LM4876 datasheet(8/9 Pages) NSC

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Application Information (Continued)

AUDIO POWER AMPLIFIER DESIGN

Design a 1W/8

Ω Audio Amplifier

Given:

Power Output

1 Wrms

Load Impedance

Ω

Input Level

1 Vrms

Input Impedance

20 k

Ω

Bandwidth

100 Hz–20 kHz ± 0.25 dB

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 Per-

formance Characteristics section, the supply rail can be

easily found. A second way to determine the minimum sup-

ply rail is to calculate the required V

opeak using Equation 3

and add the output voltage. Using this method, the minimum

supply voltage would be (V

opeak +(VODTOP +VODBOT)), where

ODBOT and VODTOP are extrapolated from the Dropout Volt-

age vs Supply Voltage curve in the Typical Performance

Characteristics section.

(3)

Using the Output Power vs Supply Voltage graph for an 8

Ω

load, the minimum supply rail is 4.6V. But since 5V is a stan-

dard voltage in most applications, it is chosen for the supply

rail. Extra supply voltage creates headroom that allows the

LM4876 to reproduce peaks in excess of 1W without produc-

ing audible distortion. At this time, the designer must make

sure that the power supply choice along with the output im-

pedance does not violate the conditions explained in the

Power Dissipation section.

Once the power dissipation equations have been addressed,

the required differential gain can be determined from Equa-

tion 4.

(4)

f/Ri = AVD/2

(5)

From Equation 4, the minimum A

VD is 2.83; use AVD =3.

Since the desired input impedance was 20 k

Ω, and with a

VD impedance of 2, a ratio of 1.5:1 of Rf to Ri results in an

allocation of R

i =20kΩ and Rf =30kΩ. The final design step

is to address the bandwidth 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

which is better than the required ±0.25 dB specified.

L = 100 Hz/5 = 20 Hz

H =20kHz*5=100 kHz

As stated in the External Components section, R

i in con-

junction with C

i create a highpass filter.

i ≥ 1/(2π*20 kΩ*20 Hz) = 0.397 µF; use 0.39 µF

The high frequency pole is determined by the product of the

desired frequency pole, f

H, and the differential gain, AVD.

With a A

VD = 3 and fH = 100 kHz, the resulting GBWP =

150 kHz which is much smaller than the LM4876 GBWP of

4 MHz. This figure displays that if a designer has a need to

design an amplifier with a higher differential gain, the

LM4876 can still be used without running into bandwidth limi-

tations.

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