LM4860M National Semiconductor, LM4860M Datasheet - Page 8

no-image

LM4860M

Manufacturer Part Number
LM4860M
Description
Audio Power Amplifier IC
Manufacturer
National Semiconductor
Datasheet

Specifications of LM4860M

No. Of Channels
1
Output Power
1.15W
Load Impedance
8ohm
Amplifier Case Style
SOIC
No. Of Pins
16
Output Power Pout
1.15W
Mounting Type
Surface Mount
Supply Voltage Min
2.7V
Lead Free Status / RoHS Status
Contains lead / RoHS non-compliant

Available stocks

Company
Part Number
Manufacturer
Quantity
Price
Part Number:
LM4860M
Manufacturer:
NS
Quantity:
3 072
Part Number:
LM4860M
Manufacturer:
NS
Quantity:
895
Part Number:
LM4860M
Quantity:
239
Part Number:
LM4860M
Manufacturer:
NS/国半
Quantity:
20 000
Company:
Part Number:
LM4860M
Quantity:
36
Company:
Part Number:
LM4860M
Quantity:
36
Part Number:
LM4860MX
Manufacturer:
SKYWELL
Quantity:
1 863
Part Number:
LM4860MX
Manufacturer:
NS/国半
Quantity:
20 000
Part Number:
LM4860MX/NOPB
Manufacturer:
TDK
Quantity:
6 221
www.national.com
Application Information
HIGHER GAIN AUDIO AMPLIFIER
The LM4860 is unity-gain stable and requires no external
components besides gain-setting resistors, an input coupling
capacitor, and proper supply bypassing in the typical appli-
cation. However if a closed-loop differential gain of greater
than 10 is required, then a feedback capacitor is needed, as
shown in Figure 2 , to bandwidth limit the amplifier. The feed-
back capacitor creates a low pass filter that eliminates un-
wanted high frequency oscillations. Care should be taken
when calculating the −3 dB frequency in that an incorrect
combination of R
typical combination of feedback resistor and capacitor that
will not produce audio band high frequency rolloff is R
100 k
point of approximately 320 kHz. Once the differential gain of
the amplifier has been calculated, a choice of R
and C
External Components Description section.
VOICE-BAND AUDIO AMPLIFIER
Many applications, such as telephony, only require a
voice-band 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 combina-
tion of R
lowpass filter. Using the typical voice-band frequency range,
with a passband differential gain of approximately 100, the
following values of R
tions stated in the External Components Description sec-
tion.
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
spectively, resulting in a flatband frequency response of bet-
ter than
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 LM4860 is a bridged
monoaural amp, it can also be used to drive a load
single-endedly in applications, such as PC cards, which re-
quire that one side of the load is tied to ground. Figure 3
shows a common single-ended application, where V
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 ca-
pacitor, designated C
forms a highpass filter. The −3 dB point of this highpass filter
is 1/(2 R
product of R
cies to the load. When driving an 8
spectrum reproduction is required, C
470 µF. V
through a 0.1 µF capacitor to a 2 k load to prevent instabil-
R
i
= 10 k , R
f
can then be calculated from the formula stated in the
and C
i
±
L
and C
0.25 dB with a rolloff of 6 dB/octave outside of the
C
O2
O
L
, the output that is not used, is connected
), so care should be taken to make sure that the
f
and C
= 5 pF. These components result in a −3 dB
i
form a highpass filter while R
f
f
= 510k, C
and C
O
i
O
, C
L
is large enough to pass low frequen-
in Figure 3 , in conjunction with R
f
and f
i
, R
will cause rolloff before 20 kHz. A
f
i
, and C
H
= 0.22 µF, and C
, are 72 Hz and 20 kHz, re-
f
load, and if a full audio
follow from the equa-
O
should be at least
(Continued)
f
and C
f
f
= 15 pF
will result,
f
form a
O1
f
is
L
=
,
8
ity. While such an instability will not affect the waveform of
V
AUDIO POWER AMPLIFIER DESIGN
Design a 500 mW/8
Given:
Power Output:
Load Impedance: 8
Input Level: 1 Vrms(max)
Input Impedance: 20 k
Bandwidth: 20 Hz-20 kHz
A designer must first determine the needed supply rail to ob-
tain the specified output power. Calculating the required sup-
ply rail involves knowing two parameters, V
dropout voltage. The latter is typically 0.7V. V
determined from equation 3.
For 500 mW of output power into an 8
V
adding V
that exists in many applications and for this reason, a supply
rail of 5V is designated. Extra supply voltage creates dy-
namic headroom that allows the LM4860 to reproduce peaks
in excess of 500 mW without 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 condi-
tions 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.
From equation 4, the minimum A
Since the desired input impedance was 20 k , and with an
A
R
capacitor is not needed. 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
a low and high frequency pole of 4 Hz and 100 kHz respec-
tively. As stated in the External Components section, R
conjunction with C
The high frequency pole is determined by the product of the
desired high frequency pole, f
With a A
100 kHz which is much smaller than the LM4860 GBWP of
7 MHz. This figure displays that if a designer has a need to
design an amplifier with a higher differential gain, the
LM4860 can still be used without running into bandwidth
problems.
O1
opeak
vd
i
= R
, it is good design practice to load the second output.
of 2, a ratio of 1:1 of R
C
i
f
is 2.83V. A minimum supply rail of 3.53V results from
= 20 k . Since the A
1/(2
opeak
vd
= 2 and f
and V
*
20 k
500 mWrms
i
create a highpass filter.
od
±
H
. But 3.53V is not a standard voltage
0.25 dB specified. This fact results in
Audio Amplifier
*
= 100 kHz, the resulting GBWP =
4 Hz) = 1.98 µF; use 2.2 µF.
±
f
vd
H
to R
0.25 dB
, and the differential gain, A
was less than 10, a feedback
vd
i
results in an allocation of
is: A
vd
load, the required
opeak
= 2
opeak
and also the
can be
i
vd
in
.

Related parts for LM4860M