LM4881M National Semiconductor, LM4881M Datasheet - Page 12

• Manufacturer Part Number: LM4881M
• Description: IC,Audio Amplifier,DUAL,BIPOLAR,SOP,8PIN,PLASTIC
• Manufacturer: National Semiconductor

Specifications of LM4881M

• Operational Class: Class-AB
• Audio Amplifier Output Configuration: 2-Channel Stereo
• Output Power (typ): 0.3x2@8OhmW
• Audio Amplifier Function: Headphone
• Input Offset Voltage: 50@5VmV
• Total Harmonic Distortion: 0.025%%
• Single Supply Voltage (typ): 3/5V
• Dual Supply Voltage (typ): Not RequiredV
• Power Supply Requirement: Single
• Rail/rail I/o Type: No
• Power Supply Rejection Ratio: 50dB
• Single Supply Voltage (min): 2.7V
• Single Supply Voltage (max): 5.5V
• Dual Supply Voltage (min): Not RequiredV
• Dual Supply Voltage (max): Not RequiredV
• Operating Temp Range: -40C to 85C
• Operating Temperature Classification: Industrial
• Mounting: Surface Mount
• Pin Count: 8
• Package Type: SOIC N
• Lead Free Status / RoHS Status: Contains lead / RoHS non-compliant
• Lead Free Status / RoHS Status: Contains lead / RoHS non-compliant

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Application Information
shutdown function should be virtually clickless and popless.
While the device will function properly, (no oscillations or
motorboating), with C
much more susceptible to turn on clicks and pops. Thus, a
value of C
but the most cost sensitive designs.
AUDIO POWER AMPLIFIER DESIGN
Design a Dual 200mW/8Ω Audio Amplifier
A designer must first determine the needed supply rail to
obtain the specified output power. Calculating the required
supply rail involves knowing two parameters, V
also the dropout voltage. The latter is typically 530 mV and
can be found from the graphs in the Typical Performance
Characteristics. V
3.
For 200 mW of output power into an 8Ω load, the required
V
from adding V
voltage in most applications, it is chosen for the supply rail.
Extra supply voltage creates headroom that allows the
LM4881 to reproduce peaks in excess of 200 mW without
clipping the signal. At this time, the designer must make sure
that the power supply choice along with the output imped-
ance does not violate the conditions explained in the Power
Given:
Power Output
Load Impedance
Input Level
Input Impedance
Bandwidth
OPEAK
is 1.79 volts. A minimum supply rail of 2.32V results
B
equal to 0.1 µF or larger is recommended in all
OPEAK
OPEAK
and V
B
equal to 0.1 µF, the device will be
can be determined from Equation
OD
. Since 5V is a standard supply
100 Hz–20 kHz
(Continued)
1 Vrms (max)
200 mWrms
±
OPEAK
0.50 dB
20 kΩ
8Ω
and
(3)
12
Dissipation section. Remember that the maximum power
dissipation point from Equation 1 must be multiplied by two
since there are two independent amplifiers inside the pack-
age.
Once the power dissipation equations have been addressed,
the required gain can be determined from Equation 4.
From Equation 4, the minimum gain is: A
Since the desired input impedance was 20 kΩ, and with a
gain of 1.26, a value of 27 kΩ is designated for R
5% tolerance resistors. This combination results in a nominal
gain of 1.35. The final design step is to address the band-
width 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 assuming a single
pole roll-off. As stated in the External Components section,
both R
order highpass filters. Thus to obtain the desired frequency
low response of 100 Hz within
taken into consideration. The combination of two single order
filters at the same frequency forms a second order response.
This results in a signal which is down 0.34 dB at five times
away from the single order filter −3 dB point. Thus, a fre-
quency of 20 Hz is used in the following equations to ensure
that the response is better than 0.5 dB down at 100 Hz.
The high frequency pole is determined by the product of the
desired high frequency pole, f
V
resulting GBWP = 135 kHz which is much smaller than the
LM4881 GBWP of 18 MHz. This figure displays that if a
designer has a need to design an amplifier with a higher
gain, the LM4881 can still be used without running into
bandwidth limitations.
. With a closed-loop gain of 1.35 and f
C
i
C
≥ 1 / (2π * 20 kΩ * 20 Hz) = 0.397 µF; use 0.39 µF.
o
i
≥ 1 / (2π * 8Ω * 20 Hz) = 995 µF; use 1000 µF.
in conjunction with C
A
V
= R
f
/R
H
i
, and C
i
±
, and the closed-loop gain, A
0.5 dB, both poles must be
(5)
o
with R
V
H
= 1.26
= 100 kHz, the
L
, create first
f
, assuming
(4)