參數(shù)資料
型號: LM4881MDA
廠商: NATIONAL SEMICONDUCTOR CORP
元件分類: 音頻/視頻放大
英文描述: 0.3 W, 2 CHANNEL, AUDIO AMPLIFIER, UUC
封裝: DIE
文件頁數(shù): 14/14頁
文件大小: 473K
代理商: LM4881MDA
Application Information (Continued)
shutdown function should be virtually clickless and popless.
While the device will function properly, (no oscillations or
motorboating), with C
B equal to 0.1 F, the device will be
much more susceptible to turn on clicks and pops. Thus, a
value of C
B equal to 0.1 F or larger is recommended in all
but the most cost sensitive designs.
AUDIO POWER AMPLIFIER DESIGN
Design a Dual 200mW/8
Audio Amplifier
Given:
Power Output
200 mWrms
Load Impedance
8
Input Level
1 Vrms (max)
Input Impedance
20 k
Bandwidth
100 Hz–20 kHz ± 0.50 dB
A designer must first determine the needed supply rail to
obtain the specified output power. Calculating the required
supply rail involves knowing two parameters, V
OPEAK and
also the dropout voltage. The latter is typically 530 mV and
can be found from the graphs in the Typical Performance
Characteristics. V
OPEAK can be determined from Equation
3.
(3)
For 200 mW of output power into an 8
load, the required
V
OPEAK is 1.79 volts. A minimum supply rail of 2.32V results
from adding V
OPEAK and VOD. 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
LM4881 to reproduce peaks in excess of 200 mW without
clipping 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 =Rf/Ri
(5)
From Equation 4, the minimum gain is: A
V = 1.26
Since the desired input impedance was 20 k
, and with a
gain of 1.26, a value of 27 k
is designated for R
f, 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 3 dB point is
0.17 dB down from passband response assuming a single
pole roll-off. As stated in the External Components section,
both R
i in conjunction with C i, and Co with RL, create first
order highpass filters. Thus to obtain the desired frequency
low response of 100 Hz within ±0.5 dB, 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 0.34 dB at five times
away from the single order filter 3 dB point. Thus, a fre-
quency of 20 Hz is used in the following equations to ensure
that the response is better than 0.5 dB down at 100 Hz.
C
i ≥ 1/(2π *20k * 20 Hz) = 0.397 F; use 0.39 F.
C
o ≥ 1/(2π *8 * 20 Hz) = 995 F; use 1000 F.
The high frequency pole is determined by the product of the
desired high frequency pole, f
H, and the closed-loop gain, A
V
. With a closed-loop gain of 1.35 and f
H = 100 kHz, the
resulting GBWP = 135 kHz which is much smaller than the
LM4881 GBWP of 18 MHz. This figure displays that if a
designer has a need to design an amplifier with a higher
gain, the LM4881 can still be used without running into
bandwidth limitations.
LM4881
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