Datasheet

LM4923, LM4923LQBD
SNAS211E –JULY 2004–REVISED MAY 2013 www.ti.com
Tolerance R
F1
R
F2
V
02
- V
01
I
LOAD
20% 0.8R 1.2R -0.500V 62.5mA
10% 0.9R 1.1R -0.250V 31.25mA
5% 0.95R 1.05R -0.125V 15.63mA
1% 0.99R 1.01R -0.025V 3.125mA
0% R R 0 0
Similar results would occur if the input resistors were not carefully matched. Adding input coupling capacitors in
between 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% tolerance or better.
AUDIO POWER AMPLIFIER DESIGN
Design a 1W/8Ω Audio Amplifier
Given:
Power Output 1Wrms
Load Impedance 8Ω
Input Level 1Vrms
Input Impedance 20kΩ
Bandwidth 100Hz–20kHz ± 0.25dB
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
Characteristics section. A second way to determine the minimum supply rail is to calculate the required V
OPEAK
using Equation 7 and add the dropout voltages. Using this method, the minimum supply voltage is (Vopeak +
(V
DO TOP
+ (V
DO BOT
)), where V
DO BOT
and V
DO TOP
are extrapolated from the Dropout Voltage vs Supply Voltage
curve in the Typical Performance Characteristics section.
(7)
Using the Output Power vs Supply Voltage graph for an 8Ω load, the minimum supply rail just about 5V. Extra
supply voltage creates headroom that allows the LM4923 to reproduce peaks in excess of 1W without producing
audible distortion. 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 Dissipation section. Once the power
dissipation equations have been addressed, the required differential gain can be determined from Equation 8.
(8)
R
f
/ R
i
= A
VD
(9)
From Equation 8, the minimum A
VD
is 2.83. Since the desired input impedance was 20kΩ, a ratio of 2.83:1 of R
f
to R
i
results in an allocation of R
i
= 20kΩ for both input resistors and R
f
= 60kΩ for both feedback resistors. The
final design 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
±0.25dB specified.
f
H
= 20kHz * 5 = 100kHz (10)
The high frequency pole is determined by the product of the desired frequency pole, f
H
, and the differential gain,
A
VD
. With a A
VD
= 2.83 and f
H
= 100kHz, the resulting GBWP = 150kHz which is much smaller than the LM4923
GBWP of 10MHz. This figure displays that if a designer has a need to design an amplifier with a higher
differential gain, the LM4923 can still be used without running into bandwidth limitations.
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