Datasheet

LM4938
www.ti.com SNAS245B –FEBRUARY 2005–REVISED MAY 2013
AUDIO POWER AMPLIFIER DESIGN
Audio Amplifier Design: Driving 1W into an 8Ω Load
The following are the desired operational parameters:
Power Output: 1 W
RMS
Load Impedance: 8Ω
Input Level: 1 V
RMS
Input Impedance: 20 kΩ
Bandwidth: 100 Hz−20 kHz ± 0.25 dB
The design begins by specifying the minimum supply voltage necessary to obtain the specified output power.
One way to find the minimum supply voltage is to use the Output Power vs Supply Voltage curve in the Typical
Performance Characteristics section. Another way, using Equation 10, is to calculate the peak output voltage
necessary to achieve the desired output power for a given load impedance. To account for the amplifier's dropout
voltage, two additional voltages, based on the Dropout Voltage vs Supply Voltage in the Typical Performance
Characteristics curves, must be added to the result obtained by Equation 10. The result is Equation 11.
(11)
V
DD
≥ (V
OUTPEAK
+ (V
OD
TOP
+ V
OD
BOT
)) (12)
The Output Power vs Supply Voltage graph for an 8Ω load indicates a minimum supply voltage of 4.6V. This is
easily met by the commonly used 5V supply voltage. The additional voltage creates the benefit of headroom,
allowing the LM4938 to produce peak output power in excess of 1W without clipping or other audible distortion.
The choice of supply voltage must also not create a situation that violates of maximum power dissipation as
explained above in the POWER DISSIPATION section.
After satisfying the LM4938's power dissipation requirements, the minimum differential gain needed to achieve
1W dissipation in an 8Ω load is found using Equation 12.
(13)
Thus, a minimum overall gain of 2.83 allows the LM4938's to reach full output swing and maintain low noise and
THD+N performance.
The last step in this design example is setting the amplifier's −6dB frequency bandwidth. To achieve the desired
±0.25dB pass band magnitude variation limit, the low frequency response must extend to at least one-fifth the
lower bandwidth limit and the high frequency response must extend to at least five times the upper bandwidth
limit. The gain variation for both response limits is 0.17dB, well within the ±0.25dB desired limit. The results are
an
f
L
= 100Hz/5 = 20Hz (14)
and an
f
H
= 20kHz x 5 = 100kHz (15)
As mentioned in the SELECTING PROPER EXTERNAL COMPONENTS section, R
i
(Right & Left) and C
i
(Right
& Left) create a highpass filter that sets the amplifier's lower bandpass frequency limit. Find the input coupling
capacitor's value using Equation 14.
C
i
≥ 1/(2πR
i
f
L
) (16)
The result is
1/(2π*20kΩ*20Hz) = 0.397μF (17)
Use a 0.39μF capacitor, the closest standard value.
The product of the desired high frequency cutoff (100kHz in this example) and the differential gain A
VD
,
determines the upper passband response limit. With A
VD
= 3 and f
H
= 100kHz, the closed-loop gain bandwidth
product (GBWP) is 300kHz. This is less than the LM4938's 3.5MHz GBWP. With this margin, the amplifier can
be used in designs that require more differential gain while avoiding performance,restricting bandwidth
limitations.
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