參數(shù)資料
型號(hào): LM4919
廠商: National Semiconductor Corporation
元件分類: 音頻放大器
英文描述: 1.5V, Mono 85mW BTL Output, 14mW Stereo Headphone Audio Amplifier
中文描述: 1.5V的,單聲道85mW BTL輸出,消耗14mW功率立體聲耳機(jī)音頻放大器
文件頁(yè)數(shù): 14/15頁(yè)
文件大小: 568K
代理商: LM4919
Application Information
(Continued)
for the supply rail. Extra supply voltage creates headroom
that allows the LM4919 to reproduce peak in excess of
10mW 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
gain can be determined from Equation 2.
(4)
From Equation 4, the minimumAV is 1; useA
V
= 1. Since the
desired input impedance is 20k, and with a A
gain of 1, a
ratio of 1:1 results from Equation 1 for R
to R. The values
are chosen with R
= 20k and R
= 20k. The final design step
is to address the bandwidth requirements which must be
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
±
0.25dB specified.
f
L
= 100Hz/5 = 20Hz
f
H
= 20kHz * 5 = 100kHz
As stated in the
External Components
section, R
i
in con-
junction with C
i
creates a
C
i
1 / (2
π
* 20k
* 20Hz) = 0.397μF; use 0.39μF.
The high frequency pole is determined by the product of the
desired frequency pole, fH, and the differential gain,A
. With
an AV
= 1 and f
= 100kHz, the resulting GBWP = 100kHz
which is much smaller than the LM4919 GBWP of 3MHz.
This example displays that if a designer has a need to design
an amplifier with higher differential gain, the LM4919 can still
be used without running into bandwidth limitations.
L
www.national.com
14
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