Datasheet

LM4808
www.ti.com SNAS051D –FEBRUARY 2000–REVISED MAY 2013
AUDIO POWER AMPLIFIER DESIGN
Design a Dual 70mW/32Ω Audio Amplifier
Given:
Power Output 70mW
Load Impedance 32Ω
Input Level 1Vrms (max)
Input Impedance 20kΩ
Bandwidth 100Hz–20kHz ± 0.50dB
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 5, 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 5. For a single-
ended application, the result is Equation 6.
(5)
V
DD
≥ (2V
OPEAK
+ (V
OD
TOP
+ V
OD
BOT
)) (6)
The Output Power vs Supply Voltage graph for a 32Ω load indicates a minimum supply voltage of 4.8V. This is
easily met by the commonly used 5V supply voltage. The additional voltage creates the benefit of headroom,
allowing the LM4808 to produce peak output power in excess of 70mW without clipping or other audible
distortion. The choice of supply voltage must also not create a situation that violates maximum power dissipation
as explained above in the POWER DISSIPATION section. Remember that the maximum power dissipation point
from Equation 1 must be multiplied by two since there are two independent amplifiers inside the package. Once
the power dissipation equations have been addressed, the required gain can be determined from Equation 7.
(7)
Thus, a minimum gain of 1.497 allows the LM4808 to reach full output swing and maintain low noise and THD+N
perfromance. For this example, let A
V
=1.5.
The amplifiers overall gain is set using the input (R
i
) and feedback (R
f
) resistors. With the desired input
impedance set at 20kΩ, the feedback resistor is found using Equation 8.
A
V
= R
f
/R
i
(8)
The value of R
f
is 30kΩ.
The last step in this design is setting the amplifier's −3db frequency bandwidth. To achieve the desired ±0.25dB
pass band magnitude variation limit, the low frequency response must extend to at lease 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 (9)
and a
f
H
= 20kHz
∗
5 = 100kHz (10)
As stated in the EXTERNAL COMPONENTS DESCRIPTION section, both R
i
in conjunction with C
i
, and C
o
with
R
L
, create first order highpass filters. Thus to obtain the desired low frequency response of 100Hz within ±0.5dB,
both poles must be taken into consideration. The combination of two single order filters at the same frequency
forms a second order response. This results in a signal which is down 0.34dB at five times away from the single
order filter −3dB point. Thus, a frequency of 20Hz is used in the following equations to ensure that the response
is better than 0.5dB down at 100Hz.
C
i
≥ 1 / (2π * 20 kΩ * 20 Hz) = 0.397µF; use 0.39µF.
C
o
≥ 1 / (2π * 32Ω * 20 Hz) = 249µF; use 330µF.
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