參數(shù)資料
型號(hào): LM4863N
廠商: NATIONAL SEMICONDUCTOR CORP
元件分類(lèi): 音頻/視頻放大
英文描述: Dual 2.2W Audio Amplifier Plus Stereo Headphone Function
中文描述: 1.5 W, 2 CHANNEL, AUDIO AMPLIFIER, PDIP16
封裝: 0.300 INCH, MDIP-16
文件頁(yè)數(shù): 11/16頁(yè)
文件大?。?/td> 513K
代理商: LM4863N
Application Information
(Continued)
NO-LOAD DESIGN CONSIDERATIONS
If the outputs of the LM4863 have a load higher than 10k
,
the LM4863 may show a small oscillation at high output lev-
els. To prevent this oscillation, place 5k
resistors from the
power outputs to ground.
AUDIO POWER AMPLIFIER DESIGN
Design a 1W/8
Bridged Audio Amplifier
Given:
Power Output:
Load Impedance:
Input Level:
Input Impedance:
Bandwidth:
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
using Equation 3
and add the dropout voltage. Using this method, the mini-
mum supply voltage would be (V
+ (2
*
V
)), where V
od
is extrapolated from the Dropout Voltage vs Supply Voltage
curve in the
Typical Performance Characteristics
section.
1 Wrms
8
1 Vrms
20 k
100 Hz20 kHz
±
0.25 dB
(4)
Using the Output Power vs Supply Voltage graph for an 8
load, the minimum supply rail is 3.9V. But since 5V is a stan-
dard supply voltage in most applications, it is chosen for the
supply rail. Extra supply voltage creates headroom that al-
lows the LM4863 to reproduce peaks in excess of 1W with-
out 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 Equa-
tion 4.
(5)
(6)
R
f
/R
i
=A
VD
/2
From equation 4, the minimum A
VD
is 2.83;
Since the desired input impedance was 20 k
, and with a
A
VD
of 3, a ratio of 1.5:1 of R
to R
results in an allocation of
R
= 20 k
and R
= 30 k
. The final design step is to ad-
dress the 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
±
0.25 dB specified.
f
L
= 100 Hz/5 = 20 Hz
f
H
= 20 kHz x 5 = 100 kHz
As stated in the
External Components
section, R
i
in con-
junction with C
i
create a highpass filter.
use A
VD
= 3
C
i
1/(2
π
*
20 k
*
20 Hz) = 0.397 μF;
The high frequency pole is determined by the product of the
desired high frequency pole, f
, and the differential gain, A
VD
. With a A
= 3 and f
= 100 kHz, the resulting GBWP =
150 kHz which is much smaller than the LM4863 GBWP of
3.5 MHz. This figure displays that if a designer has a need to
design an amplifier with a higher differential gain, the
LM4863 can still be used without running into bandwidth
problems.
use 0.33 μF
DEMOBOARD CIRCUIT LAYOUT
The demoboard circuit layout is provided here as an ex-
ample of a circuit using the LM4863. If an LM4863MTE is
used with this layout, the exposed-DAP is soldered down to
the copper pad beneath the part. Heat is conducted away
from the part by the two large copper pads in the upper cor-
ners of the demoboard.
This demoboard provides enough heat dissipation ability to
allow an LM4863MTE to output 2.2W into 4
at 25C.
DS012881-94
All Layers
DS012881-93
Silk Screen Layer
L
www.national.com
11
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