Datasheet

LM4889
SNAS157H –APRIL 2002–REVISED MAY 2013 www.ti.com
AUDIO POWER AMPLIFIER DESIGN
A 1W/8Ω Audio Amplifier
• Given:
– Power Output: 1 Wrms
– Load Impedance: 8Ω
– Input Level: 1 Vrms
– Input Impedance: 20 kΩ
– Bandwidth: 100 Hz–20 kHz ± 0.25 dB
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 Performance Characteristics section, the supply
rail can be easily found. A second way to determine the minimum supply rail is to calculate the required V
opeak
using Equation 3 and add the output voltage. Using this method, the minimum supply voltage would be (V
opeak
+
(V
OD
TOP
+ V
OD
BOT
)), where V
OD
BOT
and V
OD
TOP
are extrapolated from the Dropout Voltage vs Supply Voltage curve in
the Typical Performance Characteristics section.
(3)
5V is a standard voltage in most applications, it is chosen for the supply rail. Extra supply voltage creates
headroom that allows the LM4889 to reproduce peaks in excess of 1W without producing audible distortion. 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 Equation 4.
(4)
R
f
/R
i
= A
VD
/2 (5)
From Equation 3, the minimum A
VD
is 2.83; use A
VD
= 3.
Since the desired input impedance was 20 kΩ, and with a A
VD
impedance of 2, a ratio of 1.5:1 of R
f
to R
i
results
in an allocation of R
i
= 20 kΩ and R
f
= 30 kΩ. 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 better than the required ±0.25 dB specified.
f
L
= 100 Hz/5 = 20 Hz (6)
f
H
= 20 kHz * 5 = 100 kHz (7)
As stated in the External Components Description 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 (8)
The high frequency pole is determined by the product of the desired frequency pole, f
H
, and the differential gain,
A
VD
. With a A
VD
= 3 and f
H
= 100 kHz, the resulting GBWP = 300kHz which is much smaller than the LM4889
GBWP of 2.5MHz. This calculation shows that if a designer has a need to design an amplifier with a higher
differential gain, the LM4889 can still be used without running into bandwidth limitations.
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