LM4936MH National Semiconductor, LM4936MH Datasheet - Page 23

• Manufacturer Part Number: LM4936MH
• Description: IC,Audio Amplifier,DUAL,TSSOP,28PIN,PLASTIC
• Manufacturer: National Semiconductor

Specifications of LM4936MH

• Lead Free Status / RoHS Status: Contains lead / RoHS non-compliant

Application Information
AUDIO POWER AMPLIFIER DESIGN
Audio Amplifier Design: Driving 1W into an 8Ω Load
The following are the desired operational parameters:
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 Char-
acteristics 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 ac-
count for the amplifier’s dropout voltage, two additional volt-
ages, 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).
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
LM4936 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 LM4936’s power dissipation require-
ments, the minimum differential gain needed to achieve 1W
dissipation in an 8Ω load is found using Equation (12).
Bandwidth:
Power Output:
Load Impedance:
Input Level:
Input Impedance:
V
DD
≥ (V
OUTPEAK
+ (V
100 Hz−20 kHz
OD TOP
+ V
(Continued)
OD BOT
))
±
1 W
0.25 dB
1 V
20 kΩ
RMS
RMS
(12)
(13)
8Ω
(11)
23
Thus, a minimum overall gain of 2.83 allows the LM4936’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
pass band magnitude variation limit, the low frequency re-
sponse 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
desired limit. The results are an
As mentioned in the Selecting Proper External Compo-
nents section, R
a highpass filter that sets the amplifier’s lower bandpass
frequency limit. Find the input coupling capacitor’s value
using Equation (14).
The result is
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
upper passband response limit. With A
100kHz, the closed-loop gain bandwidth product (GBWP) is
300kHz. This is less than the LM4936’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.
and an
1/(2π*20kΩ*20Hz) = 0.397µF
i
f
H
(Right & Left) and C
f
= 20kHz x 5 = 100kHz
L
= 100Hz/5 = 20Hz
C
i
≥ 1/(2πR
i
f
L
)
i
(Right & Left) create
VD
VD
, determines the
= 3 and f
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±
±
0.25dB
0.25dB
(14)
(15)
(16)
(17)
H
=