參數(shù)資料
型號(hào): TPA0211DGN
廠商: TEXAS INSTRUMENTS INC
元件分類: 音頻/視頻放大
英文描述: 2 W, 1 CHANNEL, AUDIO AMPLIFIER, PDSO8
封裝: GREEN, PLASTIC, MSOP-8
文件頁數(shù): 5/27頁
文件大?。?/td> 770K
代理商: TPA0211DGN
TPA0211
2-W MONO AUDIO POWER AMPLIFIER
SLOS275D – JANUARY 2000 – REVISED NOVEMBER 2002
13
POST OFFICE BOX 655303
DALLAS, TEXAS 75265
APPLICATION INFORMATION
Table 1. Common Load Impedances vs Low Frequency Output Characteristics in SE Mode
RL
C(C)
Lowest Frequency
3
330
F
161 Hz
4
330
F
120 Hz
8
330
F
60 Hz
32
330
F
15 Hz
10,000
330
F
0.05 Hz
47,000
330
F
0.01 Hz
As Table 1 indicates, most of the bass response is attenuated into a 4-
load, an 8- load is adequate,
headphone response is good, and drive into line level inputs (a home stereo for example) is exceptional.
Furthermore, the total amount of ripple current that must flow through the capacitor must be considered when
choosing the component. As shown in the application circuit, one coupling capacitor must be in series with the
mono loudspeaker for proper operation of the stereo-mono switching circuit. For a 4-
load, this capacitor must
be able to handle about 700 mA of ripple current for a continuous output power of 2 W.
using low-ESR capacitors
Low-ESR capacitors are recommended throughout this applications section. A real (as opposed to ideal)
capacitor can be modeled simply as a resistor in series with an ideal capacitor. The voltage drop across this
resistor minimizes the beneficial effects of the capacitor in the circuit. The lower the equivalent value of this
resistance the more the real capacitor behaves like an ideal capacitor.
bridged-tied load versus single-ended mode
Figure 22 shows a Class-AB audio power amplifier (APA) in a BTL configuration. The TPA0211 BTL amplifier
consists of two Class-AB amplifiers driving both ends of the load. There are several potential benefits to this
differential drive configuration, but initially consider power to the load. The differential drive to the speaker
means that as one side is slewing up, the other side is slewing down, and vice versa. This, in effect, doubles
the voltage swing on the load as compared to a ground referenced load. Plugging 2
× VO(PP) into the power
equation, where voltage is squared, yields 4
× the output power from the same supply rail and load impedance.
(See equation 7.)
Power
+
V
(RMS)
2
R
L
(7)
V
(RMS
) +
V
O(PP)
22
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