參數(shù)資料
型號: LM4936
廠商: National Semiconductor Corporation
英文描述: Stereo 2W Audio Power Amplifiers with Volume Control and Selectable Control Interface (SPI or I2C)
中文描述: 2W的立體聲音頻功率放大器,帶有音量控制和可選控制接口(SPI或I2C)
文件頁數(shù): 23/28頁
文件大?。?/td> 1490K
代理商: LM4936
Application Information
(Continued)
AUDIO POWER AMPLIFIER DESIGN
Audio Amplifier Design: Driving 1W into an 8
Load
The following are the desired operational parameters:
Power Output:
Load Impedance:
Input Level:
Input Impedance:
Bandwidth:
1 W
RMS
8
1 V
RMS
20 k
100 Hz20 kHz
±
0.25 dB
The design begins by specifying the minimum supply voltage
necessary to obtain the specified output power. One way to
find the minimum supply voltage is to use the Output Power
vs Supply Voltage curve in the
Typical Performance Char-
acteristics
section. Another way, using Equation (10), is to
calculate the peak output voltage necessary to achieve the
desired output power for a given load impedance. To ac-
count for the amplifier’s dropout voltage, two additional volt-
ages, based on the Dropout Voltage vs Supply Voltage in the
Typical Performance Characteristics
curves, must be
added to the result obtained by Equation (10). The result is
Equation (11).
(11)
V
DD
(V
OUTPEAK
+ (V
OD
TOP
+ V
OD
BOT
))
(12)
The Output Power vs Supply Voltage graph for an 8
load
indicates a minimum supply voltage of 4.6V. This is easily
met by the commonly used 5V supply voltage. The additional
voltage creates the benefit of headroom, allowing the
LM4936 to produce peak output power in excess of 1W
without clipping or other audible distortion. The choice of
supply voltage must also not create a situation that violates
of maximum power dissipation as explained above in the
Power Dissipation
section.
After satisfying the LM4936’s power dissipation require-
ments, the minimum differential gain needed to achieve 1W
dissipation in an 8
load is found using Equation (12).
(13)
Thus, a minimum overall gain of 2.83 allows the LM4936’s to
reach full output swing and maintain low noise and THD+N
performance.
The last step in this design example is setting the amplifier’s
6dB frequency bandwidth. To achieve the desired
±
0.25dB
pass band magnitude variation limit, the low frequency re-
sponse must extend to at least one-fifth the lower bandwidth
limit and the high frequency response must extend to at least
five times the upper bandwidth limit. The gain variation for
both response limits is 0.17dB, well within the
±
0.25dB
desired limit. The results are an
f
L
= 100Hz/5 = 20Hz
(14)
and an
f
H
= 20kHz x 5 = 100kHz
(15)
As mentioned in the
Selecting Proper External Compo-
nents
section, R
(Right & Left) and C
(Right & Left) create
a highpass filter that sets the amplifier’s lower bandpass
frequency limit. Find the input coupling capacitor’s value
using Equation (14).
C
i
1/(2
π
R
i
f
L
)
(16)
The result is
1/(2
π
*20k
*20Hz) = 0.397μF
(17)
Use a 0.39μF capacitor, the closest standard value.
The product of the desired high frequency cutoff (100kHz in
this example) and the differential gain A
VD
, determines the
upper passband response limit. With A
= 3 and f
=
100kHz, the closed-loop gain bandwidth product (GBWP) is
300kHz. This is less than the LM4936’s 3.5MHz GBWP. With
this margin, the amplifier can be used in designs that require
more differential gain while avoiding performance,restricting
bandwidth limitations.
L
www.national.com
23
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