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參數(shù)資料
| 型號： | TC4864EUA |
| 元件分類： | 音頻／視頻放大 |
| 英文描述： | 0.74 W, 1 CHANNEL, AUDIO AMPLIFIER, PDSO8 |
| 封裝： | MSOP-8 |
| 文件頁數(shù)： | 15/18頁 |
| 文件大小： | 648K |
| 代理商： | TC4864EUA |
TC4864
6
TC4864-1 12/20/00
300mW Audio Power Amplifier with Shutdown Mode
2001 Microchip Technology Inc.
DS21483A
AUDIO POWER AMPLIFIER DESIGN
Design a 300mW/8
Audio Amplifier
Given:
Power Output
300mW
Load Impedance
8
Input Level
1 VRMS
Input Impedance
20k
Bandwidth
100Hz – 20kHz
± 0.25dB
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 Performance Characteristics section, the supply
rail can easily be found. Another way to determine the
minimum supply rail is to calculate the required VOPEAK
using Equation 4 and add the dropout voltage. Using this
method, the minimum supply voltage would be (VOPEAK
+(2*VOD )), where VOD is extrapolated from the Dropout
Voltage vs Supply Voltage curve in the Typical Perfor-
mance Characteristics section.
VOPEAK = √(2RLPO)
Equation 4.
Using the Output Power vs Supply Voltage graph for an
8
load, the minimum supply rail is 3.5V. But since 5V is a
standard supply voltage in most applications, it is chosen for
the supply rail. Extra supply voltage creates a buffer that
allows the TC4864 to reproduce peaks in excess of
500mW without producing audible distortion. At this point,
the designer must ensure that the power supply choice and
the output impedance does not violate the conditions set
forth in the Power Dissipation section.
Once the power dissipation equations have been ad-
dressed, the required differential gain can be determined
from Equation 5.
RF/Ri = AVD/2
Equation 5.
From Equation 5, the minimum AVD is 1.55; use
AVD = 2. Since the desired input impedance was 20k, and
with a AVD of 2, a ratio of 1:1 of RF to Ri results in an allocation
of Ri = RF = 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 pole gives
0.17dB down from passband response which is better than
the required
±0.25dB specified.
fL = 100Hz/5 = 20Hz
fH = 20kHz x 5 = 100kHz
As stated in the External Components section, Ri in
conjunction with Ci create a highpass filter.
Ci
≥
1
2
π Ri fC
Ci ≥ 1/(2π* 20k* 20Hz) = 0.398F; use 0.39F
The high frequency pole is determined by the product of
the desired high frequency pole, fH , and the differential gain,
AVD. With a AVD = 2 and fH = 100kHz, the resulting
GBWP = 100kHz which is much smaller than the TC4864
GBWP of 18MHz. This figure illustrates a situation in which
a designer needs to design an amplifier with a higher
differential gain. The TC4864 can still be used without
running into bandwidth problems.
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