LM4991

SNAS217A – MAY 2004 – REVISED APRIL 2013

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AUDIO POWER AMPLIFIER DESIGN

Design a 1W/8

Ω Audio Amplifier

Given:

Power Output

1 Wrms

Load Impedance

8

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 theTypical Performance Characteristics section, the supply

rail can be easily found. A second way to determine the minimum supply rail is to calculate the required Vopeak

using Equation (3) and add the output voltage. Using this method, the minimum supply voltage would be (Vopeak

+ (VODTOP + VODBOT)), where VODBOT and VODTOP are extrapolated from the Dropout Voltage vs Supply Voltage curve

in the Typical Performance Characteristics section.

(4)

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 standard voltage in most applications, it is chosen for the supply rail. Extra supply voltage creates headroom

that allows the LM4991 to reproduce peaks in excess of 1W without producing audible distortion. At this time, the

designer must make sure that the power supply choice along with the output impedance 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 Equation (4).

(5)

Rf/Ri = AVD/2

(6)

From Equation (4), the minimum AVD is 2.83; use AVD = 3.

Since the desired input impedance was 20k

Ω, and with a AVD impedance of 2, a ratio of 1.5:1 of Rf to Ri results

in an allocation of Ri = 20kΩ and Rf = 30kΩ. 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 −3dB point is 0.17dB down

from passband response which is better than the required ±0.25dB specified.

fL = 100Hz/5 = 20Hz

fH = 20kHz * 5 = 100kHz

As stated in the External Components Description section, Ri in conjunction with Ci create a highpass filter.

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

The high frequency pole is determined by the product of the desired frequency pole, fH, and the differential gain,

AVD. With a AVD = 3 and fH = 100kHz, the resulting GBWP = 150kHz which is much smaller than the LM4991

GBWP of 4MHz. This figure displays that if a designer has a need to design an amplifier with a higher differential

gain, the LM4991 can still be used without running into bandwidth limitations.

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