LM4766TF National Semiconductor, LM4766TF Datasheet - Page 17

• Manufacturer Part Number: LM4766TF
• Description: IC,Audio Amplifier,DUAL,BIPOLAR,ZIP,15PIN,PLASTIC
• Manufacturer: National Semiconductor

Specifications of LM4766TF

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

Application Information
AUDIO POWER AMPLlFIER DESIGN
Design a 30W/8Ω Audio Amplifier
A designer must first determine the power supply require-
ments in terms of both voltage and current needed to obtain
the specified output power. V
Equation (4) and I
To determine the maximum supply voltage the following
conditions must be considered. Add the dropout voltage to
the peak output swing V
current of I
the unloaded voltage which is usually about 15% higher. The
supply voltage will also rise 10% during high line conditions.
Therefore the maximum supply voltage is obtained from the
following equation.
For 30W of output power into an 8Ω load, the required
V
from adding V
supplies are
Equation (5). It should be noted that for a dual 30W amplifier
Given:
Power Output
Load Impedance
Input Level
Input Impedance
Bandwidth
OPEAK
Max supplies ≈
is 21.91V. A minimum supply rail of 25.4V results
OPEAK
±
OPEAK
32V and the required I
. The regulation of the supply determines
±
OPEAK
(V
and V
OPEAK
from Equation (5).
OPEAK
OD
OPEAK
. With regulation, the maximum
+ V
, to get the supply rail at a
OD
) (1 + regulation) (1.1)
can be determined from
OPEAK
(Continued)
20Hz−20kHz
is 2.74A from
1Vrms(max)
±
30Wrms
0.25dB
47kΩ
8Ω
(4)
(5)
17
into an 8Ω load the I
2.74A
the Power Output vs Supply Voltage to ensure that the
required output power is obtainable from the device while
maintaining low THD+N. In addition, the designer should
verify that with the required power supply voltage and load
impedance, that the required heatsink value θ
given system cost and size constraints. Once the heatsink
issues have been addressed, the required gain can be de-
termined from Equation (6).
From Equation (6), the minimum A
By selecting a gain of 21, and with a feedback resistor, R
20kΩ, the value of R
Thus with R
Since the desired input impedance was 47kΩ, a value of
47kΩ was selected for R
address the bandwidth requirements which must be stated
as a pair of −3dB frequency points. Five times away from a
−3dB point is 0.17dB down from passband response which
is better than the required
sults in a low and high frequency pole of 4Hz and 100kHz
respectively. As stated in the External Components sec-
tion, R
The high frequency pole is determined by the product of the
desired high frequency pole, f
A
which is less than the guaranteed minimum GBWP of the
LM4766 of 8MHz. This will ensure that the high frequency
response of the amplifier will be no worse than 0.17dB down
at 20kHz which is well within the bandwidth requirements of
the design.
V
= 21 and f
C
PK
i
i
in conjunction with C
≥ 1/(2π * 1kΩ * 4Hz) = 39.8µF;
or 5.48A
i
= 1kΩ a non-inverting gain of 21 will result.
H
= 100kHz, the resulting GBWP is 2.1MHz,
PK
. At this point it is a good idea to check
OPEAK
i
R
follows from Equation (7).
i
= R
IN
drawn from the supplies is twice
±
f
i
. The final design step is to
0.25dB specified. This fact re-
(A
create a high-pass filter.
H
V
, and the gain, A
− 1)
V
is: A
V
use 39µF.
≥ 15.5.
SA
www.national.com
is feasible
V
. With a
(6)
(7)
f
=