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LM1876 Datasheet(PDF) 18 Page - National Semiconductor (TI)

[Old version datasheet] Texas Instruments acquired National semiconductor. Click here to check the latest version.
Part # LM1876
Description  Dual 20W Audio Power Amplifier with Mute and Standby Modes
PDF  20 Pages
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Manufacturer  NSC [National Semiconductor (TI)]
Direct Link  http://www.national.com
Logo NSC - National Semiconductor (TI)

LM1876 Datasheet(HTML) 18 Page - National Semiconductor (TI)

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Application Information (Continued)
To achieve a transient free power-up and power-down, the
voltage seen at the input terminals should be ideally the
same. Such a signal will be common-mode in nature, and
will be rejected by the LM1876. In Figure 3, the resistor R
INP
serves to keep the inputs at the same potential by limiting the
voltage difference possible between the two nodes. This
should significantly reduce any type of turn-on pop, due to an
uneven charging of the amplifier inputs. This charging is
based on a specific application loading and thus, the system
designer may need to adjust these values for optimal perfor-
mance.
As shown in Figure 3, the resistors labeled R
BI help bias up
the LM1876 off the half-supply node at the emitter of the
2N3904. But due to the input and output coupling capacitors
in the circuit, along with the negative feedback, there are two
different values of R
BI, namely 10 k
Ω and 200 kΩ. These
resistors bring up the inputs at the same rate resulting in a
popless turn-on. Adjusting these resistors values slightly
may reduce pops resulting from power supplies that ramp
extremely quick or exhibit overshoot during system turn-on.
AUDIO POWER AMPLlFIER DESIGN
Design a 15W/8
Ω Audio Amplifier
Given:
Power Output
15 Wrms
Load Impedance
8
Ω
Input Level
1 Vrms(max)
Input Impedance
47 k
Ω
Bandwidth
20 Hz−20 kHz
±0.25 dB
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
OPEAK can be determined from
equation (4) and I
OPEAK from equation (5).
(4)
(5)
To determine the maximum supply voltage the following
conditions must be considered. Add the dropout voltage to
the peak output swing V
OPEAK, to get the supply rail at a
current of I
OPEAK. The regulation of the supply determines
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.
Max supplies
≈ ± (V
OPEAK +VOD) (1 + regulation) (1.1)
For 15W of output power into an 8
Ω load, the required
V
OPEAK is 15.49V. A minimum supply rail of 20.5V results
from adding V
OPEAK and VOD. With regulation, the maximum
supplies are ±26V and the required I
OPEAK is 1.94A from
equation (5). It should be noted that for a dual 15W amplifier
into an 8
Ω load the I
OPEAK drawn from the supplies is twice
1.94 Apk or 3.88 Apk. At this point it is a good idea to check
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
SA is feasible
given system cost and size constraints. Once the heatsink
issues have been addressed, the required gain can be de-
termined from Equation (6).
(6)
From equation 6, the minimum A
V is:
A
V
≥ 11.
By selecting a gain of 21, and with a feedback resistor, R
f =
20 k
Ω, the value of R
i follows from equation (7).
R
i =Rf (AV − 1)
(7)
Thus with R
i =1 k
Ω a non-inverting gain of 21 will result.
Since the desired input impedance was 47 k
Ω, a value of 47
k
Ω was selected for R
IN. The final design step is to address
the bandwidth requirements which must be stated as a pair
of −3 dB frequency points. Five times away from a −3 dB
point is 0.17 dB down from passband response which is
better than the required ±0.25 dB specified. This fact results
in a low and high frequency pole of 4 Hz and 100 kHz
respectively. As stated in the External Components sec-
tion, R
i in conjunction with Ci create a high-pass filter.
C
i
≥ 1/(2π * 1kΩ * 4 Hz) = 39.8 µF;
use 39 µF.
The high frequency pole is determined by the product of the
desired high frequency pole, f
H, and the gain, AV. With a
A
V = 21 and fH = 100 kHz, the resulting GBWP is 2.1 MHz,
which is less than the guaranteed minimum GBWP of the
LM1876 of 5 MHz. This will ensure that the high frequency
response of the amplifier will be no worse than 0.17 dB down
at 20 kHz which is well within the bandwidth requirements of
the design.
www.national.com
18



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