drive the speaker. This output is coupled through a 470 µF capacitor, which blocks the half-supply DC bias that exists in all single- supply amplifier configurations. This capacitor, designated C O in Figure 3, in conjunction with RL, forms a highpass filter. The −3 dB point of this high pass filter is 1/(2 πR LCO), so care should be taken to make sure that the product of R L and CO is large enough to pass low frequencies to the load. When driving an 8 Ω load, and if a full audio spectrum reproduction is required, C O should be at least 470 µF. VO2, the output that is not used, is connected through a 0.1 µF capacitor to a2k Ω load to prevent instability. While such an instability will not affect the waveform of V O1, it is good design practice to load the second output. 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 ob- tain the specified output power. By extrapolating from the Output Power vs Supply Voltage graph 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 +VOD , where VOD is typically 0.6V. (3) For 1W of output power into an 8 Ω load, the required V opeak is 4.0V. A minumum supply rail of 4.6V results from adding V opeak 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 head- room 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 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 2.83: Avd =3 Since the desired input impedance was 20 k Ω, and with a A vd of 3, a ratio of 1:1.5 of Rf to Ri results in an allocation of R i =20kΩ,Rf =30kΩ. The final design step is to address the bandwidth 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 bet- ter than the required ±0.25 dB specified. This fact results in a low and high frequency pole of 20 Hz and 100 kHz respec- tively. As stated in the External Components section, R i in conjunction 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, Avd. With a A vd = 2 and fH = 100 kHz, the resulting GBWP = 100 kHz which is much smaller than the LM4861 GBWP of 4 MHz. 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. www.national.com 7 |
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