參數(shù)資料
型號: LM4808
廠商: National Semiconductor Corporation
英文描述: Dual 105 mW Headphone Amplifier
中文描述: 雙105毫瓦耳機(jī)放大器
文件頁數(shù): 10/13頁
文件大?。?/td> 674K
代理商: LM4808
Application Information
(Continued)
For 70 mW of output power into a 32
load, the required V
O-
PEAK
is 2.12 volts. A minimum supply rail of 2.42V results
from adding V
and V
. Since 5V is a standard supply
voltage in most applications, it is chosen for the supply rail.
Extra supply voltage creates headroom that allows the
LM4808 to reproduce peaks in excess of 70 mW without clip-
ping the signal. At this time, the designer must make sure
that the power supply choice along with the output imped-
ance does not violate the conditions explained in the
Power
Dissipation
section. Remember that the maximum power
dissipation point from Equation 1 must be multiplied by two
since there are two independent amplifiers inside the pack-
age.
Once the power dissipation equations have been addressed,
the required gain can be determined from Equation 4.
(4)
A
V
= R
f
/R
i
(5)
From Equation 4, the minimum gain is: A
V
= 1.26
Since the desired input impedance was 20k
, and with a
gain of 1.26, a value of 27k
is designated for R
, assuming
5% tolerance resistors. This combination results in a nominal
gain of 1.35. The final design step is to address the band-
width requirements which must be stated as a pair of 3 dB
frequency points. Five times away from a 3dB point is
0.17dB down from passband response assuming a single
pole roll-off. As stated in the
External Components
section,
both R
in conjunction with C
, and C
with R
, create first or-
der highpass filters. Thus to obtain the desired frequency low
response of 100Hz within
±
0.5dB, both poles must be taken
into consideration. The combination of two single order filters
at the same frequency forms a second order response. This
results in a signal which is down 0.34dB at five times away
from the single order filter 3dB point. Thus, a frequency of
20Hz is used in the following equations to ensure that the re-
sponse is better than 0.5dB down at 100Hz.
C
i
1 / (2
π
* 20 k
* 20 Hz) = 0.397μF; use 0.39μF.
C
o
1 / (2
π
* 32
* 20 Hz) = 249μF; use 330μF.
The high frequency pole is determined by the product of the
desired high frequency pole, f
, and the closed-loop gain, A
V
. With a closed-loop gain of 1.35 and f
= 100kHz, the re-
sulting GBWP = 135kHz which is much smaller than the
LM4808 GBWP of 900kHz. This figure displays that if a de-
signer has a need to design an amplifier with a higher gain,
the LM4808 can still be used without running into bandwidth
limitations.
L
www.national.com
10
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