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LM4810 датащи(PDF) 13 Page - National Semiconductor (TI)

[Old version datasheet] Texas Instruments acquired National semiconductor. Click here to check the latest version.
номер детали LM4810
подробное описание детали  Dual 105mW Headphone Amplifier with Active-High Shutdown Mode
PDF  18 Pages
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производитель  NSC [National Semiconductor (TI)]
домашняя страница  http://www.national.com
Logo NSC - National Semiconductor (TI)

LM4810 датащи(HTML) 13 Page - National Semiconductor (TI)

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Application Information (Continued)
the magnitude of “clicks and pops”. Increasing the value of
C
B reduces the magnitude of turn-on pops. However, this
presents a tradeoff: as the size of C
B increases, the turn-on
time increases. There is a linear relationship between the
size of C
B and the turn-on time. Here are some typical
turn-on times for various values of C
B:
C
B
T
ON
0.1µF
80ms
0.22µF
170ms
0.33µF
270ms
0.47µF
370ms
0.68µF
490ms
1.0µF
920ms
2.2µF
1.8sec
3.3µF
2.8sec
4.7µF
3.4sec
10µF
7.7sec
In order eliminate “clicks and pops”, all capacitors must be
discharged before turn-on. Rapidly switching V
DD may not
allow the capacitors to fully discharge, which may cause
“clicks and pops”. In a single-ended configuration, the output
is coupled to the load by C
O. This capacitor usually has a
high value. C
O discharges through internal 20k
Ω resistors.
Depending on the size of C
O, the discharge time constant
can be relatively large. To reduce transients in single-ended
mode, an external 1k
Ω–5kΩ resistor can be placed in par-
allel with the internal 20k
Ω resistor. The tradeoff for using
this resistor is increased quiescent current.
AUDIO POWER AMPLIFIER DESIGN
Design a Dual 70mW/32
Ω Audio Amplifier
Given:
Power Output
70 mW
Load Impedance
32
Input Level
1 Vrms (max)
Input Impedance
20k
Bandwidth
100 Hz–20 kHz ± 0.50dB
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 (5), 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 (5). For a
single-ended application, the result is Equation (6).
(5)
V
DD
≥ (2V
OPEAK +(VODTOP +VODBOT))
(6)
The Output Power vs Supply Voltage graph for a 32
Ω load
indicates a minimum supply voltage of 4.8V. This is easily
met by the commonly used 5V supply voltage. The additional
voltage creates the benefit of headroom, allowing the
LM4810 to produce peak output power in excess of 70mW
without clipping or other audible distortion. The choice of
supply voltage must also not create a situation that violates
maximum power dissipation as explained above in the
Power Dissipation section. Remember that the maximum
power dissipation point from Equation (1) must be multiplied
by two since there are two independent amplifiers inside the
package. Once the power dissipation equations have been
addressed, the required gain can be determined from Equa-
tion (7).
(7)
Thus, a minimum gain of 1.497 allows the LM4810 to reach
full output swing and maintain low noise and THD+N perfro-
mance. For this example, let A
V=1.5.
The amplifiers overall gain is set using the input (R
i ) and
feedback (R
f ) resistors. With the desired input impedance
set at 20k
Ω, the feedback resistor is found using Equation
(8).
A
V =Rf/Ri
(8)
The value of R
f is 30k
Ω.
The last step in this design is setting the amplifier’s −3db
frequency bandwidth. To achieve the desired ±0.25dB pass
band magnitude variation limit, the low frequency response
must extend to at lease 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 ±0.25dB desired
limit. The results are an
f
L = 100Hz/5 = 20Hz
(9)
and a
f
H = 20kHz
*
5 = 100kHz
(10)
As stated in the External Components section, both R
i in
conjunction with C
i, and Co with RL, create first order high-
pass filters. Thus to obtain the desired low frequency re-
sponse of 100Hz within ±0.5dB, both poles must be taken
into consideration. The combination of two single order filters
at the same frequency forms a second order response. This
results in a signal which is down 0.34dB at five times away
from the single order filter −3dB point. Thus, a frequency of
20Hz is used in the following equations to ensure that the
response is better than 0.5dB down at 100Hz.
C
i
≥ 1/(2π * 20kΩ * 20Hz) = 0.397µF; use 0.39µF.(11)
C
o
≥ 1/(2π *32Ω * 20Hz) = 249µF; use 330µF. (12)
The high frequency pole is determined by the product of the
desired high frequency pole, f
H, and the closed-loop gain,
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