LM4928TL/NOPB National Semiconductor, LM4928TL/NOPB Datasheet - Page 14

• Manufacturer Part Number: LM4928TL/NOPB
• Description: IC AMP AUDIO PWR 2.2W AB 16USMD
• Manufacturer: National Semiconductor
• Series: Boomer®r
• Type: Class ABr

Specifications of LM4928TL/NOPB

• Output Type: 2-Channel (Stereo)
• Max Output Power X Channels @ Load: 2.2W x 2 @ 4 Ohm
• Voltage - Supply: 2.4 V ~ 5.5 V
• Features: Depop, Differential Inputs, Shutdown, Thermal Protection
• Mounting Type: Surface Mount
• Package / Case: 16-MicroSMD
• For Use With: LM4928TLBD - BOARD EVALUATION LM4928TL
• Lead Free Status / RoHS Status: Lead free / RoHS Compliant
• Other names: LM4928TLTR

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Application Information
Special care must be taken to match the values of the input
resistors (R
Because of the balanced nature of differential amplifiers,
resistor matching differences can result in net DC currents
across the load. This DC current can increase power con-
sumption, internal IC power dissipation, reduce PSRR,
CMRR, and possibly damaging the loudspeaker. The chart
below demonstrates this problem by showing the effects of
differing values between the input resistors while assuming
that the feedback resistors are perfectly matched. The re-
sults below apply to the application circuit shown in Figure 1,
and assumes that V
DC coupled inputs tied to ground.
Similar results would occur if the feedback resistors were not
carefully matched. Adding input coupling resistors in be-
tween the signal source and the input resistors will eliminate
this problem, however. To achieve best performance with
minimum component count, it is highly recommended that
both the feedback and input resistors matched to 1% toler-
ance or better for best performance.
AUDIO POWER AMPLIFIER DESIGN
Design a 1W/8Ω Audio Amplifier
Tolerance
Given:
Impedance
20%
10%
Power Output
Load Impedance
Maximum Input Level
Maximum Input
Bandwidth
5%
1%
0%
0.95R 1.05R
0.99R 1.01R
i1
0.8R
0.9R
R
R
and R
i1
1.2R
1.1R
DD
R
R
i2
i2
) and (R
= 5V, R
V
-0.500V
-0.250V
-0.125V
-0.025V
L
02
f1
= 8Ω, and the system has
100Hz–20kHz
0
- V
and R
01
f2
(Continued)
) to each other.
±
31.25mA
15.63mA
3.125mA
62.5mA
I
1Wrms
0.25dB
LOAD
1Vrms
20kΩ
0
8Ω
14
A designer must first determine the minimum supply rail to
obtain the specified output power. The supply rail can easily
be found by extrapolating from the Output Power vs Supply
Voltage graphs in the Typical Performance Characteris-
tics section. A second way to determine the minimum supply
rail is to calculate the required V
add the dropout voltages. Using this method, the minimum
supply voltage is (Vopeak + (V
BOT
vs Supply Voltage curve in the Typical Performance Char-
acteristics section.
Using the Output Power vs Supply Voltage graph for an 8Ω
load, the minimum supply rail just about 4.5V. Extra supply
voltage creates headroom that allows the LM4928 to repro-
duce peaks in excess of 1W without producing audible dis-
tortion. 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 Dissipa-
tion section. Once the power dissipation equations have
been addressed, the required differential gain can be deter-
mined from Equation 8.
From Equation 8, the minimum A
a ratio of R
step is to address the bandwidth requirement which must be
stated as a single -3dB frequency point. Five times away
from a -3dB point is 0.17dB down from passband response
which is better than the required
The high frequency pole is determined by the product of the
desired frequency pole, f
With a A
283kHz which is much smaller than the LM4928 GBWP of
10MHz. This figure displays that if a designer has a need to
design an amplifier with a higher differential gain, the
LM4928 can still be used without running into bandwidth
limitations.
and V
VD
DO TOP
f
= 2.83 and f
to R
i
f
are extrapolated from the Dropout Voltage
H
of 2.83 gives R
= 20kHz * 5 = 100kHz
R
H
f
H
/ R
= 100kHz, the resulting GBWP =
, and the differential gain, A
i
DO TOP
= A
OPEAK
VD
±
i
VD
= 14kΩ. The final design
0.25dB specified.
is 2.83. With R
+ V
using Equation 7 and
DO BOT
), where V
f
= 40kΩ,
VD
(7)
(8)
DO
.