LM4991MA National Semiconductor, LM4991MA Datasheet - Page 13

• Manufacturer Part Number: LM4991MA
• Manufacturer: National Semiconductor

Specifications of LM4991MA

• Operational Class: Class-AB
• Audio Amplifier Output Configuration: 1-Channel Mono
• Output Power (typ): 1.5x1@8OhmW
• Audio Amplifier Function: Speaker
• Total Harmonic Distortion: 0.2@8Ohm@0.5W%
• 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: 64dB
• Single Supply Voltage (min): 2.2V
• 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: Not Compliant

Application Information
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.
From Equation 4, the minimum A
Since the desired input impedance was 20kΩ, and with a
A
allocation of R
is to address the bandwidth requirements which must be
VD
impedance of 2, a ratio of 1.5:1 of R
i
= 20kΩ and R
R
f
/R
i
= A
VD
f
= 30kΩ. The final design step
/2
VD
is 2.83; use A
(5)
f
(Continued)
to R
i
results in an
VD
= 3.
(4)
13
stated as a pair of −3dB frequency points. Five times away
from a −3dB point is 0.17dB down from passband response
which is better than the required
As stated in the External Components section, R
junction with C
The high frequency pole is determined by the product of the
desired frequency pole, f
With a A
150kHz which is much smaller than the LM4991 GBWP of
4MHz. This figure displays that if a designer has a need to
design an amplifier with a higher differential gain, the
LM4991 can still be used without running into bandwidth
limitations.
f
f
C
L
H
i
= 100Hz/5 = 20Hz
= 20kHz * 5 = 100kHz
≥ 1/(2π*20kΩ*20Hz) = 0.397µF; use 0.39µF
VD
= 3 and f
i
create a highpass filter.
H
= 100kHz, the resulting GBWP =
H
, and the differential gain, A
±
0.25dB specified.
www.national.com
i
in con-
VD
.