lm4940tsx National Semiconductor Corporation, lm4940tsx Datasheet - Page 11

• Manufacturer Part Number: lm4940tsx
• Description: 6w Streo Audio Power Amplifier
• Manufacturer: National Semiconductor Corporation

Application Information
In order eliminate "clicks and pops", all capacitors must be
discharged before turn-on. Rapidly switching V
allow the capacitors to fully discharge, which may cause
"clicks and pops".
There is a relationship between the value of C
C
is applied or the shutdown mode is deactivated. Best perfor-
mance is achieved by setting the time constant created by
C
given value of C
AUDIO POWER AMPLIFIER DESIGN
Audio Amplifier Design: Driving 3W into a 4Ω load
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 Power Supply Voltage curve in the Typical Performance
Characteristics section. Another way, using Equation (8), 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 Clipping Dropout Voltage vs Power
Supply Voltage in the Typical Performance Characteris-
tics curves, must be added to the result obtained by Equa-
tion (8). The result is Equation (9).
The Output Power vs. Power Supply Voltage graph for an 8Ω
load indicates a minimum supply voltage of 11.8V. The com-
monly used 12V supply voltage easily meets this. The addi-
tional voltage creates the benefit of headroom, allowing the
LM4940 to produce an output power of 3W 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 LM4940’s power dissipation re-
quirements, the minimum differential gain needed to achieve
3W dissipation in a 4Ω BTL load is found using Equation
(10).
The following are the desired operational parameters:
Power Output
Load Impedance
Input Level
Input Impedance
Bandwidth
BYPASS
IN
and R
that ensures minimum output transient when power
i
C
V
+ R
B
DD
1.0
2.2
4.7
10
(µF)
f
= V
BYPASS
to a value less than the turn-on time for a
OUTPEAK
as shown in the table above.
+ V
ODTOP
100Hz–20kHz
+ V
T
ON
(Continued)
120
120
200
440
ODBOT
0.3V
(ms)
DD
RMS
±
may not
3W
0.25dB
IN
(max)
20kΩ
RMS
4Ω
and
(6)
(7)
11
Thus, a minimum gain of 11.6 allows the LM4940’s to reach
full output swing and maintain low noise and THD+N perfor-
mance. For this example, let A
BTL gain is set using the input (RIN
resistors of the first amplifier in the series BTL configuration.
Additionaly, A
R
the feedback resistor is found using Equation (11).
The value of R
The last step in this design example is setting the amplifier’s
-3dB 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
and an
As mentioned in the SELECTING EXTERNAL COMPO-
NENTS section, R
create a highpass filter that sets the amplifier’s lower band-
pass frequency limit. Find the coupling capacitor’s value
using Equation (14).
The result is
and
Use a 0.39µF capacitor for C
C
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
1.2mHz. This is less than the LM4940’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.
IN
OUT
and R
, the closest standard values.
f
. With the desired input impedance set at 20kΩ,
1 / (2πx20kΩx20Hz) = 0.398µF = C
1 / (2πx4Ωx20Hz) = 1989µF = C
V-BTL
f
is 240kΩ. The nominal output power is 3W.
f
L
f
is twice the gain set by the first amplifier’s
INA
L
= 20kHz x 5 = 100kHz
= 100Hz / 5 = 20Hz
C
and C
IN
R
f
= 1 / 2πR
/ R
IN
INA
IN
V
= A
= 12. The amplifier’s overall
, as well as C
and a 2000µF capacitor for
IN
V
f
L
A
) and feedback (R)
V
V
, determines the
= 12 and f
OUT
IN
OUT
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±
±
and R
0.25dB-
0.25dB
(10)
(12)
(11)
H
(8)
(9)
L
=
,