Application Information
(Continued)
AUDIO POWER AMPLIFIER DESIGN
Audio Amplifier Design: Driving 1W into an 8
Load
The following are the desired operational parameters:
Power Output
Load Impedance
Input Level
Input Impedance
Bandwidth
The design begins by specifying the minimum supply voltage
necessary to obtain the specified output power. One way to
find the minimum supply voltage is to use the Output Power
vs Supply Voltage curve in the
Typical Performance Char-
acteristics
section. Another way, using Equation (8), is to
calculate the peak output voltage necessary to achieve the
desired output power for a given load impedance. To ac-
count for the amplifier’s dropout voltage, two additional volt-
ages, based on the Dropout Voltage vs Supply Voltage in the
Typical Performance Characteristics
curves, must be
added to the result obtained by Equation (8). This results in
Equation (9).
1W
RMS
8
1V
RMS
20k
100Hz–20kHz
±
0.25dB
(8)
(9)
V
CC
≥
(V
OUTPEAK
+ (V
OD
TOP
+ V
OD
BOT
))
The Output Power vs Supply Voltage graph for an 8
load
indicates a minimum supply voltage of 4.6V. This is easily
met by the commonly used 5V supply voltage. The additional
voltage creates the benefit of headroom, allowing the
LM4876 to produce peak output power in excess of 1W with-
out clipping or other audible distortion. The choice of supply
voltage must also not create a violation of maximum power
dissipation as explained above in the
Power Dissipation
section.
After satisfying the LM4876’s power dissipation require-
ments, the minimum differential gain is found using Equation
(10).
(10)
Thus, a minimum gain of 2.83 allows the LM4876’s to reach
full output swing and maintain low noise and THD+N perfor-
mance. For this example, let A
VD
= 3.
The amplifier’s overall gain is set using the input (R
) and
feedback (R
) resistors. With the desired input impedance
set at 20k
, the feedback resistor is found using Equation
(11).
R
f
/R
i
= A
VD
/2
The value of R
f
is 30k
.
The last step in this design example is setting the amplifier’s
-3dB low frequency bandwidth. To achieve the desired
±
0.25dB pass band magnitude variation limit, the low fre-
quency response must extend to at least one-fifth the lower
bandwidth limit and the high frequency response must ex-
tend to at least five times the upper bandwidth limit. The re-
sults is an
f
L
= 100 Hz/5 = 20Hz
and an
F
H
= 20 kHz*5 = 100kHz
As mentioned in the External Components section, R
and C
i
create a highpass filter that sets the amplifier’s lower band-
pass frequency limit. Find the coupling capacitor’s value us-
ing Equation (12).
Ci
≥
1/(2
π
Rif
L
)
The result is
1/(2
π
*20k
*20Hz) = 0.398μF.
Use a 0.39μF capacitor, the closest standard value.
The product of the desired high frequency cutoff (100kHz in
this example) and the differential gain, A
VD
, determines the
upper passband response limit. With A
= 3 and f
=
100kHz, the closed-loop gain bandwidth product (GBWP) is
150kHz. This is less than the LM4876’s 4MHz GBWP. With
this margin, the amplifier can be used in designs that require
more differential gain and avoid performance-restricting
bandwidth limitations.
(11)
(12)
L
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