Application Information (Continued)
tion of R
f
and Cfwill 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 approximately 320 kHz. Once the differential gain of the amplifier has been calculated, a choice of R
f
will result, and Cfcan
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 voiceband frequency response. Such an application usually requires 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 combination of R
i
and Ciform a highpass filter while Rfand Cfform a
lowpass filter. Using the typical voice-band frequency range,
with a passband differential gain of approximately 100, the
following values of R
i,Ci,Rf
, and Cffollow from the equations stated in the External Components Description section.
R
i
=10kΩ,Rf= 510k ,Ci= 0.22 µF, and Cf=15pF
Five times away from a −3 dB point is 0.17 dB down from the
flatband response. With this selection of components, the resulting −3 dB points, f
L
and fH, are 72 Hz and 20 kHz, respectively,resulting in a flatband frequency response of better 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 singleendedly 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 singlesupply amplifier configurations. This capacitor, designated
C
O
in
Figure 3
, in conjunction with RL, forms a highpass filter.
The −3 dB point of this high pass filter is 1/(2πR
LCO
), 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. VO2, the output
that is not used, is connected through a 0.1 µF capacitor to
a2kΩload 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.
AUDIO POWER AMPLIFIER DESIGN
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 needed supply rail to obtain 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 supply rail is to calculate the required V
opeak
using Equation 3
and add the dropout voltage. Using this method, the minimum supply voltage would be (V
opeak+VOD
, where VODis
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 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 section.
Once the power dissipation equations have been addressed,
the required differential gain can be determined from Equation 4.
(4)
R
f/Ri
=
A
VD
/2 (5)
From equation 4, the minimum A
vd
is 2.83: Avd=3
Since the desired input impedance was 20 kΩ, and with a
A
vd
of 3, a ratio of 1:1.5 of Rfto Riresults in an allocation of
R
i
=20kΩ,Rf=30kΩ. 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. This fact results in
a low and high frequency pole of 20 Hz and 100 kHz respectively.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,Avd.
With a A
vd
= 2 and fH= 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.
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