LM4861
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SNAS095C –MAY 1997–REVISED MAY 2013
AUDIO POWER AMPLIFIER DESIGN
Design a 1W / 8Ω Audio Amplifier
Given:
Power Output
Load Impedance
Input Level
1 Wrms
8Ω
1 Vrms
Input Impedance
Bandwidth
20 kΩ
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
Figure 11 in Typical Performance Characteristics, the supply rail can be easily found. A second way to determine
the minimum supply rail is to calculate the required Vopeak using Equation 5 and add the dropout voltage. Using
this method, the minimum supply voltage would be (Vopeak + VOD , where VOD is typically 0.6V.
(5)
For 1W of output power into an 8Ω load, the required Vopeak is 4.0V. A minumum supply rail of 4.6V results from
adding Vopeak 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 headroom 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.
Once the power dissipation equations have been addressed, the required differential gain can be determined
from Equation 6.
(6)
Rf/Ri = AVD / 2
(7)
From Equation 6, the minimum Avd is 2.83: Avd = 3
Since the desired input impedance was 20kΩ, and with a Avd of 3, a ratio of 1:1.5 of Rf to Ri results in an
allocation of Ri = 20kΩ, Rf = 30kΩ. 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 External Components Description , Ri in conjunction with Ci create
a highpass filter.
Ci ≥ 1 / (2π*20kΩ*20Hz) = 0.397μF; use 0.39μF.
(8)
The high frequency pole is determined by the product of the desired high frequency pole, fH, and the differential
gain, Avd. With a Avd = 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.
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