Application Information
(Continued)
AUDIO POWER AMPLIFIER DESIGN
Design a 1W / 8Ω Audio Amplifier
Given:
Power Output
Load Impedance
Input Level
Input Impedance
1 Wrms
8Ω
1 Vrms
20 kΩ
tion of R
f
and C
f
will cause rolloff before 20 kHz. A typical
combination of feedback resistor and capacitor that will not
produce audio band high frequency rolloff is R
f
= 100 kΩ and
C
f
= 5 pF. These components result in a −3 dB point of ap-
proximately 320 kHz. Once the differential gain of the ampli-
fier 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 300 Hz to 3.5 kHz. By
adjusting the component values of
Figure 2,
this common
application requirement can be implemented. The combina-
tion of R
i
and C
i
form a highpass filter while R
f
and C
f
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
, C
i
, R
f
, and C
f
follow from the equa-
tions stated in the
External Components Description
sec-
tion.
R
i
= 10 kΩ, R
f
= 510k ,C
i
= 0.22 µF, and C
f
= 15 pF
Five times away from a −3 dB point is 0.17 dB down from the
flatband response. With this selection of components, the re-
sulting −3 dB points, f
L
and f
H
, are 72 Hz and 20 kHz, re-
spectively, resulting in a flatband frequency response of bet-
ter than
±
0.25 dB with a rolloff of 6 dB/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 com-
mon single-ended application, where V
O1
is used to drive the
speaker. This output is coupled through a 470 µF 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 R
L
, forms a highpass filter.
The −3 dB point of this high pass filter is 1/(2πR
L
C
O
), so care
should be taken to make sure that the product of R
L
and C
O
is large enough to pass low frequencies to the load. When
driving an 8Ω load, and if a full audio spectrum reproduction
is required, C
O
should be at least 470 µF. V
O2
, the output
that is not used, is connected through a 0.1 µF capacitor to
a 2 kΩ load to prevent instability. While such an instability will
not affect the waveform of V
O1
, it is good design practice to
load the second output.
Bandwidth
100 Hz–20 kHz
±
0.25 dB
A designer must first determine the needed supply rail to ob-
tain 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
+ V
OD
, where V
OD
is
typically 0.6V.
(3)
For 1W of output power into an 8Ω load, the required V
opeak
is 4.0V. A minumum supply rail of 4.6V results from adding
V
opeak
and V
od
. 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)
= A
VD
/2
(5)
R
f
/R
i
From equation 4, the minimum A
vd
is 2.83: A
vd
= 3
Since the desired input impedance was 20 kΩ, and with a
A
vd
of 3, a ratio of 1:1.5 of R
f
to R
i
results in an allocation of
R
i
= 20 kΩ, 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 bet-
ter than the required
±
0.25 dB specified. This fact results in
a low and high frequency pole of 20 Hz and 100 kHz respec-
tively. As stated in the
External Components
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.
The high frequency pole is determined by the product of the
desired high frequency pole, f
H
, and the differential gain, A
vd
.
With a A
vd
= 2 and f
H
= 100 kHz, the resulting GBWP =
100 kHz which is much smaller than the LM4861 GBWP of
4 MHz. 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.
7
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