LM4864MMX National Semiconductor, LM4864MMX Datasheet - Page 12

• Manufacturer Part Number: LM4864MMX
• Description: IC AMP AUDIO PWR .725W AB 8MSOP
• Manufacturer: National Semiconductor
• Series: Boomer®r
• Type: Class ABr

Specifications of LM4864MMX

• Output Type: 1-Channel (Mono)
• Max Output Power X Channels @ Load: 725mW x 1 @ 8 Ohm
• Voltage - Supply: 2.7 V ~ 5.5 V
• Features: Shutdown, Thermal Protection
• Mounting Type: Surface Mount
• Package / Case: 8-MSOP, Micro8™, 8-uMAX, 8-uSOP,
• Lead Free Status / RoHS Status: Contains lead / RoHS non-compliant
• Other names: LM4864MMXTR

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Application Information
draw will be minimized in idle mode. While the device will be
disabled with shutdown pin voltages less than V
current may be greater than the typical value of 0.7 µA. In
either case, the shutdown pin should be tied to a definite
voltage to avoid unwanted state changes.
In many applications, a microcontroller or microprocessor
output is used to control the shutdown circuitry which pro-
vides a quick, smooth transition into shutdown. Another so-
lution is to use a single-pole, single-throw switch in conjunc-
tion with an external pull-up resistor. When the switch is
closed, the shutdown pin is connected to ground and en-
ables the amplifier. If the switch is open, then the external
pull-up resistor will disable the LM4864. This scheme guar-
antees that the shutdown pin will not float, thus preventing
unwanted state changes.
PROPER SELECTION OF EXTERNAL COMPONENTS
Proper selection of external components in applications us-
ing integrated power amplifiers is critical to optimize device
and system performance. While the LM4864 is tolerant to a
variety of external component combinations, consideration
to component values must be used to maximize overall
system quality.
The LM4864 is unity-gain stable, giving a designer maximum
system flexibility. The LM4864 should be used in low gain
configurations to minimize THD+N values, and maximize the
signal to noise ratio. Low gain configurations require large
input signals to obtain a given output power. Input signals
equal to or greater than 1 Vrms are available from sources
such as audio codecs. Please refer to the section, Audio
Power Amplifier Design, for a more complete explanation
of proper gain selection.
Besides gain, one of the major considerations is the closed-
loop bandwidth of the amplifier. To a large extent, the band-
width is dictated by the choice of external components
shown in Figure 1. The input coupling capacitor, C
first order high pass filter which limits low frequency re-
sponse. This value should be chosen based on needed
frequency response for a few distinct reasons.
Selection of Input Capacitor Size
Large input capacitors are both expensive and space hungry
for portable designs. Clearly, a certain sized capacitor is
needed to couple in low frequencies without severe attenu-
ation. But in many cases the speakers used in portable
systems, whether internal or external, have little ability to
reproduce signals below 150 Hz. In this case using a large
input capacitor may not increase system performance.
In addition to system cost and size, click and pop perfor-
mance is effected by the size of the input coupling capacitor,
C
reach its quiescent DC voltage (nominally
charge comes from the output via the feedback and is apt to
create pops upon device enable. Thus, by minimizing the
capacitor size based on necessary low frequency response,
turn-on pops can be minimized.
Besides minimizing the input capacitor size, careful consid-
eration should be paid to the bypass capacitor value. Bypass
capacitor, C
turn-on pops since it determines how fast the LM4864 turns
on. The slower the LM4864’s outputs ramp to their quiescent
DC voltage (nominally
Choosing C
(in the range of 0.1 µF to 0.39 µF), should produce a click-
i
. A larger input coupling capacitor requires more charge to
B
B
, is the most critical component to minimize
equal to 1.0 µF along with a small value of C
1
⁄
2
V
DD
), the smaller the turn-on pop.
(Continued)
1
⁄
2
DD
V
DD
i
, forms a
, the idle
). This
i
12
less and popless shutdown function. While the device will
function properly, (no oscillations or motorboating), with C
equal to 0.1 µF, the device will be much more susceptible to
turn-on clicks and pops. Thus, a value of C
or larger is recommended in all but the most cost sensitive
designs.
AUDIO POWER AMPLIFIER DESIGN
Design a 300 mW/8Ω Audio Amplifier
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 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
and add the dropout voltage. Using this method, the mini-
mum supply voltage would be (V
is extrapolated from the Dropout Voltage vs Supply Voltage
curve in the Typical Performance Characteristics section.
Using the Output Power vs Supply Voltage graph for an 8Ω
load, the minimum supply rail is 3.5V. But since 5V is a
standard supply voltage in most applications, it is chosen for
the supply rail. Extra supply voltage creates headroom that
allows the LM4864 to reproduce peaks in excess of 500 mW
without producing audible distortion. At this time, the de-
signer 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 5.
From Equation 5, the minimum A
Since the desired input impedance was 20 kΩ, and with a
A
R
bandwidth requirements which must be stated as a pair of
−3 dB frequency points. Five times away from a pole gives
0.17 dB down from passband response which is better than
the required
As stated in the External Components section, R
junction with C
Given:
VD
i
Power Output
Load Impedance
Input Level
Input Impedance
Bandwidth
= R
C
of 2, a ratio of 1:1 of R
i
F
≥ 1/(2π*20 kΩ*20 Hz) = 0.397 µF; use 0.39 µF
= 20 kΩ. The final design step is to address the
±
0.25 dB specified.
i
create a highpass filter.
f
H
f
= 20 kHz x 5 = 100 kHz
R
L
F
= 100 Hz/5 = 20 Hz
/R
i
= A
F
VD
to R
/2
opeak
100 Hz–20 kHz
VD
i
results in an allocation of
is 1.55; use A
opeak
+ (2*V
(6)
using Equation 4
B
OD
equal to 1.0 µF
)), where V
300 mWrms
±
VD
0.25 dB
i
1 Vrms
in con-
20 kΩ
= 2.
8Ω
(4)
(5)
OD
B