lm4820mm-6 National Semiconductor Corporation, lm4820mm-6 Datasheet - Page 12

• Manufacturer Part Number: lm4820mm-6
• Description: Fixed Gain 1 Watt Audio Power Amplifier
• Manufacturer: National Semiconductor Corporation

www.national.com
Application Information
reproduce signals below 100 Hz to 150 Hz. Thus, using a
large input capacitor may not increase actual system perfor-
mance.
In addition to system cost and size, click and pop perfor-
mance is effected by the size of the input coupling capacitor,
C
reach its quiescent DC voltage (nominally 1/2 V
charge comes from the output via the feedback and is apt to
create pops upon device enable. Thus, by minimizing the
capacitor size based on necessary low frequency response,
turn-on pops can be minimized.
Besides minimizing the input capacitor size, careful consid-
eration should be paid to the bypass capacitor value. Bypass
capacitor, C
turn-on pops since it determines how fast the LM4820-6
turns on. The slower the LM4820-6’s outputs ramp to their
quiescent DC voltage (nominally 1/2 V
turn-on pop. Choosing C
value of C
produce a virtually clickless and popless shutdown function.
While the device will function properly, (no oscillations or
motorboating), with C
much more susceptible to turn-on clicks and pops. Thus, a
value of C
most cost sensitive designs.
AUDIO POWER AMPLIFIER DESIGN
A 1W/8Ω Audio Amplifier
A designer must first determine the minimum supply rail to
obtain the specified output power. By extrapolating from the
Output Power vs Supply Voltage graphs in the Typical Per-
formance Characteristics section, the supply rail can be
easily found. A second way to determine the minimum sup-
ply rail is to calculate the required V
Using this method, the minimum supply voltage would be
(V
extrapolated from the Dropout Voltage vs Supply Voltage
curve in the Typical Performance Characteristics section.
i.
Given:
opeak
A larger input coupling capacitor requires more charge to
Power Output
Load Impedance
Input Level
Input Impedance
Bandwidth
+ (V
B
i
OD TOP
B
equal to 1.0 µF is recommended in all but the
(in the range of 0.1 µF to 0.39 µF), should
, is the most critical component to minimize
+ V
OD BOT
B
equal to 0.1 µF, the device will be
B
equal to 1.0 µF along with a small
)), where V
100 Hz–20 kHz
opeak
OD BOT
DD
), the smaller the
using Equation 3.
(Continued)
and V
±
1 Wrms
0.25 dB
OD TOP
DD
1 Vrms
25 kΩ
). This
8Ω
are
(3)
12
2.7V
applications. Extra supply voltage creates headroom that
allows the LM4820-6 to reproduce peaks in excess of 1W
without producing audible distortion. At this time, the de-
signer must make sure that the power supply choice along
with the output impedance does not violate the conditions
explained in the Power Dissipation section.
Once the power dissipation equations have been addressed,
the differential gain is determined from Equations 4 or 5.
or
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. 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
and
As mentioned in the Selecting Proper External Compo-
nents section, R
amplifier’s lower bandpass frequency limit. To find the cou-
pling capacitor’s value, use Equation 6
The result is
Use a 0.33µf 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
200kHz. This is less than the LM4820-6’s 25MHz GBWP.
With this margin, the amplifier can be used in designs that
require more differential gain while avoiding performance,
restricting bandwidth limitations.
f
f
1/(2π*25kΩ*20kHz) = .318µf
DD
L
H
= 100Hz/5 = 20Hz
= 20kHz x 5 = 100kHz
to 5V
DD
is a standard supply voltage range for most
i
and C
A
VD
A
R
VD
f
C
= 2 ( 25kΩ/25kΩ )
i
= R
i
create a highpass filter that sets the
= 2 ( R
≥ 1/(2πR
A
VD
i
= 25kΩ
= 2
f
/R
i
f
i
L
)
)
VD
VD
, determines the
= 2 and f
±
±
0.25dB
0.25dB
H
(4)
(5)
(6)
=