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LM4861 Datasheet(PDF) 9 Page - National Semiconductor (TI) |
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LM4861 Datasheet(HTML) 9 Page - National Semiconductor (TI) |
9 / 12 page Application Information (Continued) possible high frequency oscillations. Care should be taken when calculating the −3dB frequency in that an incorrect combination of R f and Cf will cause rolloff before 20kHz. A typical combination of feedback resistor and capacitor that will not produce audio band high frequency rolloff is R f = 100k Ω and C f = 5pF. These components result in a −3dB point of approximately 320kHz. Once the differential gain of the amplifier has been calculated, a choice of R f will result, and C f 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 300Hz to 3.5kHz. 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 = 10k Ω,R f = 510k ,Ci = 0.22µF, and Cf = 15pF Five times away from a −3dB point is 0.17dB down from the flatband response. With this selection of components, the resulting −3dB points, f L and fH, are 72Hz and 20kHz, re- spectively, resulting in a flatband frequency response of better than ±0.25dB with a rolloff of 6dB/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 common single-ended application, where V O1 is used to drive the speaker. This output is coupled through a 470µF 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 −3dB 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 frequen- cies to the load. When driving an 8 Ω load, and if a full audio spectrum reproduction is required, C O should be at least 470µF. V O2, the output that is not used, is connected through a 0.1 µF capacitor to a 2k Ω load to prevent instability. While such an instability will not affect the waveform of V O1,itis 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 obtain 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 20k Ω, and with a A vd of 3, a ratio of 1:1.5 of R f to Ri results in an allocation of Ri = 20k Ω,R f = 30k Ω. 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 the External Components section, R i in conjunc- tion with C i create a highpass filter. C i ≥ 1/(2π*20kΩ*20Hz) = 0.397µF; use 0.39µF. 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 = 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. www.national.com 9 |
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