make sure that the power supply choice along with the out-
put impedance does not violate the conditions explained in
the Power Dissipation section.
Application Information (Continued)
AUDIO POWER AMPLIFIER DESIGN
Design a 500 mW/8Ω Audio Amplifier
Once the power dissipation equations have been addressed,
the required differential gain can be determined from Equa-
tion 4.
Given:
Power Output
Load Impedance
Input Level
500 mWrms
8Ω
1 Vrms
20 kΩ
(4)
=
Rf/Ri AVD/2:
(5)
Input Impedance
Bandwidth
=
From equation 4, the minimum AVD is 2; use AVD 2.
±
100 Hz–20 kHz 0.25 dB
Since the desired input impedance was 20 kΩ, and with a
AVD of 2, a ratio of 1:1 of Rf to Ri results in an allocation of Ri
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 Vopeak using equation 3
and add the dropout voltage. Using this method, the mini-
=
=
Rf 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, Ri in con-
junction with Ci create a highpass filter.
*
mum supply voltage would be (Vopeak + (2 VOD)), where
OD is extrapolated from the Dropout Voltage vs Supply Volt-
V
=
*
*
Ci ≥ 1/(2π 20 kΩ 20 Hz) 0.397 µF; use 0.39 µF.
age curve in the Typical Performance Characteristics sec-
tion.
The high frequency pole is determined by the product of the
desired high frequency pole, fH, 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
(3)
Using the Output Power vs Supply Voltage graph for an 8Ω
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
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.
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