lm4651 National Semiconductor Corporation, lm4651 Datasheet - Page 15

• Manufacturer Part Number: lm4651
• Description: Overture? Audio Power Amplifier 170w Class D Audio Power Amplifier Solution
• Manufacturer: National Semiconductor Corporation

Application Information
C
R
A lower ratio can be used if the application is for lower output
voltages than the 125Watt, 4
The feedback RC filter’s pole location should be higher than
the output filter pole. The reason for two capacitors in paral-
lel instead of one larger capacitor is to reduce the possible
EMI from the feedback traces. C
sible to the output of the LM4652 so that an audio signal is
present on the feedback trace instead of a high frequency
square wave. C
feedback inputs (pins 14, 19) of the LM4651 to filter off any
noise picked up by the feedback traces. The combination
lowers EMI and provides a cleaner audio feedback signal to
the LM4651. R
controls the bandwidth of the error signal and should be in
the range of 100pF to 470pF.
Determine the Value for C
The start-up delay is chosen to be 1 second to ensure
minimum pops or clicks when the amplifier is powered up.
Using Equation (2), the value of C
value of 10µF is used.
Determine the Value of Gain, R
The gain is set to produce a 125W output at no more than
1% distortion with a 3V
4
signal, the LM4651/LM4652 amplifier needs an overall
closed-loop gain of 22.4V
Equation (12) shows all the variables that affect the system
gain.
The values for R
the Value of the Filters section above. Therefore, R
620k , R
also found as the first step in this example to be
Inserting these values into equation (12) and reducing gives
the equation below:
The input resistance is desired to be 20k
20k . R
Lowering R
ing R
little affect on the noise level. The percent change in noise is
about what whould be expected with a higher gain. The
drawback to a lower R
sary to properly couple the lowest desired signal frequen-
cies. If a 20k
recommended values shown in Figure 1 , Typical Audio
Application Circuit should be used: with R
4.7k
Determine the Needed Heat Sink
The only remaining design requirement is a thermal design
that prevents activating the thermal protection circuitry. Use
1
fl1
Gain = [(R
should be in the range of 0.1µF to 1µF or 2 - 20% of C
load requires a 22.4V
and R
1
and R
to increase gain with the lower value for R
2
fI2
is then found to be 14k .
fl2
2
= 62k
2
are found by the ratio R
direcly affects the noise of the system. Chang-
’s value set to 3.5k
2
/ R
fl2
f
fI1
input impedance is not required, then the
should be in range of 100k
1
is then placed as close as possible to the
, R
) x ((R
and R
fI2
, and R
1
RMS
value is a larger C
fl1
RMS
R
f
RMS
input. A dissipation of 125W in a
2
= 390k . The value of V
START
+ R
= .7R
signal. To produce this output
f
/3V
were found in the Determine
solution show here.
fl2
START
fl1
1
)/ R
RMS
(Start Up Delay)
, and R
1
for a gain 7.5V/V.
is placed close as pos-
fl1
fl2
, or 7.5V/V (17.5db).
) − (R
is 11.7µF. A standard
= 10R
(Continued)
2
IN
1
so R
2
’s value set to
/ R
fl2
value, neces-
.
1
to1M . C
2
1
) + .5]. (3)
has very
is set to
CC
±
fI1
20V.
BYP
was
(4)
=
.
f
15
Equations (9) - (11) to calculate the amount of power dissi-
pation for the LM4652. The appropriate heat sink size, or
thermal resistance in ˚C/W, will then be determined.
Equation (9) determines the percentage of loss caused by
the switching. Use the typical values given in the Electrical
Characteristics for the LM4651 and Electrical Character-
istics for the LM4652 tables for the rise time, fall time and
over modulation time:
This switching loss causes a maximum power dissipation,
using Equation (10), of:
Next the power dissipation caused by the R
output FETs is found by multiplying the output current times
the R
Electrical Characteristics for the LM4652 table above.
The value for R
lating the maximum power dissipation.
The total power dissipation in the LM4652 is the sum of
these two power losses giving:
The value for Maximum Power Dissipation given in the Sys-
tem Electrical Characteristics for the LM4651 and
LM4652 is 22 watts. The difference is due to approximately
1 watt of power loss in the LM4651. The above calculations
are for the power loss in the LM4652.
Lastly, use Equation (11) to determine the thermal resistance
of the LM4652’s heat sink. The values for
found in the Operating Ratings and the Absolute Maxi-
mum Ratings section above for the LM4652. The value of
non-isolated (T) package. The value for T
value for
value when thermal grease is used. The maximum ambient
temperature from the design requirements is 50˚. The value
of
and for the non-isolated (T) package without a mica washer
to isolate the heat sink from the package:
JC
SA
is 2˚C/W for the isolated (TF) package or 1˚C/W for the
DS(ON)
for the isolated (TF) package is:
SA
SA
I
OUTRMS
CS
. Again, the value for R
%Loss = (25ns+26ns+350ns)
= [(150˚C − 50˚C)/21W] − 2˚C/W − 0.2˚C/W
= [(150˚C − 50˚C)/21W] − 1˚C/W − 0.2˚C/W
P
is set to 0.2˚C/W since this is a reasonable
DSWITCH
DS(ON)
P
P
RDS(ON)
DTOTAL
= SQRT(125watts/4 ) = 5.59 amps
at 100˚C is used since we are calcu-
P
= (5.0%
P
RDS(ON)
= 6.6W + 14.4W = 21W
%Loss = 5.0%
DSWITCH
= (5.59A)
SA
SA
= 2.5˚C/W
= 3.5˚C/W
*
= 14.4W
= 6.6W
125W) / (1−5.0%)
DS(ON)
2
*
(0.230
JMAX
is found from the
JC
*
and T
125kHz
DS(ON)
is 150˚C. The
*
2)
www.national.com
JMAX
of the
are