LM4894 datasheet(14/19 Pages) NSC

14 / 19 page

Application Information (Continued)

Tolerance R

02 -V01

LOAD

20%

0.8R

1.2R

-0.500V

62.5mA

10%

0.9R

1.1R

-0.250V

31.25mA

0.95R 1.05R -0.125V

15.63mA

0.99R 1.01R -0.025V

3.125mA

RR0

Similar results would occur if the input resistors were not

carefully matched. Adding input coupling capacitors in be-

tween the signal source and the input resistors will eliminate

this problem, however, to achieve best performance with

minimum component count it is highly recommended that

both the feedback and input resistors matched to 1% toler-

ance or better.

AUDIO POWER AMPLIFIER DESIGN

Design a 1W/8

Ω Audio Amplifier

Given:

Power Output

1Wrms

Load Impedance

Ω

Input Level

1Vrms

Input Impedance

20k

Ω

Bandwidth

–20kHz ± 0.25dB

A designer must first determine the minimum supply rail to

obtain the specified output power. The supply rail can easily

be found by extrapolating from the Output Power vs Supply

Voltage graphs in the Typical Performance Characteris-

tics section. A second way to determine the minimum supply

rail is to calculate the required VOPEAK using Equation 7

and add the dropout voltages. Using this method, the mini-

mum supply voltage is (Vopeak +(V

DO TOP+(VDO BOT )),

where V

DO BOT and VDO TOP are extrapolated from the

Dropout Voltage vs Supply Voltage curve in the Typical

Performance Characteristics section.

(7)

Using the Output Power vs Supply Voltage graph for an 8W

load, the minimum supply rail just about 5V. Extra supply

voltage creates headroom that allows the LM4894 to repro-

duce peaks in excess of 1W without producing audible dis-

tortion. 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 Dissipa-

tion section. Once the power dissipation equations have

been addressed, the required differential gain can be deter-

mined from Equation 7.

(8)

f /Ri =AVD

From Equation 7, the minimum A

VD is 2.83. Since the de-

sired input impedance was 20k

Ω, a ratio of 2.83:1 of R

f to Ri

results in an allocation of R

i = 20k

Ω for both input resistors

and R

f= 60k

Ω for both feedback resistors. The final design

step is to address the bandwidth requirement which must be

stated as a single -3dB frequency point. Five times away

from a -3dB point is 0.17dB down from passband response

which is better than the required ±0.25dB specified.

H = 20kHz * 5 =100kHz

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 = 2.83 and fH = 100kHz, the resulting GBWP =

150kHz which is much smaller than the LM4894 GBWP of

10MHz. This figure displays that if a designer has a need to

design an amplifier with a higher differential gain, the

LM4894 can still be used without running into bandwidth

limitations.

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